RS Aggarwal 2019 2020 Solutions for Class 8 Maths Chapter 17 - Construction Of Quadrilaterals

Question 1:

Question 2:

Steps of construction:
Step 1: Draw $\mathrm{AB}=4.2 \mathrm{cm}$.
Step 2: With A as the centre and radius equal to $8 \mathrm{cm}$, draw an arc.
Step 3: With B as the centre and radius equal to $6 \mathrm{cm}$, draw another arc, cutting the previous arc at C.
Step 4: Join BC.
Step 5: With A as the centre and radius equal to $5 \mathrm{cm},$ draw an arc.
Step 6: With C as the centre and radius equal to $5.2 \mathrm{cm}$, draw another arc, cutting the previous arc at D.
Step 7: Join AD and CD.

Thus, ABCD is the required quadrilateral.

Question 3:

Steps of construction:
Step 1: Draw $\mathrm{PQ}=5.4 \mathrm{cm}$.
Step 2: With P as the centre and radius equal to $4 \mathrm{cm}$, draw an arc.
Step 3: With Q as the centre and radius equal to $4.6 \mathrm{cm}$, draw another arc, cutting the previous arc at R.
Step 4: Join QR.
Step 5: With P as the centre and radius equal to $3.5 \mathrm{cm},$ draw an arc.
Step 6: With R as the centre and radius equal to $4.3 \mathrm{cm}$, draw another arc, cutting the previous arc at S.
Step 7: Join PS and RS.

Thus, PQRS is the required quadrilateral.

Question 4:

Steps of construction:
Step 1: Draw $\mathrm{AB}=3.5 \mathrm{cm}$.
Step 2: With B as the centre and radius equal to $5.6 \mathrm{cm}$, draw an arc.
Step 3: With A as the centre and radius equal to $4.5 \mathrm{cm}$, draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With D as the centre and radius equal to $4.5 \mathrm{cm},$ draw an arc.
Step 6: With B as the centre and radius equal to $3.8 \mathrm{cm}$, draw another arc, cutting the previous arc at C.
Step 7: Join BC and CD.

Thus, ABCD is the required quadrilateral.

Question 5:

Steps of construction:
Step 1: Draw $\mathrm{AB}=3.6 \mathrm{cm}$.
Step 2: With B as the centre and radius equal to $4 \mathrm{cm}$, draw an arc.
Step 3: With A as the centre and radius equal to $2.7 \mathrm{cm}$, draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With A as the centre and radius equal to $4.6 \mathrm{cm},$ draw an arc.
Step 6: With B as the centre and radius equal to $3.3 \mathrm{cm}$, draw another arc, cutting the previous arc at C.
Step 7: Join AC, BC and CD.

Thus, ABCD is the required quadrilateral.

Question 6:

Steps of construction:
Step 1: Draw $QR=7.5 \mathrm{cm}.$
Step 2: With Q as the centre and radius equal to $10 \mathrm{cm}$, draw an arc.
Step 3: With R as the centre and radius equal to $5 \mathrm{cm}$, draw another arc, cutting the previous arc at S.
Step 4: Join QS and RS.
Step 5: With S as the centre and radius equal to $6 \mathrm{cm},$ draw an arc.
Step 6: With R as the centre and radius equal to $6 \mathrm{cm}$, draw another arc, cutting the previous arc at P.
Step 7: Join PS and PR.
Step 8: PQ = 4.9 cm
Thus, PQRS is the required quadrilateral.

Question 7:

Steps of construction:
Step 1: Draw $AB=3.4 \mathrm{cm}.$
Step 2: With B as the centre and radius equal to $4 \mathrm{cm}$, draw an arc.
Step 3: With A as the centre and radius equal to $5.7 \mathrm{cm}$, draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With A as the centre and radius equal to $8 \mathrm{cm},$ draw an arc.
Step 6: With D as the centre and radius equal to $3 \mathrm{cm}$, draw another arc, cutting the previous arc at C.
Step 7: Join AC, CD and BC.

Thus, ABCD is the required quadrilateral.

Question 8:

Steps of construction:
Step 1: Draw AB= $3.5 \mathrm{cm}$.
Step 2: Make $\angle ABC={120}^{\circ }$.
Step 3: With B as the centre, draw an arc $3.5 \mathrm{cm}$ and name that point C.
Step 4: With C as the centre, draw an arc $5.2 \mathrm{cm}$.
Step 5: With A as the centre, draw another arc â€‹$5.2 \mathrm{cm}$, cutting the previous arc at D.
Step 6: Join CD and AD.
Thus, $ABCD$ is the required quadrilateral.

