Rs Aggarwal 2019 2020 for Class 9 Math Chapter 14 - Areas Of Triangles And Quadrilaterals

Answer:

We have:
Base = 24 cm
Height = 14.5 cm

Now,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times 24\times 14.5 \\ =174 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle = $\frac{\mathrm{Total} \mathrm{Cost}}{\mathrm{Rate}}=\frac{783}{58}=13.5 \mathrm{ha} =135000 {\mathrm{m}}^2$
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 135000 {\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Heigh}t =135000 \\ \Rightarrow \frac{1}{2}\times 3h\times h\mathit{ }=135000 \\ \Rightarrow h^2=\frac{135000\times 2}{3} \\ \Rightarrow h^2=90000 \\ \Rightarrow h\mathit{ }=300 \mathrm{m} \end{gathered}$$

Thus, we have:
Height = h = 300 m
Base = 3h = 900 m

Answer:

$$\begin{gathered} \mathrm{Let}: \\ a=42 \mathrm{cm}, b = 34 \mathrm{cm} \mathrm{and} c=20 \mathrm{cm} \\ \therefore s= \frac{a+b+c}{2}=\frac{42+34+20}{2}=48 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{48(48-42)(48-34)(48-20)} \\ =\sqrt{48\times 6\times 14\times 28} \\ =\sqrt{4\times 2\times 6\times 6\times 7\times 2\times 7\times 4} \\ =4\times 2\times 6\times 7 \\ =336 {\mathrm{cm}}^2 \end{gathered}$$

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 336 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 336 \\ \Rightarrow \mathrm{Height} = \frac{336\times 2}{42}=16 \mathrm{cm} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let}: \\ a=18 \mathrm{cm}, b = 24 \mathrm{cm} \mathrm{and} c=30 \mathrm{cm} \\ \therefore s= \frac{a+b+c}{2}=\frac{18+24+30}{2}=36 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{36(36-18)(36-24)(36-30)} \\ =\sqrt{36\times 18\times 12\times 6} \\ =\sqrt{12\times 3\times 6\times 3\times 12\times 6} \\ =12\times 3\times 6 \\ =216 {\mathrm{cm}}^2 \end{gathered}$$

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 216 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 216 \\ \Rightarrow \mathrm{Height} = \frac{216\times 2}{18}=24 \mathrm{cm} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let}: \\ a=91 \mathrm{m}, b = 98 \mathrm{m} \mathrm{and} c=105 \mathrm{m} \\ \therefore s= \frac{a+b+c}{2}=\frac{91+98+105}{2}=147 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{147(147-91)(147-98)(147-105)} \\ =\sqrt{147\times 56\times 49\times 42} \\ =\sqrt{7\times 3\times 7\times 2\times 2\times 2\times 7\times 7\times 7\times 7\times 3\times 2} \\ =7\times 7\times 7\times 2\times 3\times 2 \\ =4116 {\mathrm{m}}^2 \end{gathered}$$

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 4116 {\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 4116 \\ \Rightarrow \mathrm{Height} = \frac{4116\times 2}{105}=78.4 \mathrm{m} \end{gathered}$$

Answer:

Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5$\times$5 = 25 m
12$\times$5 = 60 m
13$\times$5 = 65 m

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=25 \mathrm{m}, b = 60 \mathrm{m} \mathrm{and} c=65 \mathrm{m} \\ \therefore s= \frac{150}{2}=75 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{75(75-25)(75-60)(75-65)} \\ =\sqrt{75\times 50\times 15\times 10} \\ =\sqrt{15\times 5\times 5\times 10\times 15\times 10} \\ =15\times 5\times 10 \\ =750 {\mathrm{m}}^2 \end{gathered}$$

Answer:

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25$\times$10 = 250 m
17$\times$10 = 170 m
12$\times$10 = 120 m

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=250 \mathrm{m}, b =170 \mathrm{m} \mathrm{and} c=120 \mathrm{m} \\ \therefore s= \frac{540}{2}=270 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{270(270-250)(270-170)(270-120)} \\ =\sqrt{270\times 20\times 100\times 150} \\ =\sqrt{30\times 3\times 3\times 20\times 20\times 5\times 30\times 5} \\ =30\times 3\times 20\times 5 \\ =9000 {\mathrm{m}}^2 \end{gathered}$$

Cost of ploughing 1 m² field = Rs 5
Cost of ploughing 9000 m² field = $5\times 9000=\mathrm{Rs} 45000$.

