Question

In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Answer 1

Given: ABCD is a trapezium, where AB = 7 cm, AD = BC = 5 cm, DC = x cm, and 

Distance between AB and DC = 4 cm 

Consider AL and BM are perpendiculars on DC, then 

AL = BM = 4 cm and LM = 7 cm. 

In right ΔBMC : 

Using Pythagorean Theorem, we get 

BC² = BM² + MC² 

25 = 16 + MC² 

MC² = 25 – 16 = 9 

or MC = 3 

Again, 

In right Δ ADL : 

Using Pythagorean Theorem, we get 

AD² = AL² + DL² 

25 = 16 + DL² 

DL² = 25 – 16 = 9 

or DL = 3 

Therefore, x = DC = DL + LM + MC 

= 3 + 4 + 3 

= 13 

=> x = 13 cm 

Now, Area of trapezium ABCD = 1/2(AB + CD) AL 

= 1/2(7+13)4 

= 40 

Area of trapezium ABCD is 40 cm².