Mensuration class 8 worksheets

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
$A=\frac {1}{2} [12 +20] \times 15 =240 \; cm^2$

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
$A=\frac {1}{2} [38.7 +22.3] \times 18 =549 \; cm^2$

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 cm²
Therefore
$1440=\frac {1}{2} [38.7 +22.3] \times h$
$h= 32$ cm

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 cm²
Therefore
$1586=\frac {1}{2} [84 +b] \times 26$
$h= 38$ cm

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 cm²
Therefore
$384=\frac {1}{2} [2x +6x] \times 12$
$x= 8$ cm
So parallel sides are 16 cm and 48 cm

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 m²
Therefore
$10500=\frac {1}{2} [2x +x] \times 100$
$x= 70$ m
So parallel sides are 70 m and 140 m

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 cm²
Therefore
$180=\frac {1}{2} [x +x+6] \times 9$
$x= 17$ cm
So parallel sides are 17 cm and 23 cm

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
$A=\frac {1}{2} [10 +20] \times 12$
$A= 180 \; cm^2$

(i) False
(ii) True
(iii) True
(iv) True
(v) false

(i) $S=2\pi r H= 2\pi 2b 2a= 8\pi ab$
(ii)$V=\pi r^2 H= \pi p^2 q$
(iii) 6.55m² =6.55 * 10⁴ cm²
(iv) $V_1=\pi r^2 H$ ,$V_2= \pi (2r)^2 (1/2)H=2 pi r^2 H$, So it becomes double
(v) Radius of cylinder =(1/2)b and height =b, So Surface area=2\pi b^2$

(p) -> (ii)
(q) -> (iii)
(r) -> (i)

(p) -> (iv)
(q) -> (i)
(r) -> (ii)
(s) -> (iii)