Question

In the adjacent figure, P and Q are points on the sides AB and AC respectively of a triangle ABC. PQ is parallel to BC and divides the triangle ABC into 2 parts, equal in area. The ratio of $PA:AB=$

• A. $1:1$
• B. $(\sqrt{2}-1):\sqrt{2}$
• C. $1:\sqrt{2}$
• D. $(\sqrt{2}-1):1$

Answer 1

Answer: (C)

$1:\sqrt{2}$

Given that Area of the $\Delta$ APQ $=$ Area of PQCB
That means Area $\Delta$ ABC $=$ 2 Area of $\Delta$ APQ
Since PQ $\parallel$ BC
Therefore, $\Delta$ APQ is similar to $\Delta$ ABC
We know that ratio of the areas of two triangles is equal to the square of ratio of their sides in case of similar triangles.
Therefore, $\frac{Areaof△APQ}{Areaof△ABC}$ = $\frac{{PA}^2}{{AB}^2}$
$\frac{{PA}^2}{{AB}^2}=$$\frac{Areaof△APQ}{Areaof△ABC}$ $=\frac{1}{2}$
$\frac{PA}{AB}$ $=\sqrt{\frac{1}{2}}$
Therefore, $PA:AB$ = $1:$ $\sqrt{2}$