Radha made a picture of an aeroplane with col | vedclass

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(B) Area of surface $I$:It is an isosceles triangle with sides $a = 5\, cm, b = 5\, cm, c = 1\, cm$.Semi-perimeter $s = \frac{5+5+1}{2} = 5.5\, cm$.Ar

Vedclass · Accepted Answer

$19.3$

Vedclass · Answer

(B) Area of surface $I$:It is an isosceles triangle with sides $a = 5\, cm, b = 5\, cm, c = 1\, cm$.Semi-perimeter $s = \frac{5+5+1}{2} = 5.5\, cm$.Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)} = \sqrt{5.5 	imes 0.5 	imes 0.5 	imes 4.5} = \sqrt{6.1875} \approx 2.48\, cm^{2}$.Area of surface $II$:It is a rectangle with length $6.5\, cm$ and breadth $1\, cm$.Area $= 6.5 	imes 1 = 6.5\, cm^{2}$.Area of surface $III$:It is a trapezium with parallel sides $1\, cm$ and $2\, cm$. The height $h$ is calculated using Pythagoras theorem: $h = \sqrt{1^{2} - (0.5)^{2}} = \sqrt{0.75} \approx 0.866\, cm$.Area $= \frac{1}{2} 	imes (1 + 2) 	imes 0.866 = 1.5 	imes 0.866 \approx 1.3\, cm^{2}$.Area of surface $IV$ and $V$:These are two congruent right-angled triangles with base $6\, cm$ and height $1.5\, cm$.Area of $IV = \frac{1}{2} 	imes 6 	imes 1.5 = 4.5\, cm^{2}$.Area of $V = 4.5\, cm^{2}$.Total area $= 2.48 + 6.5 + 1.3 + 4.5 + 4.5 = 19.28\, cm^{2} \approx 19.3\, cm^{2}$.

Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the total area of the paper used. (in $, cm^{2}$)

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