Let's assume the sides of the right angle triangle are a, b, and c, where c is the hypotenuse.
According to the given information,
c = a + 4 ...(1)
c = b + 8 ...(2)
We know that in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse.
So, using the Pythagorean theorem, we have:
a^2 + b^2 = c^2 ...(3)
We can substitute the values of c from equations (1) and (2) in equation (3):
a^2 + b^2 = (a + 4)^2 ...(4)
a^2 + b^2 = (b + 8)^2 ...(5)
Expanding equation (4):
a^2 + b^2 = a^2 + 8a + 16 ...(6)
Expanding equation (5):
a^2 + b^2 = b^2 + 16b + 64 ...(7)
From equations (6) and (7), we can see that 8a - 16 = 16b - 64.
This simplifies to 8a - 16b = -48. ...(8)
Now, let's solve equations (1), (2), and (8) to find the values of a, b, and c:
Subtract equation (2) from equation (1):
c - (b + 8) = a - (a + 4)
c - b - 8 = a - a - 4
c - b - 8 = -4
c - b = 4 ...(9)
Subtract equation (2) from equation (3):
c^2 - (b + 8)^2 = a^2 - (a + 4)^2
c^2 - b^2 - 16b - 64 = a^2 - a^2 - 8a - 8
(c + b)(c - b) - 16b - 64 = -8a
(c + b - 16)(c - b) = -8a
(c + b - 16)(c - b) = 8a ...(10)
Solving equations (9) and (10) simultaneously:
From equation (9), we know that c - b = 4.
Plugging this value into equation (10):
(c + b - 16)(4) = 8a
c + b - 16 = 2a
c + b = 2a + 16 ...(11)
Now let's substitute the value of c + b from equation (11) into equation (9):
2a + 16 = 4
2a = -12
a = -6
Since a cannot be negative in this context, this means that there is no solution to the given problem.
Therefore, it is not possible to find the length of the hypotenuse (x) in this case, and thus we cannot find the perimeter and area of the triangle.
The hypotenuse of a right angle triangle is 4cm longer than one side and 8cm longer than the other side if x is the hypotenuse find x and hence find the perimeter and area of the triangle
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