Understanding Shapes ICSE Class-8th Concise Selina Maths

Understanding Shapes ICSE Class-8th Concise Selina Mathematics Solutions Chapter-16 . We provide step by step Solutions of Exercise / lesson-16 Understanding Shapes ( Including Polygons ) for ICSE Class-8 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-16 A, Exe-16 B and Exe-16 C to develop skill and confidence . Visit official Website CISCE for detail information about ICSE  Board Class-8.

Understanding Shapes ICSE Class-8th Concise Selina Mathematics Solutions Chapter-16-( Including Polygons )

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–: Select Topics :–

 

Exe-16 A,

Exe-16 B,

Exe-16 C,

 

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Exercise – 16 A Understanding Shapes ( Including Polygons ) ICSE Class-8th Concise Selina Mathematics Solutions

Question 1.

State which of the following are polygons :

If the given figure is a polygon, name it as convex or concave.
Answer

Question 2.

Calculate the sum of angles of a polygon with :
(i) 10 sides
(ii) 12 sides
(iii) 20 sides
(iv) 25 sides
Answer

Question 3.

Find the number of sides in a polygon if the sum of its interior angles is :
(i) 900°
(ii) 1620°
(iii) 16 right-angles
(iv) 32 right-angles.
Answer

(iv)

Let no. of sides = n

Sum of angles of polygon = 32 right angles = 32 x 90 = 2880°

(n – 2) x 180° = 2880

n – 2 = 2880/180

n – 2 = 16

n = 16 + 2

n = 18

Question 4.

Is it possible to have a polygon ; whose sum of interior angles is :
(i) 870°
(ii) 2340°
(iii) 7 right-angles
(iv) 4500°
Answer

(i)

Let no. of sides = n

Sum of angles = 870°

(n – 2) x 180° = 870°

n – 2 = 870/180

n – 2 = 29/6

n = (29/6)+2

n = 41/6

Which is not a whole number.

Hence it is not possible to have a polygon, the sum of whose interior angles is 870°

(ii)

Let no. of sides = n

Sum of angles = 2340°

(n – 2) x 180° = 2340°

n – 2 = 2340/180

n – 2 = 13

n = 13 + 2 = 15

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.

(iii)

Let no. of sides = n

Sum of angles = 7 right angles = 7 x 90 = 630°

(n – 2) x 180° = 630°

n – 2 = 630/180

n – 2 = 7/2

n = (7/2) + 2

n = 11/2

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.

(iv)

Let no. of sides = n

(n – 2) x 180° = 4500°

n – 2 = 4500/180

n – 2 = 25

n = 25 + 2

n = 27

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

Question 5.

(i) If all the angles of a hexagon are equal ; find the measure of each angle.
(ii) If all the angles of a 14-sided figure are equal ; find the measure of each angle.
Answer

(i)

No. of sides of hexagon, n = 6

Let each angle be = x°

Sum of angles = 6x°

(n – 2) x 180° = Sum of angles

(6 – 2) x 180° = 6x°

4 x 180 = 6x

x = (4×180)/6

x = 120°

∴ Each angle of hexagon = 120°

(ii)

No.of.sides of polygon, n = 14

Let each angle = x°

∴ Sum of angles = 14x°

∴ (n – 2) × 180° = Sum of angles of polygon

∴ (14 – 2) × 180° = 14x

12 × 180° = 14x

Question 6.

Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :
(i) 7 sides
(ii) 10 sides
(iii) 250 sides.
Answer

(i)

No. of sides n = 7

Sum of interior & exterior angles at one vertex = 180°

Sum of all interior & exterior angles = 7 × 180° = 1260°

Sum of interior angles = (n – 2) × 180°

= (7 – 2) × 180°

= 900°

∴ Sum of exterior angles = 1260° – 900° = 360°

(ii)

No. of sides n = 10

Sum of interior & exterior angles at one vertex = 180°

Sum of all interior & exterior angles = 10 × 180° = 1800°

Sum of interior angles = (n – 2) × 180°

= (10 – 2) × 180°

= 1440°

∴ Sum of exterior angles = 1800° – 1440° = 360°

(iii)

No. of sides n = 250

Sum of interior & exterior angles at one vertex = 180°

Sum of all interior & exterior angles = 250 × 180° = 45000°

But Sum of interior angles = (n – 2) × 180°

= (250 – 2) × 180°

= 248 × 180°

= 44640°

∴ Sum of exterior angles = 45000° – 44640° = 360°

Question 7.

The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)° (9x – 3)°, (5x + 4)° and (12x + 6)° ; find each exterior angle.
Answer

Sum of exterior angles of hexagon formed by producing sides of order = 360°

∴ (6x – 1)° + (10x + 2)° + (8x + 2)° + (9x – 3)° + (5x + 4)°  + (12x + 6)° = 360°

50x + 10° = 360°

50x = 360° – 10°

50x = 350°

x = 350/50

x =7

∴ Angles are

(6x – 1)°; (10x + 2)°; (8x + 2)°; (9x – 3)°; (5x + 4)° and (12x + 6)°

i.e. (6 × 7 – 1)°, (10 × 7 + 2)°, (8 × 7 + 2)° (9 × 7 – 3)°, (5 × 7 + 4)° and (12 × 7 + 6)°

i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

Question 8.

The interior angles of a pentagon are in the ratio 4 : 5 : 6 : 7 : 5. Find each angle of the pentagon.
Answer

Interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.

Their sum = 4x + 5x + 6x + 7x + 5x = 21x

Sum of interior angles of a polygon = (n – 2) × 180°

= (5 – 2) × 180°

= 540°

∴ 27x = 540

⇒ x = 540/27

⇒ x = 20°

∴ Angles are 4 × 20° = 80°

5 × 20° = 100°

6 × 20° = 120°

7 × 20° = 140°

5 × 20° = 100°

Question 9.

Two angles of a hexagon are 120° and 160°. If the remaining four angles are equal, find each equal angle.
Answer

Two angles of a hexagon are 120°, 160°

Remaining four angles be x, x, x and x.

Their sum = 4x + 280°

But the sum of all the interior angles of a hexagon

= (6 – 2) × 180°

= 4 × 180° = 720°

∴ 4x + 280° = 720°

⇒ 4x = 720° – 280° = 440°

⇒ x = 110°

∴ Equal angles are 110° (each)

Question 10.

The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and ∠B : ∠C : ∠D = 5 : 6 : 7.

(i) Using formula, find the sum of interior angles of the pentagon.
(ii) Write the value of ∠A + ∠E
(iii) Find angles B, C and D.
Answer

(i)

Sum of interior angles of the pentagon

= (5 – 2) × 180°

= 3 × 180°

= 540° …

[so sum for a polygon of x sides = (x – 2) × 180°]

(ii)

Since AB || ED

∴ ∠A + ∠E = 180°

(iii)

Let ∠B = 5x , ∠C = 6x , ∠D = 7x

so 5x + 6x + 7x + 180° = 540°   …

(∠A + ∠E = 180°) (Proved in (ii))

18x = 540° – 180°

⇒ 18x = 360°

⇒ x = 20°

so ∠B = 5 × 20° = 100° ,

∠C = 6 × 20 = 120°

∠D = 7 × 20

= 140°

Question 11.

Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.
Answer

Number of sides = n

Sum of interior angles = (n – 2) × 180°

= 180n – 360°

Sum of 2 right angles = 2 × 90° = 180°

∴ Sum of other angles = 180n – 360° – 180°

= 180n – 540°

No.of vertices at which these angles are formed = n – 2

so Each interior angle = (180n – 540)/n-2

so (180n-540)/n-2 = 120°

180n – 540 = 120n – 240

180n – 120n = – 240 + 540

60n = 300

n = 300/60

n = 5

Question 12.

