Question

In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°. Find the

1. area of ΔABC
2. length of AD if AD ⊥ BC
3. area of ΔABD

Answer 1

Given: In triangle ABC,

• AB = 15 cm
• AC = 20 cm
• ∠BAC = 90°   ...(So, triangle ABC is right-angled at A)

i. Find the area of ΔABC:

Step 1: Since ∠BAC = 90°, AB and AC are the perpendicular sides legs of the right triangle. 

Step 2: Area of right-angled triangle = `1/2` × base × height

Step 3: Base = AB = 15 cm, Height = AC = 20 cm

Step 4: Area = `1/2 xx 15 xx 20`

= `1/2 xx 300`

= 150 cm²

Area of ΔABC = 150 cm².

ii. Find the length of AD if AD ⊥ BC   ...(AD is the altitude from A to BC)

Step 1: Find BC using Pythagoras theorem:

`BC = sqrt(AB^2 + AC^2)`

= `sqrt(15^2 + 20^2)`

= `sqrt(225 + 400)`

= `sqrt(625)`

= 25 cm

Step 2: Area of ΔABC can also be expressed using base BC and height AD:

`"Area" = 1/2 xx BC xx AD`

Step 3: From the area found above 150 cm², 

`150 = 1/2 xx 25 xx AD`

⇒ 150 = 12.5 × AD

⇒ `AD = 150/12.5`

⇒ AD = 12 cm

Length of AD is 12 cm.

iii. Find the area of ΔABD:

Step 1: Since AD is perpendicular to BC, ΔABD is a right triangle with base BD and height AD.

Step 2: To find BD, use the fact that in a right triangle,

`BD = (AB^2)/(BC)`

= `15^2/25`

= `225/25`

= 9 cm

Step 3: Area of ΔABD

= `1/2 xx BD xx AD`

= `1/2 xx 9 xx 12`

= `1/2 xx 108`

= 54 cm²

Area of ΔABD = 54 cm².