Q.1. What is p(–2) for the polynomial p(t) = t² – t + 1?

Solution. p(t) = t² – t + 1
⇒ p(–2) = (–2)² – (–2) + 1
= 4 + 2 + 1
= 7

Q.2. If x – 1/x = -1 then what is  

Solution. On squaring both sides we get;

⇒ 

⇒  

⇒  

Q.3. If x + y = –1, then what is the value of x³ + y³ – 3xy? 

Solution. We have x³ + y³ = (x + y)(x² – xy + y²)
⇒ x³ + y³ = (–1)(x² + 2xy + y²) + 3xy
⇒ x³ + y³ = –1(x + y)² – 3xy
⇒ x³ + y³ – 3xy = –1(–1)² = –1(1)
⇒ x³ + y³ – 3xy = –1

Q.4. Show that p(x) is not a multiple of g(x), when p(x) = x³ + x – 1 g(x) = 3x – 1

Solution. g(x) = 3x – 1 = 0 ⇒ x = 1/3

∴ Remainder 

Since remainder ≠ 0, so p(x) is not a multiple of g(x).

Q.5. (a) Find the value of ‘a’ if x – a is a factor of x³ – ax² + 2x + a – 5.
(b) Find the value of ‘a’, if (x – a) is a factor of x³ – ax² + 2x + a – 1 
[NCERT Exemplar]
(c) If x + 1 is a factor of ax³ + x² – 2x + 4a – 9 find the value of ‘a’. [NCERT Exemplar]

Solution. (a) Let p(x) = x³ – ax² + 2x + a – 5
since x – a is a factor of p(x),
so p(a) = 0
⇒ (a)³ – a(a)² + 2(a) + a – 5 = 0
⇒ (a)³ – a(a)² + 2(a) + a – 5 = 0
⇒ 3a – 5 = 0 ⇒ a = 

(b) Here, p(x) = x³ – ax² + 2x + a – 1
∵ x – a is a factor of p(x)
∴ p(a) = 0
⇒ a³ – a(a)² + 2(a) + a – 1 = 0
⇒ a³ – a³ + 2a + a – 1 = 0
⇒ 3a – 1 = 0
⇒ a = 1/3

(c) Here, x + 1 is a factor of p(x) = ax³ + x² – 2x + 4a – 9
∴ p(–1) = 0
⇒ a(–1)³ + (–1)² – 2(–1) + 4a – 9 = 0
⇒ –a + 1 + 2 + 4a – 9 = 0
⇒ 3a – 6 = 0    
⇒ a = 2

Q.6. Without finding the cubes factorise (a – b)³ + (b – c)³ + (c – a)³.

Solution. If x + y + z = 0
then x³ + y³ + z³ = 3xyz
Here, (a – b) + (b – c) + (c –a) = 0
∴ (a – b)³ + (b – c)³ + (c – a)³
= 3(a – b)(b – c)(c – a)

Q.7. What is zero of a non-zero constant polynomial?
Solution. A non-zero constant polynomial has no zero.

Q.8. What is the coefficient of a zero polynomial?
Solution. A zero polynomial has all coefficients zero.

Q.9. What is the degree of a biquadratic polynomial?
Solution.  ∵ The degree of a quadratic polynomial is 2.
∴ The degree of a biquadratic polynomial is 4.

Q.10. Is the statement: ‘0’ may be a zero of polynomial, true?
Solution. Yes, this statement is true.

Q.11. What is the value of (x + a) (x + b)?
Solution. The value of (x + a) (x + b)
= x² + (a + b) x + ab.

Q.12. What is the value of (x + y + z)² – 2[xy + yz + zx]?
Solution. ∵ (x + y + z)² 
= x² + y² + z² + 2 xy + 2 yz + 2 zx
= x² + y² + z² + 2[xy + yz + zx]
∴ (x + y + z)² – 2[xy + yz + zx]
= x² + y² + z²

Q.13. What is the value of (x + y)³ – 3xy (x + y)?
Solution. ∵ (x + y)³ = x³ + y³ + 3xy (x + y)
∴ [x³ + y³ + 3xy (x + y)] –  [3xy (x + y)] = x³ + y³ 
⇒ (x + y)³ – 3xy(x + y) = x³ + y³
Thus, value of x³ + y³ is (x + y)³ – 3xy (x + y)

Q.14. Write the value of x³ – y³.
Solution. The value of x³ – y³ is (x – y)³ + 3xy (x – y)

Q.15. Write the degree of the polynomial 4x⁴ + ox³ + ox⁵ + 5x + 7?
Solution. The degree of 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is 4.

