NCERT Solutions Class 9 Mathematics Solutions for Heron’s Formula - Exercise 12.2 in Chapter 12 - Heron’s Formula
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Radha made a picture of an airplane with colored paper as shown in Fig 12.15. Find the
the total area of the paper used.
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)]
,
For the quadrilateral II section:
This quadrilateral is a rectangl with length and breadth as 6.5 cm and 1 cm respectively.
\therefore \text { Area }=6.5 \times 1 \mathrm{cm}^{2}=6.5 \mathrm{cm}^{2}
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
For triangle IV and V:
These triangles are 2 congruent right-angled triangles having the base as 6 cm and height
1.5 cm
\begin{aligned} &\text { Area triangles } \mathrm{IV} \text { and } \mathrm{V}=2 \times(7 / 2 \times 6 \times 1.5) \mathrm{cm}^{2}=9 \mathrm{cm}^{2}\\ &\text { So, the total area of the paper used }=(2.488+6.5+1.3+9) \mathrm{cm}^{2}=19.3 \mathrm{cm}^{2} \end{aligned}
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