[4 marks sum]

Question 14 Marks

Is it possible to have a regular polygon whose interior angle is: 155°

Answer

No. of. sides = n
Each interior angle = 155°
$\therefore \frac{(2 n -4) \times 90^{\circ}}{ n }=155^{\circ}$
180n - 360° = 155n
180n - 155n = 360°
25n = 360°
$n =\frac{360^{\circ}}{25^{\circ}}$
$n =\frac{72^{\circ}}{5}$
$n =\frac{360^{\circ}}{25^{\circ}}$
$n =\frac{72^{\circ}}{5}$

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Question 24 Marks

The angles of a pentagon are in the ratio 5 : 4 : 5 : 7 : 6 ; find each angle of the pentagon.

Answer

Let the angles of the pentagon be 5x, 4x, 5x, 7x, 6x
Their sum = 5x + 4x + 5x + 7x + 6x = 27x
Sum of interior angles of a polygon
= (2n - 4) × 90°
= (2 × 5 - 4) × 90° = 540°
$\therefore 27 x=540 \Rightarrow \frac{540}{27}$
⇒ x = 20°
∴ Angles are 5 × 20° = 100°
4 × 20° = 80°
5 × 20° = 100°
7 × 20° = 140°
6 × 20° = 120°

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Question 34 Marks

If angles of quadrilateral are in the ratio 4 : 5 : 3 : 6 ; find each angle of the quadrilateral.

Answer

Let the angles of the quadrilateral be 4x, 5x, 3x and 6x.
∴ 4x + 5x + 3x + 6x = 360°
18x = 360°
$x=\frac{360^{\circ}}{18}=20^{\circ}$
∴ First angle = 4x = 4 × 20° = 180°
Second angle = 5x = 5 × 20° = 100°
Third angle = 3x = 3 × 20° = 60°
Fourth angle = 6x = 6 × 20° = 120°

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Question 44 Marks

If all the angles of an octagon are equal, find the measure of each angle.

Answer

No. of. sides of octagon n = 8
Let each angle be = x°
∴ Sum of angles = 8x°
∴ (2n - 4) × 90° = Sum of angles
(2 × 8 - 4) × 90° = 8x°
12 × 90° = 8x°
$\Rightarrow x^{\circ}=\frac{90^{\circ} \times 12^{\circ}}{8}$
⇒ x° = 135°
∴ Each angle of octagon = 135°

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Question 54 Marks

Is it possible to have a polygon, whose sum of interior angles is $1030^\circ .$

Answer

Let no. of. sides be $= n$
Sum of interior angles of polygon $= 1030^\circ$
$\therefore (2n - 4) \times 90^\circ = 1030^\circ$
$\Rightarrow 2( n -2)=\frac{1030^{\circ}}{90^{\circ}}$
$\Rightarrow( n -2)=\frac{1030^{\circ}}{2 \times 90^{\circ}}$
$\Rightarrow( n -2)=\frac{103}{18}$
$\Rightarrow n =\frac{103}{18}+2$
$\Rightarrow n =\frac{139}{18}$
Which is not a whole number.
Hence it is not possible to have a polygon,
the sum of whose interior angles is $1030^\circ .$

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Question 64 Marks

Find the number of side of a polygon, if the sum of its interior angle is: 1440°

Answer

1440°
Let no. of. sides = n
∴ Sum of interior angles of polygon = 1440°
∴ (2n - 4) × 90° = 1440°
⇒ 2n - 4 =$\frac{1440^{\circ}}{90^{\circ}}$
⇒ 2(n - 2) = $\frac{1440^{\circ}}{90^{\circ}}$
⇒ n - 2 = $\frac{1440^{\circ}}{2 \times 90^{\circ}}$
⇒ n - 2 = 8
⇒ n = 8 + 2
⇒ n = 10

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