3 Mark Question

Question 13 Marks

In the figure, find the measure of $\angle\text{MPN}$.

Answer

In the figure, OMPN is a quadrilateral in which
$\angle\text{O}=45^\circ$
$\angle\text{M}=\angle\text{N}=90^\circ$ (PM ⊥ OA and PN ⊥ OB)
Let,
$\angle\text{MPN}=\text{x}^\circ$
$\angle\text{O}+\angle\text{M}+\angle\text{N}+\angle\text{MPN}$
$=360^\circ$ (Sum of angles of a quadrilateral)
$\Rightarrow45^\circ+90^\circ+90^\circ+\text{x}^\circ$
$=360^\circ$
$\Rightarrow225^\circ+\text{x}^\circ$
$=360^\circ$
$\Rightarrow\text{x}^\circ=360^\circ-225^\circ$
$\Rightarrow\text{x}^\circ=135^\circ$
$\angle\text{MPN}=135^\circ$

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Question 23 Marks

Find the number of sides of a regular polygon when each of its angles has a measures of:-
135°

Answer

In a n-sided regular polygon, each angle $=\frac{2\text{n}-4}{\text{n}}$ right angles or $\Big[\frac{2\text{n}-4}{\text{n}}\times90^\circ\Big]$ Each interior angle = 135° Let number of sides of regular polygon = n $\frac{2\text{n}-4}{\text{n}}\times90^\circ=135^\circ$ $\Rightarrow\frac{2\text{n}-4}{\text{n}}=\frac{135}{90}=\frac{3}{2}$ By cross multiplication:$4\text{n}-8=3\text{n}$
$\Rightarrow4\text{n}-3\text{n}=8$
$\Rightarrow\text{n}=8$
$\therefore$ The Regular polygon is of 8 sided.

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Question 33 Marks

Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles?

Answer

Sum of four angles of a quadrilateral = 360°
One angle = 150°
Sum of remaining three angles = 360° - 150°
= 210°
But these three angles are equal
Measure of each angle = $\frac{210}{3}$
= 70°

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Question 43 Marks

Find the number of sides of a regular polygon when each of its angles has a measures of:-
162°

Answer

In a n-sided regular polygon, each angle $=\frac{2\text{n}-4}{\text{n}}$ right angles or $\Big[\frac{2\text{n}-4}{\text{n}}\times90^\circ\Big]$ Each interior angle = 162° Let number of sides of regular polygon = n $\frac{2\text{n}-4}{\text{n}}\times90^\circ=162^\circ$ $\Rightarrow\frac{2\text{n}-4}{\text{n}}=\frac{162}{90}=\frac{9}{5}$ By cross multiplication:$10\text{n}-20=9\text{n}$
$\Rightarrow10\text{n}-9\text{n}=20$
$\Rightarrow\text{n}=20$
$\therefore$ The Regular polygon is of 20 sides.

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Question 53 Marks

Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles?

Answer

Measures of two angles each = 65°
Sum of these two angles = 2 × 65°= 130°
But sum of four angles of a quadrilateral = 360°
Sum of the remaining two angles = 360° - 130°
= 230°
But these are equal to each other
Measure of each angle = $\frac{230}{2}$
= 115°

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Question 63 Marks

Find the number of sides of a regular polygon when each of its angles has a measures of:-
175°

Answer

In a n-sided regular polygon, each angle $=\frac{2\text{n}-4}{\text{n}}$ right angles or $\Big[\frac{2\text{n}-4}{\text{n}}\times90^\circ\Big]$ Each interior angle = 175° Let number of sides of regular polygon = n $\frac{2\text{n}-4}{\text{n}}\times90^\circ=175^\circ$ $\Rightarrow\frac{2\text{n}-4}{\text{n}}=\frac{175}{90}=\frac{35}{18}$ By cross multiplication:$36\text{n}-72=35\text{n}$
$\Rightarrow36\text{n}-35\text{n}=72$
$\Rightarrow\text{n}=72$
$\therefore$ The Regular polygon is of 72 sides.

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Question 73 Marks

Find the number of sides of a regular polygon when each of its angles has a measures of:-
160°

Answer

In a n-sided regular polygon, each angle $=\frac{2\text{n}-4}{\text{n}}$ right angles or $\Big[\frac{2\text{n}-4}{\text{n}}\times90^\circ\Big]$ when each interior angle = 160° Let number of sides of regular polygon = n $\therefore\Big[\frac{2\text{n}-4}{\text{n}}\times90^\circ\Big]=160^\circ$ $\frac{2\text{n}-4}{\text{n}}=\frac{160^\circ}{90^\circ}=\frac{16}{9}$ By cross multiplication:$18\text{n}-36=16\text{n}$
$\Rightarrow18\text{n}-16\text{n}=36$
$\Rightarrow2\text{n}=36$
$\Rightarrow\text{n}=\frac{32}{2}=18$
$\therefore$ Regular polygon is of 18 sided.

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Question 83 Marks

Find the number of sides of a regular polygon when each of its angles has a measures of:-
150°

Answer

In a n-sided regular polygon, each angle $=\frac{2\text{n}-4}{\text{n}}$ right angles or $\Big[\frac{2\text{n}-4}{\text{n}}\times90^\circ\Big]$ Each interior angle = 150° Let number of sides of regular polygon = n $\frac{2\text{n}-4}{\text{n}}\times90^\circ=150^\circ$ $\Rightarrow\frac{2\text{n}-4}{\text{n}}=\frac{150}{90}=\frac{5}{3}$ By cross multiplication:$6\text{n}-12=5\text{n}$
$\Rightarrow6\text{n}-5\text{n}=12$
$\Rightarrow\text{n}=12$
$\therefore$ The Regular polygon is of 12 sides.

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Question 93 Marks

The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.

Answer

The sum of four angles of a quadrilateral = 360°
Three angles are 110°, 50° and 40°
Let fourth angle = x
Then 110° + 50° + 40° + x° = 360°
⇒ 200° + x° = 360°
⇒ x = 360° - 200°
⇒ 160°
x = 160°

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