Question

Area of the trapezium is

• A. $\frac{75}{4}$
• B. $\frac{35}{4}$
• C. $\frac{125}{4}$
• D. $\frac{45}{4}$

Answer 1

The correct option is A $\frac{75}{4}$

Since the diagonals pass through $(1,0)$, let the equation be $y=m(x-1)$ where m is the slope of the diagonals.

To find the vertices of the trapezium, we equate the equation of diagonals to $y^2=4x$, i.e. $m^2(x-1{)}^2=4x$.

After simplifying the equation becomes, $m^2{x}^2+(-2m^2-4)x+m^2=0$.

Using the quadratic formula, we get $x=\frac{m^2+2\pm 2\sqrt{m^2+1}}{m^2}$.

Since $y=m(x-1)$, we get $y=\frac{2\pm 2\sqrt{m^2+1}}{m}$. The extremities of the diagonals can be found by taking the positive sign for the first and negative sign for the second.

Using the distance formula, we get length of the diagonal in terms of $m$ as, $l=\sqrt{(\frac{4\sqrt{m^2+1}}{m^2}{)}^2+(\frac{4\sqrt{m^2+1}}{m}{)}^2}$ $=\frac{25}{4}$.

Hence, on simplifying we get the equation $\frac{4(m^2+1)}{m^2}=\frac{25}{4}$.

That gives us, $m=\pm \frac{4}{3}$.

Hence, by putting the values of m, we get $x=4,\frac{1}{4}$ and $y=\pm 4,\pm 1$.

The four vertices of the trapezium are $A(4,4),B(4,-4),C(\frac{1}{4},1)$ and $D(\frac{1}{4}.-1)$.

So, $AB=8$ and $CD=2$, also distance between $AB$ and $CD$ is $\frac{15}{4}$.

So, area of trapezium $ABCD$ $=$ $0.5\times (AB+CD)\times \text{height}=$ $0.5\times (8+2)\times \frac{15}{4}=\frac{75}{4}$ sq. units.

So, the correct option is (A).