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.
Answers
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
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
