Perimeter & Area of Plane Figures — Exercise 17(A)

Find the area of a triangle whose base is 15 cm and the corresponding height is 9.6 cm.

Answer

Given,

Base = 15 cm

Corresponding height = 9.6 cm

We know that

Area of triangle = $\dfrac{1}{2}$ × Base × Corresponding height

= $\dfrac{1}{2}$ × 15 × 9.6

= 15 × 4.8

= 72 cm².

Hence, area of triangle = 72 cm².

Find the area of the triangle whose sides are 13 cm, 14 cm and 15 cm. Also, find the height of the triangle, corresponding to the longest side.

Answer

Let a = 13 cm, b = 14 cm and c = 15 cm.

Then,

s = $\dfrac{1}{2}(a + b + c) = \dfrac{1}{2}(13 + 14 + 15) = \dfrac{42}{2}$ = 21 cm.

⇒ (s - a) = (21 - 13) cm = 8 cm.

⇒ (s - b) = (21 - 14) cm = 7 cm.

⇒ (s - c) = (21 - 15) cm = 6 cm.

We know that,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{21 × 8 × 7 × 6} \\[1em] \Rightarrow \sqrt{7056} \\[1em] \Rightarrow 84 \text{ cm}^2.$

Length of the longest side = 15 cm.

Let the corresponding height be x cm. Then,

$\Rightarrow Area = \dfrac{1}{2} \times \text{ Base } \times \text{ Corresponding height } \\[1em] \Rightarrow 84 = \dfrac{1}{2} \times 15 \times x \\[1em] \Rightarrow x = \dfrac{168}{15} \\[1em] \Rightarrow 11.2 \text{ cm}. \\[1em]$

Hence, area of triangle = 84 cm² and the height of the triangle = 11.2 cm.

Find the area of the triangle whose sides are 30 cm, 24 cm and 18 cm. Also, find the length of the altitude corresponding to the smallest side of the triangle.

Answer

Let a = 30 cm, b = 24 cm, c = 18 cm.

Then,

s = $\dfrac{1}{2}(a + b + c) = \dfrac{1}{2}(30 + 24 + 18) = \dfrac{72}{2}$ = 36 cm.

⇒ (s - a) = (36 - 30) cm = 6 cm.

⇒ (s - b) = (36 - 24) cm = 12 cm.

⇒ (s - c) = (36 - 18) cm = 18 cm.

We know that,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{36 × 6 × 12 × 18} \\[1em] \Rightarrow \sqrt{46656} \\[1em] \Rightarrow 216 \text{ cm}^2. \\[1em]$

Length of the smallest side = 18 cm.

Let the length of altitude be x cm. Then,

$\Rightarrow Area = \dfrac{1}{2} \times \text{ Base } \times \text{ Altitude } \\[1em] \Rightarrow 216 = \dfrac{1}{2} \times 18 \times x \\[1em] \Rightarrow 216 = 9x \\[1em] \Rightarrow x = \dfrac{216}{9} \\[1em] \Rightarrow 24 \text{ cm}. \\[1em]$

Hence, area of triangle = 216 cm² and altitude = 24 cm.

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle.

Answer

It is given that the lengths of the sides of a triangle are in the ratio 3 : 4 : 5.

Let the lengths of the sides be 3x, 4x and 5x.

The perimeter of the triangle is 144 cm.

Perimeter = sum of all sides of triangle

⇒ 144 = 3x + 4x + 5x

⇒ 144 = 12x

⇒ x = $\dfrac{144}{12}$

⇒ x = 12.

So the sides of a triangle are

⇒ 3x = 3 × 12 = 36 cm

⇒ 4x = 4 × 12 = 48 cm

⇒ 5x = 5 × 12 = 60 cm

Let a = 36 cm, b = 48 cm, c = 60 cm.

s = $\dfrac{1}{2}(a + b + c) = \dfrac{1}{2}(36 + 48 + 60) = \dfrac{144}{2}$ = 72 cm.

(s - a) = (72 - 36) cm = 36 cm.