In a hexagon ABCDEF, side AB is parallel to side FE and ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3. Find ∠B and ∠D.
Answer

To find: 

∠B and ∠D

Given: 

Hexagon ABCDEF in which AB || EF and ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3.

Proof: 

No. of. sides n = 6

∴ Sum of interior angles = (n – 2) × 180°

= (6 – 2) × 180°

= 720°

∵ AB || EF (Given)

∴ ∠A + ∠F = 180°

But ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720° (Proved)

∠B + ∠C + ∠D + ∠E + 180° = 720°

∴ ∠B + ∠C + ∠D + ∠E  = 720° – 180°

Ratio = 6: 4: 2: 3

Sum of parts = 6 + 4 + 2 + 3 = 15

∴ ∠B = (6/15)×540 = 216°

∠D = (2/15)×540∘=72°

Hence proofed

∠B = 216° ; ∠D = 72°

Question 13.

the angles of a hexagon are x + 10°, 2x + 20°, 2x – 20°, 3x – 50°, x + 40° and x + 20°. Find x.
Answer

Question 14.

In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.
Answer

In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3 x 180°

= 540° – 40° – 60° = 540° – 100° = 440°

Ratio in these 3 angles

= 1 : 3 : 7

Sum of ratios =1 + 3 + 7

= 11

Biggest angle = (440 × 7)/11 = 280°

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Understanding Shapes ( Including Polygons ) of Exe -16 B for ICSE Class-8th Mathematics Selina Solutions

Question 1.

Fill in the blanks :
In case of regular polygon, with :

Answer

 

Question 2.

Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 135°
(iii)  of a right-angle
Answer

(i)

No.of.sides of regular polygon be n.

Each interior angle = 160°

so (n-2)/n × 180° = 160°

180n – 360° = 160n

180n – 160n = 360°

20n = 360°

n = 18

(ii)

No. of. sides = n

Each interior angle = 135°

(n-2)/n × 180° = 135°

180n – 360° = 135n

180n – 135n = 360°

45n = 360°

n = 8

(iii)

No. of. sides = n

Each interior angle = 1(1/5) right angles

= (6/5)×90

= 108°

so (n-2)/n × 180° = 108°

180n – 360° = 108n

180n – 108n = 360°

72n = 360°

n = 5

Question 3.

Find the number of sides in a regular polygon, if its each exterior angle is :
(i)  of a right angle
(ii) two-fifth of a right-angle.
Answer

(i)

Each exterior angle = 1/3 of a right angle

= (1/3)×90

= 30°

Let number of sides = n

∴ (360°/n) =30°

∴ n = 360°/30°

n = 12

(ii)

Question 4.

Is it possible to have a regular polygon whose each interior angle is :
(i) 170°
(ii) 138°
Answer

(i)

No. of sides = n

each interior angle = 170°

∴ (n-2)/n×180° = 170°

180n – 360° = 170n

180n – 170n = 360°

10n = 360°

n = 360°/10

n = 36

which is a whole number.

Hence it is possible to have a regular polygon

whose interior angle is 170°

(ii)

Let no. of sides = n

each interior angle = 138°

∴ (n-2)/n × 180° = 138°

180n – 360° = 138n

180n – 138n = 360°

42n = 360°

n = 360°/42

n = 60°/7

which is a whole number.

Hence it is possible to have a regular polygon

whose interior angle is 138°.

Question 5.

Is it possible to have a regular polygon whose each exterior angle is :
(i) 80°
(ii) 40% of a right angle.
Answer

(i)

No. of sides = n each exterior angle = 80°

360°/n = 80°

n = 360°/80°

n = 9/2

Which is not a whole number.
Hence

It is not possible to have a regular polygon whose each exterior angle is of 80°

(ii)

The number of sides = n

Each exterior angle = 40% of a right angle

= (40/100)×90

= 36°

n = 360°/36°

n = 10

Which is a whole number.
Hence

It is possible to have a regular polygon whose exterior angle is 40% of the right angle.

