Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

 

Solution:

For the triangle I section:

 

It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm

Perimeter = 5+5+1 = 11 cm

So, semi perimeter = 11/2 cm = 5.5 cm

Using Heron’s formula,

Area = √[s(s-a)(s-b)(s-c)]

= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm2

= √[5.5×0.5×0.5×4.5] cm2

= 0.75√11 cm2

= 0.75 × 3.317cm2

= 2.488cm2 (approx)

For the quadrilateral II section:

This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.

∴ Area = 6.5×1 cm2=6.5 cm2

For the quadrilateral III section:

It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The perpendicular height of the parallelogram will be

= 0.86 cm

And, the area of the equilateral triangle will be (√3/4×a2) = 0.43

∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm2 (approximately).

For triangle IV and V:

These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm

Area triangles IV and V = 2×(½×6×1.5) cm2 = 9 cm2

So, the total area of the paper used = (2.488+6.5+1.3+9) cm2 = 19.3 cm