Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 15 - Quadrilaterals

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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 Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 185:

Answer:

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

Page No 186:

Answer:

(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 $(∠ A, ∠ B), (∠ B, ∠ C), (∠ C, ∠ D) a n d (∠ D, ∠ A)$ .
(iv) There are two pairs of opposite angles, namely $(∠ A, ∠ C) a n d (∠ B, ∠ D)$ .
(v) There are two diagonals, namely AC and BD.

Page No 186:

Answer:

Let ABCD be a quadrilateral.
Join A and C.

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

For $△ A B C : ∠ 2 + ∠ 4 + ∠ B = 180^{o} . . . (1)$

For $△ A D C : ∠ 1 + ∠ 3 + ∠ D = 180^{o} . . . (2)$

Adding (1) and (2):
$(∠ 1 + ∠ 2 + ∠ 3 + ∠ 4) + ∠ B + ∠ D = 360^{o}$

or $∠ A + ∠ B + ∠ C + ∠ D = 360^{o}$

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

Page No 186:

Answer:

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

$Let the unknown angle be x ° . 76 + 54 + 108 + x = 360 238 + x = 360 x = 122$

The fourth angle measures 122°.

Page No 186:

Answer:

$Let the measures of the angles of the given quadrilateral be (3 x) °, (5 x) °, (7 x) ° and (9 x) ° . Sum of all the angles of a quadrilateral is 360^{o} . ∴ 3 x + 5 x + 7 x + 9 x = 360 24 x = 360 x = 15$

$Angles measure : (3 \times 15) ° = 45^{°} (5 \times 15) ° = 75^{°} (7 \times 15) ° = 105^{°} (9 \times 15) ° = 135^{°}$

Page No 186:

Answer:

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

If the unknown angle is $x$ °, then:

$75 + 75 + 75 + x = 360 x = 360 - 225 = 135$

The fourth angle measures 135°.

Page No 186:

Answer:

Let the three angles measure $x °$ each.

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

$∴ x + x + x + 120 = 360 3 x + 120 = 360 3 x = 240 x = \frac{240}{3} = 80$

Each of the equal angles measure 80°.

Page No 186:

Answer:

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

$∴ 85 + 75 + x + x = 360 160 + 2 x = 360 2 x = 360 - 160 = 200 x = 100$

Each of the equal angle measures 100°.

Page No 186:

Answer:

Sum of the angles of a quadrilateral is 360°.
$∴ ∠ A + ∠ B + 60^{o} + 100^{o} = 360 ° ∠ A + ∠ B = 360 - 100 - 60 = 200 ° o r \frac{1}{2} (∠ A + ∠ B) = 100 ° . . . (1) Sum of the angles of a triangle is 180 ° . In △ A P B : \frac{1}{2} (∠ A + ∠ B) + ∠ P = 180 ° (because A P and P B are bisectors of ∠ A and ∠ B) U sing equation (1) : 100 ° + ∠ P = 180 ° \Rightarrow ∠ P = 80 °$

∴ $∠ A P B = 80 °$

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