I need Help!!

Question

the hypotenuse length of a right angled triangle is 10cm and the length of the two sides of the right angle are x and y cm. if the perimeter =24 find the area 

i tried using pythagoras and substitution method but it was not solved help please 

Answer 1

From Pythagoras theorem x² + y² = 10²   

 

x + y + 10 = 24  (perimeter)

 

x + y = 14 ; hence x = 14 - y

 

substitute; (14 - y)² + y² = 100

 

196 -28y + y² + y² = 100

 

2y² -28y + 96 = 0; or y² -14y +48 = 0  (divide by 2)

 

(y-8)(y-6) = 0; hence y = 6, or 8

 

substituting from above x = 8 or 6

 

Therefore area = 1/2 base * height = 1/2 *8 * 6 = 24 cm²

 

check; 6 + 8 + 10 = 24; and 6² + 8² = 36 + 64 = 100 = 10² (proven)

 

 

 

 

Answer 2

x^2+y^2=10^2

x^2+y^2=100

x+y+10=24

x+y=24-10

x+y=14

we need 2 numbers that add up to 14 and...

we need 2 numbers whose squares add up to 100

these are the squares we can use:

1, 4, 9. 16, 25, 36, 49, 64, 81

36+64=100, 6^2+8^2=36+64=100 and 6+8=14

x=6 and y=8

you have a variation of a 3-4-5 right triangle (6-8-10 triangle)

A=xy/2

A=6*8/2

A=48/2

A=24 square centimeters

Answer 3

P = 10+x+y = 24
solve this for y:
y = 24-10-x = 14-x
Now you can use the Pythagorean theorem.

x² + y² = 10²
x² + (14-x)²=100
x² + 196 - 28x + x² = 100
2x²-28x+96 = 0
x²-14x+48 = 0
(x-6)(x-8) = 0
x = 6 or 8
y will wind up being 8 or 6.  (Use the perimeter equation to find it.  It's easier.)

 

Your approach sounded good.  You probably just made a simple error somewhere along the way.