Question

What is the measure of the hypotenuse of a right triangle, when its medians, drawn from the vertices of the acute angles, are 5 cm and 2V10 cm long?​

Answer 1

Step-by-step explanation:

Median AC=5 cm

Median ED=2 \sqrt 626 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^2ED2=EB2+BD2=(2x)2+y2

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

Substitute the values

(2\sqrt{10})^2=4x^2+y^2(210)2=4x2+y2

4x^2+y^2=404x2+y2=40 ....(1)

In right triangle ABC

AC^2=AB^2+BC^2AC2=AB2+BC2

5^2=(2y)^2+x^252=(2y)2+x2

x^2+4y^2=25x2+4y2=25 ....(2)

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

-15x^2=-135−15x2=−135

x^2=9x2=9

x=\sqrt{9}=3x=9=3

Length of side is always positive.

Substitute the value of x in equation (1)

then we get

4(3)^2+y^2=404(3)2+y2=40

36+y^2=4036+y2=40

y^2=40-36=4y2=40−36=4

y=2y=2

AB=2(2)=4 cm

BD=2(3)=6 cm

In triangle ABD

AD^2=AB^2+BD^2AD2=AB2+BD2

Substitute the values then we get

AD^2=(4)^2+(6)^2=52 cmAD2=(4)2+(6)2=52cm

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

Hence, the hypotenuse of the right triangle=7.2 cm

Answer 2

Given:

vertices of the acute angles, are 5 cm and $2\sqrt{10}$ cm long

In △ABC,

AE and CD are the medians of the triangle.

So, AE=5 cm and CD= $2\sqrt{10}$ cm

$\mathrm{BE}=\mathrm{EC}=\frac{1}{2} \mathrm{BC}\\\\ \text { and } \mathrm{AD}=\mathrm{DB}=\frac{1}{2} \mathrm{AB}$

$\begin{array}{l}\text { In } \triangle \mathrm{ABE} \\\\\mathrm{AE}^{2}=\mathrm{AB}^{2}+\mathrm{BE}^{2} \\\\\therefore 5^{2} =\mathrm{AB}^{2}+(\frac{1}{2})^{2} \mathrm{BC}^{2}\end{array}$

$25=\mathrm{AB}^{2}+\frac{1}{4} \mathrm{BC}^{2}$   -(1)

$\begin{array}{l}\mathrm{CD}^{2}=\mathrm{DB}^{2}+\mathrm{BC}^{2} \\\\(2\sqrt{10} )^{2} =\frac{1}{4} \mathrm{AB}^{2}+\mathrm{BC}^{2}\end{array}$

$2\times10=\frac{1}{4}AB^{2}+BC^{2}$  

$20=\frac{1}{4}AB^{2}+BC^{2}$   -(2)

Adding (1) and (2), we get

$45=\frac{5}{4}\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}\right)$

$45\times4 = 5(AC^{2} )\\\\180=5AC^{2}\\\\AC^{2}=\frac{180}{5} \\\\AC^{2}=36\\AC = \sqrt{36} \\\\AC = 6$

Hence the measure of the hypotenuse of a right triangle is 6