Q.16. What is the zero of the polynomial p(x) = 2x + 5?
Solution. ∵ p(x) = 0 ⇒ 2x + 5 = 0

⇒ x = 

∴ zero of 2x + 5 is 

Q.17. Which of the following is one of the zero of the polynomial 2x² + 7x – 4 ?
 2,  -2 ? 

Solution. ∵ 2x² + 7x – 4 = 2x² + 8x – x – 4
⇒ 2x (x + 4) – 1 (x + 4) = 0
⇒ (x + 4) (2x – 1) = 0
⇒ x = – 4, x = 1/2,
∴ One of the zero of 2x² + 7x – 4 is 1/2.

Q.18. If a + b + 2 = 0, then what is the value of a³ + b³ + 8.

Solution. ∵ x + y + z = 0
⇒ x³ + y³ + z³ = 3xyz
∴ a + b + 2 = 0
⇒ (a)³ + (b)³ + (2)³ = 3(a × b × 2) = 6ab
⇒ The value of a³ + b³ + 8 is 6ab

Q.19. If 49x² – p =, what is the value of p?
Solution. 


Q.20. If  = 1, then what is the value of  ?

Solution. On squaring both the sides we get;

Q1: Simplify: (√5 + √2)².
Ans:Here, (√5 + √2² = (√5² + 2√5√2 + (√2)²
= 5 + 2√10 + 2
= 7 + 2√10

Q2:How many rational numbers can be found between two distinct rational numbers?
(i) Two
(ii) Ten
(iii) Zero
(iv) Infinite
Ans: (iv) Infinite
Between any two distinct rational numbers, an infinite number of rational numbers can be found. This is because rational numbers are dense on the number line, meaning between any two rational numbers $$a and b ($$a<b), you can always find another rational number by calculating the average:
New rational number = a+b/2 

Q3: Identify a rational number among the following numbers :
2 + √2, 2√2, 0 and π
Ans: 0 is a rational number.

Q4:  √8 is an
(i) natural number
(ii) rational number
(iii) integer
(iv) irrational number

Ans: (D) $\mathrm{}$√8 is an irrational number

Q5: Find the value of √(3)^{– 2}.
Ans:
Q6:  Is zero a rational number? Can you write it in the form pq, where p and q are integers and q≠0?

Ans: Consider the definition of a rational number. A rational number is the one that can be written in the form $\frac{p}{q}$ , where p and q are integers and $q\ne 0$

Zero can be written as

$\frac{0}{1},\frac{0}{2},\frac{0}{3},\frac{0}{4},\frac{0}{5},\dots$

So, we arrive at the conclusion that $0$0 can be written in the form $\frac{p}{q}$, where $$p is any integer.

Therefore, zero is a rational number.

Q7: A terminating decimal is

(i) a natural number
(ii) a rational number
(iii) a whole number
(iv) an integer.

Ans: (ii) a rational number

Q8: Find 64^1/2
Ans:

Q9: Simplify
Ans:

The LCM of 3 and 4 is 12.

Q10: The sum of rational and an irrational number
(i) may be natural
(ii) may be irrational
(iii) is always irrational
(iv) is always rational

Ans: (iii) is always rational
Example: 
Rational number: $$3
Irrational number: $2$
Sum: $3+$√$2$
The sum $3+$√$2$ cannot be expressed as $\frac{p}{q}$, so it is irrational.

Multiple Choice Questions

Q1: A histogram has a class interval 40-60, and its rectangle height is 8 units. Another class interval 60-70 has rectangle height 16 units.
If the intervals are unequal, which class has higher frequency density?

(a) 40-60
(b) 60-70
(c) Both equal
(d) Cannot be determined

Ans: (b)

Explanation: In a histogram with unequal class widths, the height of each rectangle represents the frequency density (frequency ÷ class width) according to the chosen vertical scale. Here the given rectangle heights are 8 and 16 units. Comparing these heights shows that the rectangle for 60-70 is taller (16 > 8), so its frequency density is larger. If one instead treats the given numbers as raw frequencies, dividing by class widths (40-60 has width 20, 60-70 has width 10) gives densities 8/20 = 0.4 and 16/10 = 1.6, and again the class 60-70 has the higher density. Thus 60-70 has the higher frequency density.

Q2: In a frequency polygon, if two consecutive class midpoints are incorrectly taken as 2 units closer, what happens to the polygon?