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Let}: \\ a=85 \mathrm{m} \mathrm{and} b = 154 \mathrm{m} \\ \mathrm{Given}: \\ \mathrm{Perimeter} = 324 \mathrm{m} \\ or, a+b+c =324 \\ \Rightarrow c=324-85-154=85 \mathrm{m} \\ \therefore s= \frac{324}{2}=162 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{162(162-85)(162-154)(162-85)} \\ =\sqrt{162\times 77\times 8\times 77} \\ =\sqrt{1296\times 77\times 77} \\ =\sqrt{36\times 77\times 77\times 36} \\ =36\times 77 \\ =2772 {\mathrm{m}}^2 \end{gathered}$$


(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 2772 {\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 2772 \\ \Rightarrow \mathrm{Height} = \frac{2772\times 2}{154}=36 \mathrm{m} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ a=13 \mathrm{cm} \mathrm{and} b=20 \mathrm{cm} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle}=\frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{20}{4}\times \sqrt{4(13{)}^2-{20}^2} \\ =5\times \sqrt{676-400} \\ =5\times \sqrt{276} \\ =5\times 16.6 \\ =83.06 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =360 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times QR\times PX = 360 \\ \Rightarrow h =\frac{720}{80}=9 \mathrm{cm} \\ \mathrm{Now},\mathit{ } \\ QX\mathit{ }= \frac{1}{2}\times 80 = 40 \mathrm{cm} \mathrm{and} PX = 9 \mathrm{cm} \end{gathered}$$
Also,
$$\begin{gathered} PQ=\sqrt{Q{X}^2+P{X}^2} \\ a=\sqrt{{40}^2+9^2}=\sqrt{1600+81}=\sqrt{1681}=41 \mathrm{cm} \end{gathered}$$

∴ Perimeter = 80 + 41 + 41  = 162 cm

Answer:

The ratio of the equal side to its base is 3 : 2.
$\Rightarrow$Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
$$\begin{gathered} \Rightarrow 3x+3x+2x=32 \mathrm{cm} \\ \Rightarrow 8x=32 \\ \Rightarrow x=4 \mathrm{cm} \end{gathered}$$
Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, $\mathrm{S}=\frac{1}{2}\left(12+12+8\right)=\frac{32}{2}=16$
Area of the triangle will be
$$\begin{gathered} =\sqrt{S\left(S-a\right)\left(S-b\right)\left(S-c\right)} \\ =\sqrt{16\left(16-12\right)\left(16-12\right)\left(16-8\right)} \\ =\sqrt{16\times 4\times 4\times 8} \\ =4\times 4\sqrt{8}=4\times 4\times 2\sqrt{2}=32\sqrt{2} {\mathrm{cm}}^2 \end{gathered}$$
Disclaimer: The answer does not match with the answer given in the book.

Answer:

Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.

Now,
x + x + 4 + 2x − 6 = 50        (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
⇒ x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{13+17+20}{2}=\frac{50}{2}=25 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{25\left(25-13\right)\left(25-17\right)\left(25-20\right)} \\ =\sqrt{25\left(12\right)\left(8\right)\left(5\right)} \\ =20\sqrt{30} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the triangle is $20\sqrt{30} {\mathrm{cm}}^2$.

