1.

What is the area of a triangle, having perimeter 32 cm, one side 11 cm and difference of other two sides 5 cm?1). 8√30 sq cm2). 5√35 sq cm3). 6√30 sq cm4). 8√2 sq cm

Answer»

LET the sides of triangle be a, b and c respectively.

a + b + c = 32

⇒ 11 + b + c = 32

⇒ b + c = 21----(i)

And given that, b - c = 5----(II)

From eq. (i) and (ii), we get

⇒ b + c + b - c = 21 + 5

⇒ 2b = 26

∴ b = 13

∴ c = 8

Given? PERIMETER of triangle ‘2s’ = 32

∴ s = 16

And, a = 11, b = 13, c = 8

Area of triangle = √{s(s - a)(s - b)(s - c)}

⇒ Area of triangle = √{16(16 - 11)(16 - 13)(16 - 8)} = √(16 × 5 × 3 × 8)

⇒ Area of triangle = 8√30 sq cm


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