Question

Prove that the median to the hypotenuse of a right triangle is half the hypotenuse.

Answer 1

Answer:

D. 2

Step-by-step explanation:

Since midpoints will be involved, use multiples of 2 to name the coordinates for M and N.

Let

• M(0,2b)
• N(2a,0)

Then the midpoint P coordinates are

$P\left(\dfrac{2a+0}{2},\dfrac{0+2b}{2}\right)\Rightarrow P(a,b)$

Use distance formula to find OP and MN:

$OP=\sqrt{(a-0)^2+(b-0)^2}=\sqrt{a^2+b^2}\\ \\MN=\sqrt{(2a-0)^2+(2b-0)^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}$

So,

MN=2OP

or

OP=1/2 MN

Answer 2

Answer:

The correct option is D.

Step-by-step explanation:

Given: ΔMNO is a right angled triangle with right angle ∠MON, P is midpoint of MN.

To prove: $OP=\frac{1}{2}MN$

Since midpoints will be involved, use multiples of _2_ to name the coordinates for M and N.

Let the coordinates for M and N are (0,2m) and (2n,0) receptively.

Midpoint formula:

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

The coordinates of P are

$Midpoint=(\frac{2n+0}{2},\frac{2m+0}{2})$

$Midpoint=(n,m)$

The coordinates of P are (n,m).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, the distance between O(0,0) and P(n,m) is

$OP=\sqrt{(n-0)^2+(m-0)^2}=\sqrt{n^2+m^2}$

Using distance formula, the distance between M(0,2m) and N(2n,0) is

$MN=\sqrt{(2n-0)^2+(0-2m)^2}$

$MN=\sqrt{4n^2+4m^2}$

On further simplification we get

$MN=\sqrt{4(n^2+m^2)}$

$MN=2\sqrt{(n^2+m^2)}$

$MN=2(OP)$

Divide both sides by 2.

$\frac{1}{2}MN=OP$

Interchange the sides.

$OP=\frac{1}{2}MN$

Hence proved.

Therefore, the correct option is D.