NCERT Solutions Class 8 Maths Chapter 3

Q.1 Given here are some figures.

Classify each of them on the basis of the following.
(a) Simple curve
(b) Simple closed curve
(c)Polygon
(d) Convex polygon
(e) Concave polygon

Ans.

(a) Simple curve: 1, 2, 5, 6, 7
(b) Simple closed curve: 1, 2, 5, 6, 7
(c) Polygon: 1, 2
(d) Convex polygon: 2
(e) Concave polygon: 1

Q.2 How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle

Ans.

(a)

A convex quadrilateral has 2 diagonals.

(b)

A regular hexagon has 9 diagonals.

(c)

A triangle does not have any diagonal.

Q.3 What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Ans.

Let ABCD be a convex quadrilateral. Draw a diagonal BD. Now, this quadrilateral is divided into two triangles ABD and BC

\begin{array}{l}\text{We know that, the sum of interior angles of a triangle is 18}{0}^{\circ }.\\ \therefore \text{The sum of interior angles of a quadrilateral}=\text{18}{0}^{\circ }\text{+18}{0}^{\circ }\\ ={360}^{\circ }\\ \text{Hence,the sum of the measures of the angles of a convex}\\ \text{quadrilateral is}{360}^{\circ }.\end{array}

If the quadrilateral is not convex, then also this property holds.

This quadrilateral can also be divided into two triangles ABD and ACD.

\begin{array}{l}\text{We know that, the sum of interior angles of a triangle is 18}{0}^{\circ }.\\ \therefore \text{The sum of interior angles of a quadrilateral}=\text{18}{0}^{\circ }\text{+18}{0}^{\circ }\\ ={360}^{\circ }\\ \text{Hence,the sum of the measures of the angles of a concave}\\ \text{quadrilateral is}{360}^{\circ }.\end{array}

Q.4 Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

What can you say about the angle sum of a convex polygon with number of sides?

(a) 7 (b) 8 (c) 10 (d) n

Ans.

From the table it can be observed that the angle sum of a convex polygon of n sides is (n – 2)×180°.

Therefore, for given number of sides, the sum of the angles are:

$\begin{array}{l}\left(\mathrm{a}\right)7:(7-2)\times {180}^{\circ }={900}^{\circ }\\ \left(\mathrm{b}\right)8:(8-2)\times {180}^{\circ }={1080}^{\circ }\\ \left(\mathrm{c}\right)10:(10-2)\times {180}^{\circ }={1440}^{\circ }\\ \left(\mathrm{d}\right)\mathrm{n}:(\mathrm{n}-2)\times {180}^{\circ }\end{array}$

Q.5 What is a regular polygon? State the name of a regular polygon of

(i) 3 sides (ii) 4 sides (iii) 6 sides

Ans.

A polygon which is “equilateral” and “equiangular” is called a regular polygon.

(i) 3 sides: Equilateral triangle

(ii) 4 sides: Square

(iii) 6 sides: Regular Hexagon

Q.6 Find the angle measure x in the following figures.

Ans.

$\begin{array}{l}\text{We know that the sum of interior angles of a}\\ \text{quadrilateral is}{360}^{\circ }.\\ \\ \therefore \text{From the given quadrilateral ,we have}\\ {50}^{\circ }+{130}^{\circ }+{120}^{\circ }+\mathrm{x}={360}^{\circ }\\ \Rightarrow 300\mathrm{°}+\mathrm{x}={360}^{\circ }\\ \Rightarrow \mathrm{x}={360}^{\circ }-300\mathrm{°}\\ \Rightarrow \mathrm{x}=60\mathrm{°}\end{array}$