(a) Only the height changes
(b) Only the width changes
(c) Slope becomes steeper or flatter
(d) It becomes impossible to draw

Ans: (c)

Explanation: Points of a frequency polygon are plotted at class midpoints on the x-axis. If two consecutive midpoints are taken 2 units nearer than they should be, the x-coordinates of those plotted points shift. This changes the slopes of the line segments joining them, making the segment between those points either steeper or flatter compared with the correct plot. Heights (frequencies) remain the same, but the slope is distorted.

Q3: A bar graph compares profit for 12 months. If the scale is changed from 1 cm = ₹5000 to 1 cm = ₹10,000, the effect is:

(a) Bar widths double
(b) Bar heights become half
(c) Bars shift left
(d) Gaps disappear

Ans: (b)

Explanation: The vertical scale determines how much height represents a given amount. Changing the scale from 1 cm = ₹5,000 to 1 cm = ₹10,000 makes each amount occupy half the previous height on the graph. Bar widths and horizontal positions are unaffected; only the bar heights change, becoming half as tall.

Q4: A histogram is used for:
(a) Discrete data
(b) Continuous grouped data
(c) Favourite colours
(d) Names of students

Ans: (b)

Explanation: A histogram represents continuous grouped data arranged in class intervals. Adjacent bars touch, and the height of each bar gives frequency density (frequency ÷ class width). Discrete categories such as favourite colours or names use bar graphs, not histograms.

Q5: A bar graph is drawn using:

(a) Bars of equal width
(b) Bars of unequal width
(c) Bars touching each other
(d) No bars at all

Ans: (a)

Explanation: A bar graph uses bars of equal width with gaps between them to represent distinct categories; the heights show the values (frequency, amount, etc.). Histograms, by contrast, have touching bars because they represent continuous intervals.

Short Answer Questions

Q1: The following data gives the amount of manure (in tones) manufactures by a company during some years.

• Represent it with a bar graph.
• Indicate with help of bar graph in which year, the amount of manufactured by company was maximum.

Ans: The bar graph representing the given data is shown below. The year with the highest production is the year having the tallest bar in the graph.

Q2: Draw a bar graph for the following data:

Ans:

The bar graph above shows the given data; read the tallest or shortest bar to compare values across categories.

Q3: Find the class mark of 20-30.

Ans: Class mark = (Lower limit + Upper limit) ÷ 2
= (20 + 30) ÷ 2
= 25

Long Answer Questions

Q1: Draw a histogram for this data:

Ans:

When drawing the histogram, ensure that each class interval is represented by a bar whose width equals the class width and whose height equals the frequency density (frequency ÷ class width) so that areas of bars represent frequencies correctly.

Q2: You are given the following mid-points of a frequency polygon:
20, 30, 40, 50, 60
The polygon rises from (20, 6) to (40, 18), then falls to (60, 4).

(a) Reconstruct the class intervals.
(b) Draw a rough shape description of the polygon.
(c) Explain where the distribution has its “peak” and what it means.

Ans:

(a) To find class intervals from midpoints, use half the class width on either side of each midpoint.
Midpoints: 20, 30, 40, 50, 60
Difference between consecutive midpoints = 10, so class width = 10.
Each interval = midpoint ± 5, giving:

15-25
25-35
35-45
45-55
55-65

(b) Shape

• Plot points at (20, 6), (30, y₂), (40, 18), (50, y₄), (60, 4) where y₂ and y₄ are the intermediate frequencies consistent with the rise and fall described.
• The polygon rises steeply from (20, 6) up to (40, 18), then falls sharply toward (60, 4). This produces a clear single-peaked shape with a steep ascent followed by a descent.

(c) Peak at midpoint 40, frequency 18.
Meaning: The highest point of the polygon is at midpoint 40, so most observations fall in the class with midpoint 40, that is the class 35-45. This class therefore contains the largest frequency compared with the other classes.

Q3: Below is data of girls per 1000 boys in different sections of society:

(i) Draw a bar graph for the data.
(ii) Which section shows the highest value?
(iii) Comment on the trend.

Ans: (i) 

(ii) The ST (Scheduled Tribe) section shows the highest value with 970 girls per 1000 boys.

(iii) From the bar graph we observe the following trend:

• ST has the highest girls-per-1000-boys ratio.
• Backward districts and SC also show relatively better gender ratios compared with some other sections.
• Urban areas show the lowest value (910), indicating a poorer gender ratio in urban regions compared with several rural or tribal sections.
• Overall, the gender ratio varies across sections; tribal and certain backward areas perform better than urban and some other groups.

