Rd Sharma 2020 Solutions for Class 9 Math Chapter 17 Heron S Formula are provided here with simple step-by-step explanations. These solutions for Heron S Formula are extremely popular among Class 9 students for Math Heron S Formula Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 9 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 17.19:

#### Question 1:

Find the area of a quadrilateral *ABCD* is which *AB *= 3 cm, *BC* = 4 cm, *CD* = 4 cm, *DA* = 5 cm and *AC* = 5 cm.

#### Answer:

The quadrilateral ABCD having sides AB, BC, CD, DA and diagonal AC=5 cm is given, where AC divides quadrilateralABCD into two triangles ΔABC and ΔADC. We will find the area of the two triangles separately and them to find the area of quadrilateral ABCD.

In triangle ΔABC, observe that,

So the triangle ΔABC is right angled triangle.

Area of right angled triangle ΔABC,* *is given by

In ΔACD all the sides are known, so just use Heron’s formula to find out Area of triangle ΔACD,

$s=\frac{AD+DC+AC}{2}\phantom{\rule{0ex}{0ex}}=\frac{5+4+5}{2}\phantom{\rule{0ex}{0ex}}=7\mathrm{cm}$

The area of the ΔACD is:

Area of quadrilateral ABCD will be,

*Area* = Area of triangle ABC + Area of triangle ADC

#### Page No 17.19:

#### Question 2:

The sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m are 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

#### Answer:

We assume quadrilateral ABCD be the quadrangular field having sides AB, BC, CD, DA and.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC

We will find the area of two triangles ΔABC and ΔADC separately and add them to find the area of the quadrangular ABCD.

In triangle ΔADC, we have

AD = 24 m; DC = 7 m

We use Pythagoras theorem to find side AC,

AC^{2} = AD^{2 }+ DC^{2}

^{ }

Area of right angled triangle ΔADC, say *A*_{1}* *is given by

Where, Base = DA = 24 m; Height = DC = 7 m

Area of triangle ΔABC, say *A*_{2 }having sides *a,* *b*, *c* and *s* as semi-perimeter is given by

Where, *a *= AC = 25 m; *b *= AB = 26 m; *c *= BC = 27 m

Area of quadrilateral ABCD, say *A*

*A* = Area of triangle ΔADC + Area of triangle ΔABC

#### Page No 17.19:

#### Question 3:

The sides of a quadrilateral, taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.

#### Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and angle.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC. We will find the area of these two triangles and add them to find the area of the quadrilateral ABCD

In triangle ΔABC, we have

AB = 5 m; BC = 12 m

We will use Pythagoras theorem to calculate AC

AC^{2} = AB^{2}^{ }+ BC^{2}

Area of right angled triangle ΔABC, say *A*_{1}* *is given by

Where, Base = AB = 5 m; Height = BC = 12 m

Area of triangle ΔADC, say *A*_{2 }having sides *a,* *b*, *c* and *s* as semi-perimeter is given by

Where, *a *= AC = 13 m; b* *= DC = 14 m; c* *= AD = 15 m

Area of quadrilateral ABCD, say *A*

*A* = Area of triangle ΔABC + Area of triangle ΔADC

#### Page No 17.19:

#### Question 4:

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD= 5 m and AD = 8 m. How much area does it occupy?

#### Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and.

We take a diagonal DB, where DB divides ABCD into two triangles ΔBCD and ΔABD

In ΔBCD, we have

DC = 5 m; BC = 12 m

Use Pythagoras theorem

BD^{2} = DC^{2}^{ }+ BC^{2}

Area of right angled triangle ΔBCD, say *A*_{1}* *is given by

Where, Base = DC = 5 m; Height = BC = 12 m

Area of triangle ΔABD, say *A*_{2}_{ }having sides *a,* *b*, *c* and *s* as semi-perimeter is given by

Where, *a *= AD = 8 m; b* *= AB = 9 m; c* *= BD = 13 m

Area of quadrilateral ABCD, say *A*

*A* = Area of triangle DCB + Area of triangle ABD

#### Page No 17.19:

#### Question 5:

Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

#### Answer:

We assume ABCD be the given trapezium where AB is parallel to DC.

We draw CE parallel to AD from point C.

Therefore, a parallelogram ADCE is formed having AD parallel to CE and DC parallel to AE.

AE = 60 cm; CE = 25 cm; BE =

Basically we will find the area of the triangle BCE and area of the parallelogram AECD and add them to find the area of the trapezium ABCD.

Area of triangle ECB, say *A*_{1}_{ }having sides *a,* *b*, *c* and *s* as semi-perimeter is given by

Where, *a *= EB = 17 cm; b* *= EC = 25 cm; c* *= BC = 26 cm

Here we need to find the height of the parallelogram AECD which is CM to calculate area of AECD.

Where, BE = Base = 17 cm ; Height = CM = *h*

Thus area of parallelogram will be,

A_{2}= $b\times h$

$=60\times 24\phantom{\rule{0ex}{0ex}}=1440c{m}^{2}$

Total area of the trapezium will be

A = A_{1 }+ A_{2}

= 204 + 1440

= 1644 cm^{2}

#### Page No 17.19:

#### Question 6:

A rhombus, sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m^{2}. Find the cost of painting.

#### Answer:

We assume ABCD be the given rhombus having

AB = BC = CD = DA

BD and AC be the diagonals of rhombus

We need to find cost of painting both sides

Perimeter of rhombus ABCD, say P* *is 32 m

We know that BD and AC* *diagonals of rhombus and.So,

(Diagonals in rhombus intersect at right angle)

BD = 24 m; AC = *h;* side = AB = 8m

Taking square of both sides, we get

Area of rhombus, say *A*_{1}

Area of both sides of rhombus;

#### Page No 17.20:

#### Question 7:

Find the area of a quadrilateral *ABCD* in which *AD* = 24 cm, ∠*BAD* = 90° and *BCD* forms an equilateral triangle whose each side is equal to 26 cm.

#### Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA, diagonal BD and angle where BCD forms an equilateral triangle having equal sides.

We need to find area of ABCD

In triangle BAD, we have

BD^{2} = BA^{2}^{ }+ AD^{2}^{ }.So

Area of right angled triangle ABD, say *A*_{1}* *is given by

Where,

Base = BA = 10 cm; Height = AD = 24 cm

Area of equilateral triangle BCD, say *A*_{2}_{ }having sides *a,* *b*, *c* is given by

, where

*a *= BC = CD = BD = 26 cm

Area of quadrilateral ABCD, say *A*

*A* = Area of triangle BAD + Area of triangle BCD

#### Page No 17.20:

#### Question 8:

Find the area of a quadrilateral *ABCD* in which *AB* = 42 cm, *BC* = 21 cm, *CD* = 29 cm, *DA* = 34 cm and diagonal *BD* = 20 cm.

#### Answer:

The quadrilateral ABCD having sides AB,BC,CD,DA and diagonal BD is given, where BD divides ABCD into two triangles

ΔDBC and ΔDAB

In triangle DBC, we can observe that

DC^{2} = DB^{2}^{ }+ BC^{2}

Therefore, it is a right angled triangle.^{ }

Area of right angled triangle DBC, say *A*_{1}* *is given by

Where,

Base = BC = 21 cm; Height = BD = 20 cm

Area of triangle DAB, say *A*_{2}_{ }having sides *a,* *b*, *c* and *s* as semi-perimeter is given by

, where

*a *= DB = 20 cm; *b *= AD = 34 cm; *c *= AB = 42 cm

Area of quadrilateral ABCD, say *A*

*A* = Area of triangle DBC + Area of triangle DAB

#### Page No 17.20:

#### Question 9:

The adjacent sides of a parallelogram *ABCD* measure 34 cm and 20 cm, and the diagonal *AC *measures 42 cm. Find the area of the parallelogram.