(s - b) = (72 - 48) cm = 24 cm.

(s - c) = (72 - 60) cm = 12 cm.

We know that,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{72 × 36 × 24 × 12} \\[1em] \Rightarrow \sqrt{746496} \\[1em] \Rightarrow 864 \text{ cm}^2. \\[1em]$

Hence, area of triangle = 864 cm².

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of cultivating the field at ₹ 24.60 per 100 m².

Answer

It is given that the sides of a triangular field are in the ratio 25 : 17 : 12.

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

Given,

The perimeter of a triangular field is 540 m.

Perimeter = sum of all sides of triangle

⇒ 540 = 25x + 17x + 12x

⇒ 540 = 54x

⇒ x = $\dfrac{540}{54}$

⇒ x = 10.

So the sides of the triangle are

⇒ 25x = 25 × 10 = 250 m

⇒ 17x = 17 × 10 = 170 m

⇒ 12x = 12 × 10 = 120 m

Let a = 250 m, b = 170 m, c = 120 m.

s = $\dfrac{1}{2}(a + b + c) = \dfrac{1}{2}(250 + 170 + 120) = \dfrac{540}{2}$ = 270 m.

(s - a) = (270 - 250) m = 20 m.

(s - b) = (270 - 170) m = 100 m.

(s - c) = (270 - 120) m = 150 m.

We know that,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{270 × 20 × 100 × 150} \\[1em] \Rightarrow \sqrt{81000000} \\[1em] \Rightarrow 9000 \text{ m}^2. \\[1em]$

Rate = ₹ 24.60 per 100 m²

= ₹ $\dfrac{24.60}{100}$ per m².

Cost = Area of triangle × Rate

= 9000 × $\dfrac{24.60}{100}$

= 24.60 × 90 = ₹ 2,214.

Hence, area of field = 9000 m² & cost of cultivating the field = ₹ 2,214.

The base of a triangular field is twice its altitude. If the cost of cultivating the field at ₹ 14.50 per 100 m² is ₹ 52,200, find its base and altitude.

Answer

Given,

Rate of cultivation = ₹14.50 per 100 m²

= ₹ $\dfrac{14.50}{100}$ per m².

Total cost = ₹52,200

$\Rightarrow \text{Total cost} = \text{Area } \times \text{ Rate of cultivation} \\[1em] \Rightarrow 52,200 = \text{Area } \times \dfrac{14.50}{100} \\[1em] \Rightarrow Area = \dfrac{52200 × 100}{14.50} \\[1em] \Rightarrow Area = \dfrac{5220000}{14.50} \\[1em] \Rightarrow Area = 360000 \text{ m}^2. \\[1em]$

Let altitude be x meters and base be 2x meters.

We know that,

Area of a triangle = $\dfrac{1}{2}$ × base × height

⇒ 360000 = $\dfrac{1}{2}$ × 2x × x

⇒ x² = 360000

⇒ x = 600.

∴ Altitude = x = 600 m and Base = 2x = 2 × 600 = 1200 m.

Hence, altitude = 600 m and base = 1200 m.

The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the area of the triangle.

Answer

Let △ABC be the right triangle.

We know that,

Perimeter of a right-angled triangle = 60 cm

Hypotenuse = 25 cm

So, the sum of other two sides of triangle = 60 – 25 = 35 cm

Let, base (BC) = x cm

So, AB = (35 - x) cm

Using the Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 25² = (35 - x)² + x²

⇒ 625 = 1225 + x² - 70x + x²

⇒ 2x² - 70x + 600 = 0

Dividing by 2 on both sides,

⇒ x² - 35x + 300 = 0

⇒ x² - 15x - 20x + 300 = 0

⇒ x(x – 15) - 20(x - 15) = 0

⇒ (x - 15)(x - 20) = 0

⇒ x - 15 = 0 or x - 20 = 0

⇒ x = 15 or x = 20.

If x = 15, then 35 - x = 35 - 15 = 20 cm.

