# Find the lengths of the sides.?

The lengths of the hypotenuse and one leg of a right triangle have a difference of 2 feet. The other leg is 36 feet shorter than the hypotenuse. Find the lengths of the sides.

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Refer to the diagram of the triangle given below -

Let length of Hypotenuse (AC) =  x ft

BC  =  ( x - 2 ) ft  ......... Given  and

AB  =  ( x - 36 ) ft  ...... Given

According to the Pythagorus Theorem-

AC²  =  AB²  +  BC²

=>  x²  =  ( x - 36 )²  +  ( x - 2 )²

=>  x²  =  ( x² - 72 x + 1296 ) + ( x² - 4 x + 4 )

=>  x²  =  2 x² - 76 x + 1300

=>  x² - 76 x + 1300  =  0

=>  x² - 50 x - 26 x + 1300  =  0

=>   x ( x - 50 ) - 26 ( x - 50 )   =   0

=> ( x - 26 ) ( x - 50 )  =  0

=>  x  =  26 ft  OR  x  =  50 ft.

If  Length of the Hypot. ( AC )  =  26 ft,  ......

BC  =  ( x - 2 )  =  ( 26 - 2 )  =  24 ft   and

AB  =  ( x - 36 ) = ( 26 - 36 ) .... Which is negative. Hence  x ≠  26 '

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Thus the only acceptable value of x  = 50 ft.

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If  x  ( ie AC )=  50 ft. ......................... .......... Answer

BC  =  ( x - 2 )  =  ( 50 - 2 )  =  48 ft.............. Answer

AB  =  ( x - 36 )  =  ( 50 - 36 ) =  14 ft............Answer • The diagram and your explanation really helped thank you!

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• Remember Pythagoras.

h^2 = L^2 + S^2

Since the hypotenuse will be the longest side

Then h - 2 = L

h - 36 = S

Substitute

h^2 = (h - 2)^2 + (h - 36)^2

h^2 = h^2 - 4h + 4 + h^2 -72h + 1296

h^2 - 76h + 1300 = 0

Complete the Square

(h - 38)^2 - (38)^2 = -1300

(h - 38)^2 = -1300 + 1444

(h - 38)^2 = 144

h - 38 = 12

h = 50

Hence L = 48

& S = 14

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• The lengths of the hypotenuse and one leg of a right triangle

have a difference of 2 feet.

The other leg is 36 feet shorter than the hypotenuse.

h - a = 2

h - b = 36

2 + a = 36 + b

a - b = 34

a^2 + b^2 = h^2

a^2 = (h + b)(h - b)

a^2 = 36(h + b)

Find the lengths of the sides.

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• (a + 36)^2 – (a +  34)^2 = (72 – 68)a + 36^2 – 34^2 = a^2

a^2 – 4a – 140 = (a + 10)(a – 14) = 0

The sides are 14, 48 and 50

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• Right angled triangle has hypotenuse = h and 2 legs a & b.

a = h-2.

b = h-36.

a^2 + b^2 = h^2, ie., (h-2)^2 + (h-36)^2 = h^2, ie., h^2 -76h +1300 = 0.

Then 2h = 76(+/-)D, where D^2 = 76^2 -4(1300) = 4(38^2 -1300) = 24^2.

So h = 38(+/-)12. Discard root h = 38-12 = 26 since b would then = -10.

(h,a,b) = (50,48,14) ft.

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• (h - 2)² + (h - 36)²= h²

h² - 4h + 4 + h² - 72h + 1296= h²

h² - 76h + 1300= 0

h= 38 +/- 12= 26, 50.  Reject 26.

L1= 48 ft, L2= 14 ft

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