ML AGGARWAL CLASS 8 Chapter 13 Understanding Quadrilaterals Exercise 13.3

 Exercise 13.3

Question 1

Sol :

(i) Square, Rhombus

(ii) Square, Rectangle

Question 2

Sol :

(i) I. opposite sides are equal and opposite sides are parallel

    II. Adjacent angles are Supplementary

(ii) I. It has four sides

     II. Sum of all interior angles in $360^{\circ}$

(iii) I. All the sides are equal

II. All interior angles are $90^{\circ}$

III. Diagonals cut perpendicularly.

(iv) I. opposite sides are equal

II. All the interior angles are $90^{\circ}$

Question 3

Sol :

(i) Parallelogram, square, rectangle, rhombus

(ii) square,rhombus

(iii) Square, rectangle

Question 4 

Sol :

Given 

side of Rhombus = one diagonal of rhombus 

A D=D C=A C

$\therefore \triangle A D C$ is a equilateral triangle.

$\therefore \quad \angle A D C=\angle D A C=\angle D C A=60^{\circ}$

$\triangle^{\operatorname{le}} A C B$ is also a equilateral triangle

$\therefore \angle C A B=\angle A B C=\angle B C A=60^{\circ}$

$\angle A D B=\angle D A C+\angle C A B=60^{\prime}+60^{\prime}=120^{\circ}$

$\angle D C B=\angle D C A+\angle A C B=60+60=120^{\circ}$

$\therefore$ Angles of rhombus $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$

Question 5

Sol :

ABCD is a rhombus diagonals rhombus bisects each other 

$\begin{aligned} \therefore \quad & x=8 \\ & y=6 . \end{aligned}$

Diagonals of rhombus cuts orthogonally

In $\Delta^{\operatorname{le}} A O B$

$\overline{D A}^{2}+\overline{O B}^{2}=\overline{A B}^{2}(\because$ Pythagoras Theorem $)$

$x^{2}+y^{2}=z^{2}$

$8^{2}+6^{2}=z^{2}$

$z^{2}=64+36$

$z^{2}=100$

z = 10

Question 6

Sol :

ABCD is a trapezium

$\angle A: \angle D=5: 7$

$\angle B=(3 x+11)^{\circ}$

$\angle C=(5 x-31)^{\circ}$

From the property of trapezium

$\angle A+\angle D=180^{\circ}$

Let $\angle A, \angle D=5 y, 7 y$

$\begin{aligned} \therefore \quad 5 y+7 y &=180^{\circ} \\ 12 y &=180^{\circ} \\ y &=15^{\circ} \\ \angle A=5 y &=5 \times 15=75^{\circ} \\ \angle D=7 y &=7 \times 15=105^{\circ} \end{aligned}$

$\angle B+\angle C=180^{\circ}$

$3 x+11+5 x-31=180^{\circ}$

8 x-20=180

8 x=180+20

8 x=200

$x=\frac{200}{8}$

$x=25^{\circ}$

$\begin{aligned} \angle B &=3 x+11 . \\ &=3 \times 25+11 \\ &=75+11 \\ \angle B &=86^{\circ} \\ \angle C &=5 x-31 \\ &=5 \times 25-31 \\ &=125-31 \\ \angle C &=94^{\circ} \end{aligned}$

∴ $\angle A=75^{\circ}, \quad \angle B=86^{\circ}, \angle C=94^{\circ}, \angle D=105^{\circ}$

Question 7

Sol :

$\angle C E B: \angle E C B=3: 2$

$\angle C B E=90^{\circ}(\because$ angle in rectangle $)$

∴ In $\Delta$ ECB

Let

$\angle C E B=3 x, \quad \angle E C B=2 x .$

$\angle C E B+\angle E C B+\angle C B E=180^{\prime}$

3x+2 x+90=180

5 x+90=180

5 x=980-90

5 x=90

$x=\frac{90}{5}$

$x=18^{\circ}$

(ii) $\angle C E B=3 x=3 \times 18=54^{\circ}$

At ' $c^{\prime}$

$\angle C E B+\angle D C E=\angle D C B$

$54^{\circ}+\angle D C E=90^{\circ}$ (∵Angle in rectangle)