If x = 20, then 35 - x = 35 - 20 = 15 cm.

So, length of other two sides apart from hypotenuse are 15 cm and 20 cm.

Area of triangle = $\dfrac{1}{2}$ × base × height

Substituting the values we get,

A = $\dfrac{1}{2}$ × 15 × 20 = 150 cm².

Hence, area of triangle = 150 cm².

Find the length of hypotenuse of an isosceles right angled triangle, having an area of 200 cm².

Answer

Let ABC be an isosceles right-angled triangle.

Area = 200 cm²

Let, side AB = BC = x cm and hypotenuse AC = h cm

We know that,

$\Rightarrow \text{Area of triangle } = \dfrac{1}{2} \times base \times height \\[1em] \Rightarrow \text{Area of triangle } = \dfrac{1}{2} \times BC \times AB \\[1em] \Rightarrow \text{Area of triangle } = \dfrac{1}{2} \times x \times x \\[1em] \Rightarrow 200 = \dfrac{1}{2} \times x^2 \\[1em] \Rightarrow x^2 = 400 \\[1em] \Rightarrow x = \sqrt{400} \\[1em] \Rightarrow x = 20 \text{ cm}.$

Now, using the Pythagoras Theorem for the △ABC

⇒ AC² = AB² + BC²

⇒ AC² = (20)² + (20)²

⇒ AC² = 400 + 400

⇒ AC² = 800

⇒ AC = $\sqrt{800}$

⇒ AC = $\sqrt{400 × 2}$

⇒ AC = $20\sqrt{2}$

⇒ AC = 20 × 1.414

⇒ AC = 28.28 cm.

Hence, length of hypotenuse = 28.28 cm.

Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.

Answer

Given:

Perimeter = 60 cm

Let the length of each side of an equilateral triangle be 'a' cm.

Perimeter = Sum of all sides

⇒ 60 = a + a + a

⇒ 60 = 3a

⇒ a = $\dfrac{60}{3}$ = 20 cm.

$\Rightarrow \text{Area of equilateral triangle} = \dfrac{\sqrt{3}}{4} \times \text{ side}^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 20^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 400 \\[1em] = 100 \sqrt{3} \\[1em] = 173.2 \text{ cm}^2.$

By formula,

$\text{Height of an equilateral triangle} = \dfrac{\sqrt{3}}{2} \times \text{ side} \\[1em] = \dfrac{\sqrt{3}}{2} \times 20 \\[1em] = 10 \sqrt{3} \\[1em] = 17.32 \text{ cm}.$

Hence, the area of triangle = 173.2 cm² and the height = 17.32 cm.

Find the perimeter and area of an equilateral triangle whose height is 12 cm. Write your answers, correct to two decimal places.

Answer

Given,

Height (h) = 12 cm

Let the length of the side of an equilateral triangle be a cm.

We know that for an equilateral triangle,

$\Rightarrow \text{Height} = \dfrac{\sqrt{3}}{2}a \\[1em] \Rightarrow 12 = \dfrac{\sqrt{3}}{2}a \\[1em] \Rightarrow a = \dfrac{12 \times 2}{\sqrt{3}} = \dfrac{24}{\sqrt{3}} \\[1em] \Rightarrow a = \dfrac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ cm}$

Perimeter of an equilateral triangle = 3 × side

= 3 × 8 $\sqrt{3}$

= 3 × 8 × 1.732

= 41.568 ≈ 41.57 cm.

$\Rightarrow \text{Area of an equilateral triangle} = \dfrac{\sqrt{3}}{4} \times (side)^2 \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} \times (8 \sqrt{3})^2 \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} \times 64 \times 3 \\[1em] \Rightarrow \sqrt{3} \times 48 \\[1em] \Rightarrow 1.732 \times 48 \\[1em] \Rightarrow 83.136 \approx 83.14 \text{ cm}^2. \\[1em]$

Hence, the perimeter = 41.57 cm and area = 83.14 cm².

The lengths of two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24 cm², find the perimeter of the triangle.

