**Area of triangles : **

A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. There are four types of triangles. Those are,

(1) Equilateral triangle

(2) Isosceles triangle

(3) Scalene triangle

(4) Right triangle

A triangle with all three sides of equal length. All the angles are 60 |

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

Here "a" represent the length of side of a triangle.

An isosceles triangle is a triangle that has two sides of equal length. |

Area of isosceles triangle = (1/2) x b x h

Here "b" and "h " base and height of the triangle respectively.

An isosceles triangle is a triangle that has two sides of equal length. |

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

S = (a + b + c)/2

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

**Example 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

**Example 2 :**

Find the area of isosceles triangle whose side a = 7 cm and height (h) = 15 cm.

**Solution :**

a = 7 cm

h = 15 cm

Area of isosceles triangle (A) = ½ b x h square unit.

Substitute the value of b and h in Area (A),

Area (A) = ½ (7 x 15)

= ½ x 105

Area (A) = 52.5 square cm

**Example 3 :**

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

**Solution :**

In order to find the altitude on 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 (24-12) (24-16) (24-20)

= √24 (12) (8) (4)

= √2 x 12 x 12 x 4 x 2 x 4

= 12 x 2 x 4

= 96 cm²

Area of the triangle = 96 cm²

(1/2) x b x h = 96 cm²

Here the longest side is 20 cm. If we plug it instead of b we will get

(1/2) x 20 x h = 96

h = (96 x 2) /20

h = 9.6 cm

**Example 4 :**

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

- Area and polygons
- Inverse operations
- Area of square and rectangles
- Area of quadrilaterals
- Area of a parallelogram
- Finding the area of a trapezoid
- Finding the area of a rhombus
- Area of triangles
- Finding the area of a triangle
- Problems using area of a triangles
- Solving area equations
- Writing equations using the area of a trapezoid
- Solving multistep problems
- Area of polygons
- Finding areas of polygons
- Real world problems involving area and perimeter of polygon

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