$\begin{array}{l}\mathrm{From}\mathrm{the}\mathrm{figure},\\ 90\mathrm{°}+\mathrm{a}=180\mathrm{°}\left(\mathrm{Linear}\mathrm{pair}\right)\\ \Rightarrow \mathrm{a}=180\mathrm{°}-90\mathrm{°}=90\mathrm{°}\\ \\ \mathrm{Now},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{interior}\mathrm{angles}\mathrm{of}\\ \mathrm{a}\mathrm{quadrilateral}\mathrm{is}360\mathrm{°}.\\ \\ \therefore \mathrm{in}\mathrm{the}\mathrm{given}\mathrm{quadrilateral},\\ 60\mathrm{°}+70\mathrm{°}+\mathrm{x}+90\mathrm{°}=360\mathrm{°}\\ \Rightarrow 220\mathrm{°}+\mathrm{x}=360\mathrm{°}\\ \Rightarrow \mathrm{x}=140\mathrm{°}\end{array}$

From the figure, we have
70+a =180° (Linear Pair)
a = 110°
Also, 60°+b = 180° (Linear Pair)
b= 120°
Since the pentagon has 5 sides, so the sum of all the interior
angles of a pentagon is 540°.

Therefore, we have
120°+110°+30°+x+x = 540°
260° + 2x = 540°
2x= 280°
x = 140°

$\begin{array}{l}\text{We know that the sum of all the interior angles of a}\\ \text{pentagon is 54}{0}^{\circ }.\\ \text{Since, all the sides are equal,therefore, all the angles are equal.}\\ \text{Hence, it is a regular pentagon.}\\ \therefore \text{5}\mathrm{x}=\text{54}0\mathrm{°}\\ \Rightarrow \mathrm{x}=\text{1}0\text{8}\mathrm{°}\end{array}$

Q.7

(a) Find x + y + z
(b) Find x + y + z + w

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \mathrm{x}+\text{9}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{Linear pair}\right)\\ \Rightarrow \mathrm{x}=\text{9}0\mathrm{°}\\ \\ \mathrm{z}+\text{3}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{Linear pair}\right)\\ \Rightarrow \mathrm{z}=\text{15}0\mathrm{°}\\ \\ \text{We know that, the exterior angle is equal to the}\\ \text{sum of interior opposite angles.}\\ \therefore \mathrm{y}=\text{9}0\mathrm{°}+\text{3}0\mathrm{°}\\ \Rightarrow \mathrm{y}=\text{12}0\mathrm{°}\\ \\ \mathrm{Hence},\\ \mathrm{x}+\mathrm{y}+\mathrm{z}=\text{9}0\mathrm{°}+\text{12}0\mathrm{°}+\text{15}0\mathrm{°}\\ =\text{36}0\mathrm{°}\end{array}$ $\begin{array}{l}\text{(b)}\\ \text{We know that, the sum of the measures of all interior}\\ \text{angles of a quadrilateral is 3}{60}^{\circ }.\\ \\ \text{Therefore,}\\ a+\text{6}0\mathrm{°}+\text{8}0\mathrm{°}+\text{12}0\mathrm{°}=\text{36}0\mathrm{°}\\ \Rightarrow a+\text{26}0\mathrm{°}=\text{36}0\mathrm{°}\\ \Rightarrow a=\text{1}00\mathrm{°}\end{array}$ $\begin{array}{l}\mathrm{Also},\\ \mathrm{x}+\text{12}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{Linear pair}\right)\\ \Rightarrow \mathrm{x}=\text{6}0\mathrm{°}\\ \\ \mathrm{y}+\text{8}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{Linear pair}\right)\\ \Rightarrow \mathrm{y}=\text{1}00\mathrm{°}\\ \\ \mathrm{z}+\text{6}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{Linear pair}\right)\\ \Rightarrow \mathrm{z}=\text{12}0\mathrm{°}\\ \\ \mathrm{w}+\text{1}00\mathrm{°}=\text{18}0\mathrm{°}\left(\text{Linear pair}\right)\\ \mathrm{w}=\text{8}0\mathrm{°}\\ \\ \text{Hence,}\\ \mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}=\text{6}0\mathrm{°}+\text{1}00\mathrm{°}+\text{12}0\mathrm{°}+\text{8}0\mathrm{°}\\ =\text{36}0\mathrm{°}\end{array}$

Q.8 Find x in the following figures.

