Question

In ΔABC = 90° AB = 16 cm and AC = 12 cm. D is the mid-point of AC and DE ⊥ CB at E. What is the area (in cm²) of ΔCDE?

पर्याय

• 8.64

• 7.68

• 5.76

• 6.25

Answer 1

8.64

Explanation:

In ΔCAB and ΔCED, we get

∠CAB = ∠CED = 90°

And ∠C is common in both

So, ΔCAB ∼ ∠CED

CA `↔` CE

AB `↔` ED

And CB `↔`CD

⇒ `(CE)/(CA) = (ED)/(AB) = (CD)/(AB)` ......(i)

Now from question,

CB = `sqrt(CA^2 + AB^2)`

= `sqrt(12^2 + 16^2)`

= `sqrt(144 + 256)`

= `sqrt(400)`

= 20 cm

So from (i)

`(CE)/12 = (ED)/16 = 6/20`

Hence, CE = `6/20 xx 12 = 36/10` cm

And ED = `6/20 xx 16 = 48/10` cm

Required area of ΔCDE = `1/2` × CE × ED

= `1/2 xx 36/10 xx 48/10`

= 8.64 cm²