Question

A wire 84cm long is bent into a right triangle of hypotenuse 35cm.Calculate the length of the other 2 sides of the triangle.

Answer 1

Answer:

Length of other two sides is 28cm and 21cm

Step-by-step explanation:

We have perimeter,P = 84cm

Hypotenuse,c = 35cm

Let the sides be a and b

Perimeter of triangle, P = Sum of lengths of sides

P = a+b+c

84 = a+b+35

a+b = 84-35 = 49.... (1)

Also, by Pythagoras theorem

Hypotenuse^2 =base^2 +height^2

${c}^{2} = {a}^{2} + {b}^{2} \\ {35}^{2} = {a}^{2} + {b}^{2} \\ {a}^{2} + {b}^{2} = 1225.....(2)$

Squaring (1)

${(a + b)}^{2} = {49}^{2} \\ {a}^{2} + {b}^{2} + 2ab = 2401 \\ from \: (2) \\ 1225 + 2ab = 2401 \\ 2ab = 2401 - 1225 \\ ab = \frac{1176}{2} = 588 \: \: \: \: \: .....(3)$

Now, again using (1)

a+b = 49

From (3) we get b=588/a,substituting in (1)

$a + b = 49 \\ a + \frac{588}{a} = 49 \\ {a}^{2} + 588 = 49a \\ {a}^{2} - 49a + 588 = 0 \\ \\ {a}^{2} - 28a - 21a + 588 = 0 \\ a(a - 28) - 21(a - 28) = 0 \\ (a - 21)(a - 28) = 0 \\ \\ a = 28 \: and \: 21$

From (1) we get

28+b=49

b = 21

And 21+b=49

b = 28

Other two sides are 28cm and 21cm