Answer:

The sides of the triangle are of length 13 m, 14 m and 15 m.
∴ Semi-perimeter of the triangle is
$s=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{m}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\ =\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\ =84 {\mathrm{m}}^2 \end{gathered}$$

Now,
The rent of advertisements per m² per year = Rs 2000
The rent of the wall with area 84 m² per year = Rs 2000 × 84
                                                                         = Rs 168000
The rent of the wall with area 84 m² for 6 months = Rs $\frac{168000}{2}$
                                                                                = Rs 84000


Hence, the rent paid by the company is Rs 84000.

Answer:

Let the equal sides of the isosceles triangle be a cm each.
∴ Base of the triangle, b = $\frac{3}{2}$a cm
(i) Perimeter = 42 cm
or, a + a + $\frac{3}{2}$a = 42
or, 2a +$\frac{3}{2}$a= 42

$$\begin{gathered} \Rightarrow 2a+\frac{3}{2}a = 42 \\ \Rightarrow \frac{7a}{2}=42 \\ \Rightarrow a=12 \end{gathered}$$
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = $\frac{3}{2}$a = $\frac{3}{2}\times 12=18 \mathrm{cm}$
(ii)
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{18}{4}\times \sqrt{4(12{)}^2-{18}^2} (a= 12 \mathrm{cm} \mathrm{and} b=18 \mathrm{cm}) \\ =4.5\times \sqrt{576-324} \\ =4.5\times \sqrt{252} \\ =4.5\times 15.87 \\ =71.42 {\mathrm{cm}}^2 \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =71.42 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 71.42 \\ \Rightarrow \mathrm{Height} = \frac{71.42\times 2}{18}=7.94 \mathrm{cm} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 =36\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow (\mathrm{Side}{)}^2=144 \\ \Rightarrow \mathrm{Side}=12 \mathrm{cm} \end{gathered}$$

Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm

Answer:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 =81\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow (\mathrm{Side}{)}^2=324 \\ \Rightarrow \mathrm{Side}=18 \mathrm{cm} \end{gathered}$$

Now, we have:
$$\begin{gathered} \mathrm{Height} =\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ =\frac{\sqrt{3}}{2}\times 18 \\ =9\sqrt{3} \mathrm{cm} \end{gathered}$$

Answer:

Side of the equilateral triangle = 8 cm

(i)
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (8{)}^2 \\ =\frac{1.732\times 64}{4} \\ =27.71 {\mathrm{cm}}^2 \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Height} =\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ =\frac{\sqrt{3}}{2}\times 8 \\ =\frac{1.732\times 8}{2} \\ =6.93 \mathrm{cm} \end{gathered}$$

Answer:

Height of the equilateral triangle = 9 cm
Thus, we have:
$$\begin{gathered} \mathrm{Height} =\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow 9=\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow \mathrm{Side} = \frac{18}{\sqrt{3}}=\frac{18}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=6\sqrt{3 }\mathrm{cm} \end{gathered}$$

Also,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (6\sqrt{3}{)}^2 \\ =\frac{108}{4}\sqrt{3} \\ =27\sqrt{3} \\ =46.76 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,
$$\begin{gathered} PQ=\sqrt{P{R}^2-Q{R}^2} \\ =\sqrt{{50}^2-{48}^2} \\ =\sqrt{2500-2304} \\ =\sqrt{196} \\ =14 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =\frac{1}{2}\times QR\times PQ \\ =\frac{1}{2}\times 48\times 14 \\ =336 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

In right angled ∆ABD,
AB² = AD² + DB²          (Pythagoras Theorem)
⇒ AB² = 12² + 16²
⇒ AB² = 144 + 256
⇒ AB² = 400
⇒ AB = 20 cm

Area of ∆ADB = $\frac{1}{2}\times DB\times AD$
                        = $\frac{1}{2}\times 16\times 12$
                        = 96 cm²               ....(1)

In ∆ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+52+48}{2}=\frac{120}{2}=60 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACB=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{60\left(60-20\right)\left(60-52\right)\left(60-48\right)} \\ =\sqrt{60\left(40\right)\left(8\right)\left(12\right)} \\ =480 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$


Now,
Area of the shaded region = Area of ∆ACB − Area of ∆ADB
                                          = 480 − 96
                                          = 384 cm²

Hence, the area of the shaded region in the given figure is 384 cm².

