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Maths 3 Marks Question

Understanding Shapes – II (Quadrilaterals) · Jharkhand Board (JAC) STD 8 (Jharkhand - English Medium). 9 questions.

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Question 13 Marks
In the figure, find the measure of $\angle\text{MPN}$.
Answer
In the figure, $OMPN$ is a quadrilateral in which $\angle\text{O}=45^\circ$
$\angle\text{M}=\angle\text{N}=90^\circ$
$(PM \perp OA$ and $PN \perp OB)$
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^\circ $
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^\circ $
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 $150^\circ .$ What is the measure of equal angles$?$
Answer
Sum of four angles of a quadrilateral $= 360^\circ $
One angle $= 150^\circ$
Sum of remaining three angles $= 360^\circ - 150^\circ = 210^\circ $
But these three angles are equal Measure of each angle = $\frac{210}{3} = 70^\circ $
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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^\circ $
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^\circ $
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^\circ $ and the other two angles are equal. What is the measure of each of these two angles$?$
Answer
Measures of two angles each $= 65^\circ $
Sum of these two angles $= 2 \times 65^\circ = 130^\circ $
But sum of four angles of a quadrilateral $= 360^\circ $
Sum of the remaining two angles $= 360^\circ - 130^\circ = 230^\circ $
But these are equal to each other Measure of each angle $= \frac{230}{2}= 115^\circ $
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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^\circ $
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^\circ $
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^\circ $
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^\circ $
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^\circ $
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 $= 150^\circ$
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^\circ , 50^\circ $ and $40^\circ .$ Find its fourth angle.
Answer
The sum of four angles of a quadrilateral $= 360^\circ $
Three angles are $110^\circ , 50^\circ $ and $40^\circ $
Let fourth angle $= x$
Then $110^\circ + 50^\circ + 40^\circ + x^\circ = 360^\circ $
$\Rightarrow 200^\circ + x^\circ = 360^\circ $
$\Rightarrow x = 360^\circ - 200^\circ $
$\Rightarrow 160^\circ $
$x = 160^\circ $
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