S.chand books class 8 maths solution chapter 19 Special Types of Quadrilaterals exercise 19 E

 EXERCISE 19 E

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Q1 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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Question 1

Find the number of sides of a regular polygon in which each interior angle is 162°.

Sol :

Interior angle+Exterior angle=180°

Each interior angle =162°

Each exterior angle=180°-162°=18°

Number of sides$=\frac{360^{\circ}}{18^{\circ}}=20^{\circ}$

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Question 2

How many sides has a polygon the sum of whose interior angles is 1980° ?

Sol :

Sum of interior angles=(n-2)×180°

1980°=(n-2)×180°

$(n-2)=\frac{1980}{180}$

n-2=11

n=11+2=13

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Question 3

The angles of a quadrilateral taken in order are 2x , 3x , 7x ,8x. Find x and prove that two opposite sides are parallel.

Sol :

Sum of all interior angle of quadrilateral=360°

2x+3x+7x+8x=360°

20x=360°

$x=\frac{360}{20}$

x=18°

Then , angles are 

2x=2×18°=36°

3x=3×18°=54°

7x=7×18°=126°

8x=8×18°=144°

Two sides are parallel if sum of angles between them is 180°

Side AB||CD

∵∠A+∠D=36°+144°=180°

 

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Question 4

ABCDE is a regular pentagon. AB. DC are produced to meet at P. Find ∠BPC.

Sol :

Sum of interior angles=(5-2)×180°
=540°

Each interior angle$=\frac{540}{5}$=108°

Each exterior angle=180°-108°=72°

In ΔBPC,

∠PBC+∠BPC+∠BCP=180° (angle sum property of triangle)

72°+∠BPC+72°=180°

∠BPC=180°-(72°+72°)

∠BPC=36°

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Question 5

Find the sum of the interior angles of a polygon which has (i) 30 sides. (ii) 40 sides.

Sol :

(i) Sum of interior angles=(30-2)×180°

=28°×180°=5040°

(ii) Sum of interior angles=(40-2)×180°

=38°×180°=6840°

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Question 6

Prove that the sum of the interior angles of an octagon is twice the sum of the interior angles of a pentagon

Sol :

Sum of the interior angles of an octagon=(8-2)×180°
=6×180°=1080°

Sum of the interior angles of an pentagon=(5-2)×180°
=3×180°=540°

According to question,

sum of the interior angles of octagon is twice the sum of the interior angles of a pentagon

1080°=540°×2

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Question 7

In a regular pentagon ABCDE, calculate the number of degrees in the angle ABC and prove that BC||AD

Sol :

Sum of the interior angles of an pentagon=(5-2)×180°
=3×180°=540°

Each interior angle$=\frac{540}{5}$=108°

∴∠ABC=108°

In regular pentagon all sides are equal.

In ΔAED (isosceles triangle)

∠DAE+∠AED+∠EDA=180°

x+108°+x=180°

2x=180°-108°

$x=\frac{72}{2}$

x=36°

∠DAE=∠EDA=x=36°

Also,

∠DAB=∠ADC=108°-36°=72°

Two sides are parallel if angle between them is 180°.

∠DAB+∠DAE

72°+36°=180°

∴BC||AD

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Question 8

(i) Is it possible to draw a regular polygon with exterior angles
(a) 72° (b) 32° (c) 40° (d) 55° (e) 60° ? Where possible give the number of sides

Sol :

Number of sides$=\frac{360^{\circ}}{\text{exterior angle}}$

(a) $n=\frac{360^{\circ}}{72^{\circ}}=5$ (yes)

(b) $n=\frac{360^{\circ}}{32^{\circ}}=11.25$ (no)

(c) $n=\frac{360^{\circ}}{40}=9$ (yes)

(d) $n=\frac{360^{\circ}}{55}=6.54$ (no)

(e) $n=\frac{360^{\circ}}{60}=6$ (yes)

(ii) Is it possible to draw a regular polygon with interior angles
(a) 120° (b) 156° (c) 145° ? Where possible , state the number of sides.

Sol :

(a) Interior angle=120° 

Exterior angle=180°-120°=60°

Number of sides$=\frac{360^{\circ}}{\text{exterior angle}}$

$n=\frac{360^{\circ}}{60}=6$ (yes)

(b) Interior angle=156° 

Exterior angle=180°-156°=24°

$n=\frac{360^{\circ}}{24^{\circ}}=15$ (yes)

(c) Interior angle=145°

Exterior angle=180°-145°=35°

$n=\frac{360^{\circ}}{35^{\circ}}=10.28$ (No)

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Question 9

The number of sides of two regular polygons are in the ratio of 3:2 and their interior angles are in the ratio 5:3 . Find the number of their sides.