Answer:

In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle.

 

Join AC.

In right angled ∆ABC,
AC² = AB² + BC²          (Pythagoras Theorem)
⇒ AC² = 6² + 8²
⇒ AC² = 36 + 64
⇒ AC² = 100
⇒ AC = 10 cm

Area of ∆ABC = $\frac{1}{2}\times AB\times BC$
                        = $\frac{1}{2}\times 6\times 8$
                        = 24 cm²               ....(1)

In ∆ACD,
The sides of the triangle are of length 10 cm, 12 cm and 14 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+12+14}{2}=\frac{36}{2}=18 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{18\left(18-10\right)\left(18-12\right)\left(18-14\right)} \\ =\sqrt{18\left(8\right)\left(6\right)\left(4\right)} \\ =24\sqrt{6} {\mathrm{cm}}^2 \\ =24\left(2.45\right) {\mathrm{cm}}^2 \\ =58.8 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
                                             = (24 + 58.8) cm²
                                             = 82.8 cm²

Hence, the area of quadrilateral ABCD is 82.8 cm².

Answer:

We know that $△$ABD is a right-angled triangle.
∴ $A{B}^2=\sqrt{A{D}^2-D{B}^2}=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{cm}$
 $$\begin{gathered} \mathrm{Now,} \\ Area \mathrm{of} \mathrm{triangle} ABD=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ = \frac{1}{2}\times AB\times BD \\ = \frac{1}{2}\times 8\times 15 \\ =60 {\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Let}: \\ a=9 \mathrm{cm}, b = 15 \mathrm{cm} \mathrm{and} c=12 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{9+15+12}{2}=18 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} DBC = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{18(18-9)(18-15)(18-12)} \\ =\sqrt{18\times 9\times 3\times 6} \\ =\sqrt{6\times 3\times 3\times 3\times 3\times 6} \\ =6\times 3\times 3 \\ =54 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of quadrilateral ABCD  = Area of $△$ABD + Area of $△$BCD
                                           = (60 + 54) cm² =114 cm²
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm

Answer:

In right angled ∆ABC,
BC² = AB² + AC²          (Pythagoras Theorem)
⇒ BC² = 21² + 20²
⇒ BC² = 441 + 400
⇒ BC² = 841
⇒ BC = 29 cm

Area of ∆ABC = $\frac{1}{2}\times AB\times AC$
                        = $\frac{1}{2}\times 21\times 20$
                        = 210 cm²               ....(1)

In ∆ACD,
The sides of the triangle are of length 20 cm, 34 cm and 42 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+34+42}{2}=\frac{96}{2}=48 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{48\left(48-20\right)\left(48-34\right)\left(48-42\right)} \\ =\sqrt{48\left(28\right)\left(14\right)\left(6\right)} \\ =336 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
                                             = (210 + 336) cm²
                                             = 546 cm²

Also,
Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm
                                                     = 126 cm

Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 cm², respectively.

Answer:

We know that $△$BAD is a right-angled triangle.

∴ $AB=\sqrt{B{D}^2-A{D}^2}=\sqrt{{26}^2-{24}^2}=\sqrt{676-576}=\sqrt{100}=10 \mathrm{cm}$

 $$\begin{gathered} \mathrm{Now,} \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} BAD=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ = \frac{1}{2}\times AB\times AD \\ = \frac{1}{2}\times 10\times 24 \\ =120 {\mathrm{cm}}^2 \end{gathered}$$

Also, we know that $△$BDC is an equilateral triangle.
$$\begin{gathered} \therefore \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (26{)}^2 \\ =\frac{\sqrt{3}}{4}\times 676 \\ =169\sqrt{3} \\ =292.37 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of quadrilateral ABCD = Area of $△$ABD + Area of $△$BDC
                                         = (120 + 292.37) cm² = 412.37 cm²
Perimeter of ABCD = AB + BC + CD + DA = 10 + 26+ 26 + 24 = 86 cm

