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Question

In the given figure,ABCD is a trapezium such that ALDC and BMDC.If AB=7 cm,BC=AD=5 cm and AL=BM=4 cm then ar(trap.ABCD)=?

(a) 24 cm2

(b) 40 cm2

(c) 55 cm2

(d) 27.5 cm2


Solution

(b) 40 cm2

In right angled triangle MBC, we have:
MC = √5^2 - 4^2 = √9 = 3 cm
In right angled triangle ADL, we have:
DL = √5^2 - 4^2 = √9 = 3 cm
Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
∴ Area of the trapezium =1/2 × (sum of parallel sides) × distance between them
= 1/2 × ( 13 + 7) × 4
= 40 cm2


Mathematics
Secondary School Mathematics IX
Standard IX

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