#### Answer:

We are given the measure of adjacent sides of a parallelogram AB and BC that is the sides having same point of origin and the diagonal AC which divides parallelogram ABCD into two congruent triangles ABC and ADC.

Area of triangle ABC is equal to Area of triangle ADC as they are congruent triangles.

Area of parallelogram ABCD, say *A* is given by

*A =*

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangle, say *A*_{1} having sides 20 cm, 34 cm and 42 cm is given by

*a* = 20 cm ; *b* = 34 cm ; *c* = 42 cm

Area of parallelogram ABCD, say *A* is given by

#### Page No 17.20:

#### Question 10:

Find the area of the blades of the magnetic compass shown in the given figure. (Take $\sqrt{11}$ = 3.32)

#### Answer:

The blades of the magnetic compass are forming a rhombus having all equal sides measuring 5 cm each. A diagonal measuring 1 cm is given which is forming the triangular shape of the blades of the magnetic compass and diving the rhombus into two congruent triangles, say triangle ABC and triangle DBC having equal dimensions.

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangle ABC, say *A*_{1} having sides 5 cm, 5 cm and 1 cm is given by:

*a* = 5 cm ; *b* = 5 cm ; *c* = 1 cm

Area of blades of magnetic compass, say *A* is given by

*A* = Area of one of triangle ABC

#### Page No 17.20:

#### Question 11:

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

#### Answer:

It is given that the area of triangle and parallelogram are equal.

We will calculate the area of triangle with the given values and it will also give us the area of parallelogram as both are equal.

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter* *is given by,

, where

Therefore the area of a triangle; say *A, *having sides 15 cm, 13 cm and 14 cm is given by

*a *= 15 cm ; *b* = 13 cm ; *c* = 14 cm

We have to find the height of the parallelogram, say *h*

Area of parallelogram AECD say *A*_{1 }is given by

Base = 60 cm; Height = *h* cm; *A =**A*_{1}_{ }*= *84 cm^{2}

#### Page No 17.20:

#### Question 12:

Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

#### Answer:

Consider a trapezium ABCD as shown below.

Draw a line BF parallel to AD such that ABFD becomes a parallelogram. Also, draw a perpendicular BE on CD as shown in the figure.

In ΔBCF, the three sides are given as $a=\mathrm{BF}=25\mathrm{cm},b=\mathrm{BC}=26\mathrm{cm},c=\mathrm{CF}=\mathrm{CD}-\mathrm{FD}=17\mathrm{cm}$.

The semi-perimeter of ΔBCF = $\frac{25+26+17}{2}=34$.

The area of ΔBCF can be calculated using Heron's formula as

$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{34\left(34-25\right)\left(34-26\right)\left(34-17\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{34\times 9\times 8\times 17}\phantom{\rule{0ex}{0ex}}=204{\mathrm{cm}}^{2}$

Also, the area of ΔBCF = $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$\Rightarrow 204=\frac{1}{2}\times 17\times \mathrm{BE}\phantom{\rule{0ex}{0ex}}\Rightarrow 204=\frac{1}{2}\times 17\times \mathrm{BE}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BE}=24\mathrm{cm}$

Therefore, Area of Trapezium =$\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)\times \mathrm{BE}=\frac{1}{2}\left(60+77\right)\times 24=1644{\mathrm{cm}}^{2}$.

#### Page No 17.20:

#### Question 13:

Find the perimeter and area of the quadrilateral *ABCD* in which *AB* = 17 cm, *AD* = 9 cm, *CD* = 12 cm, ∠*ACB* = 90° and *AC* = 15 cm.

#### Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and ∠ACB = 90^{∘}.

We take a diagonal AC, where AC divides ABCD into two triangles ΔACB and ΔADC

Since ∆ACB is right angled at C, we have

AC = 15 cm; AB = 17 cm

AB^{2} = AC^{2}^{ }+ BC^{2}

Area of right angled triangle ABC, say *A*_{1}* *is given by

, where,

Base = BC = 8 cm; Height = AC = 15 cm

Area of triangle ADC, say *A*_{2}_{ }having sides *a,* *b*, *c* and *s* as semi-perimeter is given by

, where

*a *= AD = 9 cm; b* *= DC = 12 cm; c* *= AC = 15 cm

Area of quadrilateral ABCD, say *A*

*A* = Area of ∆ACB + Area of ∆ADC

Perimeter of quadrilateral ABCD, say *P*

#### Page No 17.20:

#### Question 14:

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in Fig. 12.28. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

#### Answer:

We have to find the area of each type of triangular strips needed for the fan.

There are 5 strips of each type having equal dimensions, so we will calculate the area of a single strip and then multiply it by 5 to ascertain the area of each type of strip needed.

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangular strip, say *A*_{1} having sides 25 cm, 25 cm and 14 cm is given by:

*a* = 25 cm ; *b* = 25 cm ; *c* = 14 cm

Area of each type of strip needed, say* A*.

_{}

#### Page No 17.24:

#### Question 1:

*Mark the correct alternative in each of the following:*

The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is

(a) 225 cm^{2}

(b) 240 cm^{2}

(c) $225\sqrt{2}$ cm^{2}

(d) 450 cm^{2}

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangle say *A, *having sides 16 cm, 30 cm and 34 cm is given by

*a* = 16 cm ; *b* = 30 cm ; *c* = 34 cm

Therefore the area of the triangle is

Hence, the correct option is (b).

#### Page No 17.24:

#### Question 2:

The base of an isosceles right triangle is 30 cm. Its area is

(a) 225 cm^{2}

(b) 225 $\sqrt{3}$ cm^{2}

(c) 225 $\sqrt{2}$ cm^{2}

(d) 450 cm^{2}

#### Answer:

$\mathrm{Let}\mathrm{ABC}\mathrm{be}\mathrm{the}\mathrm{right}\mathrm{triangle}\mathrm{in}\mathrm{which}\angle \mathrm{B}=90\xb0.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{base}=\mathrm{BC};\mathrm{perpendicular}=\mathrm{AB};\mathrm{Hypotenuse}=\mathrm{AC}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{BC}=30\mathrm{cm}\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\u2206\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{right}\mathrm{angled}\u2206\mathrm{and}\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{hypotenuse}\mathrm{is}\mathrm{the}\mathrm{longest}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{right}\u2206.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{AB}=\mathrm{BC}=30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{area}\mathrm{of}\u2206\mathrm{ABC}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 30\times 30\phantom{\rule{0ex}{0ex}}=450{\mathrm{cm}}^{2}$

Hence, the correct option is (d).

#### Page No 17.24:

#### Question 3:

The sides of a triangle are 7 cm, 9 cm and 14 cm. Its area is

(a) $12\sqrt{5}c{m}^{2}$

(b) $12\sqrt{3}c{m}^{2}$

(c) $24\sqrt{5}c{m}^{2}$

(d) $63c{m}^{2}$

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangle having sides 7 cm, 9 cm and 14 cm is given by

*a* = 7 cm ; *b* = 9 cm ; *c* = 14 cm

Therefore the answer is (a).

#### Page No 17.24:

#### Question 4:

The sides of a triangular field are 325 m, 300 m and 125 m. Its area is

(a) 18750 m^{2}

(b) 37500 m^{2}

(c) 97500 m^{2}

(d) 48750 m^{2}

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangular field, say *A* having sides 325 m, 300 m and 125 m is given by

*a* = 325 m ; *b* = 300 m ; *c* = 125 m

Therefore, the correct answer is (a).