Answer:

$$\begin{gathered} \mathrm{Let}: \\ a=26 \mathrm{cm}, b =30 \mathrm{cm} \mathrm{and} c=28 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{26+30+28}{2}=42 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} ABC = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{42(42-26)(42-30)(42-28)} \\ =\sqrt{42\times 16\times 12\times 14} \\ =\sqrt{14\times 3\times 4\times 4\times 2\times 2\times 3\times 14} \\ =14\times 4\times 2\times 3 \\ =336 {\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 336=672 {\mathrm{cm}}^2$

Answer:

$$\begin{gathered} \mathrm{Let}: \\ a=10 \mathrm{cm}, b =16 \mathrm{cm} \mathrm{and} c=14 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{10+16+14}{2}=20 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} ABC = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{20(20-10)(20-16)(20-14)} \\ =\sqrt{20\times 10\times 4\times 6} \\ =\sqrt{10\times 2\times 10\times 2\times 2\times 3\times 2} \\ =10\times 2\times 2\sqrt{3} \\ =69.2 {\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 69.2 {\mathrm{cm}}^2=138.4 {\mathrm{cm}}^2$

Answer:

$$\begin{gathered} \mathrm{Area} \mathrm{of} ABCD=\mathrm{Area} \mathrm{of} △ABD+\mathrm{Area} \mathrm{of} △BDC \\ =\frac{1}{2}\times BD\times AL+\frac{1}{2}\times BD\times CM \\ =\frac{1}{2}\times BD(AL+CM) \\ =\frac{1}{2}\times 64(16.8+13.2) \\ =32\times 30 \\ =960 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

In the given figure, ABCD is a trapezium with parallel sides AB and CD.



Let the length of CD be x.
Then, the length of AB be x + 4.

Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
$$\begin{gathered} \Rightarrow 475=\frac{1}{2}\times \left(x+x+4\right)\times 19 \\ \Rightarrow 475\times 2=19\left(2x+4\right) \\ \Rightarrow 950=38x+76 \\ \Rightarrow 38x=950-76 \\ \Rightarrow 38x=874 \\ \Rightarrow x=\frac{874}{38} \\ \Rightarrow x=23 \end{gathered}$$

∴ The length of CD is 23 cm and the length of AB is 27 cm.

Hence, the lengths of two parallel sides is 23 cm and 27 cm.                                                                                                  

Answer:

In ∆ABC,
The sides of the triangle are of length 7.5 cm, 6.5 cm and 7 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{10.5\left(10.5-7.5\right)\left(10.5-6.5\right)\left(10.5-7\right)} \\ =\sqrt{10.5\left(3\right)\left(4\right)\left(3.5\right)} \\ =21 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$

Now,
Area of parallelogram DBCE = Area of ∆ABC
                                               = 21 cm²

Also,
Area of parallelogram DBCE = base × height
$$\begin{gathered} \Rightarrow 21=BC\times DL \\ \Rightarrow 21=7\times DL \\ \Rightarrow DL=\frac{21}{7}=3 \mathrm{cm} \end{gathered}$$

Hence, the height DL of the parallelogram is 3 cm.

Answer:

In the given figure, ABCD is a trapezium having parallel sides 90 m and 30 m.



Draw DE perpendicular to AB, such that DE = BC.

In right angled ∆ADE,
AD² = AE² + ED²          (Pythagoras Theorem)
⇒ 100² = (90 − 30)² + ED²
⇒ 10000 = 3600 + ED²
⇒ ED² = 10000 − 3600
⇒ ED² = 6400
⇒ ED = 80 m

Thus, the height of the trapezium = 80 m        ...(1)

Now,
Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
                              = $\frac{1}{2}\times \left(90+30\right)\times 80$
                              = 4800 m²

The cost to plough per m² = Rs 5
The cost to plough 4800 m² = Rs 5 × 4800
                                             = Rs 24000

Hence, the total cost of ploughing the field is Rs 24000.

