Exercise 12A
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)
(c)
h=4cm ,a=3cm ,b=5cm
⇒Area=12h(a+b)
=12×4(3+5)
=2×8=16cm2
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
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
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
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
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 :
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
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+x−20)=450
2x−20=450×215
2x=60+20=80
x=802=40
∴a=40m
∴b=40-20=20m
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
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
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
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
Question 12
Find the area of trapezium EFGH having measurement as shown in the figure.
Sol :
In ΔEFG , EG2=EF2+FG2
⇒372=EF2+352⇒1369=EF2+1225⇒1369−1225=EF2⇒144=EF2⇒√144=EF
EF=12 cm
∴Area =12(23+35)×12
=12×58×12
=348 cm2
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