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06 Nov 2020, 01:15
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B
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In the figure shown, AB is the base of isosceles triangle ABR and a side of rectangle ABCD. What is the area of rectangle ABCD?
A. \(20\sqrt{7}\)
B. \(24\sqrt{7}\)
C. \(32\sqrt{7}\)
D. \(36\sqrt{7}\)
E. \(48\sqrt{7}\)
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2020-10-27_13-13-08.png [ 14.87 KiB | Viewed 2152 times ]
06 Nov 2020, 04:36
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Bunuel
Since AB is the unequal side, dropping a perpendicular from R will bisect AB. Therefore AS = SB = 6
In Triangle RSB, by Pythagoras theorem, \(RB^2 = RS^2 + SB^2\)
\(RS^2 = 8^2 - 6^2 = 28\)
RS = \(\sqrt{28} = 2\sqrt{7}\)
Area of the rectangle = \(12 * 2\sqrt{7} = 24\sqrt{7}\)
Option B
Arun Kumar
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06 Nov 2020, 23:28
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In the figure shown, AB is the base of isosceles triangle ABR and a side of rectangle ABCD. What is the area of rectangle ABCD?
Given:
Length of AB = 12 , AB is the length of the rectangle
RB = 8
If x is the bisector of AB, we know that XA=XB=6
By the pythogorean theorem \((RB)^2\) =\( (RX)^2\)+\((BX)^2\)
\(8^2\) = \(6^2\)+\((RX)^2\)
64 = 36+\(RX^2\)
\((RX)^2\) = 64-36 = 28
RX= \(2 \sqrt{7} \\
\)
Length of rectangle = 12 * \(2 \sqrt{7}\) = \(24 \sqrt{7}\\
\)
Ans: B
Given:
Length of AB = 12 , AB is the length of the rectangle
RB = 8
If x is the bisector of AB, we know that XA=XB=6
By the pythogorean theorem \((RB)^2\) =\( (RX)^2\)+\((BX)^2\)
\(8^2\) = \(6^2\)+\((RX)^2\)
64 = 36+\(RX^2\)
\((RX)^2\) = 64-36 = 28
RX= \(2 \sqrt{7} \\
\)
Length of rectangle = 12 * \(2 \sqrt{7}\) = \(24 \sqrt{7}\\
\)
Ans: B
