(a) An isosceles right-angled triangle has area 200 cm². What is the length of its hypotenuse?

(b) The perimeter of a triangle is 540 m and its sides are in the ratio 12 : 25 : 17. Find the area of the triangle.

(c) Find the area of triangle whose sides are 5 cm, 12 cm and 13 cm. Also, find the length of its altitude corresponding to the longest side.

(a) In an isosceles right-angled triangle, the two perpendicular sides (base and height) are of equal length. The area is given by:

Area = $\dfrac{1}{2} \times \text{base}\times \text{height}$

Let the base of the triangle be x. Since it is an isosceles right-angled triangle, the height is also x.

Area = $\dfrac{1}{2} \times x \times x$

⇒ 200 = $\dfrac{1}{2} \times x^2$

⇒ 200 $\times$ 2 = $x^2$

⇒ $x^2$ = 400

⇒ $x = \sqrt{400}$

⇒ $x$ = 20

Now, using the Pythagoras Theorem, the hypotenuse h is given by:

h² = base² + height²

⇒ h² = (20)² + (20)²

⇒ h² = 400 + 400

⇒ h² = 800

⇒ h = $\sqrt{800}$

⇒ h = 20 $\sqrt{2}$

Hence, the length of the hypotenuse is 20 $\sqrt{2}$ cm.

(b) The given ratio of sides is 12 : 25 : 17.

Let the sides of the triangle be 12x, 25x and 17x.

The perimeter of the triangle = Sum of all sides

⇒ 540 m = 12x + 25x + 17x

⇒ 540 m = 54x

⇒ x = $\dfrac{540}{54}$ m

⇒ x = 10 m

So, the actual lengths of the sides are:

12x = 12 $\times$ 10 = 120 m

25x = 25 $\times$ 10 = 250 m

17x = 17 $\times$ 10 = 170m

Using Heron's formula, semi-perimeter (s),

$s = \dfrac{a + b + c}{2}\\[1em] s = \dfrac{120 + 250 + 170}{2}\\[1em] = \dfrac{540}{2}\\[1em] = 270$

The area of the triangle,

$= \sqrt{s(s - a)(s - b)(s - c)}\\[1em] = \sqrt{270(270 - 120)(270 - 250)(270 - 170)}\\[1em] = \sqrt{270 \times 150 \times 20 \times 100}\\[1em] = \sqrt{81000000}\\[1em] = 9000$

Hence, the area of the triangle is 9,000 m².

(c) Using Heron's formula,

$s = \dfrac{a + b + c}{2}\\[1em] s = \dfrac{5 + 12 + 13}{2}\\[1em] = \dfrac{30}{2}\\[1em] = 15$

The area of the triangle,

$= \sqrt{s(s - a)(s - b)(s - c)}\\[1em] = \sqrt{15(15 - 5)(15 - 12)(15 - 13)}\\[1em] = \sqrt{15 \times 10 \times 3 \times 2}\\[1em] = \sqrt{900}\\[1em] = 30$

Area of the triangle = $\dfrac{1}{2}$ x base x altitude

Let the altitude corresponding to the longest side (13 cm) be h.

⇒ $\dfrac{1}{2}$ x 13 x h = 30

⇒ 13 x h = 30 x 2

⇒ 13 x h = 60

⇒ h = $\dfrac{60}{13}$

⇒ h = 4.61 cm

Hence, area of the triangle is 30 cm² and the altitude is 4.61 cm.