$\angle D C E=90-54$

$\angle D C E=36^{\circ}$

$\begin{aligned} \angle D C E+\angle D C F &=180^{\circ} \quad(\because \text { Forms Straightine }) \\ 36^{\circ}+\angle D C F &=180^{\circ} \\ \angle D C F &=180-36^{\circ} \\ \angle D C F &=144^{\circ} \end{aligned}$

Question 8

Sol :

Given ABCD is a rectangle

$\overrightarrow{A O}=\overline{O B}[\because$ intersect at 'O']

$\therefore \angle D A B=\angle O B A=x .$

(i)  In $\Delta$ AOB

$\begin{aligned} \angle A O B+\angle A B O &+\angle O A B=180^{\prime} \\ 118+x+x &=180^{\circ} \\ 2 x &=180-118 \\ 2 x &=62 \\ x &=\frac{62}{2} \\ x &=31^{'} \end{aligned}$

$\therefore \angle A B O=31^{\circ}$

(ii) $\begin{aligned} \angle A O B+\angle A O D &=180^{\circ}(\because \text { Forms straight line }\\ 118+\angle A O D &=180^{\circ} . \\ \angle A O D &=180-118 \\ \angle A O D &=62^{\circ} \end{aligned}$

$\overrightarrow{O D}=\overline{O A}(\because$ diagonals bisect each other)

$\angle D A O=\angle A D O=y$

  In $\Delta$ AOD

$\angle D A O+\angle A D O+\angle A O D=180^{\circ}$

$y+y-62=180$

$\begin{aligned} 2 y+62 &=180 \\ 2 y &=180-62 \\ 2 y &=118 \\ y &=\frac{118}{2} \\ y &=59^{\circ} \\ \therefore \quad \angle A D O &=59^{\circ} \end{aligned}$

(iii) Similarly by taking $\Delta$  BOC

We can prove $\angle O C B=59^{\circ}$

Question 9

Sol :

Give ABCD is a rhombus

$\angle A B D=50^{\circ}$

In $\triangle^{\prime \prime}$ AOB (∵In rhombus, diagonals cut perpendicularly )

$\therefore \quad \angle A O B+\angle O A B+\angle O B A=180^{\circ}$

$\begin{aligned} 90+\angle O A B+50^{\circ}=180 \\ & \angle O A B-140^{\circ}=180 \\ & \angle O A B=180-140 \\ & \angle O A B=40^{\circ} \end{aligned}$

$\angle O A B=\angle C A B=40^{\circ}$

(ii) $\begin{aligned} \angle B C D &=? \\ A B & \| D C \end{aligned}$

$\therefore \angle C A B=\angle A C D=40^{\circ}$

$\angle B C D=2 \times \angle A C D$ (∵diagonal in rhombus, bisects the angle )

$\angle B C D=2 \times 40^{\circ}$

$\angle B C D=80^{\circ}$

(iii) $\angle A D C$

$\angle A D C=\angle A B C$ (∵opposite angles are equal in rhombus )

$\angle A D C=2 \times \angle A B D$

$\angle A D C=2 \times 50^{\circ}$

$\angle A D C=100^{\circ}$

Question 10

Sol :

In a trapezium

$\begin{aligned} \angle C+\angle B &=180^{\circ} \\ 112^{\circ}+\angle B &=180^{\circ} \\ \angle B &=180-102 \\ \angle B &=78^{\circ} \end{aligned}$

Given 

$\begin{aligned} & \overline{A D}=\overline{C B} \\ \therefore & \angle C=\angle D=102^{\circ} \\ \therefore A &=\angle B=78^{\circ} \end{aligned}$

∴ Angles in trapezium $78 ; 78^{\circ}, 102^{\circ}, 102^{\circ}$

Question 11

Sol :

Given PQRS is a kite

$\begin{aligned} \therefore \quad \angle Q=\angle S &=120^{\circ} \\ & x=120^{\circ} \end{aligned}$

In a quadrilateral

$\begin{array}{rl}\angle p+\angle Q+\angle R+\angle S & =360 \\ y+120+50+120 & 2360 \\ y+290 & =360 \\ y & =360-290 \\ y & =70^{\circ}\end{array}$