Question 9:

Steps of construction:
Step 1: Draw AB= $2.9cm$
Step 2: Make $\angle A={70}^{\circ }$
Step 3: With A as the centre, draw an arc of $3.4cm$. Name that point as D.
Step 4: With D as the centre, draw an arc of $2.7cm$.
Step 5: With B as the centre, draw an arc of 3.2 cm, cutting the previous arc at C.
Step 6: Join CD and BC.
Then, $ABCD$ is the required quadrilateral.

Question 10:

Steps of construction:
Step 1: Draw BC= $5cm$
Step 2: Make $\angle B={125}^{\circ } and \angle C={60}^{\circ }$
Step 3: With B as the centre, draw an arc of $3.5 cm$. Name that point as A.
Step 4: With C as the centre, draw an arc of $4.6 cm$. Name that point as D.
Step 5: Join A and D.
Then, $ABCD$ is the required quadrilateral.

Question 11:

Steps of construction:
Step 1: Draw QR= $5.6 cm$
Step 2: Make $\angle Q={45}^{\circ } \mathrm{and} \angle R={90}^{\circ }$
Step 3: With Q as the centre, draw an arc of $6 cm$. Name that point as P.
Step 4: With R as the centre, draw an arc of $2.7cm$. Name that point as S.
Step 6: Join P and S.
Then, $PQRS$ is the required quadrilateral.

Question 12:

Steps of construction:
Step 1: Draw AB= $5.6 cm$
Step 2: Make $\angle A={50}^{\circ } and \angle B={105}^{\circ }$
Step 3: With B as the centre, draw an arc of $4cm$.
Step 3: Sum of all the angles of the quadrilateral is ${360}^{\circ }$.
$$\begin{gathered} \angle A+\angle B+\angle C+\angle D={360}^{\circ } \\ {50}^{\circ }+{105}^{\circ }+\angle C+{80}^{\circ }={360}^{\circ } \\ {235}^{\circ }+\angle C={360}^{\circ } \\ \angle C={360}^{\circ }-{235}^{\circ } \\ \angle C={125}^{\circ } \end{gathered}$$
Step 5: With C as the centre, make $\angle C \mathrm{equal} \mathrm{to} \angle {125}^{\circ }$.
Step 6: Join C and D.
Step 7: Measure $\angle D={80}^{\circ }$
Then, $ABCD$ is the required quadrilateral.

Question 13:

Steps of construction:
Step 1: Draw PQ= $5cm$
Step 2:
$$\begin{gathered} \angle P+\angle Q+\angle R+\angle S={360}^{\circ } \\ {100}^{\circ }+\angle Q+{100}^{\circ }+{75}^{\circ }={360}^{\circ } \\ {275}^{\circ }+\angle Q={360}^{\circ } \\ \angle Q={360}^{\circ }-{275}^{\circ } \\ \angle Q={85}^{\circ } \end{gathered}$$
Step 3: Make $\angle P={100}^{\circ } and \angle Q={85}^{\circ }$
Step 3: With Q as the centre, draw an arc of $6.5 cm$.
Step 4: Make $\angle R={100}^{\circ }$
Step 6: Join R and S.
Step 7: Measure $\angle S={75}^{\circ }$
Then, $PQRS$ is the required quadrilateral.

Question 1:

Steps of construction:
Step 1: Draw $AB=4cm$
Step 2: $Make \angle B={90}^{\circ }$
Step 3: $$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ {5}^2=4^2+B{C}^2 \\ 25-16=B{C}^2 \\ BC=3cm \end{gathered}$$
With B as the centre, draw an arc equal to 3 cm.
Step 4: Make $\angle C={90}^{\circ }$
Step 5: With A as the centre and radius equal to $5.5 cm$, draw an arc and name that point as D.
Then, $ABCD$ is the required quadrilateral.

