Mathematics Solutions for Class 8 Maths Chapter 18 - Circle Chord And Arc

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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.

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 = $\frac{1}{2} AB = \frac{13}{2} = 6.5$ 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, $PD = \frac{1}{2} CD = \frac{48}{2} = 24 cm$
In $△$ OPD,
We apply the Pythagoras theorem,
${OP}^{2} + {PD}^{2} = {OD}^{2} \Rightarrow {OP}^{2} + 24^{2} = 25^{2} \Rightarrow {OP}^{2} = 25^{2} - 24^{2} = 625 - 576 = 49 \Rightarrow {OP}^{2} = 49 \Rightarrow OP = 7 cm$
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, $AP = \frac{AB}{2} = \frac{24}{2} = 12 cm$
In $△$ OPA,
We apply the Pythagoras theorem,
${OP}^{2} + {AP}^{2} = {OA}^{2} \Rightarrow 9^{2} + 12^{2} = {OA}^{2} \Rightarrow {OA}^{2} = 81 + 144 = 225 \Rightarrow OA = \sqrt{225} = 15 cm$
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.
$AD = \frac{AB}{2} = \frac{12}{2} = 6 cm$
In $△$ ACD,
We apply the Pythagoras theorem
${CD}^{2} + {AD}^{2} = {AC}^{2} \Rightarrow {CD}^{2} + 6^{2} = 10^{2} \Rightarrow {CD}^{2} + 36 = 100 \Rightarrow {CD}^{2} = 64 \Rightarrow CD = 8 cm$
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.
$∠ PCS = ∠ PCQ = 90 ° \Rightarrow arc (PS) = arc (QS)$
We know that if the measures of two arcs of circle are same then two arcs are congruent
$\Rightarrow arc (PS) ≅ arc (QS)$
Similarly,
$\Rightarrow arc (PS) ≅ arc (PR) ≅ arc (RQ)$

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.
$∠ MOA + ∠ AOB + ∠ BON = 180 ° \Rightarrow 100 ° + ∠ AOB + 35 ° = 180 ° \Rightarrow ∠ AOB = 180 ° - 100 ° - 35 ° = 45 °$
Similarly, $∠ COD = 45 °$
(2) Eqaual arcs subtend equal angles at the centre.
$∠ AOB = ∠ COD = 45 °$
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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