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 =

=

=

= $\sqrt{6.1875}$ 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 $\times$ 4.5 cm= 9 cm2

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

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