Multiple Choice Questions

Q1: Surface area of bowl of radius r cm is 
(a) 4πr²
(b) 2πr²
(c) 3πr²
(d) πr² 
Ans: (c)

Sol: The area of a circle of radius r is πr² 
Thus if the hemisphere is meant to include the base then the surface area is 2πr² + πr² = 3πr² 

Q2: A conical tent is 10 m high and the radius of its base is 24 m then slant height of the tent is
(a) 26
(b) 27 
(c) 28 
(d) 29
Ans: (a)

Sol: Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l
l² = h² + r²
l² = (10)² + (24)²
l² = 100 + 576 
l² = 676 
l = √676

l = √26²
l = 26 m
Therefore, the slant height of the tent is 26 m.

Q3: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm . then curved surface area. 
(a) 155 cm² 
(b) 165 cm² 
(c) 150 cm² 
(d) none of these
Ans: 165 cm² 

Sol: Diameter of the base of the cone is 10.5 cm and slant height is 10 cm.
Curved surface area of a right circular cone of base radius, [‘r’]and slant height, l is πr.
Diameter, d = 10.5 cm
Radius, r = 10.5 / 2 cm= 5.25 cm
Slant height, l = 10 cm
Curved surface area = πrl
= 3.14 × 5.25 × 10 = 165 cm
Thus, curved surface area of the cone = 165 cm².

Q4: The surface area of a sphere of radius 5.6 cm is 
(a) 96.8π cm²
(b) 94.08π cm²
(c) 90.08π cm²
(d) none of these
Ans: (b)

Sol: Given radius of sphere = 5.6 cm
Surface area of sphere = 4πr² 
= 4 × 3.14 × (5.6)² 
Surface area of sphere = 393.88 cm² 

Q5: The height and the slant height of a cone are 21 cm and 28 cm respectively then volume of cone 
(a) 7556 cm³ 
(b) 7646 cm³ 
(c) 7546 cm³ 
(d) None of these
Ans: (c)

Sol: Volume of the cone = 1/3 πr²h
Given
Slant height = l= 28 cm
Height of cone = h= 21 cm
Let radius of cone = r cm
l² = h² + r²
28² = 21² + r² 
28² – 21² = r² 
r² = 28² – 21² 
r² = (28 – 21)(28 + 21)
r² =(7)(49)
r = √7(49)
r = √7(7)² 
r = 7√7 cm
Volume of the cone = 1/3 πr² h

Fill in the blank

Q1: Surface area of sphere of diameter 14cm is____________.
Ans: 616cm² 

Sol: Given Diameter of sphere =14cm radius =7cm
surface area of sphere = 4πr² = 4π(7)² 
= 4 × 3.14 × 49
surface area of sphere = 616cm²  

Q2: Volume of hollow cylinder is ______________.
Ans: π(R²−r²)h

Sol: The formula to calculate the volume of a hollow cylinder is given as, 
Volume of hollow cylinder =π(R²−r²)h cubic units,
where, ‘R′ is the outer radius, ‘ r ‘ is the inner radius, and, ‘ h ‘ is the height of the hollow cylinder.

Q3: Find the volume of a sphere whose surface area 154cm² is_________________.
Ans: 179.67cm³

Sol: Given surface area of sphere =154cm²
Let radius of the sphere = r cm
4πr² = 1544 × 227 × r²
=154r²
Volume of sphere =4/3πr³ 
=179.67cm³

Q4: A hemispherical bowl has a radius of 3.5cm. What would be the volume of water it would contain__________.
Ans: 89.8cm³ 

Sol: The volume of water the bowl contain =2/3πr³ 
Radius of hemisphere =r=3.5cm
The volume of water the bowl can contain =2/3πr³ 
= 2/3 × 22/7 × 3.5 × 3.5 × 3.5cm³
= 89.8cm³

Q5: The formula for the volume of a cone is __________.

Ans: 13 π r² h

Sol: The formula for the volume of a cone is:  13 π r² h

True / False

Q1: The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
Ans: True

Sol: Let the radius of the sphere = r.
According to the question,
height and diameter of cylinder = diameter of sphere.
So, the radius of the cylinder = r
And, the height of the cylinder = 2r
We know that,
Volume of sphere = 2/3 volume of cylinder
Hence, the given statement “the volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere” is true.

