NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals Exercise 3.2

Exercise 3.2

1. Find $\text{x}$ in the following figures

$\left( \text{a} \right)$

Ans: As we can see in the figure and we also know that the sum of all the exterior angles on any polygon is $ 360^{\circ}$, we got

$ \Rightarrow x + 125^{\circ} + 125^{\circ} = 360^{\circ} $

 $ \Rightarrow x + 250^{\circ} =360^{\circ} $ 

 $ \Rightarrow x = 110^{\circ} $

Therefore, we got $x = 110^{\circ}$

$\left( \text{b} \right)$

As we can see in the figure and we also know that the sum of all the exterior angles on any polygon is $360^{\circ}$, we got

$ \Rightarrow x + 60^{\circ} + 90^{\circ} + 70^{\circ} + 90^{\circ} = 360^{\circ}$

$ \Rightarrow x + 310^{\circ} = 360^{\circ}$

$ \Rightarrow x = 50^{\circ}$

Therefore, we got $x = 50^{\circ}$.

2. Find the measure of each exterior angle of a regular polygon of 

$\left( \text{i} \right)\text{9}$ sides 

Ans: Now, we know that the sum of the exterior angles of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

So, if the polygon has $\text{9}$ sides, the measure of each angle will be $\text{=}\dfrac{\text{36}{{\text{0}}^{\text{o}}}}{\text{9}}\text{=4}{{\text{0}}^{\text{o}}}$.

Therefore, the measure of each angle will be $\text{4}{{\text{0}}^{\text{o}}}$.

$\left( \text{ii} \right)\text{15}$ sides

Ans: Now, we know that the sum of the exterior angles of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

So, if the polygon has $\text{15}$ sides, the measure of each angle will be $\text{=}\dfrac{\text{36}{{\text{0}}^{\text{o}}}}{\text{15}}\text{=2}{{\text{4}}^{\text{o}}}$.

Therefore, the measure of each angle will be $\text{2}{{\text{4}}^{\text{o}}}$.

3. How many sides does a regular polygon have if the measure of an exterior angle is $\text{2}{{\text{4}}^{\text{o}}}$?

Ans: Now, we know that the sum of all the measures of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

It is given that each angle is of $\text{2}{{\text{4}}^{\text{o}}}$, so, the number of sides in regular polygon will be $\text{=}\dfrac{\text{36}{{\text{0}}^{\text{o}}}}{\text{2}{{\text{4}}^{\text{o}}}}\text{=15}$. 

Therefore, the regular polygon will have $\text{15}$ sides.

4. How many sides does a regular polygon have if the measure of an interior angle is $\text{16}{{\text{5}}^{\text{o}}}$?

Ans: Now, we have interior angle measuring $\text{16}{{\text{5}}^{\text{o}}}$, so the exterior angle will be of $\text{18}{{\text{0}}^{\text{o}}}\text{-16}{{\text{5}}^{\text{o}}}\text{=1}{{\text{5}}^{\text{o}}}$.

We got that the exterior angle is $\text{1}{{\text{5}}^{\text{o}}}$ thus we know that the sum of all the measures of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

It is given that each angle is of $\text{1}{{\text{5}}^{\text{o}}}$, so, the number of sides in regular polygon will be $\text{=}\dfrac{\text{36}{{\text{0}}^{\text{o}}}}{\text{1}{{\text{5}}^{\text{o}}}}\text{=24}$.

Therefore, the regular polygon will have $\text{24}$ sides.

5. $\left( \text{a} \right)$ Is it possible to have a regular polygon with measure of each exterior angles as $\text{2}{{\text{2}}^{\text{o}}}$?

Ans: As we know that the sum of all the exterior angles of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

Now, if we have to find that if it’s possible to have a regular polygon with a measure of exterior angle, then it is mandatory that $\text{36}{{\text{0}}^{\text{o}}}$ is a perfect multiple of the given exterior angle.

