Question

Through a point on the hypotenuse of a right triangle lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is:

• A. $\frac{1}{4m}$
• B. $\frac{1}{2m+1}$
• C. $m$
• D. $\frac{1}{8m^2}$

Answer 1

Answer: (A)

$\frac{1}{4m}$

Consider the figure,

Right angled $△ABD$ is divided into two smaller right angled triangles - $AFC$ and $CED$ by a square $BECF$

Let a side of the square be $1$

Consider triangles - $AFC$ and $CED$,

$\angle AFC=\angle CED={90}^{\circ }$

$CF=CE=1$ ------- sides of a square are equal.

$\angle FAC=\angle CDE$ ------- angles opposite to equal sides

In two triangles, if the two corresponding angles are equal, then the third corresponding angles are also equal. The triangles are similar. Thus the ratio of the sides of the triangles are the same.

Area of triangle $=\frac{1}{2}\times base\times height=\frac{1}{2}bh=\frac{h}{2}$

The height of $△CDE$ with area $m$ and base $1$ is $2m$

$\therefore \frac{2m}{1}=\frac{1}{x}$ where, $x$ - base of $AFC$ triangle

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

Area of $△AFC=\frac{1}{2}\times base\times height=\frac{1}{2}\times 1\times \frac{1}{2m}=\frac{1}{4m}$

Area of $\square BFCE=length\times breadth=1\times 1=1$

$\therefore \frac{Areaof△AFC}{Areaof\square BFCE}=\frac{\frac{1}{4m}}{1}=\frac{1}{4m}$