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S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12A

 Exercise 12A


Q1 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 1 

Find the area of the following trapeziums.

Sol :

(a)





h=6cm ,a=8cm, b=11cm

⇒Area=12h(a+b)

=12×6(8+11)

=12×6×19

=57cm2


(b)





h=10cm ,a=12cm ,b=9cm

⇒Area=12h(a+b)

=12×10(12+9)

=5×21=105cm2


(c)






h=4cm ,a=3cm ,b=5cm

⇒Area=12h(a+b)

=12×4(3+5)

=2×8=16cm2


Q2 Ex-12A Class 8 Mensuration S.chand New Learning Composite mathematics Solution

Question 2

The parallel sides of a trapezium are 9 cm and 7 cm and its area is 80 cm2. Find its altitude. 

Sol :





Area=80cm2

a=9cm ,b=7cm ,

∴We know:-

Area=12×h(a+b)

80=12×h(9+7)

h=80×216=10


Q3 Ex-12A Class 8 Mensuration S.chand New Learning Composite mathematics Solution

Question 3

The sum of the parallel sides of a trapezium is 25 cm and its area is 150 cm2. Find its altitude.

Sol :




a+b=25cm

Area=150cm2

We know:-

Area=12h(a+b)

h=area×225=150×225

=12cm


Q4 Ex-12A Class 8 Mensuration S.chand New Learning Composite mathematics Solution

Question 4

The area of a trapezium is 126 cm2 and its altitude is 6 cm. Find the sum of its parallel sides.

Sol :





Area=126cm2

h=6cm

a+b=?

We know that:-

Area=12h(a+b)

a+b=126×26

a+b=42



Q5 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 5

The area of a trapezium is 0.95 m2 and its altitude is 19 cm. Find the sum of its parallel sides.

Sol :







Area of trapezium =12×h×(a+b)

0.95=12×0.19(a+b)

0.95×20.19=a+b

95×2×10019×100=a+b

10 m=a+b

Sum of parallel sides=10 m



Q6 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 6

One of the parallel sides of a trapezium is thrice the other.The area of the trapezium is 440cm2 and its height is 22 cm. Find the length of its two parallel sides.

Sol :











Area of trapezium=12×h(a+b)

440=12×22(x+3x)

440=11×4x

44011=4x

404=x

10=x or x=10 cm

The length of parallel sides are 

x=10 cm

3x=3×10 =30 cm



Q7 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 7

The difference of the parallel sides of a trapezoidal field is 20 m. Its area is 450 m2 and its altitude is 15 m. Find the length of the parallel sides.

Sol :

 


Let, a=x

b=x-20

h=15m

Area=450m2

We know that:-

12×h(a+b)=450

=12×15(x+x20)=450

2x20=450×215

2x=60+20=80

x=802=40

∴a=40m

∴b=40-20=20m



Q8 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 8

The lengths of the parallel sides are in the ratio 7 : 9. Its area is 960 cm2 and the distance between the parallel sides is 15 cm. Find the length of each of the parallel sides.

Sol :





Let, common factor=x

a=7x

b=9x

Area=960cm2

h=15cm

ATQ,

Area=12h(a+b)

960=12×15(7x+9x)

16x=960×215

x=64×216=8

∴a=7×8=56cm

∴b=9×8=72cm



Q9 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 9

Two parallel sides of an isosceles trapezium are 10 cm and 20 cm and its non-parallel sides are each equal to 13 cm. Find the area of the trapezium.

Sol :





a=10cm=AB

b=20cm=CD

AD=BC=13cm


AD||BQ ∴AD=BQ=13cm

AB||DQ ∴AB=DQ=15cm

∴QC=DC-DQ

=20-10=10cm


∵BO bisects QC

∴QO=OC=102=5cm

∴h2=BO2=BC2-OC2=132-52

=169-25=144


or h=√144=12

A=12h(a+b)

A=12×12×(10+20)

=6×30=180cm2



Q10 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 10

Two parallel sides of an isosceles trapezium are 31 cm and 15 cm. Its non-parallel sides are each equal to 17 cm. Find the area of the trapezium.
Sol :





AB=15cm

DC=31cm

AD=BC=17cm

AD||BE  ∴AD=BE=17cm

AD=BE=17cm

EC=DC-DE=31-15=16cm

∴BO bisects EC

∴OC=EO=162=8cm


∴ΔBOC:-

BO2=BC2-OC2=172-82

=289-64=225

BO=√225=15

∴BO=h=15cm

A=12h(a+b)

=12×15(31+15)

=12×15×46

=345cm2



Q11 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 11

Find the cost of tiling the floor of a trapezium hall whose parallel sides are 25 m and 20 m and the distance between them is 16 m at the rate of  ₹40 per m2
Sol :


h=16m, a=20m, b=25m

A=12h(a+b)

=12×16(20+25)

=8×45=360m2

∴Area of floor=360m2

Rate=40 per m2

∴Cost=360×40=14400



Q12 | Ex-12A |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

Question 12

Find the area of trapezium EFGH having measurement as shown in the figure. 



Sol :
In ΔEFG , EG2=EF2+FG2

372=EF2+3521369=EF2+122513691225=EF2144=EF2144=EF

EF=12 cm

∴Area =12(23+35)×12

=12×58×12

=348 cm2

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