A maximum sized rectangular playground is made within a circular space of radius $2 \sqrt{41}$ meter. Calculate the area of the playground if its length and width are distributed in the proportion of 4: 5.
(Geometry-Rectangle)

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

Here given, radius = 2 $\sqrt{41}$ m

$∴$ Diameter of the circle will be = 2 $\times$ 2 $\sqrt{41}$ m = 4 $\sqrt{41}$

$∴$ 4 $\sqrt{41}$ will be the diagnal of the rectangular playground.

Now let, length be 4x m and width be 5x m.

According to Pythagorean theorem,

(4x)² + (5x)² = (4 $\sqrt{41}$ )²

⁼16x² + 25x² = 16 $\times$ 41

=41x² = 16 $\times$ 41

$∴$ x² = $\frac{16 \times 41}{31}$ = 16

$∴$ x = $\sqrt{16}$ = ${\sqrt{4}}^{2}$ = 4

$∴$ Area will e = 4x $\times$ 5x = 20x² = 20 $\times$ 4² = 20 $\times$ 16 = 320m²

1 year ago

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