In the given figure, $∠ A B C = 90 °$ and BM is a median, AB = 8 cm and BC = 6 cm. Then, find BM

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Solution

In triangle ABC, by applying Pythagoras rule,

$A C^{2} = A B^{2} + B C^{2}$

$A C^{2} = 8^{2} + 6^{2}$

$A C = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10 c m .$

So, AM = MC = 5cm.

We know that in a right angled triangle, mid-point of the hypotenuse is always at equidistance from all the three vertices of the triangle.

BM = AM = MC = 5 cm.

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In the given figure, ∠ABC=90° and BM is a median, AB = 8 cm and BC = 6 cm. Then, find BM

In triangle ABC, by applying Pythagoras rule, AC2=AB2+BC2 AC2=82+62 AC=√82+62=√100=10cm. So, AM = MC = 5cm. We know that in a right angled triangle, mid-point of the hypotenuse is always at equidistance from all the three vertices of the triangle. BM = AM = MC = 5 cm.

In the given figure, $∠ A B C = 90 °$ and BM is a median, AB = 8 cm and BC = 6 cm. Then, find BM