Q. 34.2( 487 Votes )

Radha made a pict

Answer :

The figure of the questions is:



Now to find area:

Sides of triangular section I = 5cm, 1cm and 5cm


Perimeter of the triangular section I = 5 + 5 + 1 = 11cm


Semi perimeter, s =   =   = 5.5cm


We use Heron’s formula to find the area of the section I,


Area of Section I =


                         =


                         =


                         =  cm2


Area of Section I = 2.48 cm2 (approx.) ~ 2.5 cm2



Length of the sides of the rectangle of section II = 6.5 cm and 1 cm


Area of section II = 6.5×1 = 6.5cm2



Section III is an isosceles trapezium which is divided into two right triangles and one rectangle. 
Now, Area of trapezium = Area (i) + Area (ii) + Area (iii)
= 1/2 × 0.5 × 1 + 1 × 1 + 1/2 × 0.5 × 1
= 1.5 cm2



Section IV and V are congruent right-angled triangles with height 1.5cm and base 6cm.


now area of triangle, 

 


Area of region IV and V = 2  4.5 cm= 9 cm2


Thus, total area is 2.5 + 6.5 + 9 + 1.5 = 19.5 sq cm

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