Rs Aggarwal 2019 2020 Solutions for Class 8 Math Chapter 15 Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Quadrilaterals are extremely popular among Class 8 students for Math Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 8 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 185:

#### Question 1:

Fill in the blanks:

(i) A quadrilateral has ......... sides.

(ii) A quadrilateral has ......... angles.

(iii) A quadrilateral has ......... vertices, no three of which are .........

(iv) A quadrilateral has ......... diagonals.

(v) A diagonal of a quadrilateral is a line segment that joins two ......... vertices of the quadrilateral.

(vi) The sum of the angles of a quadrilateral is.........

#### Answer:

(i) 4

(ii) 4

(iii) 4, co-linear

(iv) 2

(v) opposite

(vi) 360°

#### Page No 186:

#### Question 2:

In the adjoining figure, *ABCD* is a quadrilateral.

(i) How many pairs of adjacent sides are there? Name them.

(ii) How many pairs of opposite sides are there? Name them.

(iii) How many pairs of adjacent angles are there? Name them.

(iv) How many pairs of opposite angles are there? Name them.

(v) How many diagonals are there? Name them.

#### 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 $\left(\angle A,\angle B\right),\left(\angle B,\angle C\right),\left(\angle C,\angle D\right)and\left(\angle D,\angle A\right)$.

(iv) There are two pairs of opposite angles, namely $\left(\angle A,\angle C\right)and\left(\angle B,\angle D\right)$.

(v) There are two diagonals, namely *AC* and* BD*.

#### Page No 186:

#### Question 3:

Prove that the sum of the angles of a quadrilateral is 360°.

#### 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 $\u25b3ABC:\phantom{\rule{0ex}{0ex}}\angle 2+\angle 4+\angle B={180}^{o}...\left(1\right)$

For $\u25b3ADC:\phantom{\rule{0ex}{0ex}}\angle 1+\angle 3+\angle D={180}^{o}...\left(2\right)$

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°.

#### Page No 186:

#### Question 4:

The three angles of a quadrilateral are 76°, 54° and 108°. Find the measure of the fourth angle.

#### Answer:

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

$\mathrm{Let}\mathrm{the}\mathrm{unknown}\mathrm{angle}\mathrm{be}x\xb0.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}76+54+108+x=360\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}238+x=360\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}x=122$

The fourth angle measures 122°.

#### Page No 186:

#### Question 5:

The angles of a quadrilateral are in the ratio 3 : 5 : 7 : 9. Find the measure of each of these angles.

#### Answer:

$\mathrm{Let}\mathrm{the}\mathrm{measures}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{quadrilateral}\mathrm{be}\left(3x\right)\xb0,\left(5x\right)\xb0,\left(7x\right)\xb0\mathrm{and}\left(9x\right)\xb0.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{quadrilateral}\mathrm{is}{360}^{\mathrm{o}}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore 3x+5x+7x+9x=360\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}24x=360\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}x=15\phantom{\rule{0ex}{0ex}}$

â€‹

$\mathrm{Angles}\mathrm{measure}:\phantom{\rule{0ex}{0ex}}(3\times 15)\xb0={45}^{\xb0}\phantom{\rule{0ex}{0ex}}(5\times 15)\xb0={75}^{\xb0}\phantom{\rule{0ex}{0ex}}(7\times 15)\xb0={105}^{\xb0}\phantom{\rule{0ex}{0ex}}(9\times 15)\xb0={135}^{\xb0}$

#### Page No 186:

#### Question 6:

A quadrilateral has three acute angles, each measuring 75°. Find the measure of the fourth angle.

#### Answer:

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°.

#### Page No 186:

#### Question 7:

Three angles of a quadrilateral are equal and the measure of the fourth angle is 120°. Find the measure of each of the equal angles.

#### Answer:

Let the three angles measure $x\xb0$ each.

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

$\therefore x+x+x+120=360\phantom{\rule{0ex}{0ex}}3x+120=360\phantom{\rule{0ex}{0ex}}3x=240\phantom{\rule{0ex}{0ex}}x=\frac{240}{3}=80$

Each of the equal angles measure 80°.

#### Page No 186:

#### Question 8:

Two angles of a quadrilateral measure 85° and 75° respectively. The other two angles are equal. Find the measure of each of these equal angles.

#### Answer:

Let the two unknown angles measure $x\xb0$ each.

Sum of the angles of a quadrilateral is 360°.

$\therefore 85+75+x+x=360\phantom{\rule{0ex}{0ex}}160+2x=360\phantom{\rule{0ex}{0ex}}2x=360-160=200\phantom{\rule{0ex}{0ex}}x=100$

Each of the equal angle measures 100°.

#### Page No 186:

#### Question 9:

In the adjacent figure, the bisectors of ∠*A* and ∠*B** *meet in a point *P*. If ∠*C* = 100° and ∠*D* = 60°, find the measure of ∠*APB*.

#### Answer:

Sum of the angles of a quadrilateral is 360°.

$\therefore \angle A+\angle B+{60}^{o}+{100}^{o}=360\xb0\phantom{\rule{0ex}{0ex}}\angle A+\angle B=360-100-60=200\xb0\phantom{\rule{0ex}{0ex}}or\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(\angle A+\angle B\right)=100\xb0...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0.\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3APB:\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(\angle A+\angle B\right)+\angle P=180\xb0(\mathrm{because}AP\mathrm{and}PB\mathrm{are}\mathrm{bisectors}\mathrm{of}\angle A\mathrm{and}\angle B)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}U\mathrm{sing}\mathrm{equation}\left(1\right):\phantom{\rule{0ex}{0ex}}100\xb0+\angle P=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle P=80\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ $\angle APB=80\xb0$

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