Let the two legs of the right triangle be a and b.
By the Pythagorean theorem, we have:
a^2 + b^2 = 35^2.
Also, the perimeter of a triangle is the sum of all its sides. Thus, we have:
a + b + 35 = 84.
Rearranging the second equation, we get:
a + b = 84 - 35,
a + b = 49.
Solving the system of equations:
a + b = 49,
a^2 + b^2 = 35^2,
we can use the substitution method.
From the first equation, we obtain a = 49 - b.
Substituting this value of a into the second equation, we get:
(49 - b)^2 + b^2 = 35^2,
2401 - 98b + b^2 + b^2 = 1225,
2b^2 - 98b + 1176 = 0.
Dividing the equation by 2, we get:
b^2 - 49b + 588 = 0.
Factoring the equation, we have:
(b - 21)(b - 28) = 0.
Solving for b, we find two possible values:
b = 21 or b = 28.
If b = 21, substituting this value into the equation a + b = 49, we find:
a + 21 = 49,
a = 49 - 21,
a = 28.
If b = 28, substituting this value into the equation a + b = 49, we find:
a + 28 = 49,
a = 49 - 28,
a = 21.
In both cases, we obtain a = 21 and b = 28 as possible values of the legs.
Since the area of a right triangle is given by A = (1/2) * a * b, we can now calculate the area for each case:
For a = 21 and b = 28,
A = (1/2) * 21 * 28,
A = 294 square cm.
For a = 28 and b = 21,
A = (1/2) * 28 * 21,
A = 294 square cm.
Therefore, the area of the triangle is 294 square cm.
The hypotenuse of a right triangle is 35cm and the perimeter is 84cm. Find the area of the triangle
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