Area of different triangles worksheet is much useful to the students who would like to practice problems on triangles.

1) Find the area of the equilateral triangle having the length of the side equals 10 cm

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

3) Find the area of the scalene triangle whose length of sides are 12cm, 18 cm and 20 cm.

4) The sides of a triangle are 12 m, 16 m and 20 m. Find the altitude to the longest side.

5) If the area of a triangle is 1176 cm² and base : corresponding altitude is 3 : 4 the the altitude of the triangle is,

6) The sides of a triangle are in the ratio (1/2):(1/3):(1/4). If the perimeter is 52 cm,then the length of the smallest side is:

7) The area of the triangle is 216 cm² and the sides are in the ratio 3:4:5.The perimeter of the triangle is:

8) One side of a right angle triangle is twice the other,and the hypotenuse is 10 cm. The area of the triangle is:

9) If the perimeter of the isosceles right triangle is (6 + 3√2)m,then the area of the triangle is.

10) If the area of the equilateral triangle is 24√3 square cm, then find its perimeter.

**Problem 1:**

Find the area of the equilateral triangle having the length of the side equals 10 cm

**Solution:**

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

Here a = 10 cm

= (√3/4) (10)²

= (√3/4) x (10) x (10)

= (√3) x (5) x (5)

= 25 √3 cm²

**Area of the given equilateral triangle 25 √3 square cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 2:**

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

**Solution:**

Let ABC be the isosceles triangle and AD be the altitude.

Since it is isosceles triangle the length of two sides will be equal and the altitude bisects the base of the triangle.

Let AB = AC = x cm.

Perimeter of triangle ABC = 32 cm

AB + BC + CA = 32

x + BC + x = 32

BC = 32 - 2 x

DC = (1/2) **x** (32 - 2 x)

= 16 - x

length of altitude (AD) = 8 cm

In triangle ADC,

AC² = AD² + DC²

x² = 8² + (16 - x)²

x² = 64 + 16² + x² - 32 x

32 x = 64 + 256 ==> 32 x = 320 ==> x = 10 cm

DC = 16 - 10 = 6 cm

Area of triangle ADC = (1/2) **x** (bh)

= (1/2) **x** (6 x 8)

= 24 square cm

**Area of triangle ABC = 48 square cm. **

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 3:**

Find the area of the scalene triangle whose length of sides are 12cm, 18 cm and 20 cm.

**Solution:**

Length of each side as a = 12 cm, b = 18 cm and c = 20 cm respectively.

S = (a+b+c) / 2

S = (12+18+20) / 2 ==> 50/2 ==> S = 25

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

= √5 x 5 x 13 x 7 x 5

= 5√455 square cm

**Therefore area of the given scalene triangle = 5 √455 square cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 4:**

The sides of a triangle are 12 m, 16 m and 20 m. Find the altitude to the longest side.

**Solution:**

In order to find the altitude to the longest side of a triangle first we have to find the area of the triangle.

Let a = 12 m, b = 16 m and c = 20 m

S = (a+b+c) / 2 ==> (12+16+20) / 2 ==> 48/2 ==> 24 m

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

= √24 (12) (8) (4) ==> 96 cm²

Area of the triangle = 96 cm²

(1/2) x b x h = 96 cm²

Here the longest side is 20 cm.

(1/2) x 20 x h = 96

h = (96 x 2) /20 ==> 9.6 cm

**The altitude to the longest side = 9.6 cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 5:**

If the area of a triangle is 1176 cm² and base : corresponding altitude is 3 : 4 the the altitude of the triangle is,

**Solution:**

From the given information,

base of the triangle = 3x

altitude = 4x

area of the triangle = 1176

(1/2) **x** 3x x 4x = 1176 ==> 12x² = 2352 ==> x² = 196 ==> x = 14

altitude of the triangle = 4(14) = 56 cm

**Therefore altitude of the triangle = 56 cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 6:**

The sides of a triangle are in the ratio (1/2):(1/3):(1/4). If the perimeter is 52 cm,then the length of the smallest side is:

**Solution:**

From the given information,

the sides the triangle are x/2, x/3 and x/4.

Perimeter of the triangle = 52 cm

(x/2) + (x/3) + (x/4) = 52

(6x+4x+3x) / 12 = 52

13x = 624

x = 48 cm

x/2 = 24, x/3 = 16 and x/4 = 12

**Therefore the length of smallest side = 12 cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 7:**

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

**Solution:**

From the given information,

the sides the triangle are 3x,4x and 5x.

area of the triangle = 216 cm²

S = (3x + 4x + 5x) /2 ===> 6x

√s(s-a) (s-b) (s-c) = 216

√6x(3x) (2x) (x) = 216

6x² = 216 ==> x = 6

3x = 18 cm, 4x = 24 cm and 5x = 30 cm

Perimeter of triangle = 18 + 24 + 30 ==> 72 cm

**Therefore the pereimeter of the triangle = 72 cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 8:**

One side of a right angle triangle is twice the other,and the hypotenuse is 10 cm. The area of the triangle is:

**Solution:**

Let "x" be the length of one side

length of other side = 2x

10² = x² + (2x)² ==> 100 = 5x² ==> x = 2√5 cm, 2x = 4√5 cm

Area of triangle = (1/2) **x** (2√5)(4 √5) ==> 20 square cm

**Therefore area of the triangle = 20 square cm**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 9:**

If the perimeter of the isosceles right triangle is (6 + 3√2)m,then the area of the triangle is.

**Solution:**

Since the given triangle is isosceles right triangle,the length of two sides will be equal.

Let "x" be the length of one equal side, and "y" be the length of hypotenuse side

perimeter of triangle = 6 + 3√2

x + x + y = 6 + 3√2 ==> 2x + y = 6 + 3√2 ----(1)

and y² = x² + x²

y² = 2x² ==> y = √2x

by plugging y = √2x in the first equation,we get

2x + √2x = 6 + 3√2 ==> x = 3

area of triangle = (1/2)**x** (bh)

= (1/2) **x** (3)(3) ==> 4.5 square m

**Therefore the triangle = 4.5 square m**

Let us look at the next problem on "Area of different triangles worksheet".

**Problem 10:**

If the area of the equilateral triangle is 24√3 square cm, then find its perimeter.

**Solution:**

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

(√3/4) a² = 24√3

a² = 96 ==> a =4√6

**Perimeter of the equilateral triangle = 12√6 cm**

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