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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 |
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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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