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Learn/ 10th Mathematics/ Unit 8 /Numericals

Unit 8: Chords and Arcs — Exercise 8 1

10th Class Mathematics · Unit 8: Chords and Arcs

8.1.1.Calculate the length of a chord which stands at a distance of 5 cm from the centre of a circle whose radius is 13 cm.
Given
Setup OM perp AB, OM = 5 cm, OB = 13 cm
Formula
Pythagoras theorem MB2 + OM2 = OB2
Working
Substituting MB2 + 52 = 132
MB2 = 169 - 25 = 144 ⇒ MB = sqrt{144} = 12 cm
Result
Length of chord AB = 2 × MB = 2 × 12 = 24 cm
8.1.2.In construction, three steel rods are fixed at points A, B, and C (not in a straight line). A circular hoop needs to pass through all three. How many hoops can be used?
Given
Points A, B, C are non-collinear
Three non-collinear points determine exactly one unique circle.
Result
Number of hoops Number of hoops = 1
8.1.3.In a park, lamp posts are 14 m apart on the edge of a circular part of radius 10 m as shown in figure. Find the distance of the chord from the centre of the park.
Given
Given AB = 14 m ⇒ AM = MB = 7 m
Formula
In right \Delta OMA OM2 + AM2 = OA2
Working
Substituting d2 + 72 = 102
d2 = 100 - 49 = 51 ⇒ d = sqrt{51} = 7.14 m (approx.)
Result
Distance of chord from centre d = 7.14 m
8.1.4.In a circle, chords AB and CD both have length 10 cm. If the distance from the centre to AB is 6 cm, what is the distance from centre to CD?
Given
Given AB = CD = 10 cm, OM = 6 cm, let radius = r
Formula
In right \Delta OMB MB2 + OM2 = OB2
Working
Substituting MB = AB2 = 102 = 5 cm, quad 52 + 62 = r2
25 + 36 = r2 ⇒ r2 = 61
Formula
In right \Delta OND ND2 + ON2 = r2
Working
Substituting 52 + ON2 = 61
ON2 = 61 - 25 = 36 ⇒ ON = sqrt{36} = 6 cm
Result
Distance from centre to CD ON = 6 cm
8.1.5.Two holes A and B are drilled 12 cm apart on a circular tabletop of radius 10 cm. Find the perpendicular distance from the centre to AB.
Given
Given OM perp AB, AO = BO = r = 10 cm, AB = 12 cm
Working
Half chord AM = AB2 = 122 = 6 cm
Formula
Pythagoras theorem AO2 = OM2 + AM2
Working
Substituting 102 = OM2 + 62 ⇒ 100 = OM2 + 36
OM2 = 100 - 36 = 64 ⇒ OM = sqrt{64} = 8 cm
Result
Distance from centre to AB OM = 8 cm
8.1.6.A chord 8 cm long is at a distance of 3 cm from the centre. Calculate the radius of the circle.
Given
Given OM perp AB, AB = 8 cm, OM = 3 cm
Working
Half chord AM = AB2 = 82 = 4 cm
Formula
In right \Delta OMA OA2 = OM2 + AM2
r2 = 32 + 42 = 9 + 16 = 25 ⇒ r = sqrt{25} = 5 cm
Result
Radius of circle r = 5 cm
8.1.7.In the given figure, mCD = 121 units and mBC = 3x units. Find the value of x.
Given
Property If two chords are equidistant from the centre, then they are equal.
AB perp CD, so AB is perpendicular bisector of CD ⇒ CB = BD
Formula
Chord relation CD = CB + BD = 2(CB)
Working
Substituting 121 = 2(3x) = 6x
Result
Value of x x = 1216 units
8.1.8.In a circle with centre at O, the perpendicular distance of the each chord PQ and RS from the centre is 6 cm. If the length of chord PQ is 18 cm, find the length of the other chord.
Given
Given Both chords are at the same distance (6 cm) from the centre
Formula
Equal distance implies equal chords PQ = RS
Working
Given PQ = 18 cm
Result
Length of other chord RS = 18 cm
8.1.9.A line from the centre of a circle cuts a 10 cm chord at right angle where radius of the circle is 6 cm. What is the length from the centre to the chord?
Given
Given OM perp AB, AB = 10 cm, OA = 6 cm
Working
Half chord AM = AB2 = 102 = 5 cm
Formula
In right \Delta OMA OA2 = OM2 + AM2
Working
Substituting 62 = OM2 + 52 ⇒ 36 = OM2 + 25
OM2 = 36 - 25 = 11 ⇒ OM = sqrt{11} cm
Result
Distance from centre to chord OM = sqrt{11} cm
8.1.10.In a circle, a perpendicular is drawn from the centre to chord AB. If mAB = 12 cm, what is the length of each segment after bisecting?
Given
Given OM perp AB, therefore it bisects AB
Working
Substituting AM = MB = AB2 = 122 = 6 cm
Result
Length of each segment AM = MB = 6 cm