In Fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC =

1.	In Fig. 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» Draw AL perpendicular to DC And, BM perpendicular DC Then, AL = BM = 4 cm And, LM = 7 cm In Δ ADL, we have AD² = AL²⁺ DL² 25 = 16 + DL² DL = 3 cm Similarly, MC = \(\sqrt{BC^2-BM^2}\) = \(\sqrt{25-16}\) = 3 cm Therefore, x = CD = CM + ML + LD = 3 + 7 + 3 = 13 cm Area of trapezium ABCD = \(\frac{1}{2}\)(AB + CD) x AL = \(\frac{1}{2}\)(7 + 13) x 4 = 40 cm²

In Fig. 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»

Draw AL perpendicular to DC

And,

BM perpendicular DC

Then,

AL = BM = 4 cm

And,

LM = 7 cm

In Δ ADL, we have

AD² = AL²⁺ DL²

25 = 16 + DL²

DL = 3 cm

Similarly,

MC = \(\sqrt{BC^2-BM^2}\)

= \(\sqrt{25-16}\)

= 3 cm

Therefore,

x = CD

= CM + ML + LD

= 3 + 7 + 3

= 13 cm

Area of trapezium ABCD = \(\frac{1}{2}\)(AB + CD) x AL

= \(\frac{1}{2}\)(7 + 13) x 4

= 40 cm²