31-The radius of a cylindrical roller is 1.25 units and its height is 1.4 units. How much area will it cover in 1 revolution in sq. units? O a) 11
Answers
Answer:
Step-by-step explanation:
Example 1 : What is the area of the triangle ADE in the following figure?
B
D C
A
8 cm
10 cm
E
(a)45 cm2 (b) 50 cm2 (c) 55 cm2 (d) 40 cm2
Solution : The correct answer is (d).
Example 2 : What will be the change in the volume of a cube when its
side becomes 10 times the original side?
(a) Volume becomes 1000 times.
(b) Volume becomes 10 times.
(c) Volume becomes 100 times.
(d) Volume becomes
1
1000 times.
Solution : The correct answer is (a).
In examples 3 and 4, fill in the blanks to make the statements true.
Example 3 : Area of a rhombus is equal to __________ of its diagonals.
Solution : Half the product.
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Example 4 : If the area of a face of a cube is 10 cm2, then the total
surface area of the cube is __________.
Solution : 60 cm2.
In examples 5 and 6, state whether the statements are true (T) or false (F).
Example 5 : 1L = 1000 cm3
Solution : True.
Example 6 : Amount of region occupied by a solid is called its surface
area.
Solution : False.
Example 7 : 160 m3 of water is to be used to irrigate a rectangular
field whose area is 800 m2. What will be the height of the
water level in the field?
Solution : Volume of water = 160 m3
Area of rectangular field = 800 m2
Let h be the height of water level in the field.
Now, volume of water = volume of cuboid formed on the
field by water.
160 = Area of base × height
= 800 × h
h =
160
800 = 0.2
So, required height = 0.2 m
Example 8 : Find the area of a rhombus whose one side measures
5 cm and one diagonal as 8 cm.
Solution : Let ABCD be the rhombus as shown below.
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DO = OB = 4cm, since diagonals of a rhombus are
perpendicular bisectors of each other. Therefore, using
Pythagoras theorem in AOB,
AO2 + OB2 = AB2
AO = 2 2 AB OB =
2 2 5 4 = 3 cm
So, AC = 2 × 3 = 6 cm
Thus, the area of the rhombus =
1
2 × d1 × d2 =
1
2
× 8 × 6
= 24 cm2.
Example 9 : The parallel sides of a trapezium are 40 cm and 20 cm. If
its non-parallel sides are both equal, each being 26 cm,
find the area of the trapezium.
Solution : Let ABCD be the trapezium such that AB = 40 cm and
CD = 20 cm and AD = BC = 26 cm.
To become familiar with some of the vocabulary terms in the chapter,
consider the following:
1. The square root of a number is one of the two equal factors of the
number. For example, 3 is a square root because 3 3 = 9. How might
picturing plant roots help you remember the meaning of square root?
2. The word ‘perimeter’ comes from the Greek roots peri, meaning ‘all
around,’ and metron, meaning ‘measure.’ What do the Greek roots tell
you about the perimeter of a geometric figure?
3. To square a number means ‘to multiply the number by itself,’ as in
2 2. Keeping this idea of square in mind, what do you think a perfect
square might be?
4. The word ‘circumference’ comes from the Latin word circumferre,
meaning to “carry around”. How does the Latin meaning help you define
the circumference of a circle?
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Now, draw CL || AD
Then ALCD is a parallelogram
So AL = CD = 20 cm and CL = AD = 26 cm.
In ∆CLB, we have
CL = CB = 26 cm
Therefore, ∆ CLB is an isosceles triangle.
Draw altitude CM of ∆CLB.
Since ∆ CLB is an isosceles triangle. So, CM is also the
median.
Then LM = MB =
1
2 BL =
1
2 × 20 cm = 10 cm
[as BL = AB – AL = (40 – 20) cm = 20 cm].
Applying Pythagoras theorem in ∆CLM, we have
CL2 = CM2 + LM2
262 = CM2 + 102
CM2 = 262 – 102 = (26 – 10) (26 + 10) = 16 × 36 = 576
CM = 576 = 24 cm
Hence, the area of the trapezium =
1
2 (sum of parallel
sides) × Height =
1
2 (20 + 40) × 24 = 30 × 24 = 720 cm2.
Example 10 : Find the area of polygon ABCDEF, if AD = 18cm, AQ =
14 cm, AP = 12 cm, AN = 8 cm, AM = 4 cm, and FM, EP,
QC and BN are perpendiculars to diagonal AD.
Solution :
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In the figure
MP = AP – AM = (12 – 4) cm = 8 cm
PD = AD – AP = (18 – 12) cm = 6 cm
NQ = AQ – AN = (14 – 8) cm = 6 cm
QD = AD – AQ = (18 – 14) cm = 4 cm
Area of the polygon ABCDEF
= area of ∆AFM + area of trapezium FMPE + area of ∆EPD
+ area of ∆ANB + area of trapezium NBCQ + area of
∆QCD.
= 1
2 × AM × FM +
1
2 (FM + EP) × MP +
1
2 PD × EP +
1
2 ×
AN × NB +
1
2 (NB + CQ) × NQ +
1
2 QD × CQ
= 1
2 × 4 × 5 +
1
2 (5 + 6) × 8 +
1
2 × 6 × 6 +
1
2 × 8 × 5 +
1
2
(5 + 4) × 6 +
1
2 × 4 × 4.
= 10 + 44 + 18 + 20 + 27 + 8 = 127 cm2