Question

ABCD is a trapezi perpendicularum in which AB || CD AD perpendicular to DC AB = 20 cm, BC = 13 cm and DC = 25 cm. find the area of the trapezium.

Answer 1

$\mathfrak{ \huge{solution}}$

From B draw BP perpendicular DC

∴ AB = DP = 20 cm

so, PC = DC - DP

= (25 - 20) cm = 5 cm

Now, area of trapezium ABCD = Area of rectangle ABPD + Area of ∆ BPC

∆ BPC is right angled at ∠BPC

∴ Using Pythagoras theorem, we have

BC² = BP² + PC²

13² = BP² + 5²

169 = BP² + 25

169 - 25 = BP²

144 = BP²

$\implies \boxed{ \bold{BP = 12}}$

Now,

Area of trapezium ABCD = Area of rectangle ABPD + Area of ∆BPC

$= AB \: \times BP + \frac{1}{2} \times PC \times BP$

$= 20 \times 12 + \frac{1}{2} \times 5 \times 12$

= 240 + 30

${ { \boxed{270 \: \bold{cm}^{2} }}}$

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Answer 2

plz refer the attachment for the answer of your question.

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