In this page we have Worksheet for Mensurations Class 8 Maths Chapter 11 . Hope you like them and do not forget to like , social share
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Question 1
Find the area of a trapezium whose parallel sides are 12 cm and 20 cm and the distance between them is 15 cm.
Solution
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=12 cm , b=20 cm h =15 cm
Therefore
$A=\frac {1}{2} [12 +20] \times 15 =240 \; cm^2$
Question 2
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 18 cm.
Solution
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=38.7 cm , b=22.3 cm h =18 cm
Therefore
$A=\frac {1}{2} [38.7 +22.3] \times 18 =549 \; cm^2$
Question 3
The area of a trapezium is 1440 cm
^{2}. If the lengths of its parallel sides are 54.6 cm and 35.4 cm, find the distance between them.
Solution
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=54.6 cm , b=35.4 cm, h =?, A=1440 cm^{2}
Therefore
$1440=\frac {1}{2} [38.7 +22.3] \times h$
$h= 32$ cm
Question 4
The area of a trapezium is 1586 cm
^{2} and the distance between its parallel sides is 26 cm. If one of the parallel sides is 84 cm, find the other.
Solution
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=84 cm , b=?, h =26 cm., A=1586 cm^{2}
Therefore
$1586=\frac {1}{2} [84 +b] \times 26$
$h= 38$ cm
Question 5
The area of a trapezium is 384cm
^{2}. Its parallel sides are in the ratio 2: 6 and the perpendicular distance between them is 12 cm. Find the length of each of the parallel sides.
Solution
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=2x cm , b=6x, h =12 cm., A=384 cm^{2}
Therefore
$384=\frac {1}{2} [2x +6x] \times 12$
$x= 8$ cm
So parallel sides are 16 cm and 48 cm
Question 6
Mitesh wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m
^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Solution
let x be the length on the road, then 2x is the length on the river side
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=2x m , b=x, h =100 m., A=10500 m^{2}
Therefore
$10500=\frac {1}{2} [2x +x] \times 100$
$x= 70$ m
So parallel sides are 70 m and 140 m
Question 7
The area of a trapezium is 180 cm
^{2} and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
Solution
let x be one side, then x+6 is the length of another parallel side
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
$A = \frac {1}{2} [a +b] \times h$
Here a=x m , b=x+6, h =9 cm., A=180 cm^{2}
Therefore
$180=\frac {1}{2} [x +x+6] \times 9$
$x= 17$ cm
So parallel sides are 17 cm and 23 cm
Question 8
The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
Solution
Let ABCD is the trapezium with AB and CD are the parallel sides.
Now
AB=10 cm, CD=20 cm, BC=13 cm
Now draw line BE || AD and draw a perpendicular from B on EC
Now ABED is a parallelogram, then BE=13 cm
In triangle BEC, BE =BC, So Isoceles triangle,So perpendicular will bisect the EC
Hence EF=FC
Now EC= DC -DE = 20 -10 =10 cm
Therefore EF=FC= EC/2 = 5 cm
Now in Triangle BEF, it is right angle at F,So by pythagorous theorem
$BE^2 = BF^2 + EF^2$
$169 = BF^2 + 25$
$BF=12 cm$
This is also the perpendicular distance between the parallel sides.So now coming back to Area of trapezium
$A = \frac {1}{2} [a +b] \times h$
Here a=10 cm , b=20 cm, h =12 cm., A=?
Therefore
$A=\frac {1}{2} [10 +20] \times 12$
$A= 180 \; cm^2$
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