Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
A=12[a+b]×h
Here a=12 cm , b=20 cm h =15 cm
Therefore
A=12[12+20]×15=240cm2
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
A=12[a+b]×h
Here a=38.7 cm , b=22.3 cm h =18 cm
Therefore
A=12[38.7+22.3]×18=549cm2
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
A=12[a+b]×h
Here a=54.6 cm , b=35.4 cm, h =?, A=1440 cm2
Therefore
1440=12[38.7+22.3]×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=12[a+b]×h
Here a=84 cm , b=?, h =26 cm., A=1586 cm2
Therefore
1586=12[84+b]×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=12[a+b]×h
Here a=2x cm , b=6x, h =12 cm., A=384 cm2
Therefore
384=12[2x+6x]×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=12[a+b]×h
Here a=2x m , b=x, h =100 m., A=10500 m2
Therefore
10500=12[2x+x]×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=12[a+b]×h
Here a=x m , b=x+6, h =9 cm., A=180 cm2
Therefore
180=12[x+x+6]×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
BE2=BF2+EF2
169=BF2+25
BF=12cm
This is also the perpendicular distance between the parallel sides.So now coming back to Area of trapezium
A=12[a+b]×h
Here a=10 cm , b=20 cm, h =12 cm., A=?
Therefore
A=12[10+20]×12
A=180cm2
(i) False
(ii) True
(iii) True
(iv) True
(v) false
(i) S=2πrH=2π2b2a=8πab
(ii)V=πr2H=πp2q
(iii) 6.55m2 =6.55 * 104 cm2
(iv) V1=πr2H ,V2=π(2r)2(1/2)H=2pir2H, 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.