Q2: If the radius of a right circular cone is halved and height is doubled, the volume will remain unchanged.
Ans: False

Sol: Let the original radius of the cone = r
Let height of the cone = h.
The volume of cone = 1/3 πr²h
Now, when radius of a height circular cone is halved and height is doubled, then
We can observe that the new volume = half of the original volume.
Hence, the given statement “if the radius of a right circular cone is halved and height is doubled, the volume will remain unchanged” is false.

Q3: If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.
Ans: True

Sol: Let radius of the cylinder = r
Height of the cylinder = h
Then, curved surface area of the cylinder, CSA = 2πrh
According to the question,
Radius is doubled and curved surface area is not changed.
New radius of the cylinder, R = 2r
New curved surface area of the cylinder, CSA’ = 2πrh …(i)
Alternate case:
When R = 2r and CSA’ = 2πrh
But curved surface area of cylinder in this case, CSA’= 2πRh = 2π(2r)h = 4πrh …(ii)
Comparing equations (i) and (ii),
We get,
Since, 2πrh ≠ 4πrh
equation (i) ≠ equation (ii)
Thus, if h = h/2 (height is halved)
Then,
CSA’ = 2π(2r)(h/2) = 2πrh
Hence, the given statement “If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved” is true.

Q4: Doubling the radius of a sphere will double its volume.
Ans: False 

Sol: Formula for the volume of a sphere:

V = 43 π r³

If the radius is doubled (i.e., r becomes 2r), then the new volume V’ is:

V’ = 43 π (2r)³ = 43 π 8r³ = 8 × 43 π r³

Thus, doubling the radius increases the volume by a factor of 8, not 2.

Q5: The total surface area of a cone is the sum of its lateral surface area and the area of its circular base.
Ans: True

Sol: The total surface area of a cone is the sum of its lateral surface area and the area of its circular base:

Lateral Surface Area = π r l

Area of Circular Base = π r²

Total Surface Area = π r l + π r²

For a cone with radius r and slant height l, the total surface area A is given by:

A = π r l + π r2

Subjective Type Questions

Q1: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm . Find its curved surface area and its total surface area.

Ans: Diameter = 10.5 cm
Slant height of cone (l ) = 10 cm
Curved surface area of cone,
=165 cm² 
Total surface area of cone,

Q2: A Joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Ans: Radius of cap (r) = 7cm, Height of cap (h) =24 cm
Slant height of the cone (l) 

Area of sheet required to make a cap = CSA of cone = πrl

∴ Area of sheet required to make 10 caps = 10 × 550 = 5500 cm² 

Q3: Find the surface area of a sphere of diameter:
(i) 14cm

Ans: (i) Diameter of sphere = 14cm,
Therefore, Radius of sphere = 14/2 = 7cm
Surface area of sphere = 4πr² = 4 × 22/7 × 7 × 7 = 616cm² 

(ii) 21cm

Ans: Diameter of sphere = 21cm
∴ Radius of sphere =21/2cm
Surface area of sphere = 4πr² = 4 × 22/7 × 21/2 × 21/2
=1386cm² 

(iii) 3.5cm

Ans: Diameter of sphere = 3.5cm
∴ Radius of sphere =3.5/2 = 1.75cm
Surface area of sphere = 4πr² = 4 × 22/7 × 1.75 × 1.75
= 38.5cm²

Q4: A hemispherical bowl is made of steel, 0.25cm thick. The inner radius of the bowl is 5cm . Find the outer curved surface area of the bowl.

Ans:  Inner radius of bowl (r)= 5cm
Thickness of steel (t) = 0.25cm
∴ Outer radius of bowl (R) = r + t = 5 + 0.25 = 5.25cm
∴ Outer curved surface area of bowl = 2πR² = 2 × 22/7 × 5.25 × 5.25
= 2 × 22/7 × 21/4 × 21/4
= 693/4 =173.25cm² 

Q5: Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S ‘. Find the:
(i) radius r ‘ of the new sphere.

Ans: Volume of 1 sphere, V = 4/3πr³ 
Volume of 27 solid sphere
= 27 × 4/3πr³ 
Let r₁ is the radius of the new sphere.
Volume of new sphere = Volume of 27 solid sphere

(ii) ratio of S and S ‘.

Ans: S₁ : S = 9 : 1
S : S₁ = 1 : 9

Q6: A capsule of medicine is in the shape of a sphere of diameter 3.5mm . How much medicine (in mm³) is needed to fill this capsule?