Ans.

$\begin{array}{l}\text{(a)}\\ \text{We know that the sum of all exterior angles of any}\\ \text{polygon is 36}{0}^{\circ }.\\ \therefore \text{125}\mathrm{°}+\text{125}\mathrm{°}+\mathrm{x}=\text{36}0\mathrm{°}\\ \Rightarrow \text{25}0\mathrm{°}+\mathrm{x}=\text{36}0\mathrm{°}\\ \Rightarrow \mathrm{x}=\text{11}0\mathrm{°}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{We know that the sum of all exterior angles of any}\\ \text{polygon is 36}{0}^{\circ }.\\ \therefore \text{6}0°+\text{9}0°+\text{7}0°+x+\text{9}0°=\text{36}0°\\ \Rightarrow \text{31}0°+x=\text{36}0°\\ \Rightarrow x=\text{5}0°\end{array}$

Q.9 Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{We know that, the}\mathrm{sum}\mathrm{of}\mathrm{}\mathrm{all}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angles}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\\ \mathrm{given}\mathrm{polygon}={360}^{\circ }\\ \\ \mathrm{Thus},\mathrm{}\mathrm{measure}\mathrm{}\mathrm{of}\mathrm{}\mathrm{each}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{regular}\mathrm{}\\ \mathrm{polygon}\mathrm{of}9\mathrm{sides}=\frac{{360}^{\circ }}{9}={40}^{\circ }\\ \end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\\ \mathrm{We}\mathrm{know}\mathrm{that},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{}\mathrm{all}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angles}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\\ \mathrm{given}\mathrm{polygon}={360}^{\circ }\\ \\ \mathrm{Thus},\mathrm{}\mathrm{measure}\mathrm{}\mathrm{of}\mathrm{}\mathrm{each}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{regular}\mathrm{}\\ \mathrm{polygon}\mathrm{}\mathrm{of}\mathrm{}15\mathrm{}\mathrm{sides}=\frac{{360}^{\circ }}{15}={24}^{\circ }\end{array}$

Q.10 How many sides does a regular polygon have if the measure of an exterior angle is 24°?

Ans.

\begin{array}{l}\text{We know that, the sum of all exterior angles of the given}\\ \text{polygon}=\text{36}0°\\ \\ \text{It is given that the measure of each exterior angle}=\text{24}°\\ \\ \text{Thus},\text{number of sides of the regular polygon=}\frac{{360}^{\circ }}{24°}=15\end{array}

Q.11 How many sides does a regular polygon have if each of its interior angles is 165°?

Ans.

$\begin{array}{l}\text{Given,}\\ \text{Measure of each interior angle}=\text{165}°\\ \\ \therefore \text{Measure of each exterior angle}=\text{18}0°-\text{165}°=\text{15}°\\ \\ \text{Also, the sum of all exterior angles of any polygon is 36}0º.\\ \\ \mathrm{Therefore},\text{number of sides of the polygon =}\frac{{360}^{\circ }}{{15}^{\circ }}=24\end{array}$

Q.12 (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?

Ans.

$\begin{array}{l}\text{(a) Given:}\\ \text{Exterior angle}={22}^{\circ }\\ \\ \text{We know that,}\\ \text{the sum of all the exterior angles of a polygon is 3}{60}^{\circ }.\end{array}$

$\begin{array}{l}\mathrm{Also},\text{all the exterior angles are of equal measure.That}\\ {\text{means 22}}^{\circ }\text{should be exact divisor of 3}{60}^{\circ }.\\ \\ \mathrm{Since},{\text{\hspace{0.17em}22}}^{\circ }\text{\hspace{0.17em}is not an exact divisor of 3}{60}^{\circ },\mathrm{therefore},\text{such}\\ \text{a polygon is not possible.}\\ \end{array}$

\begin{array}{l}\text{(b) Given:}\\ \text{Interior angle}=\text{22}°\\ \\ \therefore \text{Exterior angle}=\text{18}0°-\text{22}°=\text{158}°\\ \\ \text{As 36}0°\text{is not a perfect multiple of 158}°,so\text{such a polygon}\\ \text{is not possible}\text{.}\end{array}

Q.13 (a) What is the minimum interior angle possible for a regular polygon? Why?

