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Question

Match the below figures with their respective areas in cm2.

A
114
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B
6+221
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C
306
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Solution

Case(1):
Quadrilateral ABCD can be divided into two triangles ABC and ADC
So, for triangle ABC
s = a+b+c2
= 3+4+52
= 6 cm
Area of ABC
=s×(sa)×(sb)×(sc)
=6×(63)×(64)×(65)
=6 cm2
Similarly,
for triangle ADC:
s =a+b+c2
=5+5+42
=7cm

Area of ADC
= s×(sa)×(sb)×(sc)
= 7×(75)×(75)×(74)
= 221 cm2

So, area of quadrilateral ABCD
= ar(ABC) + ar(ADC)
= 6+221 cm2

Case(2):
Area of trepezium ABCD
= 12×(base1+base2)×height
= 12×(28+40)×9= 306 cm2

Case(3):
Quadrilateral ABCD can be divided into two right triangles ABC and ADC
area of right triangle = 12×base×height

ar(ADC) = 12×base×height
= 12×12×9
= 54 cm2
ar(ABC) = 12×base×height
But, we dont know the value of BC
So, by applying the Pythagoreas theorem
AB2 = AC2+BC2
172 = 152+BC2
BC = 8 cm
So,
ar(ABC) = 12×8×15
= 60 cm2
So, area of quadrilateral ABCD = (60+54) cm2
= 114 cm2


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