Class 7 Maths Perimeter and Area Area of Triangle

Area of Triangle

Let ABCD is a parallelogram. Cut the parallelogram along its diagonal to get two triangles as shown in the figure.

          Class_7_Maths_Perimeter_And_Area_Traiangle_As_A_Part_Of_Parallelogram                                                     

The base and the height of the triangle are the same as the base and the height of the

parallelogram, respectively. Now, we will find that the sum of the areas of both the triangles is equal to the area of the parallelogram.

Now, Area of each triangle = (1/2) * (Area of parallelogram)

        = (1/2) * (base * height)     [Since area of a parallelogram = base * height]

          = (1/2) * (b * h)

          = bh/2

Problem: Find the area of each of the following triangles:

 Class_7_Maths_Perimeter_And_Area_Area_Of_Triangle                 

 

 

Solution:

We know that the area of triangle = (1/2) * base * height

(a) Here, base = 4 cm and height = 3 cm

So, area of triangle = (1/2) * 4 * 3 = 12/2 = 6 cm2

(b) Here, base = 5 cm and height = 3.2 cm

Area of triangle = (1/2) * 5 * 3.2 = 16/2 = 8 cm2

(c) Here, base = 3 cm and height = 4 cm

Area of triangle = (1/2) * 3 * 4 = 12/2 = 6 cm2

(d) Here, base = 3 cm and height = 2 cm

Area of triangle = (1/2) * 3 * 2 = 6/2 = 3 cm2

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