Question

In a trapezium, the two non-parallel sides are equal in length, each being of $5cm$. The parallel sides are at a distance of $3cm$. If the smaller side of the parallel side is of length $2cm$, then the length of the diagonal of the trapezium is:

• A. $3\sqrt{5}cm$
• B. $6\sqrt{5}cm$
• C. $4\sqrt{5}cm$
• D. $7cm$

Answer 1

The correct option is A $3\sqrt{5}cm$

Given - non-parallel side of trapezium are equal and measure $5cm$ and Perpendicular distance is $3cm$ with smaller $\parallel$ side equals to $2cm$.

Now let trapezium is $ABCD$

In $\Delta ACE$ & $\Delta BDF$ by Pythagoras theorem

$C{E}^2=A{C}^2-A{E}^2$ $F{D}^2=B{D}^2-B{F}^2$

$=25-9$ $=25-9$

$=16$ $=16$

$CE=\sqrt{16}$ $FD=\sqrt{16}$

$CE=4$ $FD=4$

Now $AB=EF$ because opposite sides of rectangle $ABEF$

So $EF=2cm$

Now see triangle $BCF$

In $\Delta BCF$ by Pythagoras theorems

$C{B}^2=C{F}^2+B{F}^2$

$C{B}^2=(6{)}^2+(3{)}^2$

$C{B}^2=36+9$

$C{B}^2=45$

$CB=\sqrt{45}=\sqrt{9\times 5}$

$CB=3\sqrt{5}$

CB is our diagonal.