Question

Of the Trapezium ABCD, AB ll CD. If AB=5.2cm, BC=3cm, AD=3.4cm & the dist. between the parallel sides is 2.5cm. then construct the Trapezium.

Answer 1

Of the Trapezium $ABCD$ , $AB \parallel CD$ . If $AB = 5.2 \, \text{cm}$ , $BC = 3 \, \text{cm}$ , $AD = 3.4 \, \text{cm}$ & the dist. between the parallel sides is $2.5 \, \text{cm}$ . Then construct the Trapezium.

Solution.

We know sides $AB = 5.2 \, \text{cm}$ , $BC = 3 \, \text{cm}$ , $AD$ and the distance between the parallel sides or the height $CH = DG$ . We have

$AB > BC$ & $AB > AD$

Then

$AB > CD$

We must find $CD$ for construction the Trapezium.

Consider the triangle $CHB$ . By the Pythagorean theorem:

$$B H = \sqrt {B C ^ {2} - C H ^ {2}} = \sqrt {3 ^ {2} - 2 . 5 ^ {2}} = \sqrt {2 . 7 5} (c m)$$

Consider the triangle $DGA$ . By the Pythagorean theorem:

$$G A = \sqrt {D A ^ {2} - D G ^ {2}} = \sqrt {3 . 4 ^ {2} - 2 . 5 ^ {2}} = \sqrt {5 . 3 1} (c m)$$

We have:

$$A B = A G + G H + H B$$ $$C D = H G:$$ $$A G + C D + H B = A B$$ $$C D = A B - A G - H B = 5. 2 - \sqrt {2 . 7 5} - \sqrt {5 . 3 1} \approx 1. 2 (c m)$$

Now we can construct the trapezium: