Mathematics Solutions Solutions for Class 8 Math Chapter 18 Circle Chord And Arc are provided here with simple step-by-step explanations. These solutions for Circle Chord And Arc are extremely popular among Class 8 students for Math Circle Chord And Arc Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 116:

#### Question 1:

In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find *l*(QB).

#### Answer:

P is the centre of the circle.

AB = 13 cm

seg PQ ⊥ chord AB

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, the length of QB = $\frac{1}{2}\mathrm{AB}=\frac{13}{2}=6.5$ cm.

#### Page No 116:

#### Question 2:

Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

#### Answer:

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, $\mathrm{PD}=\frac{1}{2}\mathrm{CD}=\frac{48}{2}=24\mathrm{cm}$

In $\u25b3$OPD,

We apply the Pythagoras theorem,

${\mathrm{OP}}^{2}+{\mathrm{PD}}^{2}={\mathrm{OD}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}+{24}^{2}={25}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}={25}^{2}-{24}^{2}=625-576=49\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}=49\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=7\mathrm{cm}$

Hence, the distance of the chord CD from the centre O is 7 cm.

#### Page No 116:

#### Question 3:

O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the circle.

#### Answer:

Join OA.

Let the perpendicular drawn from point O to the chord AB be P.

We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, $\mathrm{AP}=\frac{\mathrm{AB}}{2}=\frac{24}{2}=12\mathrm{cm}$

In $\u25b3$OPA,

We apply the Pythagoras theorem,

${\mathrm{OP}}^{2}+{\mathrm{AP}}^{2}={\mathrm{OA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {9}^{2}+{12}^{2}={\mathrm{OA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OA}}^{2}=81+144=225\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OA}=\sqrt{225}=15\mathrm{cm}$

Hence, the radius of the circle is 15 cm.

#### Page No 116:

#### Question 4:

C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.

#### Answer:

C is the centre of the circle.

Radius = 10 cm

Let the chord be AB.

Distance of the centre to the chord = AB.

CD is perpendiculaar to the chord AB.

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

$\mathrm{AD}=\frac{\mathrm{AB}}{2}=\frac{12}{2}=6\mathrm{cm}$

In $\u25b3$ACD,

We apply the Pythagoras theorem

${\mathrm{CD}}^{2}+{\mathrm{AD}}^{2}={\mathrm{AC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{CD}}^{2}+{6}^{2}={10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{CD}}^{2}+36=100\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{CD}}^{2}=64\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{CD}=8\mathrm{cm}$

Thus, distance of the chord from the centre is 8 cm.

#### Page No 118:

#### Question 1:

The diameters PQ and RS of the circle with centre C are perpendicular to each other at C. state, why arc PS and arc SQ are congruent. Write the other arcs which are congruent to arc PS

#### Answer:

PQ perpendicular to RS.

$\angle \mathrm{PCS}=\angle \mathrm{PCQ}=90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{arc}\left(\mathrm{PS}\right)=\mathrm{arc}\left(\mathrm{QS}\right)\phantom{\rule{0ex}{0ex}}$

We know that if the measures of two arcs of circle are same then two arcs are congruent

$\Rightarrow \mathrm{arc}\left(\mathrm{PS}\right)\cong \mathrm{arc}\left(\mathrm{QS}\right)$

Similarly,

$\Rightarrow \mathrm{arc}\left(\mathrm{PS}\right)\cong \mathrm{arc}\left(\mathrm{PR}\right)\cong \mathrm{arc}\left(\mathrm{RQ}\right)$

#### Page No 118:

#### Question 2:

In the adjoining figure O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure. hence find the following

(1) *m*∠ AOB and *m*∠ COD

(2) Show that arc AB ≌ arc CD.

(3) Show that chord AB ≌ chord CD

#### Answer:

(1) MN is the diameter of the circle with centre O.

$\angle \mathrm{MOA}+\angle \mathrm{AOB}+\angle \mathrm{BON}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 100\xb0+\angle \mathrm{AOB}+35\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{AOB}=180\xb0-100\xb0-35\xb0=45\xb0\phantom{\rule{0ex}{0ex}}$

Similarly, $\angle \mathrm{COD}=45\xb0$

(2) Eqaual arcs subtend equal angles at the centre.

$\angle \mathrm{AOB}=\angle \mathrm{COD}=45\xb0$

So, arc AB ≌ arc CD

(3) Corresponding chords of congruent arcs are congruent.

arc AB ≌ arc CD

So, chord AB ≌ chord CD

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