Mathematics Solutions Solutions for Class 8 Maths 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 Maths 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 Maths 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 Maths are prepared by experts and are 100% accurate.

#### Question 1:

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

#### 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. Perpendicular drawn from the centre of the circle to the chord bisects the chord.
So,
In $△$OPD,
We apply the Pythagoras theorem,

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

#### 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. 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,
In $△$OPA,
We apply the Pythagoras theorem,

Hence, the radius of the circle is 15 cm.

#### 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. C is the centre of the circle.
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.

In $△$ACD,
We apply the Pythagoras theorem

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

#### 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 PQ perpendicular to RS.

We know that if the measures of two arcs of circle are same then two arcs are congruent
$⇒\mathrm{arc}\left(\mathrm{PS}\right)\cong \mathrm{arc}\left(\mathrm{QS}\right)$
Similarly,
$⇒\mathrm{arc}\left(\mathrm{PS}\right)\cong \mathrm{arc}\left(\mathrm{PR}\right)\cong \mathrm{arc}\left(\mathrm{RQ}\right)$

#### 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

(1) MN is the diameter of the circle with centre O.
$\angle \mathrm{MOA}+\angle \mathrm{AOB}+\angle \mathrm{BON}=180°\phantom{\rule{0ex}{0ex}}⇒100°+\angle \mathrm{AOB}+35°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{AOB}=180°-100°-35°=45°\phantom{\rule{0ex}{0ex}}$
Similarly, $\angle \mathrm{COD}=45°$
(2) Eqaual arcs subtend equal angles at the centre.
$\angle \mathrm{AOB}=\angle \mathrm{COD}=45°$
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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