uiugolal Is 3u cm.SshĂźma divided it in two different ways (Fig 11.2

1.	uiugolal Is 3u cm.SshĂźma divided it in two different ways (Fig 11.21)J the area of this hexagon using both ways.Hende3: There is a hexagon MNOPQR of side 5 cm (Fig 11.20). Aman and11 cm5 cmRidhima' methodAman's method:Fig 11.20Fig 11.21
Answer» There are two ways of finding the area. 1) Dividing the hexagon into 6 equilateral triangles and getting the area of one triangle multiplied by 6. 2) Dividing the hexagon into two trapezium and getting the area of individual trapezium. First we need the height of the equilateral triangle. By pythagorean theorem we have : 5^2 - 2.5^2 = 18.75 Square root of 18.75 = 4.330 1) Area of one equilateral triangle : 4.330 × 0.5 × 5 = 10.825 For the six : 6 × 10.825 = 64.95 square centimeters. 2) The area of the trapezium is : The two parallel sides of the trapezium are 10cm and 5 cm. The height is 4.30cm Area : (10 + 5) /2 × 4.30 = 32.25 We have two trapezium. 2 × 32.28 = 64.5 square centimeter.

uiugolal Is 3u cm.SshĂźma divided it in two different ways (Fig 11.21)J the area of this hexagon using both ways.Hende3: There is a hexagon MNOPQR of side 5 cm (Fig 11.20). Aman and11 cm5 cmRidhima' methodAman's method:Fig 11.20Fig 11.21

Answer»

There are two ways of finding the area.

1) Dividing the hexagon into 6 equilateral triangles and getting the area of one triangle multiplied by 6.

2) Dividing the hexagon into two trapezium and getting the area of individual trapezium.

First we need the height of the equilateral triangle.

By pythagorean theorem we have :

5^2 - 2.5^2 = 18.75

Square root of 18.75 = 4.330

1) Area of one equilateral triangle :

4.330 × 0.5 × 5 = 10.825

For the six :

6 × 10.825 = 64.95 square centimeters.

2) The area of the trapezium is :

The two parallel sides of the trapezium are 10cm and 5 cm.

The height is 4.30cm

Area :

(10 + 5) /2 × 4.30 = 32.25

We have two trapezium.

2 × 32.28 = 64.5 square centimeter.