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

Exercise 12A

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Question 1

Find the area of the following trapeziums.

Sol :

(a)

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

⇒Area$=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 6(8+11)$

$=\frac{1}{2}\times 6\times 19$

=57cm²

(b)

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

⇒Area$=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 10(12+9)$

=5×21=105cm²

(c)

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

⇒Area$=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 4(3+5)$

=2×8=16cm²

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

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Question 2

The parallel sides of a trapezium are 9 cm and 7 cm and its area is $80 \mathrm{~cm}^{2}$. Find its altitude.

Sol :

Area=80cm²

a=9cm ,b=7cm ,

∴We know:-

Area$=\frac{1}{2}\times h(a+b)$

$80=\frac{1}{2}\times h(9+7)$

$h=\frac{80\times 2}{16}=10$

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

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Question 3

The sum of the parallel sides of a trapezium is 25 cm and its area is $150 \mathrm{~cm}^{2}$. Find its altitude.

Sol :

a+b=25cm

Area=150cm²

We know:-

Area$=\frac{1}{2}h(a+b)$

$h=\frac{area \times 2}{25}=\frac{150\times 2}{25}$

=12cm

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

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Question 4

The area of a trapezium is $126 \mathrm{~cm}^{2}$ and its altitude is 6 cm. Find the sum of its parallel sides.

Sol :

Area=126cm²

h=6cm

a+b=?

We know that:-

Area$=\frac{1}{2}h(a+b)$

$a+b=\frac{126\times 2}{6}$

a+b=42

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Question 5

The area of a trapezium is $0.95 \mathrm{~m}^{2}$ and its altitude is 19 cm. Find the sum of its parallel sides.

Sol :

Area of trapezium $=\frac{1}{2} \times h \times(a+b)$

$0.95=\frac{1}{2} \times 0.19(a+b)$

$\frac{0.95 \times 2}{0.19}=a+b$

$\frac{95 \times 2 \times 100}{19 \times 100}=a+b$

10 m=a+b

Sum of parallel sides=10 m

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Question 6

One of the parallel sides of a trapezium is thrice the other.The area of the trapezium is $440 cm^2$ and its height is 22 cm. Find the length of its two parallel sides.

Sol :

Area of trapezium$=\frac{1}{2}\times h (a+b)$

$440=\frac{1}{2} \times 22(x+3 x)$

$440=11 \times 4 x$

$\frac{440}{11}=4 x$

$\frac{40}{4}=x$

10=x or x=10 cm

The length of parallel sides are

x=10 cm

3x=3×10 =30 cm

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Question 7

The difference of the parallel sides of a trapezoidal field is 20 m. Its area is $450 \mathrm{~m}^{2}$ 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:-

$\frac{1}{2}\times h(a+b)=450$

$=\frac{1}{2}\times 15(x+x-20)=450$

$2x-20=\frac{450\times 2}{15}$

2x=60+20=80

$x=\frac{80}{2}=40$

∴a=40m

∴b=40-20=20m

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Question 8

The lengths of the parallel sides are in the ratio 7 : 9. Its area is $960 \mathrm{~cm}^{2}$ 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$=\frac{1}{2}h(a+b)$

$960=\frac{1}{2}\times 15(7x+9x)$

$16x=\frac{960\times 2}{15}$

$x=\frac{64\times 2}{16}=8$

∴a=7×8=56cm

∴b=9×8=72cm

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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$=\frac{10}{2}$=5cm

∴h²=BO²=BC²-OC²=13²-5²

=169-25=144

or h=√144=12

∴$A=\frac{1}{2}h(a+b)$

$A=\frac{1}{2}\times 12\times (10+20)$

=6×30=180cm²

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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$=\frac{16}{2}$=8cm

∴ΔBOC:-

BO²=BC²-OC²=17²-8²

=289-64=225

BO=√225=15

∴BO=h=15cm

$A=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 15(31+15)$

$=\frac{1}{2}\times 15\times 46$

=345cm²

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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

$A=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 16(20+25)$

=8×45=360m²

∴Area of floor=360m²

Rate=40 per m²

∴Cost=360×40=14400

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Question 12

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

Sol :
In ΔEFG , $E G^{2}=EF^{2}+FG^{2}$

$\begin{aligned}\Rightarrow 37^{2}&=E F^{2}+35^{2} \\ \Rightarrow 1369&=E F^{2}+1225 \\\Rightarrow 1369& -1225=E F^{2} \\\Rightarrow 144 &=E F^{2} \\ \Rightarrow\sqrt{144} &=E F \end{aligned}$

EF=12 cm

∴Area $=\frac{1}{2}(23+35) \times 12$

$=\frac{1}{2} \times 58 \times 12$

$=348 \mathrm{~cm}^{2}$

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

Exercise 12A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

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