# Understanding Quadrilaterals - Class 8 - Mathematics

Renews every month. Cancel anytime

### Your personal doubt-solving assistant

Chatterbot AI gives you 100% accurate answers to your questions in an instant.

## Examples - Understanding Quadrilaterals | NCERT | Mathematics | Class 8

Find measure $x$ in Fig 3.3.

In any polygon, the sum of the exterior angles is always $360^\circ$. Given that three of the exterior angles of a quadrilateral are $90^\circ$, $50^\circ$, and $110^\circ$, we can find the fourth exterior angle by subtracting the sum of these three angles from $360^\circ$:

$$ \text{Fourth exterior angle} = 360^\circ - (90^\circ + 50^\circ + 110^\circ) $$

$$ \text{Fourth exterior angle} = 360^\circ - 250^\circ $$

$$ \text{Fourth exterior angle} = 110^\circ $$

Therefore, the fourth exterior angle is $110^\circ$.

Find the number of sides of a regular polygon whose each exterior angle has a measure of $45^{\circ}$.

### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up nowFind the perimeter of the parallelogram PQRS.

### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now## Exercise 3.1 - Understanding Quadrilaterals | NCERT | Mathematics | Class 8

Given here are some figures.

Classify each of them on the basis of the following.

(a) Simple curve

(b) Simple closed curve

(c) Polygon

(d) Convex polygon

(e) Concave polygon

(a) Simple curve → This is a curve that does not cross itself. Applicable figures based on your classification are 1, 2, 5, 6, and 7.

(b) Simple closed curve →1,2,5,6,7 it forms a path which starts and ends at the same point

(c) Polygon → This is a two-dimensional closed figure with straight sides. Figures 1 and 2 fit this description,

(d) Convex polygon → This is a polygon where each interior angle is less than 180 degrees. Figure 2 is the only one that fits this definition as it has all interior angles less than 180 degrees.

(e) Concave polygon → This is a polygon where at least one interior angle is greater than 180 degrees. Figure 1 is a concave polygon because it has an interior angle greater than 180 degrees.

What is a regular polygon?

State the name of a regular polygon of

(i) 3 sides

(ii) 4 sides

(iii) 6 sides

### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now## Exercise 3.2 - Understanding Quadrilaterals | NCERT | Mathematics | Class 8

Find $x$ in the following figure

The sum of the exterior angles of any polygon, including triangles, is always 360 degrees. Given that two of the exterior angles of a triangle are 125 degrees each, we can find the third exterior angle by subtracting the sum of the two given exterior angles from 360 degrees.

Let's denote the three exterior angles of the triangle as $A_1$, $A_2$, and $A_3$. Given $A_1 = 125^\circ$ and $A_2 = 125^\circ$, we need to find $A_3$.

The formula to find the third angle, $A_3$, is:

$$ A_3 = 360^\circ - (A_1 + A_2) $$

Substituting $A_1 = 125^\circ$ and $A_2 = 125^\circ$, we get:

$$ A_3 = 360^\circ - (125^\circ + 125^\circ) = 360^\circ - 250^\circ = 110^\circ $$

Therefore, the third exterior angle of the triangle is $110^\circ$.

Find $x$ in the following figure

### Improve your grades!

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

(i) 9 sides

(ii) 15 sides

### Improve your grades!

How many sides does a regular polygon have if the measure of an exterior angle is $24^{\circ}$ ?

### Improve your grades!

How many sides does a regular polygon have if each of its interior angles is $165^{\circ}$ ?

### Improve your grades!

(a) Is it possible to have a regular polygon with measure of each exterior angle as $22^{\circ}$ ?

(b) Can it be an interior angle of a regular polygon? Why?

### Improve your grades!

(a) What is the minimum interior angle possible for a regular polygon? Why?

(b) What is the maximum exterior angle possible for a regular polygon?

### Improve your grades!

## Extra Questions - Understanding Quadrilaterals | NCERT | Mathematics | Class 8

If the side of a rhombus is $6 \, \mathrm{cm}$ and its one diagonal is $10 \, \mathrm{cm}$, then find the area of the rhombus.

A) $5 \sqrt{7} \, \mathrm{cm}^{2}$

B) $10 \sqrt{11} \, \mathrm{cm}^{2}$

C) $7 \sqrt{5} \, \mathrm{cm}^{2}$

D) $15 \sqrt{11} \, \mathrm{cm}^{2}$

The correct option is **B**) $10 \sqrt{11} , \mathrm{cm}^2$.

Given:

Side of the rhombus, $s = 6 , \mathrm{cm}$

One diagonal, $d = 10 , \mathrm{cm}$

For a rhombus, the diagonals bisect each other at right angles. Thus, each diagonal divides the rhombus into four right-angled triangles. Knowing one diagonal and the side lengths, we can calculate the area of one of these triangles and then double it to find the area of the rhombus.

Using the half diagonal (half of $10 , \mathrm{cm}$), which is $5 , \mathrm{cm}$, and the side length $6 , \mathrm{cm}$, we apply the Pythagorean theorem to find the other half diagonal ($e$):

$$ e^2 = 6^2 - 5^2 = 36 - 25 = 11 $$

$$ e = \sqrt{11} $$

The total area of the rhombus, which is made up of two congruent triangles, can be calculated by:

$$ \text{Area of one triangle} = \frac{1}{2} \times \text{First diagonal} \times \text{Second half diagonal} = \frac{1}{2} \times 10 , \mathrm{cm} \times \sqrt{11} , \mathrm{cm} $$

For one triangle, the area derived using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, where the base is the known diagonal, and the height is the second half diagonal ($\sqrt{11} , \mathrm{cm}$):

$$ \text{Area of one triangle} = 5 \sqrt{11} , \mathrm{cm}^2 $$

Since the area of the rhombus is double that of one of these triangles, it results in:

$$ \text{Area of rhombus} = 2 \times 5 \sqrt{11} , \mathrm{cm}^2 = 10 \sqrt{11} , \mathrm{cm}^2 $$

In $\triangle ABC$ and $\triangle PQR$, $\angle A = 30^{\circ}$, $\angle B = 70^{\circ}$, $\angle P = 70^{\circ}$, $\angle Q = 80^{\circ}$, and $AB = RP$, then

A) $\triangle ABC \cong \triangle PQR$

B) $\triangle ABC \cong \triangle QRP$

C) $\triangle ABC \cong \triangle RPQ$

D) $\triangle ABC \cong \triangle PRQ$