Hello sir i want to know the best solution of the question from exercise 12.2 of math of Heron’s Formula chapter of class 9^{th} give me the best and easy for solving this question how i solve it of question no.3 Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

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# Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used. Q.3

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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)] cm

^{2}= √[5.5×0.5×0.5×4.5] cm

^{2}= 0.75√11 cm

^{2 }= 0.75 × 3.317cm

^{2}= 2.488cm

^{2}(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 cm

^{2}=6.5 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

= 0.86 cm

And, the area of the equilateral triangle will be (√3/4×a

^{2}) = 0.43∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm

^{2 }(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) cm

^{2 }= 9 cm^{2}So, the total area of the paper used = (2.488+6.5+1.3+9) cm

^{2 }= 19.3 cm^{2}