Question 6.

Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
Answer

so x + x = 180°

2x = 180°

x = 90°

Now, let no. of sides = n

∵ each exterior angle = 360°/n

∴ 90° = 360°/n

n = 360°/90°

n = 4

Question 7.

The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.
Answer

Interior angle = x°

∴ x + (1/3)x = 180°

3x + x = 540

4x = 540

x = 5404

x = 135°

∴ Exterior angle = (1/3)×135°=45°

Let no.of. sides = n

∵ each exterior angle = 360°/n

∴ 45° = 360°/n

∴ n = 360°/45°

n = 8

Question 8.

The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :
(i) measure of each interior angle ;
(ii) measure of each exterior angle and
(iii) number of sides in the polygon.
Answer

Let exterior angle = x°

Interior angle = 5x°

x + 5x = 180°

6x = 180°

x = 30°

Each exterior angle = 30°

Each interior angle = 5 x 30°

= 150°

Let no. of sides = n

∵ each exterior angle = 360°/n

30°=360°/n

n = 360°/30°

n = 12

Hence

(i) 150° (ii) 30° (iii) 12

Question 9.

The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of the polygon ;
(ii) number of sides in the polygon
Answer

Interior angle : exterior angle = 2 : 1

∴ 2x° + x° = 180°

3x = 180°

x = 60°

∴ Each exterior angle = 60°

Let no.of. sides = n

360°/n = 60°

n = 360°/60°

n = 6

so (i) x = 60°

(ii) 6

Question 10.

The ratio between the exterior angle and the interior angle of a regular polygon is 1 : 4. Find the number of sides in the polygon.
Answer

Let exterior angle = x° & interior angle = 4x°

∴ 4x + x = 180°

5x = 180°

x = 36°

∴ Each exterior angle = 36°

Let no.of sides = n

∴ 360°/n=36°

n = 360°/36°

n = 10

Question 11.

The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.
Answer

Number of sides = n

Sum of exterior angles = 360°

Sum of interior angles = 360° x 2 = 720°

Sum of interior angles = (n – 2) x 180°

720° = (n – 2) x 180°

n – 2 = 720/180

n – 2 = 4

n = 4 + 2

n = 6

Question 12.

AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :
(i) its each interior angle,
(ii) its each exterior angle
(iii) the number of sides in the polygon.
Answer

∵ Polygon is regular  (Given)

∴ AB = BC

⇒ ∠BAC = ∠BCA   …[∠S opposite to equal sides]

But ∠BAC = 20°

∴ ∠BCA = 20°

i.e. In Δ ABC,

∠B + ∠BAC + ∠BCA = 180°

∠B + 20° + 20° = 180°

∠B = 180° – 40°

∠B = 140°

(i) each interior angle = 140°

(ii) each exterior angle = 180°- 140°= 40°

(iii) Let no. of. sides = n

∴ 360°/n = 40°

n = 360°/40° = 9

n = 9

Question 13.

Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.
Answer

Number of sides of regular polygon = n

AB & DC when produced meet at P such that

∠P = 90°

∵ Interior angles are equal.

∴ ∠ABC = ∠BCD

∴ 180° – ∠ABC = 180° – ∠BCD

∴ ∠PBC = ∠BCP

But ∠P = 90° (given)

∴ ∠PBC + ∠BCP = 180° – 90° = 90°

∴ ∠PBC =∠BCP

= (1/2)XX 90°=45°

∴ Each exterior angle = 45°

∴ 45° = 360°/n

n = 360°/45°

n = 8

Question 14.

