In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is :

ICSE Class 10 Mathematics PYQ · 2023 Paper · Section A

1(vi)1 Mark · 2023 · Section A · 2023 Paper

In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is :

In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is : ICSE 2023 Maths Solved Question Paper.
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Given,

∠BAP = ∠DCP = 70°

From figure,

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB // CD.

From figure,

PA = PC + AC = 6 + 4 = 10 cm.

In △ PAB and △ PCD,

⇒ ∠PAB = ∠PCD (Both equal to 70°)

⇒ ∠BPA = ∠DPC (Both are equal)

∴ △ PAB ~ △ PCD (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional to each other.

$$\begin{aligned}\Rightarrow \frac{PB}{PD} = \frac{PA}{PC} \\[1em] \Rightarrow \frac{PB}{PD} = \frac{10}{6} \\[1em] \Rightarrow \frac{PB}{PD} = \frac{5}{3}.\end{aligned}$$

Let PB = 5x and PD = 3x.

From figure,

DB = PB - PD = 5x - 3x = 2x.

$\Rightarrow \frac{PD}{DB} = \frac{3x}{2x} = \frac{3}{2}$

⇒ PD : DB = 3 : 2.

Hence, Option 3 is the correct option.

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