#### Page No 17.24:

#### Question 5:

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

(a) 20 cm

(b) 30 cm

(c) 40 cm

(d) 50 cm

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter* *is given by,

, where

Therefore the area of a triangle, say *A* having sides 50 cm, 78 cm and 112 cm is given by

The area of a triangle, having *p* as the altitude will be,

$\mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Where, *A =* 1680

We have to find the smallest altitude, so will substitute the value of the base *AC *with the length of each side one by one and find the smallest altitude distance i.e. *p*

**Case 1 **

**Case 2**

**Case 3**

Therefore, the answer is (b).

#### Page No 17.24:

#### Question 6:

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

(a) 11 m

(b) 66 m

(c) 50 m

(d) 60 m

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

We need to find the altitude to the smallest side

Therefore the area of a triangle having sides 11 m, 60 m and 61 m is given by

*a* = 11 m ; *b* = 60 m ; *c* = 61 m

The area of a triangle having base *AC* and height *p *is given by

We have to find the height *p *corresponding to the smallest side of the triangle. Here smallest side is 11 m

*AC = *11 m

Therefore, the answer is (d).

#### Page No 17.24:

#### Question 7:

The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

(a) $30\sqrt{7}cm$

(b) $\frac{15\sqrt{7}}{2}cm$

(c) $\frac{15\sqrt{7}}{4}cm$

(d) 30 cm

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

We need to find the altitude corresponding to the longest side

Therefore the area of a triangle having sides 11 cm, 15 cm and 16 cm is given by

*a* = 11 m ; *b* = 15 cm ; *c* = 16 cm

The area of a triangle having base *AC *and height *p *is given by

We have to find the height *p *corresponding to the longest side of the triangle.Here longest side is 16 cm, that is *AC=*16 cm

Therefore, the answer is (c).

#### Page No 17.24:

#### Question 8:

The base and hypotenuse of a right triangle are respectively 5 cm and 13 cm long. Its area is

(a) 25 cm^{2}

(b) 28 cm^{2}

(c) 30 cm^{2}

(d) 40 cm^{2}

#### Answer:

In right angled triangle ABC having base 5 cm and hypotenuse 13 cm we are asked to find its area

Using Pythagorean Theorem

Where, AB = hypotenuse = 13 cm, AC = Base = 5 cm, BC = Height

Area of a triangle, say *A *having base 5 cm and altitude 12 cm is given by

Where, Base = 5 cm; Height = 12 cm

Therefore, the answer is (c).

#### Page No 17.24:

#### Question 9:

The length of each side of an equilateral triangle of area $4\sqrt{3}c{m}^{2}$, is

(a) 4 cm

(b) $\frac{4}{\sqrt{3}}cm$

(c) $\frac{\sqrt{3}}{4}cm$

(d) 3 cm

#### Answer:

Area of an equilateral triangle say *A,* having each side *a* cm is given by

We are asked to find the side of the triangle

Therefore, the side of the equilateral triangle says *a, *having area is given by

Therefore, the correct answer is (a).

#### Page No 17.24:

#### Question 10:

If an isosceles right triangle has area 8 cm^{2},then the length of its hypotenuse is

(a) $\sqrt{32}\mathrm{cm}$

(b) $\sqrt{48}\mathrm{cm}$

(c) $\sqrt{24}\mathrm{cm}$

(d) 4 cm

#### Answer:

Given: Area of an isosceles right triangle is 8 cm^{2}

$\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{isosceles}\mathrm{right}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Base}\left(\because \mathrm{Base}=\mathrm{Height}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 8=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 16={\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Base}=4\mathrm{cm}=\mathrm{Perpendicular}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle},\mathrm{using}\mathrm{pythagoras}\mathrm{theorem}\phantom{\rule{0ex}{0ex}}{\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Perpendicular}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(4\right)}^{2}+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=16+16\phantom{\rule{0ex}{0ex}}=32\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Hypotenuse}=\sqrt{32}\mathrm{cm}$

Hence, the correct option is (a).

#### Page No 17.24:

#### Question 11:

The perimeter of an equilateral triangle is 60 m. The area is

(a) $10\sqrt{3}{\mathrm{m}}^{2}$

(b) $15\sqrt{3}{\mathrm{m}}^{2}$

(c) $20\sqrt{3}{\mathrm{m}}^{2}$

(d) $100\sqrt{3}{\mathrm{m}}^{2}$

#### Answer:

Given: The perimeter of an equilateral triangle is 60 m.

Let the length of the side of an equilateral triangle be *x* m.

$\mathrm{Perimeter}=60\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=60\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=20\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangl}e=\frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}{\left(20\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\left(400\right)\phantom{\rule{0ex}{0ex}}=100\sqrt{3}{\mathrm{m}}^{2}$

Hence, the correct option is (d).

#### Page No 17.24:

#### Question 12:

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

(a) $\sqrt{15}{\mathrm{cm}}^{2}$

(b) $\frac{\sqrt{15}}{2}{\mathrm{cm}}^{2}$

(c) $2\sqrt{15}{\mathrm{cm}}^{2}$

(d) $4\sqrt{15}{\mathrm{cm}}^{2}$

#### Answer:

Given:

The base of an isosceles triangle is 2 cm.

The length of one of the equal sides 4 cm.

Using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{2+4+4}{2}=\frac{10}{2}=5$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{5\left(5-2\right)\left(5-4\right)\left(5-4\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{5\left(3\right)\left(1\right)\left(1\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{15}{\mathrm{cm}}^{2}$

Hence, the correct option is (a).

#### Page No 17.24:

#### Question 13:

The length of each side of an equilateral triangle having an area of $9\sqrt{3}{\mathrm{cm}}^{2}$ is

(a) 8 cm

(b) 36 cm

(c) 4 cm

(d) 6 cm

#### Answer:

Given: The area of an equilateral triangle is $9\sqrt{3}{\mathrm{cm}}^{2}$.

Let the length of the side of an equilateral triangle be *x* cm.

$\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 9\sqrt{3}=\frac{\sqrt{3}}{4}{\left(x\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{9\sqrt{3}\times 4}{\sqrt{3}}={\left(x\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=36\phantom{\rule{0ex}{0ex}}\Rightarrow x=6\mathrm{cm}$

Hence, the correct option is (d).

#### Page No 17.25:

#### Question 14:

If the area of an equilateral triangle is $16\sqrt{3}{\mathrm{cm}}^{2}$, then its perimeter is

(a) 48 cm

(b) 24 cm

(c) 12 cm

(d) 36 cm

#### Answer:

Given: The area of an equilateral triangle is $16\sqrt{3}{\mathrm{cm}}^{2}$.

Let the length of the side of an equilateral triangle be *x* cm.

$\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 16\sqrt{3}=\frac{\sqrt{3}}{4}{\left(x\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{16\sqrt{3}\times 4}{\sqrt{3}}={\left(x\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=64\phantom{\rule{0ex}{0ex}}\Rightarrow x=8\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Perimeter}\mathrm{of}\mathrm{triangle}=3x\phantom{\rule{0ex}{0ex}}=3\left(8\right)\phantom{\rule{0ex}{0ex}}=24\mathrm{cm}$

Hence, the correct option is (b).

#### Page No 17.25:

#### Question 15:

The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. The length of its longest altitude is

(a) $16\sqrt{5}\mathrm{cm}$

(b) $10\sqrt{5}\mathrm{cm}$

(c) $24\sqrt{5}\mathrm{cm}$

(d) 28 cm

#### Answer:

Given:

The sides of a triangle are 35 cm, 54 cm and 61 cm respectively.

Let *CD* be the longest altitude of the triangle.

