Question

In the given figure, ΔABC is right angled triangle with ∠A = 90°. AD is perpendicular to BC.

Prove that:

1. ΔDBA ∼ ΔDAC
2. DA² = DB × DC
3. Find the area of ΔABC when DB = 9 cm and DC = 16 cm.

Answer 1

Given:

ΔABC is right angled at A (i.e., ∠A = 90°)

AD is perpendicular to BC

To prove:

1. ΔDBA ∼ ΔDAC
2. DA² = DB × DC
3. Find the area of ΔABC when DB = 9 cm and DC = 16 cm.

i. Prove ΔDBA ∼ ΔDAC

In ΔDBA and ΔDAC:

∠ADB = ∠ADC = 90°   ...(Since AD ⊥ BC)

∠DBA = ∠DAC  ...(Common angles in the triangles)

By AA (Angle-Angle) similarity criterion,

ΔDBA ∼ ΔDAC

ii. Prove DA² = DB × DC

Since ΔDBA ∼ ΔDAC, corresponding sides are proportional:

`(DA)/(DB) = (DC)/(DA)`

⇒ DA² = DB × DC

iii. Find the area of ΔABC when DB = 9 cm and DC = 16 cm

From part (ii),

DA² = DB × DC

= 9 × 16

= 144

⇒ DA = 12 cm

Since ΔABC is right angled at A, AD is the height from A to hypotenuse BC.

Length of BC = DB + DC

= 9 + 16

= 25 cm

Area of ΔABC is given by:

Area = `1/2 xx BC xx AD`

= `1/2 xx 25 xx 12`

= 150 cm²