Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17A Solutions
EXERCISE – 17A
(Q1) A chord of length 6cm is drawn in a circle of radius 5cm calculate its distance from the centre of the circle.
Solution:
By Pythagoras theorem,
AO2 = OD2 + AD2
(5)2 = OD2 + (3)2
25 = OD2 + 9
25 – 9 = OD2
16 = OD2
Taking square root on both sides,
√16 = √OD2
4cm = OD
(Q2) A chord of length 8cm is drawn at a distance of 3cm from the centre of a circle. Calculate the radius of a circle.
Solution:
In △ODA
By Pythagoras Theorem,
(Hypotenuse)2 = (Base)2 + (height)2
(AO)2 = AD2 + OD2
AO2 = (4)2 + (3)2
AO2 = 16 + 9
AO2 = 25
Taking square root on both sides,
√AO2 = √25
AO = 5
(Q3) The Radius of a circle is 17.0 cm and the length of perpendicular is drawn from its centre to a chord is 8.0 cm. Calculate the length of the chord.
Solution:
In △ODA,
By Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
AO2 = AD2 + OD2
(17)2 = AD2 + (8)2
289 = AD2 + 64
289 – 64 = AD2
Taking square root on both sides,
√225 = √AD2
215 = AD
∴ AB = 2×15 = 30cm
(Q4) A chord of length 24 cm is at a distance of 5cm from the centre of the circle find the length of the chord of the same circle which is at a distance of 12m from the centre.
Solution:
In △OPA, by Pythagoras theorem,
OA2 = OP2 + AP2
OA2 = (5)2 + (12)2
OA2 = 25 + 144
OA2 = 169
Taking square root on both sides,
√OA2 = √169
OA = 13 cm
In △CQO, by Pythagoras theorem,
CQ2 = QO2 + CQ2
(13)2 = (12)2 + CQ2
169 = 144 + CQ2
169 – 144 = CQ2
25 = CQ2
Taking square root on both sides,
√25 = √CQ2
5 = CQ
CD = 2×5
CD = 10cm
(Q5) In the following figure, AD is a straight line. OP⊥AD and O is the centre of both circles. If OA = 34 cm, OB = 20 cm and OP = 16cm, find the length of AB
Solution:
In △OPA, by Pythagoras theorem,
OA2 = OP2 + AP2
(34)2 = (16)2 + AP2
(34)2 – (16)2 = AP2
By using formula,
A2 – b2 = (a + b) (a – b)
(34 + 16) (34 – 16) = AP2
(50) (18) = AP2
Taking square root on both sides,
√50×18 = AP2
√2×25×2×3×3 = AP
√2×5×5×2×3×3 = AP
2×5×3 = AP
30 cm = AP
In △OPB, By Pythagoras theorem,
OB2 = OP2 + BP2
(20)2 = (16)2 + BP2
(20)2 – (16)2 = BP2
(20 + 16) (20 – 16) = BP2
36 × 4 = BP2
Taking square root on both sides,
√36×4 = BP
√144 = BP
12 cm = BP
AB = AP – BP
= 30 – 12
AB = 18 cm
∴ The length of AB = 18cm
Here is your solution of Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17A
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