Using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{35+54+61}{2}=\frac{150}{2}=75$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{75\left(75-35\right)\left(75-54\right)\left(75-61\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{75\left(40\right)\left(21\right)\left(14\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(5\times 5\times 3\right)\left(5\times 2\times 2\times 2\right)\left(3\times 7\right)\left(2\times 7\right)}\phantom{\rule{0ex}{0ex}}=\left(5\times 3\times 7\times 2\times 2\right)\sqrt{5}\phantom{\rule{0ex}{0ex}}=420\sqrt{5}{\mathrm{cm}}^{2}$

We know,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}\Rightarrow 420\sqrt{5}=\frac{1}{2}\times 35\times CD\phantom{\rule{0ex}{0ex}}\Rightarrow 12\sqrt{5}=\frac{1}{2}CD\phantom{\rule{0ex}{0ex}}\Rightarrow CD=24\sqrt{5}$

Thus, the length of its longest altitude is $24\sqrt{5}\mathrm{cm}$.

Hence, the correct option is (c).

#### Page No 17.25:

#### Question 16:

The sides of a triangle are 56 cm, 60 cm and 52 cm. Area of the triangle is

(a) 1322 cm^{2}

(b) 1311 cm^{2}

(c) 1344 cm^{2}

(d) 1392 cm^{2}

#### Answer:

Given:

The sides of a triangle are 56 cm, 60 cm and 52 cm.

Using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{56+60+52}{2}=\frac{168}{2}=84$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{84\left(84-56\right)\left(84-60\right)\left(84-52\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{84\left(28\right)\left(24\right)\left(32\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{3\times 7\times 2\times 2\left(7\times 2\times 2\right)\left(3\times 2\times 2\times 2\right)\left(2\times 2\times 2\times 2\times 2\right)}\phantom{\rule{0ex}{0ex}}=\left(3\times 7\times 2\times 2\times 2\times 2\times 2\times 2\right)\phantom{\rule{0ex}{0ex}}=1344{\mathrm{cm}}^{2}$

Hence, the correct option is (c).

#### Page No 17.25:

#### Question 17:

The edges of a triangular board are 6 cm, 8 cm and 10 cm long. The cost of painting it at the rate of 9 paise per cm^{2 }is

(a) ₹ 2

(b) ₹ 2.16

(c) ₹ 2.48

(d) ₹ 3

#### Answer:

Given:

The edges of a triangular board are 6 cm, 8 cm and 10 cm long.

The cost of painting per cm^{2} is 9 paise.

Using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{6+8+10}{2}=\frac{24}{2}=12$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{12\left(12-6\right)\left(12-8\right)\left(12-10\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{12\left(6\right)\left(4\right)\left(2\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(3\times 2\times 2\right)\left(3\times 2\right)\left(2\times 2\right)\left(2\right)}\phantom{\rule{0ex}{0ex}}=\left(3\times 2\times 2\times 2\right)\phantom{\rule{0ex}{0ex}}=24{\mathrm{cm}}^{2}$

The cost of painting per cm^{2} = 9 paise.

The cost of painting 24 cm^{2} = 24 × 9 paise

= 216 paise

= ₹ 2.16

Hence, the correct option is (b).

#### Page No 17.25:

#### Question 18:

The area of an equilateral triangle with side $2\sqrt{3}$ cm is

(a) 5.196 cm^{2}

(b) 0.866 cm^{2}

(c) 3.496 cm^{2}

(d) 1.732 cm^{2}

#### Answer:

Given: The side of an equilateral triangle is $2\sqrt{3}$ cm.

$\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangl}e=\frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}{\left(2\sqrt{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\left(12\right)\phantom{\rule{0ex}{0ex}}=3\sqrt{3}\phantom{\rule{0ex}{0ex}}=5.196{\mathrm{cm}}^{2}$

Hence, the correct option is (a).

#### Page No 17.25:

#### Question 19:

If the area of a regular hexagon is $54\sqrt{3}{\mathrm{cm}}^{2}$, then the length of its each side is

(a) 3 cm

(b) $2\sqrt{3}\mathrm{cm}$

(c) 6 cm

(d) $6\sqrt{3}\mathrm{cm}$

#### Answer:

Given: The area of a regular hexagon is $54\sqrt{3}{\mathrm{cm}}^{2}$.

We know, a hexagon is formed by joining 6 equilateral triangles together.

Therefore, Area of 6 equilateral triangles = $54\sqrt{3}{\mathrm{cm}}^{2}$

$\mathrm{Area}\mathrm{of}6\mathrm{equilateral}\mathrm{triangles}=6\times \frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 54\sqrt{3}=6\times \frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{54\sqrt{3}\times 4}{6\times \sqrt{3}}={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 9\times 4={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 36={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=6\mathrm{cm}$

Hence, the correct option is (c).

#### Page No 17.25:

#### Question 20:

If the length of each edge of a regular tetrahedron is '*a*', then its surface area is

(a) $\sqrt{3}{a}^{2}\mathrm{sq}.\mathrm{units}$

(b) $3\sqrt{2}{a}^{2}\mathrm{sq}.\mathrm{units}$

(c) $2\sqrt{3}{a}^{\mathit{2}}\mathrm{sq}.\mathrm{units}$

(d) $\sqrt{6}{a}^{2}\mathrm{sq}.\mathrm{units}$

#### Answer:

Given: The length of each edge of a regular tetrahedron is '*a*' units.

We know, the surface area of tetrahedron = area of 4 equilateral triangles

$\mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{Surface}\mathrm{Area}=4\times \frac{\sqrt{3}}{4}\times {a}^{2}\phantom{\rule{0ex}{0ex}}=\sqrt{3}{a}^{2}\mathrm{sq}.\mathrm{units}$

Hence, the correct option is (a).

#### Page No 17.25:

#### Question 21:

If the area of an isosceles right triangle is 8 cm^{2}, what is the perimeter of the triangle?

(a) 8 + $\sqrt{2}$ cm^{2}

(b) 8 + 4$\sqrt{2}$ cm^{2}

(c) 4 + 8$\sqrt{2}$ cm^{2}

(d) 12$\sqrt{2}$ cm^{2}

#### Answer:

We are given the area of an isosceles right triangle and we have to find its perimeter.

Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. We are asked to find the perimeter of the triangle

Let us take the base and height of the triangle be *x* cm.

Area of a isosceles right triangle, say *A* having base *x* cm and height *x* cm is given by

*A* = 8 cm^{2}; Base = Height = *x* cm

Using Pythagorean Theorem we have;

Let ABC be the given triangle

Perimeter of triangle ABC, say *P* is given by

AB = 4 cm; BC = 4 cm; AC =

Therefore, the answer is (b).

#### Page No 17.25:

#### Question 22:

The lengths of the sides of Δ *ABC* are consecutive integers. It Δ *ABC* has the same perimeter as an equilateral triangle with a side of length 9 cm, what is the length of the shortest side of Δ*ABC**?*

(a) 4

(b) 6

(c) 8

(d) 10

#### Answer:

We are given that triangle ABC has equal perimeter as to the perimeter of an equilateral triangle having side 9 cm. The sides of triangle ABC are consecutive integers. We are asked to find the smallest side of the triangle ABC

Perimeter of an equilateral triangle, say *P *having side 9 cm is given by

Let us assume the three sides of triangle ABC be *x,* *x+*1, *x−*1

Perimeter of triangle ABC, say *P*_{1 }is given by

*P*_{1}_{ }*=* AB + BC + AC

AB = *x*; BC =* x* +1; AC = *x−*1. Since *P _{1}*

_{ }=

*P*. So

By using the value of *x,* we get the sides of triangle as 8 cm, 9 cm and 10 cm

Therefore, the answer is (c).

#### Page No 17.25:

#### Question 23:

In the given figure, the ratio *AD* to *DC* is 3 to 2. If the area of Δ ABC is 40 cm^{2}, what is the area of Δ* BDC*?

#### Answer:

Area of triangle ABC is given 40 cm^{2}.

Also

We are asked to find the area of the triangle BDC

Let us take BE perpendicular to base AC in triangle ABC.

