Construct a line passing through AD
Now, AD and CD are tangents to the circle with centre O from the external point D.
So, AD = CD (Tangents drawn from an external point to a circle are equal.) .....(1)
Also, AB and AD are the tangents to the circle with centre O' from the external point A.
So, AD = AB (Tangents drawn from an external point to a circle are equal.) .....(2)
From (1) and (2)
AB = CD
Hence Proved.
As indicated by the inquiry,
Abdominal muscle = CD
Development: Produce AB and CD, to converge at P.
Confirmation:
Think about the circle with more noteworthy span.
Digressions drawn from an outside highlight a circle are equivalent
AP = CP … (1)
Too,
Think about the circle with more modest sweep.
Digressions drawn from an outer highlight a circle are equivalent
\[BP\text{ }=\text{ }BD\text{ }\ldots \text{ }\left( 2 \right)\]
Deduct Equation (2) from (1). We Get
\[AP\text{ }\text{ }BP\text{ }=\text{ }CP\text{ }\text{ }BD\]
Abdominal muscle = CD
Thus Proved.
Text Solution
Solution : Given AB and CD are common tangent to two circles of unequal radius <br> To prove AB=CD <br> <img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_X_C09_S01_025_S01.png" width="80%"> <br> Construction Produce AB and CD, to intersect at P. <br> Proof PA=PC <br> [the length of tangents drawn from an internal point to a circle are equal] <br> Also, PB=PD <br> [the lengths of tangents drawn from an internal point to a circle are equal] <br> `:.PA-PB=PC-PD` <br> AB=CD Hence proved.
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Question 5In figure AB and CD are common tangents to two circles of unequal radii prove that AB = CD
2
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
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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