WORD PROBLEMS ON AREA AND PERIMETER OF TRIANGLE

Problem 1 :

The semi perimeter of a triangle having the length of its sides as 20 cm, 15 cm and 9 cm is

(a) 44 cm     (b) 21 cm     (c) 22 cm     (d) None

Solution :

Let a = 20 cm, b = 15 cm and c = 9 cm

Semi perimeter S = (a + b + c) / 2

S = (20 + 15 + 9) / 2

S = 44/2

S = 22 cm

Therefore, semi perimeter of triangle is 22 cm.

So, option (c) is correct.

Problem 2 :

The angles of a triangle are in the ratio 5:3:7, then the triangle is

(a) Acute angled     (b) Obtuse angled

(c) Right triangle     (d) Isosceles triangle

Solution :

Angles of triangle are in the ratio 5: 3: 7

Let the angles be 5x, 3x and 7x

5x + 3x + 7x = 180°

15x = 180°

x = 12°

5x = 5 × 12 = 60°

3x = 3 × 12 = 36°

7x = 7 × 12 = 42°

Since, all angles are less than 90°.

Therefore, the triangle is an acute angled triangle.

So, option (a) is correct.

Problem 3 :

The base of an isosceles triangle whose area is 12 cm² and one of the equal sides is 5 cm is

(a) 6 cm     (b) 8 cm     (c) Both a and b     (d) None of these

Solution:

Let equal sides be (a) = 5 cm 

Let b be the base of isosceles triangle.

Area of an isosceles triangle = 12 cm²

12 = (b/4) √4a² - b²

12 = (b/4) √4(5)² - b²

48 = b√100 - b²

Squaring both sides,

2304 = b² (100 - b²)

b⁴ - 100b² + 2304 = 0

b⁴ - 64b² - 36b² + 2304 = 0

b² (b² - 64) - 36 (b² - 64) = 0

(b² - 64) (b² - 36) = 0

(b² - 64) = 0

b² = 64

b =  ± 8

(b² - 36) = 0

b² = 36

b =  ± 6

Hence, base = 8 cm or 6 cm.

So, option (c) is correct.

Problem 4 :

The length of each side of an equilateral triangle having an area of 9√3 cm² is

(a) 8 cm     (b) 36 cm     (c) 4 cm     (d) 6 cm

Solution:

The area of equilateral triangle is 9√3 cm²

Since, the area of equilateral triangle is = (√3/4) a²

(√3/4) a² = 9√3

a² = 9 × 4

a² =  36

a = 6 cm

Hence, the side of the triangle is 6 cm.

So, option (d) is correct.

Problem 5 :

The length of the perpendicular drawn on the longest side of a scalene triangle is

(a) Largest     (b) Smallest     (c) No relation     (d) None

Solution :

In a scalene triangle the length of perpendicular drawn on smallest side is largest and the length of perpendicular drawn on largest side is smallest.

So, option (b) is correct.

Problem 6 :

Semi perimeter of scalene triangle of side k, 2k and 3k is

(a) k     (b) 2k     (c) 3k     (d) None

Solution:

Let a = k, b = 2k and c = 3k

Semi perimeter S = (a + b + c) / 2

S = k + 2k + 3k / 2

S = 6k/2

S = 3k

Therefore, the semi perimeter of given triangle is 3k.

So, option (c) is correct.

Problem 7 :

If the area of an equilateral triangle is 16√3 cm², then the perimeter of the triangle is

(a) 48 cm     (b) 24 cm     (c) 12 cm     (d) 36 cm

Solution:

Given, area of an equilateral triangle = 16√3 cm²

Area of an equilateral triangle = (√3/4) (side)²

(√3/4) (side)² = 16√3

(side)² = 64

Side = 8 cm

Perimeter of an equilateral triangle = 3 × side

= 3 × 8

= 24 cm

Hence, the perimeter of an equilateral triangle is 24 cm.

So, option (b) is correct.

Problem 8:

An isosceles right triangle has an area 8 cm². The length of its hypotenuse is 

(a) √32 cm     (b) √16 cm     (c) √48 cm     (d) √24 cm

Solution :

Let height of triangle = h

Let base = height = h

Area of triangle = 8 cm²

 1/2 × Base × Height = 8

1/2 × h × h = 8

h² = 16

h = 4 cm

Base = Height = 4 cm

Since the triangle is right angles,

Hypotenuse² = Base² + Height²

Hypotenuse² = 4² + 4²

= 32

Hypotenuse = √32 cm

So, option (a) is correct.

Problem 9 :

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at 0.09 per cm² is 

(a) 2.00     (b) 2.16     (c) 2.48     (d) 3.00

Solution:

s = (a + b + c) / 2

= 6 + 8 + 10 / 2

s = 12

By Heron's formula,

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

= √12(12-6) (12 - 8) (12 - 10)

= √12(6)(4)(2)

= √576

= 24 cm²

Cost of painting = 9 × 24 paise

= 216 paise

= 2.16

So, option (b) is correct.