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Find the area of a trapezium whose parallel sides are 12 cm and 20 cm and the distance between them is 15 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=12 cm , b=20 cm h =15 cm

Therefore

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

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.

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

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

The area of a trapezium is 1440 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=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

The area of a trapezium is 1586 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 cm^{2}

Therefore

$1586=\frac {1}{2} [84 +b] \times 26$

$h= 38$ cm

The area of a trapezium is 384cm

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

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

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

The area of a trapezium is 180 cm

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

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.

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) Amount of region occupied by a solid is called its surface area

(ii)The areas of any two faces of a cube are equal.

(iii)The areas of two oppossite faces of a cuboid are equal.

(iv) 2.5 litres is equal to 0.0025 cubic meters

(v) Ratio of area of a circle to the area of a square whose side equals radius of circle is 1 : π.

(i) False

(ii) True

(iii) True

(iv) True

(v) false

(i) Curved surface area of a cylinder of radius 2b and height 2a is _______.

(ii) Volume of a cylinder with radius p and height q is __________.

(iii) 6.55m

(iv)The volume of a cylinder becomes __________ the original volume if its radius becomes doubles of the original radius and height becomes half of its original values

(v)The surface area of a cylinder which exactly fits in a cube of side b is____

(i) $S=2\pi r H= 2\pi 2b 2a= 8\pi ab$

(ii)$V=\pi r^2 H= \pi p^2 q$

(iii) 6.55m^{2} =6.55 * 10^{4} cm^{2}

(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)

This mensuration class 8 worksheets is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

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