Question

If the sides AB, BC, CD and DA of a trapezium ABCD measure 10 cm, 20 cm, 18 cm and 16 cm respectively then find the length of the longer diagonal given that AB is parallel to CD

• A. $\sqrt{760}$cm
• B. $\sqrt{231}$cm
• C. $\sqrt{54}$cm
• D. $\sqrt{930}$cm

Answer 1

The correct option is A $\sqrt{760}$cm

$\Rightarrow$ Draw two altitudes from A and B onto CD and call the intersection P and Q.

$\Rightarrow$ Then Two rigjht triangles will be formed, $△$APD and $△$BQC.

$\Rightarrow$ $△$BQC has sides $x,h,20$.

$\Rightarrow$ and $△$APD has sides $x-8,h,16$.

$\Rightarrow$ Applying Pythagoras theorem in $△$APD and $△$BQC,

$\Rightarrow$ $x^2+h^2={20}^2$

$\Rightarrow$ $x^2+h^2=400$ ------- (1)

$\Rightarrow$ $(x-8{)}^2+h^2={16}^2$

$\Rightarrow$ $x^2-16x+64=256$ -------- (2)

Now subtracting equation (1) and (2) we get,

$\Rightarrow$ $16x-64=400-256$

$\Rightarrow$ $16x=144+64$

$\Rightarrow$ $16x=208$

$\Rightarrow$ $x=13$

This shows $QC=13$ and $QD=5$

So that the longer diagonal must be AC rather than BD.

Now to find the length of AC, look at the right triangle APC, with sides $23,h,d$.

$\Rightarrow$ $d^2=h^2+{23}^2$

$\Rightarrow$ $d^2={20}^2-{13}^2+{23}^2$

$\Rightarrow$ $d^2=400-169+529$

$\Rightarrow$ $d^2=760$

$\Rightarrow$ $d=\sqrt{760}cm$