**Introduction of solving geometry examples:-**

In geometry, we deal with the problems in triangles, circle, and square are solved using certain formulas is called Solving geometry examples. Here the formulas are very important to solve any examples problems. From this we can find area, volume, and perimeter etc.

To solving the geometry examples are,

Area of triangle formula =½(bh)

Area of square formula = a²

Area of circle formula = πr²

Area of rectangle formula = l*b

**Examples for solving Geometry Problems**

Example 1:

Find the area of square given that a= 35?

Solution:

Area of square = a²

= 35² (calculate the area square )

= 1225

Area of square =1225

Example 2:

Find the area of triangle given that base = 21, height = 18?

Solution:

Area of triangle = ½(bh)

= ½(21*18) ( multiply the values)

= ½(378)

= 189

Area of triangle = 189

Example 3:

Find the area circle given that diameter = 26?

Solution:

Area of circle = πr²

But the radius is not given here; we have found the radius from diameter.

Radius = diameter /2

Radius = 26/2 ( dived the values)

Radius = 13

Area of circle = πr²

Area of circle = π*13²

=3.14*169 ( multiply the values)

Area of circle = 530.66sq.m.

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**Examples for solving Geometry Problems**

Example 4:

Find the length of the rectangle given that area = 155 and base = 23?

Solution:

Area of rectangle = l*b

155 = l*23

l= 155/23

l = 6.7

l = 7

Example 5:

Find the dimension of 3rd side. When the perimeter is fifty and the two sides are same. The equal sides have five greater the 3rd side?

Solution:

Let us assume s be the unknown length of the triangle, then

We know that perimeter of triangle is

P = a+b+c

50 = s + s + s+ 5

50 = 3s + 5

Then solving the value of variable s, we get

3s = 50 – 5

3s = 45

s =15

The value of third side = 15 + 5 =20

The value of the third side is 20

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