Question

the perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm .find the area of the triangle​

Answer 1

Answer

The required area of the triangle is 60cm²

$\bf\large\underline{Given}$

• The perimeter of a right triangle is 40cm
• The hypotenuse of the triangle measures 17cm

$\large\bf\underline{To \: Find}$

• The area of the triangle

$\large\bf\underline{Solution}$

Let us consider the base and the height of the right angled triangle be x cm and y cm respectively

$\sf\underline{According \ to \ question}$

$\sf\implies x + y +17=40\\\\ \sf\implies x + y = 23 \\\\ \sf\implies x = 23-y \longrightarrow(1)$

$\sf\underline{Again \: question}$

$\sf\implies x^{2}+y^{2}=17^{2} \\\\ \sf\implies (23-y)^2+y^{2}=17^{2} \\\\ \sf\implies 529+y^{2} -46y + y^{2} = 289 \\\\ \sf\implies 2y^{2} - 46y + 240 = 0 \\\\ \sf\implies y^{2} - 23y + 120 = 0 \\\\ \sf\implies y^{2} - 15y - 8y + 120 = 0 \\\\ \sf\implies y (y - 15)-8 (y - 15).= 0 \\\\ \sf\implies (y - 15)(y-8)=0$

Therefore ,

$\sf\implies y - 15 = 0 \: \: and \implies y - 8 = 0 \\\\ \sf\implies y = 15 \: \: and y = 8$

Using the value of y in (1)

$\sf\implies x = 23-15 \: \: and \: \implies x = 23 - 8 \\\\ \sf\implies x = 8 \: \: and \: \implies x = 15$

Thus when the height is 15cm , the base is 8cm

and when the height is 8cm , the base is 15cm

Hence , required area of triangle :

$\sf\implies Area \ of \ triangle = \dfrac{1}{2} \times base \times height \\\\ \sf\implies Ar. \: of \triangle = \dfrac{1}{2}\times 8cm\times 15cm \\\\ \sf\implies Ar. \: of \triangle = 60cm^{2}$