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Midpoint Theorem

In the Δ ABC,...

Question

In the $Δ A B C, ∠ B$ is a right angle, $D$ and $E$ are the mid-points of the sides $A B = 3 c m$ and $A C = 5 c m$ respectively. Then, the length of $D E$ is

$4 c m$

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$2 c m$

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$5 c m$

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$3 c m$

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Solution

The correct option is B
$2 c m$

In $△ A B C$ ,

$A C^{2} = A B^{2} + B C^{2}$ [Pythagoras theorem]
$\Rightarrow B C^{2} = 5^{2} - 3^{2} = 25 - 9 = 16$
$\Rightarrow B C = \sqrt{16} = 4 c m$

From mid-point theorem, $D E = \frac{B C}{2} = \frac{4}{2} = 2 c m$

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In the ΔABC, ∠B is a right angle, D and E are the mid-points of the sides AB=3 cm and AC=5 cm respectively. Then, the length of DE is

The correct option is B 2 cm In △ABC, AC2=AB2+BC2 [Pythagoras theorem] ⇒BC2=52−32=25−9=16 ⇒BC=√16=4 cm From mid-point theorem, DE=BC2=42=2 cm

In the $Δ A B C, ∠ B$ is a right angle, $D$ and $E$ are the mid-points of the sides $A B = 3 c m$ and $A C = 5 c m$ respectively. Then, the length of $D E$ is

The correct option is B
$2 c m$

In $△ A B C$ ,

$A C^{2} = A B^{2} + B C^{2}$ [Pythagoras theorem]
$\Rightarrow B C^{2} = 5^{2} - 3^{2} = 25 - 9 = 16$
$\Rightarrow B C = \sqrt{16} = 4 c m$

From mid-point theorem, $D E = \frac{B C}{2} = \frac{4}{2} = 2 c m$