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Let ABC be a triangle of area 16 cm2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is

(a) 3.5 cm2 

(b) 3.7 cm2 

(c) 3.75 cm2 

(d) 4.0 cm2 

1 Answer

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(c) 3.75 cm2

Let the areas of ΔAXY and ΔBXY be A1 and A2 respectively. The heights of the ΔAXY and ΔBXY are equal, so 

\(\frac{A_1}{A_2}\) = \(\frac{\frac{1}{2}\times{AX}\times{height}}{\frac{1}{2}\times{BX\times{height}}}\)

⇒ \(\frac{A_1}{A_2}\) = \(\frac{AX}{BX} = \frac{3}{5}\)        .......(i)

Also, ΔAXY ~ ΔABC

⇒ \(\frac{\text{Area of ΔAXY}}{\text{Area of ΔABC}}\) = \(\big(\frac{AX}{AB}\big)^2\)

⇒ \(\frac{A_1}{16}\) = \(\bigg(\frac{3x}{3x+5x}\bigg)^2\) ⇒ \(\frac{A_1}{16}\) = \(\frac{9}{64}\)

⇒ A1\(\frac{9}{64}\) x 16 = \(\frac{9}{4}\)

∴ From (i) and (ii) A2\(\frac{5}{3}\) x A\(\frac{5}{3}\) x \(\frac{9}{4}\) = \(\frac{15}{4}\) = 3.75 cm2.

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