So, as we can see that $\text{36}{{\text{0}}^{\text{o}}}$ is not a perfect multiple of $\text{2}{{\text{2}}^{\text{o}}}$.

Therefore, it is not possible to have a regular polygon with measure of each exterior angles as $\text{2}{{\text{2}}^{\text{o}}}$.

$\left( b \right)$ Is it possible to have a regular polygon with measure of each interior angles as $\text{2}{{\text{2}}^{\text{o}}}$?

Ans: Now, we have interior angle measuring $\text{2}{{\text{2}}^{\text{o}}}$, so the exterior angle will be of $\text{18}{{\text{0}}^{\text{o}}}\text{-2}{{\text{2}}^{\text{o}}}\text{=15}{{\text{8}}^{\text{o}}}$.

We got that the exterior angle is $\text{15}{{\text{8}}^{\text{o}}}$.

As we know that the sum of all the exterior angles of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

Now, if we have to find that if it’s possible to have a regular polygon with measure of exterior angle, then it is mandatory that $\text{36}{{\text{0}}^{\text{o}}}$ is a perfect multiple of the given exterior angle.

So, as we can see that $\text{36}{{\text{0}}^{\text{o}}}$ is not a perfect multiple of $\text{15}{{\text{8}}^{\text{o}}}$.

Therefore, it is not possible to have a regular polygon with measure of each exterior angles as $\text{15}{{\text{8}}^{\text{o}}}$.

6. $\left( \text{a} \right)$ What is the minimum interior angle possible for a regular polygon?

Ans: As we needed to find out the minimum interior angle possible for a regular polygon, we will consider a regular polygon with the lowest sides.

The regular polygon with the lowest sides is an equilateral triangle with $\text{3}$ sides

Now, we know that the sum of all the measures of any polygon is $\text{36}{{\text{0}}^{\text{o}}}$.

So, the maximum exterior angle in equilateral triangle will be $\dfrac{\text{36}{{\text{0}}^{\text{o}}}}{\text{3}}\text{=12}{{\text{0}}^{\text{o}}}$.

We got the maximum exterior angle is $\text{12}{{\text{0}}^{\text{o}}}$.

As we know, the sum of all the interior angles in a triangle is $\text{18}{{\text{0}}^{\text{o}}}$.

So, the minimum interior angle will be $\text{18}{{\text{0}}^{\text{o}}}\text{-12}{{\text{0}}^{\text{o}}}\text{=6}{{\text{0}}^{\text{o}}}$, we get the minimum angle as $\text{6}{{\text{0}}^{\text{o}}}$

Therefore, the minimum interior angle is possible for a regular polygon is $\text{6}{{\text{0}}^{\text{o}}}$.

$\left( \text{b} \right)$ What is the maximum exterior angle possible for a regular polygon?

Ans: Now, we know that the maximum exterior angle of a regular polygon is possible if the interior angle of the same polygon is minimum.

Now, we know that the minimum interior angles possible is $\text{6}{{\text{0}}^{\text{o}}}$,

So, the maximum exterior angle possible is $\text{18}{{\text{0}}^{\text{o}}}\text{-6}{{\text{0}}^{\text{o}}}\text{=12}{{\text{0}}^{\text{o}}}$.

Therefore, the maximum exterior angle is $\text{12}{{\text{0}}^{\text{o}}}$.

Conclusion

Exercise 3.2 focuses on exploring various properties and types of quadrilaterals. The exercise provides a series of problems that help students understand the classification and characteristics of different quadrilaterals, such as parallelograms, rhombuses, rectangles, and squares. This chapter is designed to reinforce students' knowledge of quadrilaterals, helping them to distinguish between different types and understand their properties thoroughly. This exercise is essential for building a strong base in geometry, enabling students to tackle more complex geometrical concepts in future studies.

Class 8 Maths Chapter 3: Exercises Breakdown

CBSE Class 8 Maths Chapter 3 Other Study Materials

Other than Maths Class 8 Chapter 3 Exercise 3.2  you can also check on the additional study materials provided for Class 8 Maths Chapter 3.

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