Answer

ABC is a right angled triangle with a right angle at B.

The area of the triangle is 24 cm².

Let the lengths of BC and AB be x and y, respectively.

Given,

The difference between the two perpendicular sides is 2 cm.

x - y = 2

∴ y = x - 2

Area = $\dfrac{1}{2}$ × base × height

⇒ 24 = $\dfrac{1}{2}$ × BC × AB

⇒ 24 = $\dfrac{1}{2}$ × x × (x - 2)

⇒ x × (x - 2) = 48

⇒ x² - 2x = 48

⇒ x² - 2x - 48 = 0

⇒ x² - 8x + 6x - 48 = 0

⇒ x(x - 8) + 6(x - 8) = 0

⇒ (x - 8)(x + 6) = 0

⇒ (x - 8) = 0 or (x + 6) = 0

⇒ x = 8 or x = -6

Since length cannot be negative, ∴ x = 8 cm.

y = x - 2 = 8 - 2 = 6 cm.

Thus, AB = 6 cm and BC = 8 cm.

By using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ AC² = 8² + 6²

⇒ AC² = 64 + 36

⇒ AC² = 100

⇒ AC = $\sqrt{100}$

⇒ AC = 10 cm.

Perimeter of a triangle = Sum of all the sides of a triangle

= AB + BC + AC

= 10 + 8 + 6

= 24 cm.

Hence, perimeter of a triangle = 24 cm.

The sides of a right-angled triangle containing the right angle are (5x) cm and (3x - 1) cm. If its area is 60 cm², find its perimeter.

Answer

Let ABC be a right-angled triangle,

AB = 5x cm and BC = (3x – 1) cm

We know that,

Area of △ ABC = $\dfrac{1}{2}$ × base × height

Substituting the values we get,

⇒ 60 = $\dfrac{1}{2}$ × (3x - 1) × 5x

⇒ 120 = 5x(3x – 1)

⇒ 120 = 15x² - 5x

⇒ 15x² - 5x - 120 = 0

⇒ 5(3x² - x - 24) = 0

⇒ 3x² - x - 24 = 0

⇒ 3x² – 9x + 8x – 24 = 0

⇒ 3x(x – 3) + 8(x - 3) = 0

⇒ (3x + 8)(x - 3) = 0

⇒ 3x + 8 = 0 or x - 3 = 0

⇒ 3x = -8 or x = 3

⇒ x = $\dfrac{-8}{3}$ or x = 3.

Since, x cannot be negative. So, x = 3.

⇒ AB = 5 × 3 = 15 cm

⇒ BC = (3 × 3 – 1) = 9 – 1 = 8 cm

In right angled △ABC,

Using Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 cm.

Perimeter of a triangle = Sum of all the sides of a triangle

= AB + BC + AC

= 15 + 8 + 17

= 40 cm.

Hence, perimeter of the triangle = 40 cm.

Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.

Answer

Let ABC be an isosceles triangle in which AB = AC, AD ⊥ BC and BC is the base.

Given,

Each of the equal sides of the isosceles triangle is 2 cm more than its height.

Let the height of the triangle be h cm.

Equal sides: AB = AC = h + 2

Base: BC = 12 cm

In Δ ABD and Δ ACD,

AD = AD [Common Side]

∠ADB = ∠ADC [Both equal to 90°]

AB = AC [Δ ABC is an isosceles triangle]

∴ Δ ABD ≅ Δ ACD [By R.H.S. axiom]

∴ BD = CD [C.P.C.T.C.]

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{12}{2}$ = 6 cm.

By using the Pythagoras theorem in Δ ABD,

⇒ BD² + AD² = AB²

⇒ 6² + h² = (h + 2)²

⇒ 36 + h² = h² + 4 + 2 × h × 2

⇒ 36 = 4 + 4h

⇒ 4h = 32

⇒ h = $\dfrac{32}{4}$

⇒ h = 8 cm.

