KVPY 2016 SA Question 2

Let the height of the trapezium be $h$ . By Pythagorean theorem, we have:

$\begin{aligned} \sqrt{AD^2-h^2} + \sqrt{CB^2-h^2} & = AB-CD \\ \sqrt{3^2-h^2} + \sqrt{4^2-h^2} & = 11-6 \\ \sqrt{9-h^2} + \sqrt{16-h^2} & = 5 & \small \color{#3D99F6} \text{Squaring both sides.} \\ 9-h^2 + 2 \sqrt {(9-h^2)(16-h^2)} + 16-h^2 & = 25 \\ 2 \sqrt {(9-h^2)(16-h^2)} & = 2h^2 \\ \sqrt {(9\times 16 - 25h^2 + h^4} & = h^2 & \small \color{#3D99F6} \text{Squaring both sides.} \\ 9\times 16 - 25h^2 + h^4 & = h^4 \\ h^2 & = \frac {9 \times 16}{25} \\ \implies h & = \frac {3\times 4}5 \\ & = \boxed{2.4} \end{aligned}$

Through $D$ , draw a line parallel to $BC$ . Let it meet $AB$ in $E$ . Then, quadrilateral $BCDE$ is a parallelogram because its opposite sides are parallel. Therefore, $DE=4\\$ . And, $\\AE=AB-BE=11-DC=5\\$ .

Thus, $\Delta ADE$ is a 3-4-5 right triangle. Thus, if $h$ is the height(length of perpendicular from A) of the triangle then, $\operatorname{ar}{(\Delta ADE)}=\frac{1}{2}3\times4=\frac{1}{2}5\times h\\ \boxed{h=2.4}$

My approach is Algebraic.

For this one important formula is to be applied, which is that area of a quadrilateral is $\large\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi perimeter of the quadrilateral and $\large a,b,c,d$ are the sides. And for obtaining the height we have to equate this with the normal formula for calculating the area of a Trapezium which is $\large\frac{1}{2}(sum\hspace{1mm}of \hspace{1mm}parallel\hspace{1mm}sides)*height$ .

We get $\large s=24$ and the sum of the parallel sides as $\large AB+CD=17$ .

so $\large12\sqrt{3}$ = $\large\frac{1}{2}*17*height$ . and the height comes out to be

$\large=\frac{24\sqrt{3}}{17}$ $\large\approx2.44$