Circle : Chord And Arc Class 8th Mathematics (new) MHB Solution

Circle : Chord And Arc

Class 8th Mathematics (new) MHB Solution


Class 8th Mathematics (new) MHB Solution

Practice Set 17.1

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:

We know that,


The perpendicular from the centre of a circle to a chord bisects the chord.


Therefore, it is given that,


AB=13 cm


PQ perpendicular to AB


|(QB) = AB/2


| (QB) = 13/2


| (QB) = 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 the length of the chord is 48 cm.




Answer:

As we know that, the perpendicular from the centre of a circle to a chord bisects the chord.


Therefore, OP perpendicular to CD and OP bisects the CD. Therefore, it makes a right angle triangle, which is ∆OPD. We have OD=25 cm and PD=48/2=24 cm.


By Pythagoras theorem,


OD2 = OP2 + PD2


OP2 = OD2 - PD2


OP2= (25)2 - (24)2


OP2 = 625 – 576


OP2 = 49


OP = 7 cm


Therefore, distance of the chord from the centre is 7 cm.



Question 3.

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




Answer:


As we know that, the perpendicular from the centre of a circle to a chord bisects the chord.


So let P is the point, which bisects chord AB. So OP is perpendicular, it makes a right angle triangle ∆OPA.


Now we have OP = 9cm and AP as 12 cm


So by Pythagoras theorem,


AO2 = AP2 + PO2


AO2 = (12)2 + (9)2


AO2 = 144+81


AO2 = 225


AO = 15 cm


Length of radius 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.


Answer:


As we know that, the perpendicular from the centre of a circle to a chord bisects the chord.


So here we have C as a centre where CP is perpendicular on AB which bisects the chord AB and radius as CA = 10 cm and chord length = 12 cm, so AP=6cm.


It makes a right angle triangle ∆CPA.


Therefore, by using Pythagoras theorem, we have,


AC2 = CP2 + AP2


We have to find CP so


CP2 = AC2 - AP2


CP2 = (10)2 – (6)2


CP2 = 100 – 36


CP2= 64


CP = 8 cm


Therefore, a distance of the chord from the centre is 8 cm.




Practice Set 17.2

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:

As we know that, according to the theorem of the circle, two arcs are congruent, if their central angles are congruent, so arc PS and arc SQ are congruent because the angles between the chords are same and both are at 90° of the centre.

The other arcs, which are congruent to arcs PS, are


arc PS ≅ arc PR ≅ arc RQ because if two arcs of a circle are congruent, then their corresponding arcs are also congruent.



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) In given figure, we can see that


∠NOC + ∠COD +∠DOM = 180° (linear pair)


35° +∠COD+100° =180°


∠COD = 180° - 135°= 45°


So ∠COD and ∠AOB = 45°


(2) arc AB ≅ arc CD because the arcs are of equal measure 45° each angle and equal angle made equal sector.


(3) chord AB ≅ chord CD because corresponding chords of congruent arcs are congruent.


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