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Q.

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164. The larger and smaller parts of the squares are [[1]], [[2]] respectively.

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answer is 6, 10.

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Detailed Solution

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Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164. The larger and smaller parts of the squares are 10, 6 respectively.
Given that 16 is divided into two parts.
Let’s assume that the larger and smaller parts are

p

and

q

respectively.
Then,

⟹ p + q = 16

⟹ p = 16 - q

According to the given condition,
Twice the square of the larger part exceeds the square of the smaller part by 164.

⟹ 2 {(16 - q)}^{2} = q^{2} + 164

⟹ 2 {(16 - q)}^{2} = q^{2} + 164

⟹ 2 (q^{2} - 32 q + 256) = q^{2} + 164

⟹ 2 q^{2} - 64 q + 512 = q^{2} + 164

⟹ q^{2} - 64 q + 512 - 164 = 0

⟹ q^{2} - 64 q + 348 = 0

⟹ q^{2} - 58 q - 6 q + 348 = 0

⟹ q (q - 58) - 6 (q - 58) = 0

⟹ (q - 6) (q - 58) = 0

⟹ q - 6 = 0 or q - 58 = 0

∴ q = 6 or q = 58

We will choose the value

q = 6

because

58 > 16

⟹ q = 6 p = 16 - q = 16 - 6 = 10

.
Therefore, the larger and smaller part is 10 and 6 respectively.

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