Answer:

Let ABCD be a rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front.



According to the laws, the length of the inner rectangle = 40 − 3 − 3 = 34 m and the breath of the inner rectangle = 15 − 2 − 2 = 11 m.

∴ Area of the inner rectangle PQRS = Length × Breath
                                                          = 34 × 11
                                                          = 374 m²

Hence, the largest area where house can be constructed is 374 m².

Answer:

Let the sides of rhombus be of length x cm.



Perimeter of rhombus = 4x
⇒ 40 = 4x
⇒ x = 10 cm

Now,
In ∆ABC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+10+12}{2}=\frac{32}{2}=16 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)} \\ =\sqrt{16\left(6\right)\left(6\right)\left(4\right)} \\ =48 {\mathrm{cm}}^2 ...\left(1\right) \end{gathered}$$

In ∆ADC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+10+12}{2}=\frac{32}{2}=16 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ADC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)} \\ =\sqrt{16\left(6\right)\left(6\right)\left(4\right)} \\ =48 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC
                                      = 48 + 48
                                      = 96 cm²

The cost to paint per cm² = Rs 5
The cost to paint 96 cm² = Rs 5 × 96
                                        = Rs 480
The cost to paint both sides of the sheet = Rs 2 × 480
                                                                = Rs 960

Hence, the total cost of painting is Rs 960.

Answer:

Let the semi-perimeter of the triangle be s.
Let the sides of the triangle be a, b and c.
Given: s − a = 8, s − b = 7 and s − c = 5      ....(1)

Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20
⇒ 3s − 2s = 20                       $\left(\because s=\frac{a+b+c}{2}\right)$
⇒ s = 20 cm                  ...(2)

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{20\left(8\right)\left(7\right)\left(5\right)} \left(\mathrm{from} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =20\sqrt{14} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the triangle is $20\sqrt{14} {\mathrm{cm}}^2$.

Answer:

Area of one triangular-shaped tile can be found in the following manner:

$$\begin{gathered} \mathrm{Let}: \\ a=16 \mathrm{cm}, b = 12 \mathrm{cm} \mathrm{and} c=20 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{16+12+20}{2}=24 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{24(24-16)(24-12)(24-20)} \\ =\sqrt{24\times 8\times 12\times 4} \\ =\sqrt{6\times 4\times 4\times 4\times 4\times 6} \\ =6\times 4\times 4 \\ =96 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of 16 triangular-shaped tiles = $16\times 96=1536 {\mathrm{cm}}^2$
Cost of polishing tiles of area 1 cm² = Rs 1
Cost of polishing tiles of area 1536 cm² = $1\times 1536 =\mathrm{Rs} 1536$

Answer:

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{20}{4}\times \sqrt{4(50{)}^2-{20}^2} (a= 50 \mathrm{cm} \mathrm{and} b=20 \mathrm{cm}) \\ =5\times \sqrt{10000-400} \\ =5\times \sqrt{9600} \\ =5\times 40\sqrt{6} \\ =200\sqrt{6} \\ =490 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of 1 triangular piece of cloth = 490 cm²
Area of 12 triangular pieces of cloth = $12\times 490 =5880 {\mathrm{cm}}^2$

Answer:

In the given figure, ABCD is a square with diagonal 44 cm.
∴ AB = BC = CD = DA.        ....(1)

In right angled ∆ABC,
AC² = AB² + BC²          (Pythagoras Theorem)
⇒ 44² = 2AB²
⇒ 1936 = 2AB²
⇒ AB² = $\frac{1936}{2}$
⇒ AB² = 968
⇒ AB = $22\sqrt{2}$ cm      ...(2)

∴ Sides of square = AB = BC = CD = DA = $22\sqrt{2}$ cm

Area of square ABCD = (side)²
                                    = ($22\sqrt{2}$)²
                                    = 968 cm²          ...(3)

