The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
Answer
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Hint: In a Right-angled triangle ,the square of the hypotenuse side is equal to the sum of the other two sides and the area of the triangle is half of its product of base and Height.
Let, ABC be the given right angled triangle such that base $ = BC = x{\text{ }}cm$ and hypotenuse $AC = 25cm$ . Perimeter is given to us as $60cm$ . Then,
$
AB + BC + AC = 60 \\
\Rightarrow AB + x + 25 = 60 \\
\Rightarrow AB = 35 - x \\
$
By Pythagoras theorem, we know that,
$
A{B^2} + B{C^2} = A{C^2} \\
\Rightarrow {(35 - x)^2} + {x^2} = {25^2} \\
\Rightarrow 1225 + {x^2} - 70x + {x^2} = 625 \\
\Rightarrow 2{x^2} - 70x + 600 = 0 \\
\Rightarrow {x^2} - 35x + 300 = 0 \\
\Rightarrow {x^2} - (15 + 20)x + 300 = 0 \\
\Rightarrow {x^2} - 15x - 20x + 300 = 0 \\
\Rightarrow x(x - 15) - 20(x - 15) = 0 \\
\Rightarrow (x - 20)(x - 15) = 0 \\
\Rightarrow x = 15,20 \\
$
If, $x = 20{\text{ then }}AB = 35 - x = 35 - 20 = 15{\text{ and }}BC = x = 20$ .
Area of a triangle can be calculated by $\frac{1}{2}(base \times height)$ so, $\frac{1}{2}(BC \times AB) = \frac{1}{2}(15 \times 20) = 150c{m^2}$ .
Note: Pythagoras theorem is only applicable for right angle triangles.
Let, ABC be the given right angled triangle such that base $ = BC = x{\text{ }}cm$ and hypotenuse $AC = 25cm$ . Perimeter is given to us as $60cm$ . Then,
$
AB + BC + AC = 60 \\
\Rightarrow AB + x + 25 = 60 \\
\Rightarrow AB = 35 - x \\
$
By Pythagoras theorem, we know that,
$
A{B^2} + B{C^2} = A{C^2} \\
\Rightarrow {(35 - x)^2} + {x^2} = {25^2} \\
\Rightarrow 1225 + {x^2} - 70x + {x^2} = 625 \\
\Rightarrow 2{x^2} - 70x + 600 = 0 \\
\Rightarrow {x^2} - 35x + 300 = 0 \\
\Rightarrow {x^2} - (15 + 20)x + 300 = 0 \\
\Rightarrow {x^2} - 15x - 20x + 300 = 0 \\
\Rightarrow x(x - 15) - 20(x - 15) = 0 \\
\Rightarrow (x - 20)(x - 15) = 0 \\
\Rightarrow x = 15,20 \\
$
If, $x = 20{\text{ then }}AB = 35 - x = 35 - 20 = 15{\text{ and }}BC = x = 20$ .
Area of a triangle can be calculated by $\frac{1}{2}(base \times height)$ so, $\frac{1}{2}(BC \times AB) = \frac{1}{2}(15 \times 20) = 150c{m^2}$ .
Note: Pythagoras theorem is only applicable for right angle triangles.
Last updated date: 21st Sep 2023
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