(b) What is the maximum exterior angle possible for a regular polygon?

Ans. For a regular polygon, as the number of sides decreases, interior angle also decreases and exterior angle increases.

Therefore, polygon with maximum exterior angle and minimum interior angle is an equilateral triangle as it has the minimum number of sides possible for a polygon.

Consider an equilateral triangle. The exterior angle of this triangle will be the maximum exterior angle possible for any regular polygon.

Therefore,

$\mathrm{Exterior}\mathrm{angle}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{{360}^{\circ }}{3}=120\mathrm{°}$

$\begin{array}{l}\mathrm{Hence},\mathrm{maximum}\mathrm{possible}\mathrm{measure}\mathrm{of}\mathrm{exterior}\mathrm{angle}\mathrm{for}\mathrm{any}\\ \mathrm{polygon}\mathrm{is}120\mathrm{°}.\\ \\ \mathrm{Also},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{an}\mathrm{exterior}\mathrm{angle}\mathrm{and}\mathrm{an}\mathrm{interior}\mathrm{angle}\\ \mathrm{are}\mathrm{always}\mathrm{in}\mathrm{a}\mathrm{linear}\mathrm{pair}.\\ \mathrm{Hence},\mathrm{minimum}\mathrm{interior}\mathrm{angle}=180\mathrm{°}-120\mathrm{°}=60\mathrm{°}\end{array}$

Q.14 Given a parallelogram ABCD. Complete each statement along with the definition or property used.

1. AD = ……
2. ∠ DCB = ……
3. OC = ……
4. m ∠DAB + m ∠CDA = …..

Ans.

1. AD = BC

(Opposite sides are of equal length in a parallelogram.)

2. ∠ DCB = ∠ DAB

(Opposite angles are equal in measure.)

3. OC = OA

(In a parallelogram, diagonals bisect each other)

4. m ∠DAB + m ∠CDA = 180°

(In a parallelogram, adjacent angles are supplementary to each other.)

Q.15 Consider the following parallelograms. Find the values of the unknowns x, y, z.

Ans.

(i)

$\begin{array}{l}\mathrm{x}+\text{1}00\mathrm{°}=\text{18}0\mathrm{°}\left(\begin{array}{l}\text{Adjacent}\mathrm{}\text{angles of a parallelogram}\\ \text{are supplementary}\end{array}\right)\\ \Rightarrow \mathrm{x}=\text{8}0\mathrm{°}\\ \\ \text{Since opposite angles of a parallelogram are equal}\\ \therefore \angle \mathrm{A}=\angle \mathrm{C}\\ \Rightarrow \mathrm{z}=\mathrm{x}=\text{8}0\mathrm{º}\\ \\ \mathrm{Also},\angle \mathrm{B}=\angle \mathrm{D}\\ \mathrm{Hence},\mathrm{y}=\text{1}00\mathrm{°}\\ \\ \therefore \mathrm{x}=\text{8}0\mathrm{°},\mathrm{y}=100\mathrm{°}\mathrm{and}\mathrm{z}=\text{8}0\mathrm{°}\end{array}$

(ii)

$\begin{array}{l}\mathrm{Since}\mathrm{adjacent}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{parallelogram}\mathrm{are}\\ \mathrm{supplementary},\mathrm{therefore},\\ 50\mathrm{°}+\mathrm{y}=180\mathrm{°}\\ \Rightarrow \mathrm{y}=130\mathrm{°}\\ \mathrm{Also},\text{opposite angles of a parallelogram are equal}\\ \therefore \mathrm{x}=\mathrm{y}=130\mathrm{°}\\ \\ \text{Now,’z’ and ‘x’ are the corresponding angles.}\\ \therefore \mathrm{z}=\mathrm{x}=130\mathrm{°}\\ \\ \therefore \mathrm{x}=130\mathrm{°},\mathrm{y}=130\mathrm{°}\mathrm{and}\mathrm{z}=\text{13}0\mathrm{°}\end{array}$

