# Worksheet on Area and Perimeter of Triangle | Area and Perimeter of Triangle Worksheet with Answers

Students looking for different problems on Perimeter and Area of the Triangle can get them in one place. Use the Worksheet on Area and Perimeter of Triangle and kick start your preparation. Students can assess their strengths and weaknesses by solving all questions from Triangle Area and Perimeter Worksheet. Make use of these worksheets and understand different formulas and ways of solving isosceles triangle, equilateral triangle, scalene, and right-angled triangle areas and perimeter.

The Questions covered in the Area and Perimeter of Triangle Worksheet include various triangles area and perimeter. Get those formulas and step by step process to solve all questions. Here, we are offering a detailed solution for each and every problem for a better understanding of concepts.

1. Find the Area, Perimeter of the following triangles:

(a) (b) (c) Solution:

(a)

Sides of the triangle are a = 12, b = 16, c = 20

s = (a + b + c)/2

s = (12 + 16 + 20)/2 = 48/2

s = 24

Area of the triangle formula is A = √[s (s – a) (s – b) (s – c)]

A = √[24 (24 – 12) (24 – 16) (24 – 20)]

= √[24 (12) (8) (4)]

= √(9216) = 96 sq units.

Perimeter of the traingle = (a + b + c)

= 12 + 16 + 20 = 48 units

∴ The area, perimeter of the given triangle is 96 sq units, 48 units.

(b)

The sides of the given triangle are a = 5, b = 13, c = 12

Semiperimeter of the traingle is s = (a + b + c)/2

s = (5 + 13 + 12)/2 = 30/2 = 15

Area of the triangle formula is A = √[s (s – a) (s – b) (s – c)]

A = √[15 (15 – 5) (15 – 13) (15 – 12)]

= √[15 (10) (2) (3)]

= √ = 30

Perimeter of the triangle formula is P = (a + b + c)

P = (5 + 13 + 12) = 30

∴ The perimeter and area of the triangle is 30 units, 30 sq units.

(c)

Sides of the triangle are a = 30 cm, b = 35 cm, c = 55 cm

Perimeter of the triangle = (a + b + c)

= (30 + 35 + 55) = 120 cm

Semiperimeter of the triangle s = (a + b + c)/2

= (30 + 35 + 55) = 120/2 = 60 cm

Area of the triangle formula is A = √[s (s – a) (s – b) (s – c)]

A = √[60 (60 – 30) (60 – 35) (60 – 55)]

= √[60 (30) (25) (5)]

= √[225,000] = 474.34 sq. cm

2. Find the area of the triangle whose dimensions are:

(i) base = 20 cm, height = 5 cm

(ii) base = 6.5 m, height = 8 m

(iii) base = 28 m, height = 7 m

Solution:

(i)

Given that,

base = 20 cm, height = 5 cm

Area of the triangle formula is A = ½ x base x height

= ½ x 20 x 5

Area = 10 x 5 = 50 cm²

(ii)

Given that,

base = 6.5 m, height = 8 m

Area of the triangle formula is A = ½ x base x height

A = ½ x 6.5 x 8

Area = 6.5 x 4 = 26 m²

(iii)

Given that,

base = 28 m, height = 7 m

The area of the triangle formula is A = ½ x base x height

= ½ x 28 x 7 = 14 x 7

Area = 98 m²

3. Find the base of a triangle whose

(i) Area = 612 cm², height = 6.2 cm

(ii) Area = 528 cm², height = 10 cm

(iii) Area = 126 m², height = 8 m

Solution:

(i)

Given that,

Area = 612 cm², height = 6.2 cm

Area A = ½ x base x height

612 = ½ x base x 6.2

base = (2 x 612)/6.2 = 1224/6.2

base = 197.41 cm

(ii)

Given that,

Area = 528 cm², height = 10 cm

Area A = ½ x base x height

528 = ½ x base x 10

base = 528/5 = 105.6 cm

(iii)

Given that,

Area = 126 m², height = 8 m

Area A = ½ x base x height

126 = ½ x base x 8

base = 126/4 = 31.5 m

4. Find the height of the triangle, whose

(i) Area = 1500 m², Base = 7.5 m

(ii) Area = 480 cm², Base = 8 cm

(iii) Area = 2580 cm², Base = 12 cm

Solution:

