Question

If the area of a rhobus is 24cm² and perimeter is 20cm. find the length of its diagonals?

Answer 1

Answer:

$\bold{8\ cm\ and\ 6\ cm}$

Step-by-step explanation:

$Let\ the\ diagonals\ be\ x\ and\ y\ and\ side\ be\ a. \\ \\ \\ Area = \frac{1}{2}xy = 24\ cm^2 \\ \\ xy = 24 \div \frac{1}{2} = 24 \times 2 = 48\ cm^2 \\ \\ \\ Perimeter = 4a = 20\ cm \\ \\ a = \frac{20}{4} = 5\ cm$

$\\ \\ \\ Here,\ a^2 = (\frac{x}{2})^2 + (\frac{y}{2})^2 \\ \\ = (\frac{x}{2})^2 + (\frac{y}{2})^2 = 5^2 \\ \\ = \frac{x^2}{4} + \frac{y^2}{4} = 25 \\ \\ = \frac{x^2 + y^2}{4} = 25 \\ \\ x^2 + y^2 = 25 \times 4 = 100 \\ \\ \\ We\ have, \\ \\ x^2 + y^2 = 100 \to (1) \\ \\ \& \\ \\ xy = 48 \to (2)$

$\\ \\ \\ (1) + 2 \times (2) \\ \\ = x^2 + y^2 + 2xy = 100 + 2 \times 48 \\ \\ = (x + y)^2 = 100 + 96 = 196 \\ \\ x + y = \sqrt{196} = 14 \to (3) \\ \\ \\ (1) - 2 \times (2) \\ \\ = x^2 + y^2 - 2xy = 100 - 2 \times 48 \\ \\ = (x - y)^2 = 100 - 96 = 4 \\ \\ x - y = \sqrt{4} = 2 \to (4)$

$\\ \\ \\ (3) + (4) \\ \\ = (x + y) + (x - y) = 14 + 2 \\ \\ = x + y + x - y = 16 \\ \\ = 2x = 16 \\ \\ x = \frac{16}{2} = \bold{8\ cm} \\ \\ \\ (3) - (4) \\ \\ = (x + y) - (x - y) = 14 - 2 \\ \\ = x + y - x + y = 12 \\ \\ = 2y = 12 \\ \\ y = \frac{12}{2} = \bold{6\ cm} \\ \\ \\ \therefore\ The\ length\ of\ diagonals\ are\ \bold{8\ cm}\ and\ \bold{6\ cm}.$

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