Question

Radha made a picture of an aeroplane with coloured paper as shown. Find the total area of the paper used.
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Answer 1

$\sf \underline{\large{Solution - }}$

Area of first region :

➢ In the first region, sides of triangle are given as 5 cm, 5 cm and 1 cm.

$\sf Area \: of \: triangle = \sqrt{S(S - a)( S- b)( S- c)}$

Where,

➢ S = Semi-perimeter

➢ a, b, c = Sides of triangle

Let us take : a = 5 cm, b = 5 cm, c = 1 cm

$\sf : \implies \: S = \dfrac{a + b + c}{2}$

$\sf : \implies \: S = \dfrac{5+ 5 + 1}{2}$

$\sf : \implies \: S = \dfrac{11}{2}$

Now,

$\sf Area \: of \: triangle = \sqrt{S(S - a)( S- b)( S- c)}$

$\sf : \implies \: \sqrt{ \dfrac{11}{2} \bigg( \dfrac{11}{2} - 5\bigg ) \bigg( \dfrac{11}{2} - 5\bigg) \bigg( \dfrac{11}{2} - 1 \bigg)}$

$\sf : \implies \: \sqrt{ \dfrac{11}{2} \bigg( \dfrac{1}{2} \bigg ) \bigg( \dfrac{1}{2} \bigg) \bigg( \dfrac{9}{2} \bigg)}$

$\sf : \implies \: \sqrt{ \dfrac{11 \times 9}{2 \times 2 \times 2 \times 2} }$

$\sf : \implies \: \sqrt{ \dfrac{99}{16}} = \dfrac{ \sqrt{99} }{ \sqrt{16} } = \dfrac{3 \sqrt{11} }{4} = \dfrac{3 (3.31) }{4} = \dfrac{9.93 }{4} = \bf2.48 \: cm^{2}$

Hence, area of first region = 2.48 cm²

$\quad$

Area of second region :

➢ In the second region, a rectangle is given with, Length = 6.5 cm and Breadth = 1 cm.

$\sf Area \: of \: rectangle = Length \times Breadth$

So,

➢ Area of rectangle = 6.5 × 1 = 6.5 cm²

Hence, area of second region = 6.5 cm²

$\quad$

Area of third region :

➢ In the third region, a trapezium is given as shown in the attachment.

Area of △ BOC =

$\sf Area \: of \: triangle = \sqrt{S(S - a)( S- b)( S- c)}$

$\sf : \implies \: S = \dfrac{1 + 1 + 1}{2}$

$\sf : \implies \: S = \dfrac{3}{2}$

Now,

$\sf Area \: of \: triangle = \sqrt{S(S - a)( S- b)( S- c)}$

$\sf : \implies \: \sqrt{ \dfrac{3}{2} \bigg( \dfrac{3}{2} - 1\bigg ) \bigg( \dfrac{3}{2} - 1\bigg) \bigg( \dfrac{3}{2} - 1 \bigg)}$

$\sf : \implies \: \sqrt{ \dfrac{3}{2} \bigg( \dfrac{1}{2}\bigg ) \bigg( \dfrac{1}{2}\bigg) \bigg( \dfrac{1}{2} \bigg)}$

$\sf : \implies \: \sqrt{ \dfrac{ 3}{2 \times 2 \times 2 \times 2} }$

$\sf : \implies \: \sqrt{ \dfrac{3}{16}} = \dfrac{ \sqrt{3} }{ \sqrt{16} } = \dfrac{ \sqrt{3} }{4} = \bf0.43 \: cm^{2}$

Also,

$\sf : \implies \: Area \: of \: triangle = \dfrac{1}{2} \times Base \times Height$

$\sf : \implies \: 0.43 = \dfrac{1}{2} \times 1 \times Height$

$\sf : \implies \: Height \: (h) = \bf0.866 \: cm$

So,

$\sf : \implies \: Area \: of \: trapezium =\dfrac{1}{2} \:( sum\:of \: parallel \: sides)\:\times height$

$\sf : \implies \: \dfrac{1}{2} \:(1+2)\:\times 0.866$

$\sf : \implies \:\dfrac{3}{2} \times 0.866$

$\sf : \implies \:\dfrac{2.598}{2} = \bf 1.299 \: cm^2$

Hence, area of third region = 1.299 cm²

$\quad$

Area of fourth and fifth region :

➢ Fourth and fifth region are similar right angled triangle.

So, Area of fourth region = Area of fifth region

$\sf : \implies \: Area \: of \: triangle = \dfrac{1}{2} \times Base \times Height$

$\sf : \implies \: \dfrac{1}{2} \times 1.5 \times 6$

$\sf : \implies \: 3 \times 1.5= \bf 4.5 \: cm ^2$

Hence, area of fourth and fifth region = 4.5 cm²

$\quad$

Now,

The total area of the paper used :

$\sf : \implies \: Area\: (I) + Area \:(II)+ Area \:(III) +Area \:(IV)+ Area\: (V)$

$\sf : \implies \: 2.48 + 6.5+ 1.299+4.5+ 4.5$

$\bf : \implies \: 19.279 \: cm^2$

Therefore, total area of the paper used = 19.279 cm²