Answer
Verified
479.1k+ views
Hint- Here, we will be dividing the figure of the aeroplane into simpler figures like rectangle, triangle and trapezium whose formulas for areas we already know and then using those formulas we will find the area of the whole figure.
“Complete step-by-step answer:”
From the figure, we can see that the picture of the aeroplane can be divided into five regions as shown.
As we know that area of the triangle$ = \dfrac{1}{2}\left( {{\text{Base}}} \right)\left( {{\text{Height}}} \right)$.
Also, area of the rectangle\[ = \left( {{\text{Length}}} \right)\left( {{\text{Breadth}}} \right)\]
Also, area of the trapezium$ = \dfrac{{\left( {{\text{Height of trapezium}}} \right)}}{2} \times \left( {{\text{Sum of parallel sides}}} \right)$
In $\vartriangle {\text{GFN}}$, ${\text{NF}} = \dfrac{{{\text{CF}}}}{2} = \dfrac{{{\text{KJ}}}}{2} = \dfrac{1}{2}$ cm and $GF = 5$ cm
According to Pythagoras Theorem, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
$
\Rightarrow {\left( {{\text{GF}}} \right)^2} = {\left( {{\text{GN}}} \right)^2} + {\left( {{\text{NF}}} \right)^2} \\
\Rightarrow {\left( {\text{5}} \right)^2} = {\left( {{\text{GN}}} \right)^2} + {\left( {\dfrac{{\text{1}}}{2}} \right)^2} \\
\Rightarrow {\left( {{\text{GN}}} \right)^2} = 25 - \dfrac{{\text{1}}}{4} = \dfrac{{99}}{4} \\
\Rightarrow {\text{GN}} = \sqrt {\dfrac{{99}}{4}} = \dfrac{{3\sqrt {11} }}{2}{\text{ cm}} \\
$
So, height of the $\vartriangle {\text{GCF}}$ is \[{\text{GN}} = \dfrac{{3\sqrt {11} }}{2}{\text{ cm}}\]
Area of region I = Area of $\vartriangle {\text{GCF}}$ $ = \dfrac{1}{2}\left( {{\text{CF}}} \right)\left( {{\text{GN}}} \right) = \dfrac{1}{2}\left( {\text{1}} \right)\left( {\dfrac{{3\sqrt {11} }}{2}} \right) = \dfrac{{3\sqrt {11} }}{4}{\text{ c}}{{\text{m}}^2}$
Area of region II = Area of rectangle KJFC\[ = \left( {{\text{KJ}}} \right)\left( {{\text{CK}}} \right) = \left( {\text{1}} \right)\left( {{\text{6}}{\text{.5}}} \right) = 6.5{\text{ c}}{{\text{m}}^2}\].
Since, the height of the trapezium KHIJ = KL = JM
In $\vartriangle {\text{JMI}}$, IJ = 1 cm and MI = HL = $\dfrac{{{\text{HI}} - {\text{KJ}}}}{2} = \dfrac{{{\text{2}} - {\text{1}}}}{2} = \dfrac{1}{2}$ cm
According to Pythagoras Theorem, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
$
\Rightarrow {\left( {{\text{IJ}}} \right)^2} = {\left( {{\text{JM}}} \right)^2} + {\left( {{\text{MI}}} \right)^2} \\
\Rightarrow {\left( {\text{1}} \right)^2} = {\left( {{\text{JM}}} \right)^2} + {\left( {\dfrac{{\text{1}}}{2}} \right)^2} \\
\Rightarrow {\left( {{\text{JM}}} \right)^2} = 1 - \dfrac{{\text{1}}}{4} = \dfrac{3}{4} \\
\Rightarrow {\text{JM}} = \sqrt {\dfrac{3}{4}} = \dfrac{{\sqrt 3 }}{2}{\text{ cm}} \\
$
So, the height of trapezium KHIJ is ${\text{JM}} = \dfrac{{\sqrt 3 }}{2}{\text{ cm}}$
Area of region III = Area of trapezium KHIJ$ = \dfrac{{\left( {{\text{JM}}} \right)}}{2} \times \left( {{\text{KJ}} + {\text{HI}}} \right) = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{2} \times \left( {1 + 2} \right) = \dfrac{{3\sqrt 3 }}{4}{\text{ c}}{{\text{m}}^2}$
Area of region IV = Area of region V [because both the triangles ABC and DEF have equal areas due to their same dimensions]
Area of region IV = Area of region V = Area of $\vartriangle {\text{DEF}}$$ = \dfrac{1}{2}\left( {{\text{EF}}} \right)\left( {{\text{DF}}} \right) = \dfrac{1}{2}\left( {{\text{1}}{\text{.5}}} \right)\left( {\text{6}} \right) = \dfrac{9}{2}{\text{ c}}{{\text{m}}^2}$
Therefore, the total area of the paper used = Area of region I + Area of region II + Area of region III + Area of region IV + Area of region V \[ = \dfrac{{3\sqrt {11} }}{4} + 6.5 + \dfrac{{3\sqrt 3 }}{4} + \dfrac{9}{2} + \dfrac{9}{2} = \dfrac{{3\sqrt {11} + 26 + 3\sqrt 3 + 18 + 18}}{4} = 19.2865{\text{ c}}{{\text{m}}^2}\].
So, the total area of the coloured paper used is 19.2865 \[{\text{c}}{{\text{m}}^2}\].
Note- In this particular problem, from symmetry point of view we can say that the perpendicular GN drawn from point G to the line CF divides the line into two equal parts which are CN and NF i.e., ${\text{NF}} = {\text{CN}} = \dfrac{{{\text{CF}}}}{2} = \dfrac{{{\text{KJ}}}}{2}$ and the triangles KLH and JMI are symmetrically same that’s why MI = HL = $\dfrac{{{\text{HI}} - {\text{KJ}}}}{2}$.