We assume AC equal to *y* and BE equal to *x* in triangle ABC

Area of triangle ABC, say *A* is given by

We are given the ratio between AD to DC equal to 3:2

So,

In triangle BDC, we take BE as the height of the triangle

Area of triangle BDC, say *A*_{1}* *is given by

Therefore, the answer is (a).

#### Page No 17.26:

#### Question 24:

If the length of a median of an equilateral triangle is *x* cm, then its area is

(a) x^{2}

(b) $\frac{\sqrt{3}}{2}{x}^{2}$

(c) $\frac{{x}^{2}}{\sqrt{3}}$

(d) $\frac{{x}^{2}}{2}$

#### Answer:

We are given the length of median of an equilateral triangle by which we can calculate its side. We are asked to find area of triangle in terms of *x*

Altitude of an equilateral triangle say *L,* having equal sides of *a *cm* *is given by, where, *L = x *cm

Area of an equilateral triangle, say *A*_{1} having each side *a* cm is given by

Since * *.So

Therefore, the answer is (c).

#### Page No 17.26:

#### Question 25:

If every side of a triangle is doubled, then increase in the area of the triangle is

(a) $100\sqrt{2}\%$

(b) 200%

(c) 300%

(d) 400%

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

We take the sides of a new triangle as 2*a*, 2*b*, 2*c* that is twice the sides of previous one

Now, the area of a triangle having sides 2*a*, 2*b,* and 2c* *and as semi-perimeter is given by,

Where,

Now,

Therefore, increase in the area of the triangle

Percentage increase in area

Therefore, the answer is (c).

#### Page No 17.26:

#### Question 26:

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $12\sqrt{2}$ cm, then area of the triangle is

(a) $24\sqrt{2}c{m}^{2}$

(b) $24\sqrt{3}c{m}^{2}$

(c) $48\sqrt{3}c{m}^{2}$

(d) $64\sqrt{3}c{m}^{2}$

#### Answer:

It is given the perimeter of a square ABCD is equal to the perimeter of triangle PQR.

The measure of the diagonal of the square is given .We are asked to find the area of the triangle

In square ABCD, we assume that the adjacent sides of square be *a.*

Since, it is a square then* *

By using Pythagorean Theorem

Therefore, side of the square is 12 cm.

Perimeter of the square ABCD say *P* is given by

Side = 12 cm

Perimeter of the equilateral triangle PQR say *P*_{1} is given by

The side of equilateral triangle PQR is equal to 16 cm.

Area of an equilateral triangle say *A,* having each side *a* cm is given by

Area of the given equilateral triangle having each equal side equal to 4 cm is given by

*a* = 16 cm

Therefore, the answer is (d).

#### Page No 17.26:

#### Question 1:

The side and altitude of an equilateral triangle are in the ratio __________.

#### Answer:

Let *ABC* be an equilateral triangle of side '*a*' units and *AD* be the altitude of the triangle.

In ∆*ABD*,

*AB* = *a* units

*BD *= $\frac{a}{2}$ units

Using pythagoras theorem,

$A{B}^{2}=B{D}^{2}+A{D}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}={\left(\frac{a}{2}\right)}^{2}+A{D}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A{D}^{2}={a}^{2}-\frac{{a}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow A{D}^{2}=\frac{4{a}^{2}-{a}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow A{D}^{2}=\frac{3{a}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow AD=\frac{\sqrt{3}}{2}a...\left(1\right)\phantom{\rule{0ex}{0ex}}$

Now,

$\frac{\mathrm{Side}}{\mathrm{Altitude}}=\frac{AB}{AD}\phantom{\rule{0ex}{0ex}}=\frac{a}{{\displaystyle \frac{\sqrt{3}}{2}}a}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{3}}$

Hence, the side and altitude of an equilateral triangle are in the ratio $\overline{)2:\sqrt{3}}$.

#### Page No 17.26:

#### Question 2:

If the area of an isosceles right-angled triangle is 72 cm^{2}, then its perimeter is _________.

#### Answer:

Given: Area of an isosceles right-angled triangle is 72 cm^{2}

$\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{isosceles}\mathrm{right}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Base}\left(\because \mathrm{Base}=\mathrm{Height}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 72=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 144={\left(\mathrm{Base}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Base}=12\mathrm{cm}=\mathrm{Perpendicular}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\mathrm{a}\mathrm{right}-\mathrm{angled}\mathrm{triangle},\mathrm{using}\mathrm{pythagoras}\mathrm{theorem}\phantom{\rule{0ex}{0ex}}{\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Perpendicular}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(12\right)}^{2}+{\left(12\right)}^{2}\phantom{\rule{0ex}{0ex}}=144+144\phantom{\rule{0ex}{0ex}}=288\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Hypotenuse}=\sqrt{288}\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Hypotenuse}=12\sqrt{2}\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{Perimeter}=12+12+12\sqrt{2}=12\left(2+\sqrt{2}\right)\mathrm{cm}$

Hence, its perimeter is $\overline{)12\left(2+\sqrt{2}\right)\mathrm{cm}}.$

#### Page No 17.26:

#### Question 3:

The height of an equilateral triangle is $\sqrt{3}$ *a* units. Then its area is _________.

#### Answer:

Given: The height of an equilateral triangle is $\sqrt{3}$*a* units.

Let *ABC* be an equilateral triangle of side '*x*' units and *AD* be the altitude of the triangle of length $\sqrt{3}$*a* units.

In ∆*ABD*,

*AB* = *x* units

*BD* = $\frac{x}{2}$ units

Using pythagoras theorem,

$A{B}^{2}=B{D}^{2}+A{D}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}={\left(\frac{x}{2}\right)}^{2}+{\left(\sqrt{3}a\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{a}^{2}={x}^{2}-\frac{{x}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{a}^{2}=\frac{4{x}^{2}-{x}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{a}^{2}=\frac{3{x}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=\frac{{x}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{x}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=2a...\left(1\right)\phantom{\rule{0ex}{0ex}}$

Now,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times {\left(2a\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times 4{a}^{2}\phantom{\rule{0ex}{0ex}}=\sqrt{3}{a}^{2}$

Hence, its area is $\overline{)\sqrt{3}{a}^{2}\mathrm{sq}.\mathrm{units}}$.

#### Page No 17.26:

#### Question 4:

The area of a triangle with base 4 cm and height 6 cm is __________.

#### Answer:

Given:

Base of the triangle = 4 cm

Height = 6 cm

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 4\times 6\phantom{\rule{0ex}{0ex}}=\frac{24}{2}\phantom{\rule{0ex}{0ex}}=12{\mathrm{cm}}^{2}$

Hence, the area is $\overline{)12{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 5:

Δ*ABC *is an isosceles right triangle right-angled at *A*. If *AB *= 4 cm, then its area is _________.

#### Answer:

Given:

Δ*ABC *is an isosceles right triangle right-angled at *A
AB *= 4 cm

Since, Δ

*ABC*is an isosceles right triangle right-angled at

*A*

Therefore,

*AB = AC*= 4 cm

In ∆

*ABC*,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times AB\times AC\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 4\times 4\phantom{\rule{0ex}{0ex}}=8{\mathrm{cm}}^{2}$

Hence, its area is $\overline{)8{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 6:

If the side of a rhombus is 10 cm and one diagonal is 16 cm, then its area is _________.

#### Answer:

Given:

The side of a rhombus is 10 cm

One diagonal is 16 cm

Let *ABCD* be a rhombus.

*AB* = 10 cm ...(1)

*AC* = 16 cm ...(2)

We know, the diagonals of a rhombus bisects each other at right angles.

Let the point of intersection of the diagonals be *O*.

Then,

*AO = OC* = $\frac{16}{2}=8$ cm ...(3)

$\mathrm{In}\u2206AOB,\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{pythagoras}\mathrm{theorem},\phantom{\rule{0ex}{0ex}}A{B}^{2}=A{O}^{2}+B{O}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {10}^{2}={8}^{2}+B{O}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 100=64+B{O}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B{O}^{2}=100-64\phantom{\rule{0ex}{0ex}}\Rightarrow B{O}^{2}=36\phantom{\rule{0ex}{0ex}}\Rightarrow BO=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},BD=2\times BO\phantom{\rule{0ex}{0ex}}=2\times 6\phantom{\rule{0ex}{0ex}}=12\mathrm{cm}...\left(4\right)$

$\mathrm{Area}\mathrm{of}\mathrm{rhombus}=\frac{1}{2}\times \left(\mathrm{product}\mathrm{of}\mathrm{diagonals}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 12\times 16\left(\mathrm{from}\left(2\right)\mathrm{and}\left(4\right)\right)\phantom{\rule{0ex}{0ex}}=6\times 16\phantom{\rule{0ex}{0ex}}=96{\mathrm{cm}}^{2}$

Hence, the area is $\overline{)96{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 7:

The area of a regular hexagon of side 6 cm is __________.

#### Answer:

Given: The side of a regular hexagon is 6 cm.

We know, a hexagon is formed by joining 6 equilateral triangles together.

Therefore,

$\mathrm{Area}\mathrm{of}\mathrm{hexagon}=\mathrm{Area}\mathrm{of}6\mathrm{equilateral}\mathrm{triangles}\phantom{\rule{0ex}{0ex}}=6\times \frac{\sqrt{3}}{4}\times {x}^{2}\phantom{\rule{0ex}{0ex}}=6\times \frac{\sqrt{3}}{4}\times {\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=6\times \frac{\sqrt{3}}{4}\times 36\phantom{\rule{0ex}{0ex}}=6\times 9\times \sqrt{3}\phantom{\rule{0ex}{0ex}}=54\sqrt{3}{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

Hence, the area of a regular hexagon of side 6 cm is $\overline{)54\sqrt{3}{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 8:

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Then its area is _________.

#### Answer:

Given:

The base of a right triangle is 8 cm

hypotenuse is 10 cm

$\mathrm{Using}\mathrm{pythagoras}\mathrm{theorem},\phantom{\rule{0ex}{0ex}}{\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Height}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {10}^{2}={8}^{2}+{\left(\mathrm{Height}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 100=64+{\left(\mathrm{Height}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\mathrm{Height}\right)}^{2}=100-64\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\mathrm{Height}\right)}^{2}=36\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Height}=6$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \left(\mathrm{base}\right)\times \left(\mathrm{height}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 8\times 6\phantom{\rule{0ex}{0ex}}=4\times 6\phantom{\rule{0ex}{0ex}}=24{\mathrm{cm}}^{2}$

Hence, the area is $\overline{)24{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 9:

Each side of a triangle is multiplied by with the sum of the squares of the other two sides. The sum of all such possible results is 6 times the product of sides. The triangle must be _________.

#### Answer:

Let the sides of the triangle be *a*, *b* and *c*.

According to the question,

$a\left({b}^{2}+{c}^{2}\right)+b\left({a}^{2}+{c}^{2}\right)+c\left({b}^{2}+{a}^{2}\right)=6abc\phantom{\rule{0ex}{0ex}}\Rightarrow a{b}^{2}+a{c}^{2}+b{a}^{2}+b{c}^{2}+c{b}^{2}+c{a}^{2}=6abc\phantom{\rule{0ex}{0ex}}\Rightarrow a{b}^{2}+a{c}^{2}+b{a}^{2}+b{c}^{2}+c{b}^{2}+c{a}^{2}-6abc=0\phantom{\rule{0ex}{0ex}}\Rightarrow a{b}^{2}+a{c}^{2}-2abc+b{a}^{2}+b{c}^{2}-2abc+c{b}^{2}+c{a}^{2}-2abc=0\phantom{\rule{0ex}{0ex}}\Rightarrow a\left({b}^{2}+{c}^{2}-2bc\right)+b\left({a}^{2}+{c}^{2}-2ac\right)+c\left({b}^{2}+{a}^{2}-2ab\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow a{\left(b-c\right)}^{2}+b{\left(a-c\right)}^{2}+c{\left(b-a\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(b-c\right)}^{2}=0\mathrm{and}{\left(a-c\right)}^{2}=0\mathrm{and}{\left(b-a\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(b-c\right)=0\mathrm{and}\left(a-c\right)=0\mathrm{and}\left(b-a\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow b=c\mathrm{and}a=c\mathrm{and}b=a\phantom{\rule{0ex}{0ex}}\Rightarrow a=b=c$

Hence, the triangle must be __equilateral__.

#### Page No 17.26:

#### Question 10:

In a scalene triangle, one side exceeds the other two sides by 4 cm and 5 cm respectively and the perimeter of the triangle is 36 cm. The area of the triangle is _________.

#### Answer:

Given: The perimeter of the triangle is 36 cm.

Let the sides of the triangle be *a*, *b* and *c*.

According to the question,

*b = a* − 4 and *c = a* − 5

Perimeter of the triangle =* a + b + c*

$\Rightarrow 36=a+a-4+a-5\phantom{\rule{0ex}{0ex}}\Rightarrow 36=3a-9\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=36+9\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=45\phantom{\rule{0ex}{0ex}}\Rightarrow a=15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow b=a-4=15-4=11\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow c=a-5=15-5=10\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{sides}\mathrm{are}15\mathrm{cm},11\mathrm{cm}\mathrm{and}10\mathrm{cm}.$

Now, using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{15+11+10}{2}=\frac{36}{2}=18$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{18\left(18-15\right)\left(18-11\right)\left(18-10\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{18\left(3\right)\left(7\right)\left(8\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(3\times 3\times 2\right)\left(3\right)\left(7\right)\left(2\times 2\times 2\right)}\phantom{\rule{0ex}{0ex}}=\left(3\times 2\times 2\right)\sqrt{21}\phantom{\rule{0ex}{0ex}}=12\sqrt{21}{\mathrm{cm}}^{2}$

Hence, the area of the triangle is $\overline{)12\sqrt{21}{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 11:

Among an equilateral triangle, an isosceles triangle and a scalene triangle, ________ has the maximum area if the perimeter of each triangle is same.

#### Answer:

We know, when the triangles have same perimeter, then the equilateral triangle have the greatest area.

Hence, among an equilateral triangle, an isosceles triangle and a scalene triangle, __an equilateral triangle__ has the maximum area if the perimeter of each triangle is same.

#### Page No 17.26:

#### Question 12:

Area of an isosceles triangle, one of whose equal side is 5 units and base 6 units is __________.

#### Answer:

Given:

Base of isosceles triangle is 6 units.

equal sides of isosceles triangle is 5 units.

Now, using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{6+5+5}{2}=\frac{16}{2}=8$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{8\left(8-5\right)\left(8-5\right)\left(8-6\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{8\left(3\right)\left(3\right)\left(2\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(2\times 2\times 2\right)\left(3\right)\left(3\right)\left(2\right)}\phantom{\rule{0ex}{0ex}}=\left(3\times 2\times 2\right)\phantom{\rule{0ex}{0ex}}=12{\mathrm{cm}}^{2}$

Hence, the area of the triangle is $\overline{)12{\mathrm{cm}}^{2}}$.

#### Page No 17.26:

#### Question 13:

The sides of a scalene triangle are 11 cm, 12 cm and 13 cm. The length of the altitude corresponding to the side having length 12 cm is ________.

#### Answer:

Given:

The sides of a triangle are 11 cm, 12 cm and 13 cm respectively.

Let *ABC* be a triangle with sides

*AB* = 11 cm

*BC* = 12 cm

*AC* = 13 cm

Let *AD* be the altitude corresponding to the side having length 12 cm.

Using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{11+12+13}{2}=\frac{36}{2}=18$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{18\left(18-11\right)\left(18-12\right)\left(18-13\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{18\left(7\right)\left(6\right)\left(5\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(2\times 3\times 3\right)\left(7\right)\left(3\times 2\right)\left(5\right)}\phantom{\rule{0ex}{0ex}}=\left(2\times 3\right)\sqrt{105}\phantom{\rule{0ex}{0ex}}=6\sqrt{105}{\mathrm{cm}}^{2}$

We know,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}\Rightarrow 6\sqrt{105}=\frac{1}{2}\times 12\times AD\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{105}=AD\phantom{\rule{0ex}{0ex}}\Rightarrow AD=\sqrt{105}\mathrm{cm}$

Hence, the length of the altitude corresponding to the side having length 12 cm is $\overline{)\sqrt{105}\mathrm{cm}}$.

#### Page No 17.26:

#### Question 14:

A ground is in the form of a triangle having sides 51 m, 37 m and 20 m. The cost of levelling the ground at the rate of ₹ 3 per m^{2} is __________.

#### Answer:

Given:

A ground is in the form of a triangle having sides 51 m, 37 m and 20 m.

The cost of levelling the ground per m^{2} is ₹ 3.

Using Heron's formula:

If *a*, *b *and *c* are three sides of a triangle, then

area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.

Here, $s=\frac{51+37+20}{2}=\frac{108}{2}=54$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{54\left(54-51\right)\left(54-37\right)\left(54-20\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{54\left(3\right)\left(17\right)\left(34\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(2\times 3\times 3\times 3\right)\left(3\right)\left(17\right)\left(2\times 17\right)}\phantom{\rule{0ex}{0ex}}=\left(3\times 3\times 2\times 17\right)\phantom{\rule{0ex}{0ex}}=306{\mathrm{m}}^{2}$

The cost of painting per m^{2} = ₹ 3

The cost of painting 306 m^{2} = ₹ 306 × 3

= ₹ 918

Hence, the cost of levelling the ground at the rate of ₹ 3 per m^{2} is __₹ 918__.

#### Page No 17.26:

#### Question 15:

If each side of a triangle is doubled, then its area is __________ times the area of the original triangle.

#### Answer:

Let the base of the original triangle be *a* units and height be *b* units.

Then, the area of original triangle is $\frac{1}{2}ab$ ...(1)

According to the question,

Each side of a triangle is doubled

Thus, base of the new triangle is 2*a* units and height is 2*b* units.

Area of new triangle = $\frac{1}{2}\times 2a\times 2b$

= $\frac{1}{2}\left(4ab\right)$

= $4\times \frac{1}{2}ab$

= 4 × area of original triangle (from (1))

Hence, if each side of a triangle is doubled, then its area is __4__ times the area of the original triangle.

#### Page No 17.27:

#### Question 16:

In a triangle, the sum of any two sides exceeds the third by 6 cm. The area of the triangle is __________.

#### Answer:

Given:

In a triangle, the sum of any two sides exceeds the third by 6 cm.

Let the sides of the triangle be *a*, *b *and *c.*

According to the question,

*$a+b-c=6...\left(1\right)\phantom{\rule{0ex}{0ex}}b+c-a=6...\left(2\right)\phantom{\rule{0ex}{0ex}}a+c-b=6...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}2a+2b+2c-a-b-c=18\phantom{\rule{0ex}{0ex}}\Rightarrow a+b+c=18...\left(4\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(1\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a+b+c-a-b+c=18-6\phantom{\rule{0ex}{0ex}}\Rightarrow 2c=12\phantom{\rule{0ex}{0ex}}\Rightarrow c=6...\left(5\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(2\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a+b+c-b-c+a=18-6\phantom{\rule{0ex}{0ex}}\Rightarrow 2a=12\phantom{\rule{0ex}{0ex}}\Rightarrow a=6...\left(6\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(3\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a+b+c-a-c+b=18-6\phantom{\rule{0ex}{0ex}}\Rightarrow 2b=12\phantom{\rule{0ex}{0ex}}\Rightarrow b=6...\left(7\right)$*

From (5), (6) and (7),

*a = b = c *= 6 cm

⇒ triangle is equilateral

$\mathrm{Area}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times {\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times 36\phantom{\rule{0ex}{0ex}}=9\sqrt{3}{\mathrm{cm}}^{2}$

Hence, the area of the triangle is $\overline{)9\sqrt{3}{\mathrm{cm}}^{2}}.$

#### Page No 17.27:

#### Question 17:

If the circumradius of a right triangle is 10 cm and one of the two perpendicular sides is 12 cm, then the area of the triangle is ________.

#### Answer:

Given:

The circumradius of a right triangle is 10 cm

One of the two perpendicular sides is 12 cm

We know, the circumradius of the right-angled triangle is half of its hypotenuse.

Thus,

$\mathrm{Hypotenuse}=2\times \mathrm{circumradius}\phantom{\rule{0ex}{0ex}}=2\times 10\phantom{\rule{0ex}{0ex}}=20\mathrm{cm}$

Using Pythagoras theorem,

${\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Perpendicular}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(20\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(12\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 400={\left(\mathrm{Base}\right)}^{2}+144\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\mathrm{Base}\right)}^{2}=400-144\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\mathrm{Base}\right)}^{2}=256\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Base}=16\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{right}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 16\times 12\phantom{\rule{0ex}{0ex}}=8\times 12\phantom{\rule{0ex}{0ex}}=96{\mathrm{cm}}^{2}$

Hence, the area of the triangle is $\overline{)96{\mathrm{cm}}^{2}}.$

#### Page No 17.27:

#### Question 1:

Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.

#### Answer:

Given, base = 5 cm; height = 4 cm

Area of the triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

#### Page No 17.27:

#### Question 2:

Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where

Therefore the area of a triangle, say having sides 3 cm, 4 cm and 5 cm is given by

*a* = 3 cm ; *b* = 4 cm ; *c* = 5 cm

Now, area

$=\sqrt{6(6-3)(6-4)(6-5)}\phantom{\rule{0ex}{0ex}}=\sqrt{6\times 3\times 2\times 1}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6c{m}^{2}$

#### Page No 17.27:

#### Question 3:

Find the area of an isosceles triangle having the base x cm and one side y cm.

#### Answer:

Let us assume triangle ABC be the given isosceles triangle having sides AB = AC and base BC. The area of a triangle ABC, say *A* having given sides AB and AC equals to *y *cm and given base BC equals to *x* cm is given by

Where,

Base = BC = *x* cm; Height = $\sqrt{{y}^{2}-\frac{{x}^{2}}{4}}$

$A=\frac{1}{2}\left(Base\times Height\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times x\left(\sqrt{{y}^{2}-\frac{{x}^{2}}{4}}\right)\phantom{\rule{0ex}{0ex}}=\frac{x}{2}\left(\sqrt{{y}^{2}-\frac{{x}^{2}}{4}}\right)$

#### Page No 17.27:

#### Question 4:

Find the area of an equilateral triangle having each side 4 cm.

#### Answer:

Area of an equilateral triangle having each side *a* cm is given by

Area of the given equilateral triangle having each equal side equal to 4 cm is given by

*a* = 4 cm

#### Page No 17.27:

#### Question 5:

Find the area of an equilateral triangle having each side* x* cm.

#### Answer:

Area of an equilateral triangle, say *A* having each side *a* cm is given by

Area of the given equilateral triangle having each equal side equal to x cm is given by

*a* = x cm

#### Page No 17.27:

#### Question 6:

The perimeter of a triangullar field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

, where,

* *

It is given the sides of a triangular field are in the ratio 3:4:5 and perimeter=144 m

Therefore, *a*: *b*: *c* = 3:4:5

We will assume the sides of triangular field as

* *

Substituting the value of *x *in, we get sides of the triangle as

Area of a triangular field, say *A *having sides *a*, *b *, *c *and *s *as semi-perimeter is given by

#### Page No 17.27:

#### Question 7:

Find the area of an equilateral triangle having altitude *h* cm.

#### Answer:

Altitude of a equilateral triangle, having side *a *is given by

Substituting the given value of altitude *h *cm, we get

Area of a equilateral triangle, say *A* having each side *a* cm is given by

Area of the given equilateral triangle having each equal side equal to is given by;* *

#### Page No 17.27:

#### Question 8:

Let Δ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.

#### Answer:

We are given assumed value is the area of a given triangle *ABC*

We assume the sides of the given triangle *ABC* be *a,* *b,* c

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

Where,

We take the sides of a new triangle as 2*a*, 2b, 2c that is twice the sides of previous one

Now, the area of a triangle having sides 2*a**,* 2*b**,* and 2c* *and as semi-perimeter is given by,

, where

Now,

#### Page No 17.27:

#### Question 9:

If each side of a triangle is doubled, the find percentage increase in its area.

#### Answer:

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

Where,

We take the sides of a new triangle as 2*a*, 2b, 2*c* that is twice the sides of previous one

Now, the area of a triangle having sides 2*a,* 2*b,* and 2*c *and as semi-perimeter is given by,

Where,

Now,

Therefore, increase in the area of the triangle

Percentage increase in area

#### Page No 17.27:

#### Question 10:

If each side of a equilateral triangle is tripled then what is the percentage increase in the area of the triangle?

#### Answer:

Area of an equilateral triangle having each side *a* cm is given by

Now, Area of an equilateral triangle, say if each side is tripled is given by

*a *= 3*a*

Therefore, increase in area of triangle

Percentage increase in area

#### Page No 17.8:

#### Question 1:

Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

#### Answer:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by *A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given:

*a* =150 cm

*b=*120 cm

*c *=200 cm

Here we will calculate *s,*

So the area of the triangle is:

#### Page No 17.8:

#### Question 2:

Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.

#### Answer:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by *A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given:

*a* = 9 cm, *b = *12 cm, *c *= 15 cm

Here we will calculate *s,*

So the area of the triangle is:

#### Page No 17.8:

#### Question 3:

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

#### Answer:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by *A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given:

*a* = 18 cm

*b = *10 cm, and perimeter = 42 cm

We know that perimeter = 2*s**, *

So 2*s** = *42

Therefore *s *= 21 cm

We know that, so

So the area of the triangle is:

#### Page No 17.8:

#### Question 4:

In a Δ*ABC*, *AB* = 15 cm, *BC* = 13 cm and *AC* = 14 cm. Find the area of Δ*ABC* and hence its altitude on *AC*.

#### Answer:

If we denote area of the triangle by ‘Area’*,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given:

AB = 15 cm, BC *= *13 cm, AC = 14 cm

Here we will calculate *s,*

So the area of the triangle is:

Now draw the altitude from point B on AC which intersects it at point D.BD is the required altitude. So if you draw the figure, you will see,

Here . So,

#### Page No 17.8:

#### Question 5:

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle.

#### Answer:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by *A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given, and

Here,

Using these data we will find the sides of the triangle. Suppose the sides of the triangle are as follows,

Since, so

Now we know each side that is,

Now we know all the sides. So we can use Heron’s formula.

The area of the triangle is;

#### Page No 17.8:

#### Question 6:

The perimeter of a triangle is 300 m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.

#### Answer:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by *A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given, and

Here,

Using these data we will find the sides of the triangle. Suppose the sides of the triangle are as follows,

Since, so

Now we know each side that is,

Now we know all the sides. So we can use Heron’s formula.

The area of the triangle is;

$\mathrm{A}=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{150\left(150-60\right)\left(150-100\right)\left(150-140\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{150\left(90\right)\left(50\right)\left(10\right)}\phantom{\rule{0ex}{0ex}}=100\sqrt{15\times 9\times 5}\phantom{\rule{0ex}{0ex}}=100\sqrt{5\times 3\times 3\times 3\times 5}\phantom{\rule{0ex}{0ex}}=100\times 3\times 5\sqrt{3}\phantom{\rule{0ex}{0ex}}=\overline{)1500\sqrt{3}{\mathrm{m}}^{2}}$

#### Page No 17.8:

#### Question 7:

The perimeter of a triangular field is 240 dm. If two of its sides are 78 cm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

#### Answer:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by *A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where, * *

We are given two sides of the triangle and.

That is *a *= 78 dm, *b *= 50 dm

We will find third side *c* and then the area of the triangle using Heron’s formula.

Now,

Use Heron’s formula to find out the area of the triangle. That is* *

Consider the triangle ΔPQR in which

PQ=50 dm, PR=78 dm, QR=120 dm

Where RD is the desired perpendicular length

Now from the figure we have

#### Page No 17.8:

#### Question 8:

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.

#### Answer:

*A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given: *a *= 35 cm; *b *= 54 cm; *c *= 61 cm

The area of the triangle is:

Suppose the triangle is ΔPQR and focus on the triangle given below,

In which PD1, QD2 and RD3 are three altitudes

Where PQ=35 cm, QR=54 cm, PR=61 cm

We will calculate each altitude one by one to find the smallest one.

**Case 1 **

In case of ΔPQR:

**Case 2**

**Case 3**

The smallest altitude is QD2.

The smallest altitude is the one which is drawn on the side of length 61 cm from apposite vertex.

#### Page No 17.8:

#### Question 9:

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.

#### Answer:

*A,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

We are given, and

Here,

Using these data we will find the sides of the triangle. Suppose the sides of the triangle are as follows,

Since 2*s*=144, so

Now we know each side that is,

Now we know all the sides. So we can use Heron’s formula.

The area of the triangle is;

We are asked to fin out the height corresponding to the longest side of the given triangle. The longest side is *c *and supposes the corresponding height is H then,

#### Page No 17.8:

#### Question 10:

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.

#### Answer:

We are given that and its base is (3/2) times each of the equal sides. We are asked to find out the length of each side, area of the triangle and height of the triangle. In this case ‘height’ is the perpendicular distance drawn on the base from the apposite vertex.

In the following triangle ΔABC

BC = *a,* AC = *b*, AB = *c* and AB = AC

Let the length of each of the equal sides be *x* and *a, b* and *c* are the side of the triangle. So,

Since .This implies that,

Therefore all the sides of the triangle are:

All the sides of the triangle are 18 cm, 12 cm, and 12 cm.

If we denote area of the triangle by *Area,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

To calculate area of the triangle we need to find *s:*

The area of the triangle is:

Now we will find out the height, say H. See the figure, in which AD = H

So,

#### Page No 17.8:

#### Question 11:

Find the area of the shaded region in the given figure.

#### Answer:

We are given the following figure with dimensions.

Figure:

Let the point at which angle is be D.

AC = 52 cm, BC = 48 cm, AD = 12 cm, BD = 16 cm

We are asked to find out the area of the shaded region.

Area of the shaded region=Area of triangle ΔABC−area of triangle ΔABD

In right angled triangle ABD, we have

Area of the triangle ΔABD is given by

If we denote area of the triangle by *Area,* then the area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by;

Where,

Here *a *= 48 cm,* b *=* 52 *cm, *c = *20 cm and

Therefore the area of a triangle ΔABC is given by,

Now we have all the information to calculate area of shaded region, so

Area of shaded region = Area of ΔABC − Area of ΔABD

The area of the shaded region is 384 cm^{2}.

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