$\Rightarrow \text{Area of triangle ABC} = \dfrac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \dfrac{1}{2} \times BC \times AD \\[1em] = \dfrac{1}{2} \times 12 \times 8 \\[1em] = 6 \times 8 \\[1em] = 48 \text{ cm}^2.$

Hence, area of the triangle = 48 cm².

Find the area of an isosceles triangle, each of whose equal sides is 13 cm and base 24 cm.

Answer

Each of equal sides (a) = 13 cm and base (b) = 24 cm

By formula,

$\Rightarrow \text{Area of an isosceles triangle} = \dfrac{1}{4}b\sqrt{4a^2 - b^2} \\[1em] \Rightarrow \dfrac{1}{4} \times 24 \times \sqrt{4 × 13^2 - 24^2} \\[1em] \Rightarrow 6 \times \sqrt{676 - 576} \\[1em] \Rightarrow 6 \times \sqrt{100} \\[1em] \Rightarrow 6 \times 10 \\[1em] \Rightarrow 60 \text{ cm}^2.$

Hence, area of the triangle = 60 cm².

The base of an isosceles triangle is 18 cm and its area is 108 cm². Find its perimeter.

Answer

Given,

Base (b) = 18 cm

Area = 108 cm².

Let the length of equal sides of the isosceles triangle be a cm.

By formula,

$\Rightarrow \text{Area of an isosceles triangle} = \dfrac{1}{4}b\sqrt{4a^2 - b^2} \\[1em] \Rightarrow 108 = \dfrac{1}{4} \times 18 \times \sqrt{4a^2 - 18^2} \\[1em] \Rightarrow 432 = 18 \times \sqrt{4a^2 - 324} \\[1em] \Rightarrow \sqrt{4a^2 - 324} = \dfrac{432}{18} \\[1em] \Rightarrow \sqrt{4a^2 - 324} = 24 \\[1em] \Rightarrow 4a^2 - 324 = 24^2 \\[1em] \Rightarrow 4a^2 - 324 = 576 \\[1em] \Rightarrow 4a^2 = 576 + 324 \\[1em] \Rightarrow 4a^2 = 900 \\[1em] \Rightarrow a^2 = \dfrac{900}{4} \\[1em] \Rightarrow a^2 = 225 \\[1em] \Rightarrow a = \sqrt{225} \\[1em] \Rightarrow a = 15 \text{ cm}.$

Perimeter = Sum of all sides of triangle

= 15 + 15 + 18

= 48 cm.

Hence, perimeter of the isosceles triangle = 48 cm.

In the given figure, △ABC is an equilateral triangle having each side equal to 10 cm and △PBC is right angled at P in which PB = 8 cm. Find the area of the shaded region.

Answer

Given,

△ABC is an equilateral triangle.

Each side = 10 cm

By formula,

$\Rightarrow \text{Area of equilateral △ABC} = \dfrac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} \times 10^2 \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} \times 100 \\[1em] \Rightarrow 25\sqrt{3} \\[1em] \Rightarrow 43.3 \text{ cm}^2.$

Given,

PB = 8 cm and BC = 10 cm

By using the Pythagoras theorem in △PBC,

⇒ BC² = PB² + PC²

⇒ 10² = 8² + PC²

⇒ 100 = 64 + PC²

⇒ PC² = 36

⇒ PC = $\sqrt{36}$

⇒ PC = 6 cm.

Area of triangle = $\dfrac{1}{2}$ × base × height

Area of △PBC = $\dfrac{1}{2}$ × PB × PC

= $\dfrac{1}{2}$ × 8 × 6

= 4 × 6 = 24 cm².

Shaded region = Area of △ABC − Area of △PBC

= 43.3 - 24

= 19.3 cm².

Hence, area of shaded region = 19.3 cm².

If the area of an equilateral triangle is $81\sqrt{3}$ cm², find its perimeter.

Answer

Let the length of the side of an equilateral triangle be a cm.

