Question

In the given figure, AX : XB = 3 : 5 and XY || BC.

Then, the ratio of the areas of trapezium XBCY and triangle ABC is

• A. 64:45
• B. 45:64
• C. 55:64
• D. 64:55

Answer 1

The correct option is C

55:64

We are given in the $\Delta ABC$, AX : XB =3 : 5 and XY || BC.

Let AX = 3x and XB = 5x

$\therefore AB=3x+5x=8x.$

Now in $\Delta ABC$ and $\Delta AXY,$

$\angle ABC=\angle AXY$ (corresponding angles)

$\angle A=\angle A$ (common)

$\therefore \Delta ABC\sim \Delta AXY$ (by AA similarity citerion)

So, we have

$\frac{area\left(\Delta ABC\right)}{area\left(\Delta AXY\right)}=\frac{A{B}^2}{A{X}^2}={\left(\frac{8x}{3x}\right)}^2=\frac{64x^2}{9x^2}$ or $\frac{64}{9}$ ....... (i)

Subtracting 1 from each side we get,

$\frac{area\left(\Delta ABC\right)}{area\left(\Delta AXY\right)}-1=\frac{64}{9}$ - 1

$\Rightarrow \frac{area\left(\Delta ABC\right)-area\Delta AXY}{area\Delta AXY}=\frac{64-9}{9}$

$\Rightarrow \frac{\text{area of trapezium XBCY}}{\text{area}\Delta AXY}=\frac{55}{9}$ ....... (ii)

Dividing equation (ii) by equation (i) we get,

$\therefore$ area of trapezium XBCY : area of $\Delta ABC=55:64$