In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD
Solution:
Given, AB and CD are the common tangents to two circles of unequal radii.
We have to prove that AB = CD.
Extend AB and CD such that it intersects at P.
We know that the tangents to a circle through an external point are equal.
Considering smaller circle,
The tangents are PB and PD
So, PB = PD ---------- (1)
Considering the larger circle,
The tangents are PA and PC
So, PA = PC ---------- (2)
Subtracting (1) and (2),
PA - PB = PC - PD
From the figure,
PA - PB = AB
PC - PD = CD
Therefore, AB = CD
✦ Try This: In the given figure, the length of tangents PA and PD are 8 cm and 3 cm respectively. Find the length of CD and AB.
Given, the length of tangent PA = 8 cm
The length of tangent PD = 3 cm
We have to find the length of CD and AB
We know that the tangents to a circle through an external point are equal.
So, PA = PC
Now, PC = 8 cm
Also, PD = PB
So, PB = 3 cm
From the figure,
AB = PA - PB
AB = 8 - 3 = 5 cm
CD = PC - PD
CD = 8 - 3 = 5 cm
Therefore, the length of AB and CD 5 cm and 5 cm.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 5
In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD
Summary:
In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. It is proven that AB = CD
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