In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Answer

(i) Since number of sides in the pentagon = 5

Each exterior angle = 360/5 = 72°

∠BAE = 180° – 72°= 108°

(ii) In Δ ABE, AB = AE

∴ ∠ABE = ∠AEB

But ∠BAE + ∠ABE + ∠ AEB = 180°

∴ 108° + 2 ∠ABE = 180° – 108° = 72°

⇒ ∠ABE = 36°

(iii) Since ∠AED = 108° ..

[∵ each interior angle = 108°]

⇒ ∠AEB = 36°

⇒ ∠BED = 108° – 36° = 72°

Question 15.

The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.
Answer

We know that sum of exterior angles of a polynomial is 360°

(i) If sides of a regular polygon = n – 1

Then each angle = 360°/(n-1)

and if sides are n + 1, then

each angle = 360°/(n+1)

According to the condition,

⇒ n² – 81 = 0

⇒ (n)² – (9)² = 0

⇒ (n + 9)(n – 9) = 0

Either n + 9 = 0. then n = -9 which is not possible being negative,

or n – 9 = 0, then n = 9

∴ n = 9

∴ No. of. sides of a regular polygon = 9

Question 16.

If the difference between the exterior angle of a n sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.
Answer

We know that sum of exterior angles of a polygon = 360°

Each exterior angle of a regular polygon of 360°

n sides = 360°/n

and exterior angle of the regular polygon of

⇒ n² + n = 36    (Dividing by 12)

⇒ n² + n – 30 = 0

⇒ n² + n – 30 = 0

⇒ n² + 6n – 5n – 30 = 0  …

{∵ -30 = 6 × (-5), 1 = 6 – 5}

⇒ n(n + 6) – 5(n + 6) = 0

⇒ (n + 6)(n + 5) = 0

Either n + 6 = 0, then n = -6

which is not possible being negative

or – 5 = 0

then n = 5

Hence

n = 5.

Question 17.

The ratio between the number of sides of two regular polygons is 3 : 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.
Answer

Ratio of sides of two regular polygons = 3 : 4

Let sides of first polygon = 3n

and sides of second polygon = 4n

Sum of interior angles of first polygon

= (2 × 3n – 4) × 90° = (6n – 4) × 90°

and sum of interior angle of second polygon

= (2 × 4n – 4) × 90° = (8n – 4) × 90°

⇒ 18n – 12 = 16n – 8

⇒ 18n – 16n = – 8 + 12

⇒ 2n = 4

⇒ n = 2

∴ No. of sides of first polygon

= 3n = 3 × 2 = 6

and no. of sides of second polygon

= 4n = 4 × 2 = 8

Question 18.

Three of the exterior angles of a hexagon are 40°, 51 ° and 86°. If each of the remaining exterior angles is x°, find the value of x.
Answer

Sum of exterior angles of a hexagon = 4 x 90° = 360°

Three angles are 40°, 51° and 86°

Sum of three angle = 40° + 51° + 86° = 177°

Sum of other three angles = 360° – 177° = 183°

Each angle is x°

3x = 183°

x = 183/3

Hence

x = 61

Question 19.

Calculate the number of sides of a regular polygon, if:
(i) its interior angle is five times its exterior angle.
(ii) the ratio between its exterior angle and interior angle is 2 : 7.
(iii) its exterior angle exceeds its interior angle by 60°.
Answer

(i)

(ii)

(iii)

Question 20.

The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.
Answer

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ICSE Class-8 Maths Exe-16 C Understanding Shapes ( Including Polygons ) Concise Selina Solutions

Question 1.

Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Answer

The other angle = x°

According to question,

89° + 113° + x° + x° = 360°

2x° = 360° – 202°

2x° = 158°

x° = 158/2

other two angles = 79° each

Question 2.

Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
Answer

Two angles are 68° and 76°

Other two angles be 5x and 7x

68° + 76°+ 5x + 7x = 360°

12x + 144° = 360°

12x = 360° – 144°

12x = 216°

x = 18°

angles are 5x and 7x

i.e. 5 x 18° and 7 x 18° i.e. 90° and 126°

Question 3.

