In the given figure (not drawn to scale) chords AD and BC intersect

Q) In the given figure (not drawn to scale) chords AD and BC intersect at P, where AB = 9 cm, PB = 3 cm and PD = 2 cm.

(a) Prove that ∆APB ~ ∆CPD.

(b) Find the length of CD.

(c) Find area ∆APB : area ∆CPD.

ICSE Specimen Question Paper (SQP)2026

Ans:

a) Prove Δ AΡΒ ~ Δ CPD:

Let’s compare Δ APB and Δ CPD :

Here, ∠BAP = ∠ BCD (angles subtended by same segment)

similarly, ∠ ABP = ∠ CDP (angles subtended by same segment)

∴ by AA similarity rule,

Δ APB ~ Δ CPD .…….. Hence Proved !

b) Length of CD:

Let’s look at Δ APB and Δ CPD:

We just proved that these are similar triangles,

∴ $\frac{AB}{CD} = \frac{BP}{PD} = \frac{AP}{CP}$ … (i)

By substituting the given values, we get:

$\frac{9}{CD} = \frac{3}{2}$

∴ CD = $\frac{9 \times 2}{3}$

∴ CD = 6 cm

Therefore, length of CD is 6 cm.

(c) Area ∆APB : Area ∆CPD:

Since Area of a triangle is given by 1/2 x Height x Base

∴ Area ∆APB = $\frac{1}{2}$ x AP x BP

Similarly, Area ∆CPD = $\frac{1}{2}$ x CP x PD

∴ Area ∆APB : Area ∆CPD = $\frac{\frac{1}{2} \times AP \times BP}{\frac{1}{2} \times CP \times PD}$

= $\frac{AP \times BP}{CP \times PD}$

= $\frac{AP}{CP} \times\frac{BP}{PD}$

From equation (i), we have $\frac{BP}{PD} = \frac{AP}{CP}$

∴ Area ∆APB : Area ∆CPD = $\frac{BP}{PD} \times\frac{BP}{PD}$ = $\frac{BP^2}{PD^2}$

By substituting the given values, we get:

Area ∆APB : Area ∆CPD = $\frac{3^2}{2^2}$ = $\frac{9}{4}$

Therefore, Area ∆APB : Area ∆CPD = 9:4

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