(iii)

\begin{array}{l}\end{array} $\begin{array}{l}\mathrm{x}=\text{9}0\mathrm{°}\left(\text{Vertically opposite angles}\right)\\ \\ \mathrm{Also},\\ \mathrm{x}+\mathrm{y}+\text{3}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{By angle sum property of triangles}\right)\\ \Rightarrow \text{12}0\mathrm{°}+\mathrm{y}=\text{18}0\mathrm{°}\\ \Rightarrow \mathrm{y}=\text{6}0\mathrm{°}\\ \\ \mathrm{Now},\mathrm{z}=\mathrm{y}=\text{6}0\mathrm{°}\left(\text{Alternate interior angles}\right)\\ \\ \therefore \mathrm{x}=90\mathrm{°},\mathrm{y}=60\mathrm{°}\mathrm{and}\mathrm{z}=60\mathrm{°}\end{array}$

(iv)

$\begin{array}{l}\text{In the given figure},\\ \mathrm{z}=\text{8}0\mathrm{°}\left(\text{Corresponding angles}\right)\\ \\ \mathrm{y}=\text{8}0\mathrm{°}\left(\text{Opposite angles are equal in a parallelogram}\right)\\ \mathrm{Also},\\ \mathrm{x}+\mathrm{y}=\text{18}0\mathrm{°}\left(\text{Adjacent angles are supplementary}\right)\\ \Rightarrow \mathrm{x}=\text{18}0\mathrm{°}-\text{8}0\mathrm{°}\\ =\text{1}00\mathrm{°}\\ \\ \therefore \mathrm{x}=100\mathrm{°},\mathrm{y}=80\mathrm{°}\mathrm{and}\mathrm{z}=80\mathrm{°}\end{array}$

(v)

\begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{y}=\text{112}\mathrm{°}\left(\text{Opposite angles are equal}\right)\\ \\ \mathrm{In}\mathrm{a}\text{}\mathrm{triangle},\\ \mathrm{x}+\mathrm{y}+\text{4}0\mathrm{°}=\text{18}0\mathrm{°}\left(\text{By angle sum property}\right)\\ \Rightarrow \mathrm{x}+\text{112}\mathrm{°}+\text{4}0\mathrm{°}=\text{18}0\mathrm{°}\\ \Rightarrow \mathrm{x}+\text{152}\mathrm{°}=\text{18}0\mathrm{°}\\ \Rightarrow \mathrm{x}=\text{28}\mathrm{°}\\ \\ \mathrm{Also},\\ \mathrm{z}=\mathrm{x}=\text{28}\mathrm{°}\left(\text{Alternate interior angles}\right)\\ \\ \therefore \mathrm{x}=\text{28}\mathrm{°},\mathrm{y}=\text{112}\mathrm{°}\mathrm{and}\mathrm{z}=\text{28}\mathrm{°}\end{array}$

Q.16 Can a quadrilateral ABCD be a parallelogram if

(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?

Ans.

A parallelogram has opposite sides equal and parallel. The opposite angles are also equal. The adjacent angles are supplementary.

1. So, if ∠D + ∠B = 180°, a quadrilateral ABCD may or may not be a parallelogram as all the other conditions of the parallelogram should also be fulfilled by a quadrilateral ABCD.
2. In a parallelogram the opposite sides should be of equal length. Here, AB is equal to CD but AD is not equal to BC. Hence, ABCD is not a parallelogram.
3. In a parallelogram the opposite angles are of equal length. Here, ∠A is not equal to ∠C. Hence, ABCD is not a parallelogram.

Q.17 Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Ans.

