If the perimeter of a right angled triangle is 56 cm and the area of the triangle is 84 $$\text{cm}^2$$​, then the length of the hypotenuse is:

Correct option is A

Given:

Perimeter of the right-angled triangle = 56 cm

Area of the triangle = 84 cm²

Formula Used:

Let the two perpendicular and base are a and b, and hypotenuse = c.
For a right-angled triangle:

Area$$= \frac{1}{2}ab$$​​

Perimeter = a + b + c

Pythagoras Theorem: c $$= \sqrt{a^2 + b^2}$$​​

Solution:

Let a, b be the perpendicular and base, and c be the hypotenuse of a right-angled triangle.

Given that a + b + c = 56 (Equation 1)

Also, the area of the triangle is 84.

​$$\frac12\times$$​ ab = 84

ab = 2 $$\times$$​ 84 = 168

Pythagorean Theorem:

c² = a² + b²

From Equation 1, a + b = 56 - c

(a + b)² = a² + b² + 2ab

(56 - c)² = c² + 2(168)

(56 - c)² = c² + 336

3136 - 112c + c² = c² + 336

3136 - 112c = 336

112c = 3136 - 336

112c = 2800

c = $$\frac{2800 }{ 112}$$​​

c = 25

Alternate Method:

​Area = $$\frac{1}{2}ab = 84$$

ab = 168

Perimeter = a + b $$+ \sqrt{a^2 + b^2}$$​ = 56
Let’s solve using values satisfying:

ab = 168

​$$a + b + \sqrt{a^2 + b^2} = 56$$​​

Try a = 12, b = 14:

ab = 12 × 14 = 168

​$$\sqrt{a^2 + b^2} = \sqrt{144 + 196} = \sqrt{340} \approx 18.44$$​​

Perimeter = 12 + 14 + 18.44$$\approx$$ 44.44 not possible

Try a = 8, b = 21:

ab = 168

​$$\sqrt{64 + 441} = \sqrt{505} \approx 22.47$$​​

Perimeter = 8 + 21 + 22.$$47 \approx 51.47$$ not possible

Try a = 7, b = 24:

ab = 168

​$$\sqrt{49 + 576} = \sqrt{625} = 25$$ 

Perimeter = 7 + 24 + 25 = 56 right

Thus, hypotenuse = 25 cm