RS Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 15 Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Quadrilaterals are extremely popular among class 8 students for Maths Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2020 2021 Book of class 8 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

(i) 4
(ii) 4
(iii) 4, co-linear
(iv) 2
(v) opposite
(vi) 360°

#### Question 2:

(i) 4
(ii) 4
(iii) 4, co-linear
(iv) 2
(v) opposite
(vi) 360°

(i) There are four pairs of adjacent sides, namely (AB,BC), (BC,CD), (CD,DA) and (DA,AB).
(ii) There are two pairs of opposite sides, namely (AB,DC) and (AD,BC).
(iii) There are four pairs of adjacent angles, namely .
(iv) There are two pairs of opposite angles, namely .
(v) There are two diagonals, namely AC and BD.

#### Question 3:

(i) There are four pairs of adjacent sides, namely (AB,BC), (BC,CD), (CD,DA) and (DA,AB).
(ii) There are two pairs of opposite sides, namely (AB,DC) and (AD,BC).
(iii) There are four pairs of adjacent angles, namely .
(iv) There are two pairs of opposite angles, namely .
(v) There are two diagonals, namely AC and BD.

Let ABCD be a quadrilateral.
Join A and C.

Now, we know that the sum of the angles of a triangle is 180°.

For

For

Adding (1) and (2):
$\left(\angle 1+\angle 2+\angle 3+\angle 4\right)+\angle B+\angle D={360}^{o}\phantom{\rule{0ex}{0ex}}$

or $\angle A+\angle B+\angle C+\angle D={360}^{o}\phantom{\rule{0ex}{0ex}}$

Hence, the sum of all the angles of a quadrilateral is 360°.

#### Question 4:

Let ABCD be a quadrilateral.
Join A and C.

Now, we know that the sum of the angles of a triangle is 180°.

For

For

Adding (1) and (2):
$\left(\angle 1+\angle 2+\angle 3+\angle 4\right)+\angle B+\angle D={360}^{o}\phantom{\rule{0ex}{0ex}}$

or $\angle A+\angle B+\angle C+\angle D={360}^{o}\phantom{\rule{0ex}{0ex}}$

Hence, the sum of all the angles of a quadrilateral is 360°.

Sum of all the four angles of a quadrilateral is 360°.

The fourth angle measures 122°.

#### Question 5:

Sum of all the four angles of a quadrilateral is 360°.

The fourth angle measures 122°.

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#### Question 6:

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Sum of the four angles of a quadrilateral is 360°.

If the unknown angle is $x$°, then:

$75+75+75+x=360\phantom{\rule{0ex}{0ex}}x=360-225=135\phantom{\rule{0ex}{0ex}}$

The fourth angle measures 135°.

#### Question 7:

Sum of the four angles of a quadrilateral is 360°.

If the unknown angle is $x$°, then:

$75+75+75+x=360\phantom{\rule{0ex}{0ex}}x=360-225=135\phantom{\rule{0ex}{0ex}}$

The fourth angle measures 135°.

Let the three angles measure $x°$ each.

Sum of all the angles of a quadrilateral is 360°.

Each of the equal angles measure 80°.

#### Question 8:

Let the three angles measure $x°$ each.

Sum of all the angles of a quadrilateral is 360°.

Each of the equal angles measure 80°.

Let the two unknown angles measure $x°$ each.
Sum of the angles of a quadrilateral is 360
°.

Each of the equal angle measures 100
°.

#### Question 9:

Let the two unknown angles measure $x°$ each.
Sum of the angles of a quadrilateral is 360
°.

Each of the equal angle measures 100
°.

Sum of the angles of a quadrilateral is 360°.

$\angle APB=80°$

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