Here, ABCD is a quadrilateral that is not a parallelogram but ∠B=∠D.

Q.18 The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.

Ans.

$\begin{array}{l}\text{The two angles are in the ratio 3 : 2.}\\ \text{Let}\angle \text{A = 3x and}\angle \text{B = 2x}\\ \text{We know that , the adjacent angles are supplementary in a parallelogram.}\end{array}$ $\begin{array}{l}\therefore \angle \mathrm{A}+\angle \mathrm{B}=180\mathrm{°}\\ \Rightarrow 3\mathrm{x}+2\mathrm{x}=180\mathrm{°}\\ \Rightarrow 5\mathrm{x}=180\mathrm{°}\\ \Rightarrow \mathrm{x}=\frac{180\mathrm{°}}{5}=36\mathrm{°}\\ \mathrm{Hence},\\ \angle \mathrm{A}=3\mathrm{x}=3\times 36\mathrm{°}=108\mathrm{°}\\ \angle \mathrm{B}=2\mathrm{x}=2\times 36\mathrm{°}=72\mathrm{°}\\ \\ \mathrm{Since},\text{opposite angles of a paralle}\log \text{ra}\mathrm{m}\mathrm{are}\\ \mathrm{equal},\mathrm{therefore},\\ \angle \text{A=}\angle \text{C and}\angle \text{B=}\angle \text{D}\\ \therefore \angle \mathrm{A}=\angle \mathrm{C}=108\mathrm{°}\mathrm{and}\angle \mathrm{B}=\angle \mathrm{D}=72\mathrm{°}\\ \end{array}$

Thus, the measures of the angles of the parallelogram are 108°, 72°, 108°, and 72°.

Q.19 Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Ans.

In a parallelogram the adjacent angles are supplementary

$\begin{array}{l}\therefore \angle \text{A}+\angle \text{B}=\text{18}0\mathrm{°}\\ \text{Since}\angle \text{A}=\angle \text{B},\\ \Rightarrow \text{2}\angle \text{A}=\text{18}0\mathrm{°}\\ \Rightarrow \angle \text{A}=\text{9}0\mathrm{°}\\ \Rightarrow \angle \text{B}=\text{9}0\mathrm{°}\\ \\ \mathrm{Now},\angle \text{C}=\angle \text{A}\left(\text{Opposite angles}\right)\\ \Rightarrow \angle \text{C}=\text{9}0\mathrm{°}\\ \angle \text{D}=\angle \text{B}\left(\text{Opposite angles}\right)\\ \Rightarrow \angle \text{D}=\text{9}0\mathrm{°}\\ \\ \text{Hence},\text{all the angles of the parallelogram are 9}0\mathrm{º}.\end{array}$

Q.20 The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Ans.

$\begin{array}{l}\text{7}0\mathrm{°}=\mathrm{z}+\text{4}0\mathrm{°}\left(\text{Corresponding angles of a parallelogram}\right)\\ \Rightarrow \text{7}0\mathrm{°}-\text{4}0\mathrm{°}=\mathrm{z}\\ \Rightarrow \text{z}=\text{3}0\mathrm{°}\\ \\ \mathrm{Also},\\ \mathrm{y}=\text{4}0\mathrm{°}\left(\text{Alternate interior angles}\right)\\ \\ \mathrm{Now},\angle \mathrm{E}+\angle \mathrm{H}=180\mathrm{°}\left(\text{Adjacent angles are supplementary}\right)\\ \Rightarrow \mathrm{x}+(\mathrm{z}+\text{4}0\mathrm{º})=\text{18}0\mathrm{°}\\ \Rightarrow \mathrm{x}+(30\mathrm{°}+\text{4}0\mathrm{º})=\text{18}0\mathrm{°}\\ \Rightarrow \mathrm{x}+\text{7}0\mathrm{°}=\text{18}0\mathrm{°}\\ \Rightarrow \mathrm{x}=\text{11}0\mathrm{°}\\ \\ \therefore \mathrm{x}=\text{11}0\mathrm{°},\mathrm{y}=40\mathrm{°}\mathrm{and}\mathrm{z}=30\mathrm{°}\end{array}$

Q.21 The following figures GUNS and RUNS are parallelograms . Find x and y. (Lengths are in cm)

Ans.

