Question

Let ∆xoy be a right angled triangle. Measure of angle xoy = 90°. Let M and N be the midpoints of ox and oy respectively. Given that XN = 19 and YM = 22, find XY.

Answer 1

Let OM = x, ON = y. By the Pythagorean Theorem on \triangle XON, MOY respectively, (2x)^2 + y^2 = 19^2

x^2 + (2y)^2 = 22^2

Summing these gives 5x^2 + 5y^2 = 845

x^2 + y^2 = 169.

By the Pythagorean Theorem again, we have

[(2x)^2 + (2y)^2 = XY^2

XY = sqrt{4(x^2 + y^2)} = sqrt{4(169)} = sqrt{676} =  26

Alternatively, we could note that since we found x^2 + y^2 = 169, segment MN=13.

Right triangles triangle MON and triangle XOY are similar by Leg-Leg with a ratio of frac{1}{2}, so XY=2(MN) = 26

Answer 2

See attachment

Let OM = XM = A

Let ON = NY = B

ΔMAY is a right angle triangle:

a² + b² = c²

OM² + OY² = MY²

A² + (2B)² = (22)²

A² + 4B² = 484

ΔXON is a right angle triangle:

a² + b² = c²

XO² + ON² = XN²

(2A)² + B² = 19²

4A² + B² = 361

Put the 2 equations together:

A² + 4B² = 484 ----------------- [ 1 ]

4A² + B² = 361  ----------------- [ 2 ]

[1 ] x 4 :

4A² + 16B² = 1936  ----------------- [ 3 ]

Find B:

[ 3 ] - [ 2 ]:

15B²  = 1575

B² = 105

B = √105

Find A:

Sub B = √105 into [ 1 ]:

A² + 4(√105)² = 484

A² = 484  - 4(√105)²

A² = 484  - 420

A² = 64

A = √64

A = 8

Find the length OX:

OX = 2A = 2(8) = 16 units

Find the length OY:

OY = 2B = 2(√105) = 2√105 units

ΔXOY is a right angle triangle:

a² + b² = c²

XO² + OY² = XY²

16² + (2√105)² = XY²

XY² = 256 + 420

XY² = 676

XY = √676

XY = 26 units

Answer: XY = 26 units