In a $Δ A B C$ , $A C = 3$ , $B C = 4$ and median AD and BE are at right angles. Then $A B =$ ?

\sqrt{5}

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5

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\frac{5}{2}

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\sqrt{\frac{5}{2}}

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Solution

The correct option is A $\sqrt{5}$
$A D = \frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}}$
$A E = \frac{1}{2} \sqrt{2 a^{2} + 2 c^{2} - b^{2}}$
$(A B)^{2} = (A D \times \frac{2}{3})^{2} + (B E \times \frac{2}{3})^{2}$
$(A B)^{2} = (\frac{2}{3})^{2} (\frac{1}{4} (2 b^{2} + 2 c^{2} - a^{2} + 2 a^{2} + 2 c^{2} - b^{2})$
$9 c^{2} = b^{2} + 4 c^{2} + a^{2}$
$45 c^{2} = 16 + 9$
$c^{2} = 5$
$c = \sqrt{5}$

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Similar questions

In the given figure, $Δ A B C$ is a right triangle right angled at B. AD and CE are the two medians drawn from A and C respectivley. If AC =5 cm and AD $= \frac{3 \sqrt{5}}{2}$ cm, then CE =??

Q. Consider

Δ A B C

the medians AD & CF intersect at right angles at G. If

B C = 3 c m

and

A B = 4 c m

, then the length of

A C

is (in cm):

$Δ A B C$ is a right triangle right angled at B. AD and CE are the two medians drawn from A and C respectivley. If AC =5cm and AD $= \frac{3 \sqrt{5}}{2}$ cm, then CE =??