    Question

# Match the below figures with there 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×(s−a)×(s−b)×(s−c) =√6×(6−3)×(6−4)×(6−5) =6 cm2 Similarly, for triangle ADC: s =a+b+c2 =5+5+42 =7cm Area of ADC = √s×(s−a)×(s−b)×(s−c) = √7×(7−5)×(7−5)×(7−4) = 2√21 cm2 So, area of quadrilateral ABCD = ar(ABC) + ar(ADC) = 6+2√21 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  Suggest Corrections  0      Similar questions
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