Q) In the given figure PT is a tangent to the circle.

Given PT-20cm and PA-16cm.
(a) Prove AΡΤΒ ~ΔΡΑΤ
(b) Find the length of AB.

Chord BA produced meets the tangent PT at P. Given PT = 20cm and PA = 16cm.
(a) Prove Δ ΡΤΒ ~ Δ ΡΑΤ
(b) Find the length of AB.

ICSE Specimen Question Paper (SQP)2025

Ans:

a) Prove Δ ΡΤΒ ~ Δ ΡΑΤ:

Let’s compare Δ  PTB and Δ  PAT :

Here, ∠PTA = ∠ PBT   (by alternate segment theorem)

also ∠ APT = ∠ BPT    (common angle)

∴ by AA similarity  rule,

Δ  PTB ~ Δ  PAT .…….. Hence Proved !

b) Length of AB:

Let’s look at Δ  PTB and Δ  PAT  – we get these 2 triangles separately:

Given PT-20cm and PA-16cm.
(a) Prove AΡΤΒ ~ΔΡΑΤ
(b) Find the length of AB.

For better understanding, let’s rotate Δ  PAT, we get these 2 triangles:

Given PT-20cm and PA-16cm.
(a) Prove AΡΤΒ ~ΔΡΑΤ
(b) Find the length of AB.

Now, \frac{PA}{PT} = \frac{PT}{PB}

∴ PT = PA x PB

By substituting the values, we get:

(20) = 16 x PB

∴ PB = \frac{400}{16} = 25 cm

Now, from the original diagram, we can see that: Given PT-20cm and PA-16cm.
(a) Prove AΡΤΒ ~ΔΡΑΤ
(b) Find the length of AB.

PB = PA + AB

∴ 25 = 16 + AB

∴ AB= 25 = 16 = 9 cm

Therefore, length of AB is 9 cm.

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