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Standard VIII

Mathematics

Area of Trapezium by Division into Shapes of Known Area

From the adjo...

Question

From the adjoining sketch, calculate
i) the length $A D$
ii) the area of trapezium $A B C D$
iii) the area of triangle $B C D$

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Solution

Area of $A B C D$
area of $△ B C D$
$A D^{2} = (41)^{2} - (40)^{2} = (41 + 40) (41 - 40)$
$A D^{2} = 81$
$A D = 9$
area of $A B C D = \frac{1}{2} | 40 + 15 | \times 9 = \frac{1}{5} \times 55 \times 9 \frac{495}{2}$
$247.5$
area of $△ B C D =$ area of $A B C D -$ area of $△ A D B$
$= 247.5 - \frac{1}{2} \times 9 \times 20$
$= 247.5 - 180 = 16.75 = 167.5$ unit

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From the adjoining sketch, calculate i) the length ADii) the area of trapezium ABCDiii) the area of triangle BCD

Area of ABCDarea of △BCDAD2=(41)2−(40)2=(41+40)(41−40)AD2=81AD=9area of ABCD=12|40+15|×9=15×55×94952247.5area of △BCD= area of ABCD− area of △ADB =247.5−12×9×20=247.5−180=16.75=167.5 unit

From the adjoining sketch, calculate
i) the length $A D$
ii) the area of trapezium $A B C D$
iii) the area of triangle $B C D$