“Complete step-by-step answer:”
From the figure, we can see that the picture of the aeroplane can be divided into five regions as shown.
As we know that area of the triangle$ = \dfrac{1}{2}\left( {{\text{Base}}} \right)\left( {{\text{Height}}} \right)$.
Also, area of the rectangle\[ = \left( {{\text{Length}}} \right)\left( {{\text{Breadth}}} \right)\]
Also, area of the trapezium$ = \dfrac{{\left( {{\text{Height of trapezium}}} \right)}}{2} \times \left( {{\text{Sum of parallel sides}}} \right)$
In $\vartriangle {\text{GFN}}$, ${\text{NF}} = \dfrac{{{\text{CF}}}}{2} = \dfrac{{{\text{KJ}}}}{2} = \dfrac{1}{2}$ cm and $GF = 5$ cm
According to Pythagoras Theorem, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
$
\Rightarrow {\left( {{\text{GF}}} \right)^2} = {\left( {{\text{GN}}} \right)^2} + {\left( {{\text{NF}}} \right)^2} \\
\Rightarrow {\left( {\text{5}} \right)^2} = {\left( {{\text{GN}}} \right)^2} + {\left( {\dfrac{{\text{1}}}{2}} \right)^2} \\
\Rightarrow {\left( {{\text{GN}}} \right)^2} = 25 - \dfrac{{\text{1}}}{4} = \dfrac{{99}}{4} \\
\Rightarrow {\text{GN}} = \sqrt {\dfrac{{99}}{4}} = \dfrac{{3\sqrt {11} }}{2}{\text{ cm}} \\
$
So, height of the $\vartriangle {\text{GCF}}$ is \[{\text{GN}} = \dfrac{{3\sqrt {11} }}{2}{\text{ cm}}\]
Area of region I = Area of $\vartriangle {\text{GCF}}$ $ = \dfrac{1}{2}\left( {{\text{CF}}} \right)\left( {{\text{GN}}} \right) = \dfrac{1}{2}\left( {\text{1}} \right)\left( {\dfrac{{3\sqrt {11} }}{2}} \right) = \dfrac{{3\sqrt {11} }}{4}{\text{ c}}{{\text{m}}^2}$
Area of region II = Area of rectangle KJFC\[ = \left( {{\text{KJ}}} \right)\left( {{\text{CK}}} \right) = \left( {\text{1}} \right)\left( {{\text{6}}{\text{.5}}} \right) = 6.5{\text{ c}}{{\text{m}}^2}\].
Since, the height of the trapezium KHIJ = KL = JM
In $\vartriangle {\text{JMI}}$, IJ = 1 cm and MI = HL = $\dfrac{{{\text{HI}} - {\text{KJ}}}}{2} = \dfrac{{{\text{2}} - {\text{1}}}}{2} = \dfrac{1}{2}$ cm
According to Pythagoras Theorem, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
$
\Rightarrow {\left( {{\text{IJ}}} \right)^2} = {\left( {{\text{JM}}} \right)^2} + {\left( {{\text{MI}}} \right)^2} \\
\Rightarrow {\left( {\text{1}} \right)^2} = {\left( {{\text{JM}}} \right)^2} + {\left( {\dfrac{{\text{1}}}{2}} \right)^2} \\
\Rightarrow {\left( {{\text{JM}}} \right)^2} = 1 - \dfrac{{\text{1}}}{4} = \dfrac{3}{4} \\
\Rightarrow {\text{JM}} = \sqrt {\dfrac{3}{4}} = \dfrac{{\sqrt 3 }}{2}{\text{ cm}} \\
$
So, the height of trapezium KHIJ is ${\text{JM}} = \dfrac{{\sqrt 3 }}{2}{\text{ cm}}$
Area of region III = Area of trapezium KHIJ$ = \dfrac{{\left( {{\text{JM}}} \right)}}{2} \times \left( {{\text{KJ}} + {\text{HI}}} \right) = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{2} \times \left( {1 + 2} \right) = \dfrac{{3\sqrt 3 }}{4}{\text{ c}}{{\text{m}}^2}$
Area of region IV = Area of region V [because both the triangles ABC and DEF have equal areas due to their same dimensions]
Area of region IV = Area of region V = Area of $\vartriangle {\text{DEF}}$$ = \dfrac{1}{2}\left( {{\text{EF}}} \right)\left( {{\text{DF}}} \right) = \dfrac{1}{2}\left( {{\text{1}}{\text{.5}}} \right)\left( {\text{6}} \right) = \dfrac{9}{2}{\text{ c}}{{\text{m}}^2}$
Therefore, the total area of the paper used = Area of region I + Area of region II + Area of region III + Area of region IV + Area of region V \[ = \dfrac{{3\sqrt {11} }}{4} + 6.5 + \dfrac{{3\sqrt 3 }}{4} + \dfrac{9}{2} + \dfrac{9}{2} = \dfrac{{3\sqrt {11} + 26 + 3\sqrt 3 + 18 + 18}}{4} = 19.2865{\text{ c}}{{\text{m}}^2}\].
So, the total area of the coloured paper used is 19.2865 \[{\text{c}}{{\text{m}}^2}\].
Note- In this particular problem, from symmetry point of view we can say that the perpendicular GN drawn from point G to the line CF divides the line into two equal parts which are CN and NF i.e., ${\text{NF}} = {\text{CN}} = \dfrac{{{\text{CF}}}}{2} = \dfrac{{{\text{KJ}}}}{2}$ and the triangles KLH and JMI are symmetrically same that’s why MI = HL = $\dfrac{{{\text{HI}} - {\text{KJ}}}}{2}$.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE