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ABCD is a square. Draw a triangle QBC on side BC considering BC as base and draw a triangle PAC on AC as its base such that ∆QBC ~ ∆PAC. Then,
Area of ∆QBC is equal to : Area of ∆PAC
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- 1/2
- 2/1
- 1/3
- 2/3
- 1/2
Correct Option: A
From ∆ABC
AC = √AB² + BC²
= √BC² + BC²
= √2BC
∆QBC ~ ∆PAC
| ∴ | = | ||
| Area of ∆PAC | AC² |
| = | ||
| (√2BC)² |
| = | ||
| (√2BC)² |