$\Rightarrow \text{Area of triangle} = \dfrac{\sqrt{3}}{4} \times a^2 \\[1em] \Rightarrow 81\sqrt{3} = \dfrac{\sqrt{3}}{4} \times a^2 \\[1em] \Rightarrow a^2 = 81 \times 4 \\[1em] \Rightarrow a^2 = 324 \\[1em] \Rightarrow a = \sqrt{324} \\[1em] \Rightarrow a = 18. \\[1em]$

Perimeter = Sum of all sides of a triangle

= 3a

= 3 × 18

= 54 cm.

Hence, perimeter of the triangle = 54 cm.

The base of a right-angled triangle is 24 cm and its hypotenuse is 25 cm. Find the area of the triangle.

Answer

Given,

Hypotenuse = 25 cm

Base = 24 cm

By using the Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ 25² = 24² + Height²

⇒ 625 = 576 + Height²

⇒ Height² = 625 - 576

⇒ Height² = 49

⇒ Height = $\sqrt{49}$

⇒ Height = 7 cm.

Area = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × 24 × 7

= 12 × 7 = 84 cm².

Hence, area of the triangle = 84 cm².

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

Answer

Let ABC be an isosceles triangle with AB = AC = a cm and BC = b cm.

Altitude (AD) = 8 cm

Perimeter = 32 cm

Perimeter = sum of all sides of a triangle

⇒ 32 = a + a + b

⇒ 32 = 2a + b

⇒ b = 32 - 2a .........(1)

In an isosceles triangle, the altitude drawn from the common vertex bisects the base.

Thus, AD bisects BC.

So, BD = DC = $\dfrac{b}{2}$

∴ ∠ADC = ∠ADB = 90°.

In triangle ADB,

By pythagorean theorem,

$\Rightarrow AB^2 = AD^2 + DB^2 \\[1em] \Rightarrow a^2 = 8^2 + \Big(\dfrac{b}{2}\Big)^2 \\[1em] \Rightarrow a^2 = 64 + \dfrac{b^2}{4} \\[1em] \Rightarrow a^2 = \dfrac{256 + b^2}{4} \\[1em] \Rightarrow 4a^2 = 256 + b^2$

Substituting the value of b from equation (1) in above equation, we get :

$\Rightarrow 4a^2 = 256 + (32 - 2a)^2 \\[1em] \Rightarrow 4a^2 = 256 + 1024 -128a + 4a^2 \\[1em] \Rightarrow 0 = 1280 - 128a \\[1em] \Rightarrow 128a = 1280 \\[1em] \Rightarrow a = \dfrac{1280}{128} \\[1em] \Rightarrow a = 10 \text{ cm}.$

∴ b = 32 - 2(10)

⇒ b = 32 - 20

⇒ b = 12 cm.

Area of triangle ABC = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × BC × AD

= $\dfrac{1}{2}$ × 12 × 8

= 6 × 8 = 48 cm².

Hence, area of triangle = 48 cm².

The area of a triangle is 216 cm² and its sides are in the ratio 3 : 4 : 5. Find the perimeter of the triangle.

Answer

Given,

Area = 216 cm²

Sides = 3 : 4 : 5

Let the sides of a triangle be 3x, 4x and 5x.

Since 3 : 4 : 5 is a pythagorean triplet (3² + 4² = 5²)

Thus, the triangle is a right angled triangle, and sides containing right angle are 3x cm and 4x cm.

$\Rightarrow \text{Area of right angled triangle} = \dfrac{1}{2} \times \text{ (product of sides containg right angle)} \\[1em] \Rightarrow 216 = \dfrac{1}{2} \times 3x \times 4x \\[1em] \Rightarrow 432 = 12x^2 \\[1em] \Rightarrow x^2 = 36 \\[1em] \Rightarrow x = \sqrt{36} \\[1em] \Rightarrow x = 6.$

∴ Sides of a triangle are

⇒ 3x = 3 × 6 = 18 cm

⇒ 4x = 4 × 6 = 24 cm

⇒ 5x = 5 × 6 = 30 cm

Perimeter = Sum of all sides of a triangle

= 18 + 24 + 30

= 72 cm.

Hence, perimeter of the triangle = 72 cm.