# AREA OF DIFFERENT TYPES OF TRIANGLES

## About "Area of different types of triangles"

Area of different types of triangles :

Finding area of a triangle is depending on the type of it.

To have better understanding on area of different types of triangles, first we have to know the types of triangles.

In geometry, triangles can be classified using various properties related to their angles and sides.

There are six different types of triangles.

Here we are going to see, "How triangles in geometry can be classified"

## Equilateral Triangle

An equilateral triangle is a triangle in which all the three sides will be equal.

Each angle will be 60°.

## Isosceles triangle

A triangle with two equal sides is called as isosceles triangle.

The angles corresponding to the equal sides will always be equal.

## Scalene triangle

In a scalene triangle the length of all the three sides will be different.

## Right triangle

A right triangle is the triangle in which one of the angles will be 90°.

## Acute triangle

An acute triangle is a triangle with all three angles are less than 90 degree.

## Obtuse triangle

An obtuse triangle is a triangle in which one of the angles are obtuse (greater than 90 degree).

## Area of different types of triangles - Practice problems

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 types of triangles".

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 types of triangles".

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 types of triangles".

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 types of triangles".

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 types of triangles".

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 types of triangles".

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 types of triangles".

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 types of triangles".

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 types of triangles".

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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