Unit 6
Measurement
PRACTICE QUESTIONS: UNIT 6
1. In the trapezium DEFG below, not drawn to scale, DE = 10 cm, DG = 13 cm and GX =
5 cm. Angle EFX and DXF are right angles.
D
E
Calculate:
a. The length of DX
b. The area of trapezium DEFG
G
X
F
Ans:
DE = 10cm
DG = 13cm and GX = 5cm
So by Pythagoras theorem
DX² =
 DX = 12cm
Area of a trapezium = area of triangle + area of quadrilateral
={
}
=(
)
(
)
2. The figure OAB below, not drawn to scale, represent a flower bed in the shape of a
sector of a circle with centre O, and radius 10 m. The triangular region OAB is
planted in rose and the arc AB is planted in marigold. Given that angle AOB is 36 o,
calculate the area of the segment assigned to marigolds. (Take
)
A
O
B
Ans:
Total area under the curve =
Angel in radian =
= 36o
Total area under the curve = (
)
= 31.4cm²
1
Unit 6
Measurement
Area of a triangle OAB =
=
Area of a marigold = Total area
Area of a marigold = 31.4cm²
Area of a triangle OAB
= 2.01cm²
3. A rectangular steel pyramid of height 6 cm and base dimensions 11 cm by 16 cm, is
melted down and rolled into a cylinder of height 7 cm, calculate:
i.
The radius of the cylinder in cm.
Ans: as one shape is transformed to another shape so volume will remain same
Volume of cylinder = volume of pyramid
ii.
Ans:
The mass of the cylinder in kg, if the density of the steel is 5g/cm3
Volume of cylinder = 352cm³
Mass of cylinder = density
Mass of cylinder = 5 g/cm3
=1760g = 1.76kg
4. The diagram below represents the net of a solid.
2
Unit 6
Measurement
i.
Draw a sketch of the solid represented.
Ans:
ii.
Write down the number of edges for the solid.
Ans:
It has 10 sides
iii.
State the name of the solid.
Ans:
It is hexagonal shape
5. A closed cylinder has a base diameter 14 cm and a vertical height of 20 cm. Take
Calculate
i.
Ans:
The area, in cm2, of the base.
Base Area =
Radius = 14/2 = 7cm
Area =
ii.
Ans:
= 154cm²
The volume, in cm3, of the cylinder.
Volume of cylinder = Base Area
= 154
iii.
Ans:
Height
20 = 3080cm³
Show that the total surface area is 1188 cm2
Total surface Area = {circumference of base
height} + {2
Base Area}
=
=
+ 308
= 1188cm²
3
Unit 6
Measurement
6. A piece of wire is bent in the form of a circle and it encloses an area of 154 cm 2.
Calculate
i.
The radius of the circle
Ans:
Area =
154cm²=
ii.
Ans :
The circumference of the circle.
circumference =
=2
= 44cm
The same piece of wire is then bent in the form of a square.
iii.
Calculate the area enclosed by the square.
Ans:
area will be same as of circle i.e 154cm²
7. A circular drain pipe, in the diagram below, not drawn to scale, is 1 metre long with
outer and inner radii of 20 cm and 15 cm, respectively.
i.
Draw a cross –sectional view of the drain pipe, showing the measurement of
the outer and inner radii.
Ans:
Material area
Calculate:
ii.
Ans:
The area, in cm2, of the cross –section of the drain pipe.
Cross sectional area =
=
iii.
Ans:
The amount, in cm3, of the material required to construct the drain pipe.
Amount of material = Total Volume
(r2² r1²)
Hollow Volume
h
(10² 7.5²)
1 = 137.44cm³
4
Unit 6
Measurement
iv.
Ans:
The capacity, in litres, of the hollow space of the drain pipe.
As
1cm³ = 0.001 liter
So capacity in hollow area = 176.7cm³
v.
0.001 = 0.1767 liter
The volume, in litres, of the water passing through the pipe in 1 minute, if
water flows through the pipe at a speed of 1 metre per second. Take
Ans:
Volume flow Rate = Area hollow
= 176.7cm²
velocity
100cm/s = 17670cm³/s
Volume flow in one Minute = 17670cm³/s 60 s = 1060200cm³
Volume flow in one Minute in liter = 1060200 0.001 = 1060.2 liter
1m
5