Among all the parallelograms whose diagonals | vedclass

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(C) Let the diagonals of the parallelogram be $d_1 = 10$ and $d_2 = 4$. Let the angle between the diagonals be $	heta$.The area of the parallelogram

Vedclass · Accepted Answer

$(21, 22]$

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(C) Let the diagonals of the parallelogram be $d_1 = 10$ and $d_2 = 4$. Let the angle between the diagonals be $	heta$.The area of the parallelogram is given by $A = \frac{1}{2} d_1 d_2 \sin 	heta = \frac{1}{2} 	imes 10 	imes 4 	imes \sin 	heta = 20 \sin 	heta$.The area is maximum when $\sin 	heta = 1$,i.e.,$	heta = \frac{\pi}{2}$.When the diagonals intersect at right angles,the parallelogram is a rhombus.The sides of the rhombus $s$ are given by $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$.The perimeter of the rhombus is $P = 4s = 4\sqrt{29}$.Since $5 More precisely,$\sqrt{29} \approx 5.385$,so $P \approx 4 	imes 5.385 = 21.54$.Thus,the perimeter lies in the interval $(21, 22]$.

Among all the parallelograms whose diagonals are $10$ and $4$,the one having maximum area has its perimeter lying in the interval

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