Question 123

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 $$\triangle QBC\sim\triangle PAC$$, Then $$\frac{Area\ of\ \triangle QBC}{Area\ of\ \triangle PAC}$$ is equal to:

Solution

Let side of square ABCD = 1 unit

=> Diagonal AC = $$\sqrt{1^2+1^2}=\sqrt2$$ units

It is given that $$\triangle QBC\sim\triangle PAC$$

Ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides.

=> $$\frac{Area\ of\ \triangle QBC}{Area\ of\ \triangle PAC}=\frac{(BC)^2}{(AC)^2}$$

= $$\frac{1^2}{(\sqrt2)^2}$$

= $$\frac{1}{2}$$

=> Ans - (C)

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