In ΔABC, ∠A = 90º, AB = 16 cm and AC = 12 cm. D is the midpoint of AC and DE ⊥ CB at E. What is the area (in cm2) of ΔCDE?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is Option 1 i.e. 8.64
| Understanding | Application |
| Tricks |
Here
Given in the figure, ∠DCE = ∠ACB ∠CED = ∠CAB So, ΔCED ~ ΔCAB (AA rule) So CA/AB = CE/ED = 12/16 = 3/4 So Let, CE = 3x and ED = 4x Given D is mid-point of AC. so, DC = 6 ⇒ CE2 + ED2 = DC2 ⇒ (3x)2 + (4x)2 = 62 = 36 ⇒ 25x2 = 36 ⇒ x = 6/5 ⇒ arΔCED = 1/2 × 3x × 4x = 1/2 × 12x2 ⇒ arΔCED = 1/2 × 12 × 36/25 ⇒ arΔCED = 216/25 |
| Hence | arΔCDE = 8.64 |