Sol :

Ratio of sides=3:2

Ratio of interior angle=5:3

Sides of 1st polygon=3x

Sides of 2nd polygon=2x

Sum of interior angles of 1st polygon=(3x-2)×180°

Sum of interior angles of 2nd polygon=(2x-2)×180°

According to question,

$\frac{(3 x-2) \times 180^{\circ}}{(2 x-2) \times 180^{\circ}}=\frac{5}{3}$

$\frac{(3 x-2) }{(2 x-2) }=\frac{5}{3}$

3(3x-2)=5(2x-2)

9x-6=10x-10

10-6=10x-9x

4=x or x=4

Number of sides are

3x=3×4=12

2x=2×4=8

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Question 10

The sides of a pentagon are produced in order and the exterior angles so obtained measure 2x+20° , x-10° , 3x+30° , 4x-15° , 5x-10° . Find the measure of x and find all the exterior angles of the pentagon.

Sol :

Sum of exterior angles of polygon is 360°

2x+20°+x-10°+3x+30°+4x-15°+5x-10°=360°

2x+x+3x+4x+5x+20°-10°+30°-15°-10°=360°

15x+50°-35°=360°

15x+15=360°

15x=360°-15°

15x=345°

$x=\frac{345}{15}$=23°

All angles are 

2x+20°=2(23°)+20°=46°+20°=66°

x-10°=23°-10°=13°

3x+30°=3(23°)+30°=69°+30°=99°

4x-15°=4(23°)-15°=92°-15°=77°

5x-10°=5(23°)-10°=115°-10°=105°

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Question 11

The ratio between the interior and exterior angles of a regular polygon is 8:1. Find
(i) The number of sides of the regular polygon
(ii) each exterior angle of the polygon

Sol :

Let the common ratio be x

Interior angle : Exterior angle=8x : 1x

Also,

Interior angle+Exterior angle=180°

8x+1x=180°

9x=180°

$x=\frac{180}{9}$=20°

So, each exterior angle is 20°

Number of sides$=\frac{360^{\circ}}{20^{\circ}}=18^{\circ}$

(i) n=18

(ii) 20°

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Question 12

An octagon has two pairs of equal angles , one measuring 72° and the other 56° and 4 equal angles . Find the size of equal angles.

Sol :

An octagon has 8 sides:

Find the sum of all the interior angles:

Sum of interior angles = (n - 2)×180 where n is the number of sides

Sum of interior angles = (8 - 2)×180 = 1080º

Find the total of the 4 equal angles:

4 equal angles = 1080 - [(2×72) +(2×56)]

=1080 -(144 + 112)

=1080 - 256= 824º

Find one equal angle:

4 equal angles = 824º

1 equal angle = 824 ÷ 4 =206º

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Question 13

Each exterior angle of a regular polygon is two-thirds of its interior angle. Find the number of sides in the regular polygon

Sol :

Let the interior angle be x

then exterior angle be $\frac{2x}{3}$

Also,

Interior angle+Exterior angle=180º

$x+\frac{2 x}{3}=180$

$\frac{3x+2x}{3}=180$

5x=180×3

$x=\frac{180\times 3}{5}$

x=36×3=108º

Each exterior angle be $\frac{2x}{3}=\frac{2\times 108}{3}$

=2×36=72º

Number of sides (n)$=\frac{360}{\text{exterior angle}}=\frac{360}{72}$

=5

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Question 14

Each interior angle of a regular polygon is 160° . Find the interior angle of regular polygon which has double the number of sides as the given polygon.

Sol :

Each interior angle=160°

Each exterior angle=180°-160°=20°

Number of sides (n)$=\frac{360}{\text{exterior angle}}=\frac{360}{20}$

=18

ATQ,

New Polygon has double number of sides=2×18=36

Each Exterior angle$=\frac{360}{\text{Number of sides}}=\frac{360}{36}$

=10°

Interior angle+Exterior angle=180°

Interior angle=180°-10°=170°

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Question 15

Two interior angles of a polygon are right angles and sum of remaining angles is 720°. Find the number of sides of the polygon.

Sol :

90°+90°+720°=(n-2)×180°

900=(n-2)×180°

$(n-2)=\frac{900}{180}$

n-2=5

n=5+2=7