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Q.

In a triangle ABC, a = 7, b = 8, c = 9, BD is the median and BE the altitude from the vertex B, then
COLUMN-I COLUMN-II
A) BD p) 2
B) BE q) 7
C) ED r) $\sqrt{45}$
D) AE s) 6

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a

$A - s; B - r; C - q; D - p$

b

$A - r; B - p; C - q; D - s$

c

$A - q; B - r; C - p; D - s$

d

$A - r; B - s; C - q; D - p$

answer is C.

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Detailed Solution

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$Δ A B D \cos A = \frac{4^{2} + 9^{2} - B D^{2}}{2 \times 4 \times 9}$
= $\frac{8^{2} + 9^{2} - 7^{2}}{2 \times 8 \times 7}$ (From $Δ A B C$ ) \
$\Rightarrow B D^{2} = 49 \Rightarrow B D = 7 \Rightarrow Δ B C D$ is isosceles
$\Rightarrow E D = 1 / 2 C D = 2$
And BE = $\sqrt{7^{2} - 2^{2}} = \sqrt{45}, A E = 6$
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In a triangle ABC, a = 7, b = 8, c = 9, BD is the median and BE the altitude from the vertex B, then COLUMN-I COLUMN-IIA)BDp)2B)BEq)7C)EDr)45D)AEs)6