Area of red portion = $\frac{968}{4}=242 {\mathrm{cm}}^2$
Area of yellow portion = $\frac{968}{2}=484 {\mathrm{cm}}^2$
Area of green portion = $\frac{968}{4}=242 {\mathrm{cm}}^2$

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+14}{2}=\frac{54}{2}=27 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆AEF=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{27\left(27-20\right)\left(27-20\right)\left(27-14\right)} \\ =\sqrt{27\left(7\right)\left(7\right)\left(13\right)} \\ =21\sqrt{39} \\ =131.04 {\mathrm{cm}}^2 ...\left(4\right) \end{gathered}$$

Total area of the green portion = 242 + 131.04 = 373.04 cm²

Hence, the paper required of each shade to make a kite is red paper 242 cm², yellow paper 484 cm² and green paper 373.04 cm².

Answer:

Area of rectangle ABCD = Length × Breath
                                        = 75 × 4
                                        = 300 m²

Area of rectangle PQRS = Length × Breath
                                       = 60 × 4
                                       = 240 m²

Area of square EFGH = (side)²
                                    = (4)²
                                    = 16 m²

∴ Area of the footpath = Area of rectangle ABCD + Area of rectangle PQRS − Area of square EFGH
                                     = 300 + 240 − 16
                                     = 524 m²

The cost of gravelling the road per m² = Rs 50
The cost of gravelling the roads 524 m² = Rs 50 × 524
                                                                = Rs 26200

Hence, the total cost of gravelling the roads at Rs 50 per m² is Rs 26200.

Answer:

The top and the bottom of the canal are parallel to each other.
Let the height of the trapezium be h.

Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
⇒ 640 = $\frac{1}{2}\times \left(10+6\right)\times h$
⇒ 640 = $8\times h$
⇒ h = $\frac{640}{8}$
⇒ h = 80 m

Hence, the depth of the canal is 80 m.

Answer:

In the given figure, ABCD is the trapezium.

 

Draw a line BE parallel to AD.

In ∆BCE,
The sides of the triangle are of length 15 m, 13 m and 14 m.
∴ Semi-perimeter of the triangle is
$s=\frac{15+13+14}{2}=\frac{42}{2}=21 \mathrm{m}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆BCE=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-15\right)\left(21-13\right)\left(21-14\right)} \\ =\sqrt{21\left(6\right)\left(8\right)\left(7\right)} \\ =84 {\mathrm{m}}^2 ...\left(1\right) \end{gathered}$$

Also,
Area of ∆BCE = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$
$$\begin{gathered} \Rightarrow 84=\frac{1}{2}\times 14\times \mathrm{Height} \\ \Rightarrow 84=7\times \mathrm{Height} \\ \Rightarrow \mathrm{Height}=\frac{84}{7} \\ \Rightarrow \mathrm{Height}=12 \mathrm{m} \end{gathered}$$

∴ Height of ∆BCE = Height of the parallelogram ABED = 12 m

Now,
Area of the parallelogram ABED = Base × Height
                                                     = 11 × 12
                                                     = 132 m²                     ...(2)

∴ Area of the trapezium = Area of the parallelogram ABED + Area of the triangle BCE
                                        = 132 + 84
                                        = 216 m²

Hence, the area of a trapezium is 216 m².

Answer:

Let the length of the parallel sides be x and x − 8.
The height of the trapezium = 24 cm

Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
⇒ 312 = $\frac{1}{2}\times \left(x+x-8\right)\times 24$
⇒ 312 = 12(2x − 8)
⇒ 2x − 8 = $\frac{312}{12}$
⇒ 2x − 8 = 26
⇒ 2x = 26 + 8
⇒ 2x = 34
⇒ x = 17 cm

Hence, the lengths of the parallel sides are 17 cm and 9 cm.

Answer:

Diagonals d₁ and d₂ of the rhombus measure 120 m and 44 m, respectively.