Question 2:

Steps of construction:
Step 1: Draw AB = $5.2cm$
Step 2: With B as the centre, draw an arc of $4.7 cm$.
Step 3: With A as the centre, draw another arc of $7.6 cm$, cutting the previous arc at C.
Step 4: Join A and C.
Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with C as the centre, draw an arc of $5.2cm$.
Step 6: With A as the centre, draw another arc of $4.7 cm$, cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

Question 3:

Steps of construction:
Step 1: Draw AB= $4.3cm$
Step 2: With B as the centre, draw an arc of $6.8 cm$.
Step 3: With A as the centre, draw another arc of $4cm$, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: We know that the opposite sides of a parallelogram are equal.
            Thus, with D as the centre, draw an arc of $4.3cm$.
Step 6: With B as the centre, draw another arc of $4 cm$, cutting the previous arc at C.
Step 7: Join CD and BC.
​then, ABCD is the required parallelogram.

Question 4:

Steps of construction:
Step 1: Draw PQ= 4 cm
Step 2: Make $\angle PQR={60}^{\circ }$
Step 2: With Q as the centre, draw an arc of 6 cm and name that point as R.
Step 3: With R as the centre, draw an arc of 4 cm and name that point as S.
Step 4: Join SR and PS.
Then, PQRS is the required parallelogram.

Question 5:

Steps of construction:
Step 1: Draw BC= $5cm$
Step 2: Make an $\angle BCD={120}^{\circ }$
Step 2: With C as centre draw an arc of $4.8 \mathrm{cm}$, name that point as D
Step 3: With D as centre draw an arc $5\mathrm{cm}$, name that point as A
Step 4: With B as centre draw another arc $4.8 \mathrm{cm}$ cutting the previous arc at A.
Step 5: Join AD and AB
​then, ABCD is a required parallelogram.

Question 6:

We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw AB= $4.4cm$
Step 2: With A as the centre and radius $2.8cm$, draw an arc.
Step 3: With B as the centre and radius $3.5cm$, draw another arc, cutting the previous arc at point O.
Step 4: Join OA and OB.
Step 5: Produce OA to C, such that OC= AO. Produce OB to D, such that OB=OD.
Step 5: Join AD, BC, and CD.
Thus, ABCD is the required parallelogram. The other side is 4.5 cm in length.

Question 7:

Steps of construction:
Step 1: Draw AB= 6.5cm
Step 2: Draw a perpendicular at point A. Name that ray as AX. From point A, draw an arc of length 2.5 cm on the ray AX and name that point as L.
Step 3: On point L, make a perpendicular. Draw a straight line YZ passing through L, which is perpendicular to the ray AX.
Step 4: Cut an arc of length 3.4 cm on the line YZ and name it as C.
Step 5: From point C, cut an arc of length 6.5 cm on the line YZ. Name that point as D.
 Step 6: Join BC and AD.

Therefore, quadrilateral ABCD is a parallelogram.

The altitude from C measures 2.5 cm in length.

Question 8:

 We know that the diagonals of a parallelogram bisect each other.

Steps of construction:
Step 1: Draw AC= $3.8\mathrm{cm}$
Step 2: Bisect AC at O.
Step 3: Make $\angle COX={60}^{\circ }$ 
Produce XO to Y.
Step 4:
$$\begin{gathered} OB = \frac{1}{2}(4.6) \mathrm{cm} \\ OB=2.3 \mathrm{cm} \\ \mathrm{and} OD =\frac{1}{2}(4.6) \mathrm{cm} \\ OD=2.3 \mathrm{cm} \end{gathered}$$
Step 5: Join AB, BC, CD and AD.
​Thus, ABCD is the required parallelogram.
                 

Question 9:

Steps of construction:
Step 1: Draw AB = $11cm$ 
Step 2: Make $$\begin{gathered} \angle A={90}^{\circ } \\ \angle B={90}^{\circ } \end{gathered}$$
Step 3: Draw an arc of 8.5 cm from point A and name that point as D.
Step 4: Draw an arc of 8.5 cm from point B and name that point as C.
Step 5: Join C and D.
Thus, ABCD is the required rectangle.

Question 10:

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB = $6.4cm$ 
Step 2: Make $$\begin{gathered} \angle A={90}^{\circ } \\ \angle B={90}^{\circ } \end{gathered}$$
Step 3: Draw an arc of length 6.4 cm from point A and name that point as D.
Step 4: Draw an arc of length 6.4 cm from point B and name that point as C.
Step 5: Join C and D.
​Thus, ABCD is a required square.

Question 11:

We know that the diagonals of a square bisect each other at right angles.
Steps of construction:
Step 1: Draw AC= $5.8 \mathrm{cm}$
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:
: $$\begin{gathered} From O: \\ OB=\frac{1}{2}(5.8) \mathrm{cm} = 2.9 \mathrm{cm} \\ OD=\frac{1}{2}(5.8) \mathrm{cm}= 2.9 \mathrm{cm} \end{gathered}$$
Step 4: Join AB, BC, CD and DA.
ABCD is the required square.