Angles of a quadrilateral are (4x)°, 5(x + 2)°, (7x – 20)° and 6(x + 3)°. Find :
(i) the value of x.
(ii) each angle of the quadrilateral.
Answer

Angles of the quadrilateral are,

(4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°

∴ 4x + 5(x + 2) + (7x – 20) + 6(x + 3)

= 360°

4x + 5x + 10 + 7x – 20 + 6x + 18

= 360°

22x + 8 = 360°

22x = 360° – 8°

22x = 352°

x = 16°

Hence angles are,

(4x)° = (4 × 16)°

= 64°

5(x + 2)° = 5(16 + 2)°

= 90°

(7x – 20)° = (7 × 16 – 20)°

= 92°

6(x + 3)° = 6(16 + 3)

= 114°

Question 4.

Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C

Answer

∵ ∠A = 90°    (it’s Given)

∠B = (2x + 4°)

∠C = (3x – 5°)

∠D = (8x – 15°)

∠A + ∠B + ∠C + ∠D = 360°

90° + (2x + 4°) + (3x – 5°) + (8x – 15°)

= 360°

90° + 2x + 4° + 3x – 5° + 8x – 15°

= 360°

⇒ 74° + 13x

= 360°

⇒ 13x = 360° – 74°

⇒ 13x = 286°

⇒ x = 22°

∵ ∠B = 2x  4 = 2 × 22° + 4

= 48°

∠C = 3x – 5 = 3 × 22° – 5 = 61°

Hence

(i) 22°

(ii) ∠B = 48°,

∠C = 61°

Question 5.

In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD
Answer

∵ ∠A : ∠D = 1 : 2

Let ∠A = x

and ∠B = 2x

∵ ∠C : ∠B = 4 : 5

Let ∠C = 4y

and ∠B = 5y

∵ AB || DC

∴ ∠A + ∠D

= 180°

x + 2x = 180°

3x = 180°

x = 60°

∴ A = 60°

∠D = 2x = 2 × 60

= 120°

∠B + ∠C = 180°

5y + 4y = 180°

9y = 180°

y = 20°

∴ ∠B = 5y = 5 20

= 100°

∠C = 4y = 4  20

= 80°

Hence

∠A = 60°; ∠B = 100°; ∠C = 80°

and ∠D = 120°

Question 6.

From the following figure find ;
(i) x
(ii) ∠ABC
(iii) ∠ACD
Answer

(i) In Quadrilateral ABCD,

x + 4x + 3x + 4x + 48° = 360°

12x = 360° – 48°

12x = 312

x = 31212

= 26°

(ii) ∠ABC = 4x

4 × 26 = 104°

(iii) ∠ACD = 180° – 4x – 48°

= 180° – 4 × 26° – 48°

= 180° – 104° – 48°

= 180° – 152°

= 28°

Question 7.

Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a + 2)° and ∠B = 2(2a + 7)°.
Calculate ∠A.
Answer

∵ ∠C = 64° (Given)

∴ ∠D = ∠C – 8° = 64°- 8°

= 56°

∠A = 5(a+2)°

∠B = 2(2a+7)°

Now ∠A + ∠B + ∠C + ∠D = 360°

5(a+2)° + 2(2a+7)° + 64° + 56°

= 360°

5a + 10 + 4a + 14° + 64° + 56°

= 360°

9a + 144°

= 360°

9a = 360° – 144°

9a = 216°

a = 24°

∴ ∠A = 5 (a + 2) = 5(24+2)

= 130°

Question 8.

In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.

Answer

∵ Sum of the angles of quadrilateral = 360°

70° + a + 2a + 15 + 3a + 5

= 360°

6a + 90°

= 360°

6a = 270°

a = 45°

∴ b = 2a + 15 = 2 x 45 + 15

= 105°

c = 3a + 5 = 3 x 45 + 5

= 140°

Hence :

∠b and ∠c are 105°

and 140°

Question 9.

Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
Answer

Each equal angle be x°

x + x + x + 69° = 360°

3x = 360°- 69

3x = 291

x = 97°

Each, equal angle

= 97°

Question 10.

In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other
(i) Is PS also parallel to QR?
(ii) Assign a special name to quadrilateral PQRS.
Answer

∵ ∠P : ∠Q : ∠R : ∠S

= 3: 4: 6: 7

Let ∠P = 3x

∠Q = 4x

∠R = 6x

∠S = 7x

∴ ∠P + ∠Q + ∠R + ∠S

= 360°

3x + 4x + 6x + 7x

= 360°

20x = 360°

x = 18°

∴ ∠P = 3x = 3 × 18

= 54°

∠Q = 4x = 4 × 18

= 72°

∠R = 6x = 6 × 18

= 108°

∠S = 7x = 7 × 18

= 126°

∠Q + ∠R = 72° + 108°

= 180°

or ∠P + ∠S = 54° + 126°

= 180°

Hence 

PQ || SR

As ∠P + ∠Q = 72° + 54°

= 126°

Which is ≠ 180°

∴ PS and QR are not parallel.

PQRS is a Trapezium as its one pair of opposite side is parallel.

Question 11.

Use the information given in the following figure to find the value of x.

Answer

A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).

Since ∠EAD = 70°

∴ ∠DAB = 180° – 70°

= 110°

[EAB is a straight line and AD stands on it ∠EAD+ ∠DAB = 180°]

∴ 110° + 80° + 56° + 3x – 6° = 360°
[∵ sum of interior angles of a quadrilateral = 360°]

∴ 3x = 360° – 110° – 80° – 56° + 6°

3x = 360° – 240°

= 120°

∴ x = 40°

Question 12.

The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.

Answer

Let the ∠A = 4x

∠D = 5x

Since ∠A + ∠D = 180°

 [AB||DC]

4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 20°

∠A = 4 (20)

= 80°,

∠D = 5 (20) = 100°

Again ∠B + ∠C = 180°

[ AB||DC]

3x – 15° + 4x + 20° = 180°

7x = 180° – 5°

⇒ 7x = 175°

⇒ x = 25°

∠B = 75° – 15°

= 60°

and ∠C = 4 (25) + 20 = 100°+ 20°

= 120°

Question 13.

Use the following figure to find the value of x

Answer

Sum of the exterior angles of a quadrilateral

Question 14.

ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
Answer

Given:

ABCDE is a regular pentagon.

The bisector ∠A of the pentagon meets the side CD at point M.

To prove : 

∠AMC = 90°

Proof:

We know that the measure of each interior angle of a regular pentagon is 108°.

∠BAM =  x 108°

= 54°

Since, we know that the sum of a quadrilateral is 360°

In quadrilateral ABCM, we have

∠BAM + ∠ABC + ∠BCM + ∠AMC

= 360°

54° + 108° + 108° + ∠AMC

= 360°

∠AMC = 360° – 270°

∠AMC = 90°

Question 15.

In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:
∠AOB =  (∠C + ∠D)
Answer

Given:

 AO and BO are the bisectors of ∠A and ∠B respectively.
∠1 = ∠4

and ∠3 = ∠5 ……..(i)

To prove : ∠AOB = (1/2) (∠C + ∠D)

Proof: In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°

(1/2) (∠A + ∠B + ∠C + ∠D) = 180° …………(ii)

Now in ∆AOB

∠1 + ∠2 + ∠3 = 180° ………(iii)

Equating equation (ii) and equation (iii), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + (1/2)  (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 + (1/2)  (∠C + ∠D)

∠2 = (1/2)  (∠C + ∠D)

∠AOB = (1/2)  (∠C + ∠D)

Hence proved.

 

— End of Understanding Shapes Solutions :–

 

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