Question

in the figure the area of trapezium ABCD is 100 square.cm and the area of triangle ADC is 40sq. cm. find the areq of triangle DEC​

Answer 1

Solution :-

since AB || DC in given trapezium ABCD .

In ∆AEB and ∆CED we have,

→ ∠BAE = ∠DCE (Alternate angles.)

→ ∠ABE = ∠CDE (Alternate angles.)

so,

→ ∆AEB ~ ∆CED (By AA similarity.)

we know that,

• when two triangles are similar, ratio of their areas is equal to square of their corresponding sides .

Let AB = a cm and CD = b cm .

then,

→ Area ∆AEB / Area ∆CED = a² / b²

and,

→ Area ∆AED / Area ∆BEC = ab / ab = 1

given that,

→ Area trapezium ABCD = 100 cm²

→ Area ∆ADC = 40 cm²

so,

→ a² + b² + ab + ab = 100

→ (a + b)² = 100

→ (a + b) = 10 ------- Eqn.(1)

and,

→ b² + ab = 40 -------- Eqn.(2)

then,

→ a² + ab = 60 --------- Eqn.(3)

possible values of (a , b) from Eqn.(1) which satisfy eqn.(2) and eqn.(3) are (6 , 4) .

as,

→ 4² + 4 * 6 = 16 + 24 = 40

→ 6² + 4 * 6 = 36 + 24 = 60

therefore,

→ Area ∆DEC = b² = 4² = 16 cm² (Ans.)

Hence, area of ∆DEC is equal to 16 cm² .

Learn more :-

ABCD is a rhombus with A = 60° , BC = (3x+5)cm , CD =(6x-10)cm and AC =(3y-1)cm. Find

x and y.

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3.

In the fig, AB || CD,FIND x.(Hint:Prove that AOB-COD).

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