Question 12:

Steps of construction:
Step 1: Draw QR = $3.6cm$ 
Step 2: Make  $$\begin{gathered} \angle Q={90}^{\circ } \\ \angle R={90}^{\circ } \end{gathered}$$
Step 3:
 $$\begin{gathered} P{R}^2=P{Q}^2+Q{R}^2 \\ {6}^2=P{Q}^2+3.6^2 \\ P{Q}^2=36-12.96 \\ P{Q}^2=23.04 \\ PQ=4.8 \mathrm{cm} \end{gathered}$$

Step 3: Draw an arc of length 4.8 cm from point Q and name that point as P.
​Step 4: Draw an arc of length 6 cm from point R, cutting the previous arc at P.
​Step 5: Join PQ
Step 6: Draw an arc of length 4.8 cm from point R.
From point P, draw an arc of length 3.6 cm, cutting the previous arc. Name that point as S.
Step 7: Join P and S.
Thus, PQRS is the required rectangle. The other side is 4.8 cm in length.

Question 13:

We know that the diagonals of a rhombus bisect each other.
.Steps of construction:
Step 1: Draw AC= $6cm$
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:
$$\begin{gathered} OB = \frac{1}{2}\left(8\right) \mathrm{cm} \\ OB=4\mathrm{cm} \\ \mathrm{and} OD =\frac{1}{2}\left(8\right) \mathrm{cm} \\ OD=4\mathrm{cm} \end{gathered}$$
Draw an arc of length 4 cm on OX and name that point as B.
Draw an arc of length 4 cm on OY and name that point as D.
Step 4 : Join AB, BC, CD and AD.
​Thus, ABCD is the required rhombus, as shown in the figure.

Question 14:

Steps of construction:
Step 1: Draw AB = $4cm$
Step 2: With B as the centre, draw an arc of $4 \mathrm{cm}$.
Step 3: With A as the centre, draw another arc of $6.5 \mathrm{cm}$, cutting the previous arc at C.
​Step 4: Join AC and BC.
Step 5: With C as the centre, draw an arc of 4 cm.
Step 6: â€‹With A as the centre, draw another arc of $4 \mathrm{cm}$, cutting the previous arc at D.
Step 7: Join AD and CD.
ABCD is the required rhombus.

Question 15:

Steps of construction:
Step1: Draw AB = $7.2 \mathrm{cm}$
Step2: Draw $$\begin{gathered} \angle ABY=60° \\ \angle BAX=120° \end{gathered}$$
Sum of the adjacent angles is 180°.
$$\begin{gathered} \angle BAX+\angle ABY=180° \\ =>\angle BAX=180°-60°=120° \end{gathered}$$
Step 3:  
$\mathrm{Set} \mathrm{off} AD (7.2 \mathrm{cm}) \mathrm{along} AX \mathrm{and} BC ( 7.2 \mathrm{cm}) \mathrm{along} BY.$
Step 4: Join C and D.
Then, ABCD is the required rhombus.

Question 16:

Steps of construction:
Step 1: Draw AB=$6 \mathrm{cm}$
Step 2: Make  $\angle ABX={75}^{\circ }$
Step 3: With B as the centre, draw an arc at $4cm$. Name that point as C.
Step 4:  $AB\parallel CD$

$$\begin{gathered} \therefore \angle ABX+\angle BCY=180° \\ \Rightarrow \angle BCY=180° -75°=105° \end{gathered}$$
Make  $\angle BCY=105°$
At C, draw an arc of length $3.2 \mathrm{cm}$.
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

Question 1:

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle, $\angle ABX, \mathrm{equal} \mathrm{to} 60°.$
Step3: With B as the centre, draw an arc of $5 \mathrm{cm}$. Name that point as C. Join B and C.
Step4:
$$\begin{gathered} AB\parallel DC \\ \therefore \angle ABX+\angle BCY=180° \\ \Rightarrow \angle BCY=180°-60°=120° \end{gathered}$$

Draw an angle,  $\angle BCY, \mathrm{equal} \mathrm{to} 120°.$

Step4: With A as the centre, draw an arc of length $6.5 \mathrm{cm}$, which cuts CY. Mark that point as D.