(i) Given that,

Area = 1500 m², Base = 7.5 m

Area A = ½ x base x height

1500 = ½ x 7.5 x height

height = 3000/7.5

height = 400 m

(ii) Given that,

Area = 480 cm², Base = 8 cm

Area A = ½ x base x height

480 = ½ x 8 x height

height = 480/4

height = 120 cm

(iii) Given that,

Area = 2580 cm², Base = 12 cm

Area A = ½ x base x height

2580 = ½ x 12 x height

height = 2580/6

height = 430 cm.

5. The three sides of the triangle are 10 cm, 14 cm, 18 cm. Find its semi perimeter, perimeter, and area.

Solution:

Given that,

Sides of the triangle are a = 10 cm, b = 14 c, c = 18 cm

Perimeter = (a + b + c)

= 10 + 14 + 18 = 42 cm

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

= (10 + 14 + 18)/2 = 42/2

= 21 cm

Area of the triangle = √[s (s – a) (s – b) (s – c)]

= √[21 (21 – 10) (21 – 14) (21 – 18]

= √[21 (11) (7) (3)]

= √ = 69.64 cm²

6. Find the area, the perimeter of the triangle, whose sides are 12 m, 15 m, and the semiperimeter is 17.5 m.

Solution:

Given that,

The sides of the triangle are a = 12 m, b = 15 m

Triangle semiperimeter s = 17.5

s = (a + b + c)/2

17.5 = (12 + 15 + c)/2

35 = 27 + c

c = 35 – 27

c = 8 m

Perimeter of the triangle = (a + b + c)

= 12 + 15 + 8 = 35 m

Triangle area = √[s (s – a) (s – b) (s – c)]

= √[17.5 (17.5 – 12) (17.5 – 15) (17.5 – 8)]

= √[17.5 (5.5) (2.5) (9.5)] = √[2,285.9375]

= 47.81 m².

∴ The Triangle perimeter is 35 m, another side is 8 m, and the area is 47.81 m².

7. The three sides of the triangle are in the ratio 5: 2: 6 and its perimeter is 78 units. Find its area?

Solution:

Given that,

The ratio of three sides of the triangle = 5: 2: 6

Triangle perimeter = 78 units

Let us take x as the side of the triangle.

Then, the three sides of a triangle are 5x, 2x, 6x.

Triangle perimeter = 78 units

5x + 2x + 6x = 78

13x = 78

x = 78/13

x = 6 units

So, three sides of the triangle are 30, 12, 36

Semi perimeter s = (30 + 12 + 36)/2 = 78/2 = 39

Triangle area = √[s (s – a) (s – b) (s – c)]

= √[39 (39 – 30) (39 – 12) (39 – 36)]

= √[39 (9) (27) (3)]

= √[28,431] = 168.61 sq units.

8. Find the base of the triangle, whose area is 120 m², altitude is 15 m?

Solution:

Given that,

Altitude = 15 m

Triangle area A = 120 m²

½ x base x height = 120

½ x base x 15 = 120

base = 240/15 = 16 m

9. Find the height of the triangle, whose base is 9 units, and the area is 144 sq units?

Solution:

Given that,

Base = 9 units

Triangle Area = 144 sq units

½ x base x height = 144

½ x 9 x height = 144

height = 288/9 = 32 units.

10. Find the area of an equilateral triangle, the length of whose sides is 11 cm?

Solution:

Given that,

Side of the equilateral triangle a = 11 cm

Area of triangle = (√3/4) a²

Area = (√3/4) 11² = (√3/4) x 11 x 11

= 52.39 cm²

∴ The area of an equilateral triangle is 52.39 cm².

11. The base and height of a triangle are in the ratio 15: 14 and its area is 320 m². Find the height and base of the triangle?

Solution:

Given that,

The ratio of base and height of triangle = 15: 14

Let us take 15x, 14x as the triangle base and height.

