Question

The medians of a right triangle which are drawn from the vertices of the acute angles are 5 cm and 2√10 cm. Then
its hypotenuse is
​

Answer 1

Answer:

2√13

Step-by-step explanation:

Use Pythagoras twice on medians and add.

Answer 2

The hypotenuse of the right triangle=7.2 cm

Step-by-step explanation:

Median AC=5 cm

Median ED=$2 \sqrt 6$ cm

We know that median is that segment which is drawn from the vertex of triangle and bisect opposite side.

Let BC=x and AE=y,BD=2x

AE=EB=y,AB=2y

In triangle EBD

$ED^2=EB^2+BD^2=(2x)^2+y^2$

By using Pythagoras theorem $(Hypotenuse)^2=(Perpendicular\;side)^2+(Base)^2}$

Substitute the values

$(2\sqrt{10})^2=4x^2+y^2$

$4x^2+y^2=40$....(1)

In right triangle ABC

$AC^2=AB^2+BC^2$

$5^2=(2y)^2+x^2$

$x^2+4y^2=25$....(2)

Equation (1) multiply by 4 and then subtract from equation (2) then, we get

$-15x^2=-135$

$x^2=9$

$x=\sqrt{9}=3$

Length of side is always positive.

Substitute the value of x in equation (1)

then we get

$4(3)^2+y^2=40$

$36+y^2=40$

$y^2=40-36=4$

$y=2$

AB=2(2)=4 cm

BD=2(3)=6 cm

In triangle ABD

$AD^2=AB^2+BD^2$

Substitute the values then we get

$AD^2=(4)^2+(6)^2=52 cm$

$AD=\sqrt{52}=7.2 cm$

Hence, the hypotenuse of the right triangle=7.2 cm

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