Step5: Join A and D.

​Thus, ABCD is the required trapezium.

Question 2:

( i) Open curve: An open curve is a curve where the beginning and end points are different.
Example:     Parabola

(ii) Closed Curve: A curve that joins up so there are no end points.
Example: Ellipse

(iii) Simple closed curve:  A closed curve that does not intersect itself.

Question 3:

Let the angles be $\left(x\right)°, \left(2x\right)°, \left(3x\right)° \mathrm{and} \left(4x\right)°.$
Sum of the angles of a quadrilateral is ${360}^{\circ }$.
$$\begin{gathered} x+2x+3x+4x=360 \\ 10x=360 \\ x=\frac{360}{10} \\ x=36 \end{gathered}$$

$$\begin{gathered} \left(2x\right)°=(2\times 36{)}^{\circ }={72}^{\circ } \\ (3x)°=(3\times 36{)}^{\circ }={108}^{\circ } \\ \left(4x\right)°=(4\times 36{)}^{\circ }={144}^{\circ } \end{gathered}$$

The angles of the quadrilateral are ${36}^{\circ }, {72}^{\circ }, {108}^{\circ } \mathrm{and} {144}^{\circ }.$

Question 4:

$\mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{b}\mathrm{e} \left(2x\right)° \mathrm{and} \left(3x\right)°.$
Sum of any two adjacent angles of a parallelogram is ${180}^{\circ }$.

$$\begin{gathered} \therefore 2x+3x =180 \\ \Rightarrow 5x=180 \\ \Rightarrow x=36 \end{gathered}$$

$$\begin{gathered} \left(2x\right)°=(2\times 36)°={72}^{\circ } \\ \left(3x\right)°=(3\times 36)°={108}^{\circ } \end{gathered}$$

Measures of the angles are ${72}^{\circ } and {108}^{\circ }$.

Question 5:

Let the length be $4x$ cm and the breadth be $5x$ cm.
Perimeter of the rectangle =180 $cm$
Perimeter of the rectangle=$2(l+b)$

 $$\begin{gathered} 2(l+b)=180 \\ \Rightarrow 2(4x+5x)=180 \\ \Rightarrow 2\left(9x\right)=180 \\ \Rightarrow 18x=180 \\ \Rightarrow x=10 \end{gathered}$$
 
$$\begin{gathered} \therefore \mathrm{Length}=4x \mathrm{cm}=4\times 10=40 \mathrm{cm} \\ \mathrm{Breadth}=5x \mathrm{cm}=5\times 10=50 \mathrm{cm} \end{gathered}$$

Question 6:

Rhombus is a parallelogram.
 
 

Question 7:

All the sides of a rhombus are equal in length.
The diagonals of a rhombus intersect at ${90}^{\circ }$.
The diagonal and the side of a rhombus form right triangles.

In $△AOB$:
$$\begin{gathered} A{B}^2=AO{ }^2+O{B}^2 \\ =8^2+6^2 \\ =64+36 \\ =100 \\ AB=10 \mathrm{cm} \end{gathered}$$
Therefore, the length of each side of the rhombus is 10 cm.

Question 8:

(b) 37^o, 143^o, 37^o 143^o

Opposite angles of a parallelogram are equal.
 $$\begin{gathered} \therefore 3x-2=50-x \\ \Rightarrow 3x+x=50+2 \\ \Rightarrow 4x=52 \\ \Rightarrow x=13 \end{gathered}$$

Therefore, the first and the second angles are:
$$\begin{gathered} {\left(3x-2\right)}^{\circ }={\left(2\times 13-2\right)}^{\circ }={37}^{\circ } \\ {\left(50-x\right)}^{\circ }={\left(50-13\right)}^{\circ }={37}^{\circ } \end{gathered}$$
Sum of adjacent angles in a parallelogram is ${180}^{\circ }$.
Adjacent angles = ${180}^{\circ }-{37}^{\circ }={143}^{\circ }$

Question 9:

(d) none of the these

Let the angles be $\left(x\right)°, \left(3x\right)°, \left(7x\right)° \mathrm{and} \left(9x\right)°$.

Sum of the angles of the quadrilateral is ${360}^{\circ }$.

$$\begin{gathered} x+3x+7x+9x=360 \\ 20x=360 \\ x=18 \end{gathered}$$

$$\begin{gathered} \mathrm{Angles}: \\ \left(3x\right)°=(3\times 18)={54}^{\circ } \\ \left(7x\right)°=(7\times 18)°={126}^{\circ } \\ \left(9x\right)°=(9\times 18)°={162}^{\circ } \end{gathered}$$

Question 10:

(b) 6 cm
Let the breadth of the rectangle be x cm.
Diagonal =10 cm
Length= 8 cm
The rectangle is divided into two right triangles.
$$\begin{gathered} Diagona{l}^2= Lengt{h}^2+ Breadt{h}^2 \\ {10}^2=8^2+x^2 \\ 100-64=x^2 \\ {x}^2=36 \\ x=6 cm \end{gathered}$$

Breadth of the rectangle = 6 cm

Question 11:

(d) x = 8
All sides of a square are equal.


$$\begin{gathered} PQ=QR \\ (2x+3)=(3x-5) \\ =>2x-3x=-5-3 \\ =>x=8 \mathrm{cm} \end{gathered}$$

Question 12:

(d) 90°

We know that the opposite sides and the angles in a parallelogram are equal.
Also, its adjacent sides are supplementary, i.e. sum of the sides is equal to 180.
Now, the bisectors of these angles form a triangle, whose two angles are:
 $$\begin{gathered} \frac{A}{2}\mathrm{and}\frac{B}{2} \mathrm{or} \frac{A}{2}=(90-\frac{A}{2}) \\ \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \frac{\angle A}{2}+90-\frac{\angle A}{2}+\angle O={180}^{\circ } \\ \angle O=180-90 \\ \angle O={90}^{\circ } \end{gathered}$$

Hence, the two bisectors intersect at right angles.

Question 13:

(c) 9
Hexagon has six sides.
$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{diagonals} = \frac{n(n-3)}{2} (\mathrm{where} n \mathrm{is} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{side}s) \\ =\frac{6(6-3)}{2} \\ =9 \end{gathered}$$

Question 14:

(b) 8
$$\begin{gathered} \mathrm{Interior} \mathrm{angle}=\frac{180(n-2)}{n} \\ \Rightarrow 135=\frac{180(n-2)}{n} \\ \Rightarrow 135n=180n-360 \\ \Rightarrow 360=180n-135n \\ \Rightarrow n=8 \end{gathered}$$

It has 8 sides.

Question 15:

(i) Sum of all exterior angles =  ${360}^{\circ }$

(ii) Sum of all interior angles = $(n-2)\times 180°$

(iii) Number of diagonals = $\frac{n(n-3)}{2}$

Question 16:

(i) Sum of all exterior angles of a regular polygon is ${360}^{\circ }$.

(ii) Sum of all interior angles of a polygon is $(n-2)\times 180°, \mathrm{where} n \mathrm{is} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{sides}.$

Question 17:

(i) Octagon has 8 sides.
$$\begin{gathered} \therefore \mathrm{Interior} \mathrm{angle} =\frac{180°n-360°}{n} \\ \mathrm{Int}\mathrm{erior} \mathrm{angle} =\frac{(180°\times 8)-360°}{8} =135° \end{gathered}$$
(ii) Sum of the interior angles of a regular hexagon = $(6-2)\times {180}^{\circ }={720}^{\circ }$

(iii) Each exterior angle of a regular polygon is ${60}^{\circ }$.
$\therefore \frac{360}{60}=6$
Therefore, the given polygon is a hexagon.

(iv) If the interior angle is ${108}^{\circ }$, then the exterior angle will be ${72}^{\circ }$.                (interior and exterior angles are supplementary)
Sum of the exterior angles of a polygon is 360°.

Let there be n sides of a polygon.

$$\begin{gathered} 72n=360 \\ n=\frac{360}{72} \\ n=5 \end{gathered}$$

Since it has 5 sides, the polygon is a pentagon.

(v) A pentagon has 5 diagonals.

 $$\begin{gathered} \mathrm{If} n \mathrm{is} \mathrm{the} \mathrm{number} \mathrm{of} s\mathrm{ides,} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{diagonals} = \frac{n(n-3)}{2} \\ \frac{5(5-3)}{2} \\ =5 \end{gathered}$$

Question 18:

(i) F
The diagonals of a parallelogram need not be equal in length.

(ii) F
The diagonals of a rectangle are not perpendicular to each other.

(iii) T

(iv) T

Adjacent sides of a kite are equal and this is also true for a rhombus. Additionally, all the sides of a rhombus are equal to each other.