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Find the Area Enclosed by Each of the Following Figures [Fig. 20.49 (I)-(Iii)] as the Sum of the Areas of a Rectangle and a Trapezium: - Mathematics

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ConceptArea of Trapezium

Question

Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:

Solution

\[(i)\]
The given figure can be divided into a rectangle and a trapezium as shown below:
From the above firgure:
Area of the complete figure = (Area of square ABCF)+(Area of trapezium CDEF)
\[=(AB\times BC)+[\frac{1}{2}\times(FC+ED)\times(\text{ Distance between FC and ED })]\]
\[=(18\times18)+[\frac{1}{2}\times(18+7)\times(8)]\]
\[=324+100\]
\[ {=424 cm}^2\]
\[(ii)\]
The given figure can be divided in the following manner:\]
From the above figure:
AB = AC-BC=28-20=8 cm
So that area of the complete figure = (area of rectangle BCDE)+(area of trapezium ABEF)
\[=(BC\times CD)+[\frac{1}{2}\times(BE+AF)\times(AB)]\]
\[=(20\times15)+[\frac{1}{2}\times(15+6)\times(8)]\]
\[=300+84\]
\[ {=384 cm}^2\]

\[(iii)\]
The given figure can be divided in the following manner:

From the above figure:
EF = AB = 6 cm
Now, using the Pythagoras theorem in the right angle triangle CDE:
\[ 5^2 {= 4}^2 {+CE}^2 \]
\[ {CE}^2 = 25-16=9\]
\[CE =\sqrt{9}= 3 cm\]
\[\text{ And, }GD=GH+HC+CD=4+6+4=14 cm\]
\[ \therefore\text{ Area of the complete figure }= (\text{ Area of rectangle ABCH })+(\text{ Area of trapezium GDEF })\]
\[=(AB\times BC)+[\frac{1}{2}\times(GD+EF)\times(CE)]\]
\[=(6\times4)+[\frac{1}{2}\times(14+6)\times(3)]\]
\[=24+30\]
\[ {=54 cm}^2\]

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APPEARS IN

 RD Sharma Solution for Mathematics for Class 8 by R D Sharma (2019-2020 Session) (2017 to Current)
Chapter 20: Mensuration - I (Area of a Trapezium and a Polygon)
Ex. 20.3 | Q: 2 | Page no. 28

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Solution Find the Area Enclosed by Each of the Following Figures [Fig. 20.49 (I)-(Iii)] as the Sum of the Areas of a Rectangle and a Trapezium: Concept: Area of Trapezium.
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