(i) We know that, the lengths of opposite sides of a parallelogram are equal to each other.

Therefore, GU = SN and SG =NU

$\begin{array}{l}\Rightarrow \text{3}\mathrm{y}-\text{1}=\text{26}\\ \Rightarrow \text{3}\mathrm{y}=\text{27}\\ \Rightarrow \mathrm{y}=\text{9}\\ \\ \text{Also, SG}=\text{NU}\\ \Rightarrow \text{3}\mathrm{x}=\text{18}\\ \Rightarrow \mathrm{x}=\text{6}\end{array}$

Hence, the measures of x and y are 6 cm and 9 cm.

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

$\begin{array}{l}\mathrm{y}+\text{7}=\text{2}0\mathrm{and}\mathrm{x}+\mathrm{y}=\text{16}\\ \\ \mathrm{Now},\mathrm{y}+\text{7}=\text{2}0\\ \Rightarrow \mathrm{y}=\text{13}\\ \\ \mathrm{Also},\mathrm{x}+\mathrm{y}=\text{16}\\ \Rightarrow \mathrm{x}+\text{13}=\text{16}\\ \Rightarrow \mathrm{x}=\text{3}\end{array}$

Hence, the measures of x and y are 3 cm and 13 cm.

Q.22

In the above figure both RISK and CLUE are parallelograms. Find the value of x.

Ans.

$\begin{array}{l}\text{In parallelogram RISK},\\ \angle \text{RKS}+\angle \text{ISK}=\text{18}0°\begin{array}{l}[\text{Adjacent angles of a parallelogram}\\ \text{are supplementary.}]\end{array}\\ \Rightarrow \text{12}0°+\angle \text{ISK}=\text{18}0°\\ \Rightarrow \angle \text{ISK}=\text{6}0°\\ \\ \\ \text{In parallelogram CLUE},\\ \angle \text{ULC}=\angle \text{CEU}=\text{7}0°\left(\begin{array}{l}\text{opposite angles of}\\ \text{a parallelogram are equal}.\end{array}\right)\\ \\ \text{Now, in a triangle,}\\ x+\text{6}0°+\text{7}0°=\text{18}0°\left(\text{By angle sum property}\right)\\ \Rightarrow x=\text{5}0°\end{array}$

Q.23 Explain how this figure is a trapezium. Which of its two sides are parallel?

Ans.

Here, ∠NML + ∠MLK = 180°

Hence, NM||LK and ML is a transversal.(If a transversal intersects the two given lines such that the sum of the angles on the same side of transversal is 180º, then the given two lines will be parallel to each other.)

Therefore, KLMN is a trapezium as it has one pair of parallel lines.

Q.24

$\mathrm{Find}\mathrm{m}\angle \mathrm{C}\mathrm{in}\mathrm{Fig}.\mathrm{if}\overline{\mathrm{AB}}\left|\right|\overline{\mathrm{DC}}.$

Ans.

$\begin{array}{l}\mathrm{}\text{Given}:\mathrm{AB}\left|\right|\mathrm{BC}\\ \Rightarrow \angle \text{B}+\angle \text{C}=\text{18}0\mathrm{°}\begin{array}{l}[\text{Angles on the same side of transversal}\\ \text{are supplementary}.]\end{array}\\ \Rightarrow \text{12}0\mathrm{°}+\angle \text{C}=\text{18}0\mathrm{°}\\ \Rightarrow \angle \text{C}=\text{6}0\mathrm{°}\end{array}$

Q.25

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{measure}\mathrm{of}\angle \mathrm{P}\mathrm{and}\angle \mathrm{S}\mathrm{if}\mathrm{SP}\left|\right|\mathrm{RQ}\mathrm{in}\mathrm{Fig}.(\mathrm{If}\mathrm{you}\mathrm{find}\\ \mathrm{m}\angle \mathrm{R},\mathrm{is}\mathrm{there}\mathrm{more}\mathrm{than}\mathrm{one}\mathrm{method}\mathrm{to}\mathrm{find}\mathrm{m}\angle \mathrm{P}?)\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{SP}\left|\right|\mathrm{RQ}\\ \therefore \angle \text{P}+\angle \text{Q}=\text{18}0\mathrm{°}\left(\text{Angles on the same side of transversal}\right)\\ \Rightarrow \angle \text{P}+\text{13}0\mathrm{°}=\text{18}0\mathrm{°}\\ \Rightarrow \angle \text{P}=\text{5}0\mathrm{°}\\ \\ \mathrm{Also},\\ \angle \text{R}+\angle \text{S =18}0\mathrm{°}\left(\text{Angles on the same side of transversal}\right)\\ \Rightarrow \text{9}0\mathrm{°}+\angle \text{R}=\text{18}0\mathrm{°}\\ \Rightarrow \angle \text{S}=\text{9}0\mathrm{°}\\ \end{array}$

$\begin{array}{l}\text{Yes,}\text{there is one more method to find the measure of}\angle \text{P}.\\ \text{We can apply the angle sum property of a quadrilateral}\\ \text{to find the}\mathrm{m}\angle \text{P}.\end{array}$

Q.26 State whether True or False.
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles (d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.

Ans.

(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Q.27 Identify all the quadrilaterals that have.

(a) four sides of equal length

(b) four right angles

Ans.

(a) four sides of equal length: Rhombus and Square

(b) four right angles: Square and rectangle

Q.28 Explain how a square is.
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle

Ans.

$\begin{array}{l}\text{(i) Quadrilateral: A square has four sides}.\mathrm{Therefore},\text{it is a}\\ \text{quadrilateral}\text{.}\\ \\ \left(\text{ii}\right)\text{Parallelogram:The opposite sides of a square are parallel}\\ \text{to each other and diagonals bisect each other}\text{.}\mathrm{Hence},\text{it is a}\\ \text{parallelogram}\text{.}\\ \\ \left(\text{iii}\right)\text{Rhombus:All the four sides of a square are of the same}\\ \text{length and diagonals bisect each other at}{90}^{\circ }.\mathrm{Hence},\text{it is a}\\ \text{rhombus}\text{.}\\ \\ \left(\text{iv}\right)\text{Rectangle:A square is a rectangle since each interior}\\ \text{angle measures 9}0°\text{and diagonals are of equal length}\text{.}\\ \mathrm{Hence},\text{it is a rectangle}\text{.}\end{array}$

Q.29 Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal

Ans.

(i) The quadrilaterals whose diagonals bisect each other are: parallelogram, square, rhombus and rectangle.
(ii) The quadrilaterals whose diagonals are perpendicular bisectors of each other are: square and rhombus.
(iii) The quadrilaterals having equal diagonals are: square and rectangle.

Q.30 Explain why a rectangle is a convex quadrilateral.

Ans. A convex quadrilateral is a quadrilateral whose diagonals lie in the interior of it and its all interior angles are less than 180°.

We know that, a rectangle has two diagonals. Both the diagonals lie in the interior of the rectangle. Also, each interior angle of a rectangle measures 90°.

Hence, it is a convex quadrilateral.

Q.31 ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A,B and C. (The dotted lines are drawn additionally to help you).

Ans.

Given: ABC is a right angled triangle and O is the midpoint of AC.

Draw lines AD and DC such that AD||BC, AB||DC and join OD.

Therefore, AD = BC, AB = DC.

Since opposite sides are equal and parallel to each other and all the interior angles are of 90º, therefore, ABCD is a rectangle.

Also, diagonals bisect each other in a rectangle.

Hence, AO= OC = BO = OD

Thus, O is equidistant from A, B, and C.