Question

Q.2. Find the length of the side BC of the given
trapezium ABCD.
​

Answer 1

Answer

The length of the side BC is 8cm

Given

ABCD is a trapezium where :

• AB = 9cm
• AD = 10cm
• CD = 3cm

To Find

• The side BC

Construction

DE ⏊ AB is drawn so that BCDE become a rectangle.

Solution

We have from construction ,

BCDE is a rectangle ,

∴ BC = DE and CD = BE

Now , from figure

∆AED is right angled triangle since DE ⏊ AB,

From Pythagorean triplet we get ,

$\sf\implies AD^{2} = DE^{2} + AE^{2} \\\\ \sf\implies 10^{2} = BC^{2} + (AB - BE)^{2} \: \: \: \{ \because BC = DE \} \\\\ \sf\implies AD^{2}= BC^{2} + (AB - CD)^{2} \: \: \: \: \{\because CD = BE\} \\\\ \sf\implies BC^{2} = AD^{2} - (AB - CD)^{2} \\\\ \sf\implies BC^{2} = 10^{2} - (9-3)^{2} \\\\ \sf\implies BC^{2} = 10^{2} - 6^{2} \\\\ \sf\implies BC^{2} = 100 - 36 \\\\ \sf\implies BC^{2} = 64 \\\\ \sf\implies BC^{2} = 8^{2} \\\\ \sf\implies BC = 8$

Therefore , the length of the side BC is 8cm