Base of the parallelogram = 66 m

Now,
Area of the rhombus = Area of the parallelogram
$$\begin{gathered} \Rightarrow \frac{1}{2}\times d_1\times d_2=\mathrm{Base}\times \mathrm{Height} \\ \Rightarrow \frac{1}{2}\times 120\times 44=66\times \mathrm{Height} \\ \Rightarrow 60\times 44=66\times \mathrm{Height} \\ \Rightarrow 2640=66\times \mathrm{Height} \\ \Rightarrow \mathrm{Height}=\frac{2640}{66} \\ \Rightarrow \mathrm{Height}=40 \mathrm{m} \end{gathered}$$

Hence, the measure of the altitude of the parallelogram is 40 m.

Answer:

It is given that,
Sides of the square = 40 m
Altitude of the parallelogram = 25 m

Now,
Area of the parallelogram = Area of the square
$$\begin{gathered} \Rightarrow \mathrm{Base}\times \mathrm{Height}={\left(\mathrm{side}\right)}^2 \\ \Rightarrow \mathrm{Base}\times 25={\left(40\right)}^2 \\ \Rightarrow \mathrm{Base}\times 25=1600 \\ \Rightarrow \mathrm{Base}=\frac{1600}{25} \\ \Rightarrow \mathrm{Base}=64 \mathrm{m} \end{gathered}$$

Hence, the length of the corresponding base of the parallelogram is 64 m.

Answer:

It is given that,
The sides of rhombus = 20 cm.
One of the diagonal = 24 cm.

 

In ∆ABC,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+24}{2}=\frac{64}{2}=32 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{32\left(32-20\right)\left(32-20\right)\left(32-24\right)} \\ =\sqrt{32\left(12\right)\left(12\right)\left(8\right)} \\ =192 {\mathrm{cm}}^2 ...\left(1\right) \end{gathered}$$

In ∆ACD,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+24}{2}=\frac{64}{2}=32 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{32\left(32-20\right)\left(32-20\right)\left(32-24\right)} \\ =\sqrt{32\left(12\right)\left(12\right)\left(8\right)} \\ =192 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ACD
                                      = 192 + 192
                                      = 384 cm²

Hence, the area of a rhombus is 384 cm².

Answer:

It is given that,
Area of rhombus = 480 cm².
One of the diagonal = 48 cm.

(i) Area of the rhombus = $\frac{1}{2}\times d_1\times d_2$
$$\begin{gathered} \Rightarrow 480=\frac{1}{2}\times 48\times d_2 \\ \Rightarrow 480=24\times d_2 \\ \Rightarrow d_2=\frac{480}{24} \\ \Rightarrow d_2=20 \mathrm{cm} \end{gathered}$$

Hence, the length of the other diagonal is 20 cm.

(ii) We know that the diagonals of the rhombus bisect each other at right angles.

 

In right angled ∆ABO,
AB² = AO² + OB²          (Pythagoras Theorem)
⇒ AB² = 24² + 10²
⇒ AB² = 576 + 100
⇒ AB² = 676
⇒ AB = 26 cm

Hence, the length of each of the sides of the rhombus is 26 cm.

(iii) Perimeter of the rhombus = 4 × side
                                                = 4 × 26
                                                = 104 cm

Hence, the perimeter of the rhombus is 104 cm.

Answer:

(b) 30 cm²

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ \mathrm{Area} \mathrm{of}\mathit{ }\mathit{∆}ABC=\frac{1}{2}\times 12\times 5=30 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

(a) 96 cm²

$$\begin{gathered} \mathrm{Let}: \\ a=20 \mathrm{cm}, b = 16 \mathrm{cm} \mathrm{and} c=12 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{20+16+12}{2}=24 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{24(24-20)(24-16)(24-12)} \\ =\sqrt{24\times 4\times 8\times 12} \\ =\sqrt{6\times 4\times 4\times 4\times 4\times 6} \\ =6\times 4\times 4 \\ =96 {\mathrm{cm}}^2 \end{gathered}$$