ABCD is a rhombus. The circumradii of Delta A | vedclass

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(A) Let the diagonals of the rhombus $ABCD$ intersect at $O$. Let $AO = y$ and $BO = x$. Since the diagonals of a rhombus bisect each other at right a

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(A) Let the diagonals of the rhombus $ABCD$ intersect at $O$. Let $AO = y$ and $BO = x$. Since the diagonals of a rhombus bisect each other at right angles,$\Delta AOB$ is a right-angled triangle.In $\Delta ABD$,the sides are $AB, AD, BD$. Let $\angle OAB = 	heta$. Then $AO = y, BO = x$. The circumradius $R_1$ of $\Delta ABD$ is given by $R_1 = \frac{BD}{2 \sin \angle BAD} = \frac{2x}{2 \sin 2	heta} = \frac{x}{\sin 2	heta} = \frac{25}{2}$.Thus,$x = \frac{25}{2} \sin 2	heta = 25 \sin 	heta \cos 	heta$.In $\Delta ACD$,the sides are $AC, AD, CD$. The circumradius $R_2$ of $\Delta ACD$ is given by $R_2 = \frac{AC}{2 \sin \angle ADC} = \frac{2y}{2 \sin(\pi - 2	heta)} = \frac{y}{\sin 2	heta} = 25$.Thus,$y = 25 \sin 2	heta = 50 \sin 	heta \cos 	heta$.Dividing the two equations,we get $\frac{x}{y} = \frac{25 \sin 	heta \cos 	heta}{50 \sin 	heta \cos 	heta} = \frac{1}{2}$,so $y = 2x$.Using the property of the rhombus,the area is $2xy$. From the circumradius formula $R = \frac{abc}{4K}$,for $\Delta ABD$,$R_1 = \frac{AB \cdot AD \cdot BD}{4 \cdot 	ext{Area}(\Delta ABD)} = \frac{AB^2 \cdot 2x}{4 \cdot (\frac{1}{2} \cdot 2x \cdot y)} = \frac{AB^2}{2y} = \frac{x^2+y^2}{2y} = \frac{25}{2}$.Substituting $y=2x$,we get $\frac{x^2+4x^2}{4x} = \frac{5x^2}{4x} = \frac{5x}{4} = \frac{25}{2} \Rightarrow x = 10$.Then $y = 20$. The area of the rhombus is $2xy = 2(10)(20) = 400 \, sq. \, unit$.

$ABCD$ is a rhombus. The circumradii of $\Delta ABD$ and $\Delta ACD$ are $\frac{25}{2}$ and $25$ respectively. Then the area of the rhombus is .............. $sq. \, unit$.

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