AREA OF TRIANGLE

Area of triangle with coordinates :

Here, we are going to see, how to find area of a triangle when coordinates of the three vertices are given.

Let us consider the triangle given below.

In the above triangle, A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are the vertices.

To find area of the triangle ABC, now we have take the vertices A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) of the triangle ABC in order (counter clockwise direction) and write them column-wise as shown below.

And the diagonal products x₁y₂, x₂y₃ and x₃y₁ as shown in the dark arrows.

Also add the diagonal products xy, xy and xy as shown in the dotted arrows.

Now, subtract the latter product from the former product to get area of the triangle ABC.

Hence, area of the triangle ABC.

Area of triangle using vertices - Practice problems

Problem 1 :

Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)

Solution :

Plot the given points in a rough diagram as given below and take them in order (counter clock wise)

Let the vertices be A(1, 2), B(-3, 4) and C(-5, -6)

Then, we have

(x₁, y₁)  =  (1, 2)

(x₂, y₂)  =  (-3, 4)

(x₃, y₃)  =  (-5, -6)

Area of triangle ABC is

=  (1/2) x  { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }

=  (1/2) x  { [1.4 + (-3).(-6) + (-5).2] - [(-3).2 + (-5).4 + 1.(-6)] }

=  (1/2) x  { [4 + 18 - 10] - [-6 - 20 -6] }

=  (1/2) x  { [12] - [-32] }

=  (1/2) x  { 12 + 32 }

=  (1/2) x  {  44 }

=  22 square units.

Hence, are of triangle ABC is 22 square units.

Problem 2 :

If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".

Solution :

Let (x₁, y₁)  =  (6, 7),  (x₂, y₂)  =  (-4, 1) and (x₃, y₃)  =  (a, -9)

Given :  Area of triangle ABC   =  68 square units

(1/2) x  { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }  =  68

Multiply by 2 on both sides,

{ [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] }  =  136

{ [6 + 36 + 7a] - [-28 + a - 54] }  =  136

[42 + 7a] - [a - 82]  =  136

42 + 7a -a +82  =  136

6a + 124  =  136

6a  =  12

a  =  2

Hence, the value of "a" is 2.

Area of different types of triangle - 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 triangle"

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

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 triangle"

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 triangle"

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 triangle"

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 triangle"

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 triangle"

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 triangle"

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 triangle"

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

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

After having gone through the stuff given above, we hope that the students would have understood "Area of triangle".

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