EXERCISE 19 E
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$=\frac{360^{\circ}}{18^{\circ}}=20^{\circ}$
Q2 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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
Q3 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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°
Q4 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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°
Q5 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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°
Q6 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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
Q7 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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°
Q8 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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$=\frac{360^{\circ}}{\text{exterior angle}}$
(b) Interior angle=156°
Exterior angle=180°-156°=24°
(c) Interior angle=145°
Exterior angle=180°-145°=35°
Q9 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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
Q10 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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°
Q11 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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°
Q12 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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
Q13 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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
Q14 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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)$=\frac{360}{\text{exterior angle}}=\frac{360}{20}$
Q15 | Ex-19E | Class 8 | S.Chand | Composite maths | chapter 19 |Special Types of Quadrilaterals | myhelper
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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