Area of triangle = 320

½ x base x height = 320

½ x 15x x 14x = 320

105x² = 320

x² = 320/105

x = √(3.047) = 1.74 m

Height = 14 x 1.74 = 24.44 m

Base = 15 x 1.74 = 26.1 m

12. Find the area of an isosceles right-angled triangle of equal sides 15 cm each?

Solution:

Given that,

Right-angles triangle sides = 15 cm

Triangle area = ½ x side x side

= ½ x 15 x 15 = 225/2 = 112.5 cm

∴ The area of the triangle is 112.5 cm.

13. Find the area of a triangle, whose sides are 5 m, 4 m and angle is 120°.

Solution:

Given that,

The sides of a triangle are a = 5 m, b = 4 m

Angle c = 180°.

Area of the triangle = ½ab sin (A)

= ½ x 5 x 4 x sin(120°)

= 10 x (√3/2)

= 5√3

∴ Area of the triangle = 5√3 sq units.

14. Find the other side and perimeter of the right-angled triangle, whose sides are 8 cm, 15 cm?

Solution:

Given that,

The sides of the triangle are a = 8 cm, b = 15 cm

As per the Pythagorean theorem,

c² = a² + b²

c² = 8² + 15²

c² = 64 + 225

c² = 289

c = √289 = 17 cm

Perimeter = (8 + 15 + 17) = 40

∴ Another side, the perimeter of the right-angled triangle is 17 cm, 40 cm.

15. Find the area of a right-angled triangle whose hypotenuse is 13 cm and one of its sides containing the right angle is 12 cm. Find the length of the other side and perimeter.

Solution:

Given that,

Hypotenuse = 13 cm

Side = 12 cm

Hypotenuse² = Side² + Side²

13² = 12² + Side²

Side² = 169 – 144

Side² = 25

Side = √25 = 5 cm

Semi perimeter s = (13 + 5 + 12)/2 = 30/2 = 15

Perimeter = (13 + 5 + 12) = 30 cm

Triangle Area = √[s (s – a) (s – b) (s – c)]

= √[15 (15 – 5) (15 – 12) (15 – 13)]

= √[15 (10) (3) (2)] = √

= 30 cm²

∴ Area is 30 cm², other side is 5 cm.

16. The area of the triangle is equal to that of the square whose each side measures 30 cm. Find the side of the triangle whose corresponding altitude is 36 cm.

Solution:

Given that,

Area of Triangle = Area of square

Side of square = 30 cm

Altitude of triangle = 36 cm

Area of square = 30 x 30 = 900 cm²

Area of Triangle = Area of square

Area of Triangle = 900

½ x base x height = 900

½ x base x 36 = 900

base = 1800/36 = 50 cm

∴ Side of the triangle is 50 cm.

17. The length of one of the diagonals of a field in the form of a quadrilateral is 52 m. The perpendicular distance of the other two vertices from the diagonal is 15 m and 10 m, find the area of the field.

Solution:

Given that,

Diagonal = 52 m

Altitudes h1 = 15, h2 = 10

Area of quadrilateral = ½ x diagonal (h1 + h2)

= ½ x 52 x (15 + 10)

= 26(25)

= 650 m²

Therefore, the quadrilateral area is 650 m².

18. The area of a right triangle is 156 sq units and one of its legs is 13 units, find the other leg?

Solution:

Given that,

Base = 13 units

Area of triangle = 156 sq units

½ x base x height = 156

½ x 13 x height = 156

height = 312/13 = 24.

19. ∆ ABC is isosceles with AB = AC = 18 cm, BC = 12 cm. The height AD from A to BC is 7 cm. Find the area, perimeter of ∆ABC.

Solution:

Given that,

Sides of the triangle are AB = AC = 18 cm, BC = 12 cm

Area of ∆ABD = ½bh

= ½ x 18 x 7

= 9 x 7 = 63 cm²

Area of ∆ABC = Area of ∆ABD + ∆ADC

= 2 x 63 = 126 cm².

20. ∆ ABC is an equilateral triangle with sides 28 cm, Find its perimeter and area?

Solution:

Given that,

Side of triangle = 28 cm

Perimeter = (28 + 28 + 28) = 84 cm

Area = (√3/4)a²

= (√3/4) x 28² = (√3/4) x 784

= √3 x 196 = 339.4

Therefore, perimeter, area of the equilateral trinagle is 84 cm, 339.4 cm².