A wire 84 cm bend into right angled triangle of hypotenuse 35 cm calculate the length of other two sides
Answers
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
\begin{gathered}{c}^{2} = {a}^{2} + {b}^{2} \\ {35}^{2} = {a}^{2} + {b}^{2} \\ {a}^{2} + {b}^{2} = 1225.....(2)\end{gathered}
c
2
=a
2
+b
2
35
2
=a
2
+b
2
a
2
+b
2
=1225.....(2)
Squaring (1)
\begin{gathered}{(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)\end{gathered}
(a+b)
2
=49
2
a
2
+b
2
+2ab=2401
from(2)
1225+2ab=2401
2ab=2401−1225
ab=
2
1176
=588.....(3)
Now, again using (1)
a+b = 49
From (3) we get b=588/a,substituting in (1)
\begin{gathered}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\end{gathered}
a+b=49
a+
a
588
=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=28and21
From (1) we get
28+b=49
b = 21
And 21+b=49
b = 28
Other two sides are 28cm and 21cm