AREA AND PERIMETER OF TRIANGLES WORD PROBLEMS

Problem 1 :

Find the area of an equilateral triangle with side 10 cm.

Solution:

Given, side length = 10 cm

Area of an equilateral triangle = (√3/4) a²

= (√3/4) × 10²

= (√3/4) × 100

A = 25√3 cm²

Problem 2 :

If the angles of a triangle are in the ratio 2: 3: 4 find the angles of the triangle.

Solution:

Let us consider the angles as 2x, 3x and 4x.

2x + 3x + 4x = 9x

Angle sum of triangle = 180˚

9x = 180˚

x = 180/9

x = 20˚

So, 2x = 2 × 20 = 40˚

3x = 3 × 20 = 60˚

4x = 4 × 20 = 80˚

Therefore, the angles are 40˚, 60˚, 80˚.

Problem 3 :

The area of a triangle is 48 cm². If its base is 12 cm, find its altitude.

Solution:

Given, area of a triangle = 48 cm²

base = 12 cm

Area of triangle = 1/2 × base × altitude

48 = 1/2 × 12 × altitude

altitude = 48/6

altitude = 8 cm

Problem 4 :

The hypotenuse of an isosceles right triangle is 10 cm. Find its area.

Solution:

Let sides are x.

Base² + Height² = Hypotenuse²

x² + x² = 10²

2x² = 100

x² = 50

Area of triangle = 1/2 × base × height

= 1/2 × x × x

= x²/2

= 50/2

A = 25 cm²

Hence, the area of isosceles triangle is 25 cm².

Problem 5 :

A rhombus has perimeter 120 m and one of its diagonal is 50 m. Find the area of the rhombus.

Solution:

Perimeter of rhombus = 120 m

d₁ + d₂ = 120 m

50 + d₂ = 120 m

d₂ = 120 - 50

d₂ = 70 m

Area of rhombus = 1/2 × d₁ × d₂

= 1/2 × 50 × 70

= 1750 m²

So, the area of rhombus is 1750 m² .

Problem 6 :

The area of an equilateral triangle is 2√3 cm². Find its perimeter.

Solution:

Given, Area of an equilateral triangle = 2√3 cm²

Area of an equilateral triangle = √3/4 a²

√3/4 a² = 2√3

a² = 2 × 4

a²  = 8

a = 2√2

Perimeter of an equilateral triangle = 3a

= 3 × 2√2

= 6√2 cm

Problem 7 :

The area of an equilateral triangle is 81√3 cm², find its height.

Solution:

Given, Area of an equilateral triangle = 2√3 cm2

Area of an equilateral triangle = √3/4 a²

√3/4 a² = 81√3

a² = 81 × 4

a²  = 324

a = 18 cm

Height of an equilateral triangle = √3/2 (a)

= √3/2 (18)

= 9√3 cm

So, height of an equilateral triangle is 9√3 cm.

Problem 8 :

Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm respectively.

Solution:

Let sides are 13 cm, 14 cm and 15 cm.

s = (a + b + c) / 2

= (13 + 14 + 15) / 2

= 42/2

s = 21

Area of triangle = √ (s(s - a) (s - b) (s - c))

= √ 21(21 - 13) (21 - 14) (21 - 15)

= √ 21 × 8 × 7 × 6

= √7056

A = 84 cm²

So, the area of triangle is 84 cm².

Problem 9 :

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

Solution:

Area of an isosceles triangle = (1/4) b√(4a² - b²)

= 1/4 × 24√(4(13)² - (24)²)

= 6√ (4 × 169 - 576)

= 6√676 - 576

= 6√100

= 6 × 10

= 60 cm²

So, Area of an isosceles triangle is 60 cm².

Problem 10 :

Prove that the area of a quadrilateral ABCD is 4(√3 + 2√2)m². AB = 6 cm, BC = 6 cm, CD = 4 cm and AD = 4 cm and diagonal AC = 4 cm.

Solution:

∆ ABC, AB = 6 cm, BC = 6 cm, AC = 4 cm

s = (a + b + c) / 2

= (6 + 6 + 4) / 2

= 16/2

s = 8

Area of triangle = √ (s(s - a) (s - b) (s - c))

= √ 8(8 - 6) (8 - 6) (8 - 4)

= √ 8 × 2 × 2 × 4

= √128

A = 8√2 cm²

∆ ADC, AC = 4 cm, AD = 4 cm, CD = 4 cm

s = (a + b + c) / 2

= (4 + 4 + 4) / 2

= 12/2

s = 6

Area of triangle = √ (s(s - a) (s - b) (s - c))

= √ 6(6 - 4) (6 - 4) (6 - 4)

= √ 6 × 2 × 2 × 2

= √48

A = 4√3 cm²

Total area of quadrilateral ABCD = 8√2 + 4√3 cm²

= 4(√3 + 2√2) cm²

Hence, it is proved.

Problem 11 :

The perimeter of a right angles is 144 cm and its hypotenuse measures 65 cm. Find the length of other sides and calculate its area. Verify the result using Heron's formula.

Solution:

x + y + 65 = 144

x + y = 144 - 65

x + y = 79 ---> (1)

x² + y² = (65)²

x² + y² = 4225

Now, 

(x + y)² = x² + y² + 2xy

2xy = (x + y)² - (x² + y²)

= (79)² - 4225 

= 6241 - 4225

= 2016

Then, x - y = ± √ (x² + y² - 2xy)

x - y = ± √ (4225 - 2016)

= ± √2209

x - y = ± 47 ---> (2)

Solving (1) and (2), we get

x = 63 cm, y = 16 cm (or)

x = 16 cm, y = 63 cm

Area of triangle = 1/2 × b × h

= 1/2 × 63 × 16

A = 504 cm²

Now, verify using Heron's formula,

s = (a + b + c) / 2

= (65 + 63 + 16) / 2

= 144/2

s = 72

Area of triangle = √ (s(s - a) (s - b) (s - c))

= √ 72(72 - 65) (72 - 63) (72 - 16)

= √ 72 × 7 × 9 × 56

= √254016

A = 504 cm²

So, it is verified.

Problem 12 :

Find the perimeter of an isosceles right triangle having an area equal to 200 cm².

Solution:

Let ABC be an isosceles right angled triangle with right angle at B such that AB = BC = a cm

Area of triangle = 1/2 × b × h

200 = 1/2 × a × a

200 = a²/2

a² = 400

a = 20 cm 

By using Pythagorean Theorem,

AC² = AB² + BC²

AC² = a² + a²

AC² = 2a²

AC = a√2 cm

= 20√2 cm

Perimeter = AB + BC + AC

= 20 + 20 + 20√2

= 68.2 cm

Problem 13 :

The area of an isosceles right triangle is 128 cm². Find the length of its hypotenuse.

Solution:

Let the base and height of the isosceles triangle be x.

Area = 1/2 × base × height

128 = 1/2 × x × x

x² = 256

x = 16

Using Pythagorean Theorem,

AC² = AB² + BC²

AC² = 16² + 16²

AC² = 256 + 256

AC² = 512

AC = 22.62 cm

So, the length of hypotenuse is 22.62 cm.