Ans: Diameter of spherical capsule = 3.5mm
∴ Radius of spherical capsule (r) = 3.5/2 = 35/20 = 7/4mm
Medicine needed to fill the capsule = Volume of sphere

Multiple Choice Questions

Q1: Surface area of bowl of radius r cm is 
(a) 4πr²
(b) 2πr²
(c) 3πr²
(d) πr² 

Q2: A conical tent is 10 m high and the radius of its base is 24 m then slant height of the tent is
(a) 26
(b) 27
(c) 28
(d) 29

Q3: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm . then curved surface area. 
(a) 155 cm² 
(b) 165 cm² 
(c) 150 cm² 
(d) None of these

Q4: The surface area of a sphere of radius 5.6 cm is 
(a) 96.8π cm²
(b) 94.08π cm²
(c) 90.08π cm²
(d) None of these

Q5: The height and the slant height of a coneare 21 cm and 28 cm respectively then volume of cone 
(a) 7556 cm³ 
(b) 7646 cm³ 
(c) 7546 cm³ 
(d) None of these

Fill in the blank

Q1: Surface area of sphere of diameter 14cm is____________.

Q2: Volume of hollow cylinder is ______________.

Q3: Find the volume of a sphere whose surface area 154cm² is_________________.

Q4: A hemispherical bowl has a radius of 3.5cm. What would be the volume of water it would contain__________.

Q5: The formula for the volume of a cone is __________.

True / False

Q1: The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.

Q2: If the radius of a right circular cone is halved and height is doubled, the volume will remain unchanged.

Q3: If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.

Q4: Doubling the radius of a sphere will double its volume.

Q5: The total surface area of a cone is the sum of its lateral surface area and the area of its circular base.

Subjective Type Questions

Q1: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm . Find its curved surface area and its total surface area.

Q2: A Joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Q3: Find the surface area of a sphere of diameter:
(i) 14cm
(ii) 21cm
(iii) 3.5cm

Q4: A hemispherical bowl is made of steel, 0.25cm thick. The inner radius of the bowl is 5cm . Find the outer curved surface area of the bowl.

Q5: Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S ‘. Find the:
(i) radius r ‘ of the new sphere.
(ii) ratio of S and S ‘.

Q6: A capsule of medicine is in the shape of a sphere of diameter 3.5mm . How much medicine (in mm³) is needed to fill this capsule?

Multiple Choice Questions

Q1: The difference between sides at right angles in a right-angled triangle is 14 cm. The area of the triangle is 120 cm². The perimeter of the triangle is
(a) 80
(b) 45
(c) 60
(d) 64

Q2: ABCD is a trapezium with AB  = 10cm, AD = 5 cm, BC = 4 cm and DC = 7 cm. Find the area of the ABCD?
(a) 34 cm²
(b) 28cm²
(c) 20 cm²
(d) None of these

Q3: Find the area and perimeter of the right angle triangle whose hypotenuse is 5 cm and Base is 4 cm.
(a) 6 cm² ,12 cm
(b) 12 cm² ,14 cm
(c) 4 cm², 6 cm
(d) 12 cm² ,6 cm

Q4: In an isosceles triangle ABC with AB = AC = 13 cm. D is mid point on BC. Also BC=10 cm. Which of the following is true?
(a) Area of Triangle ABD and ADC are equal
(b) Area of triangle ABD is 30 cm²
(c) Area of triangle ABC is 60 cm²
(d) All the above

Q5: A triangle and a parallelogram have the same base and the same area. The sides of the triangle are 26 cm and 30 cm and parallelogram stands on the base 28 cm. Calculate the height of the parallelogram.
(a) 12 cm
(b) 14 cm
(c) 10cm
(d) 13 cm

True/False

(i) Heron formula for area of triangle is not valid for all triangles.
(ii) If each side of the triangles is tripled, the area will becomes 9 times.
(iii) Base and corresponding altitude of the parallelogram are 8 and 5 cm respectively. The area of the parallelogram is 40 cm².
(iv) If each side of triangle is doubled, the perimeter will become 4 times.
(v) If p is the perimeter of the triangle of sides a,b,c ,the area of triangle is

(vi) When two triangles are congruent, there areas are same.
(vii) Heron’s belongs to America.
(viii) If the side of the equilateral triangle is a rational number, the area would always be irrational number.

Conceptual Questions

Q1: Calculate the area in each case
(i) Triangle have sides as a=5 cm ,b=4 cm, c=3 cm.
(ii) Equilateral triangle having side a=2 cm.
(iii) Right angle triangle have base=4 cm and Height =3 cm.
(iv) Square whose diagonal is 10 cm.
(v) Rectangle whose length and breath are 6 and 4 cm.
(vi) Parallelogram whose two sides are 10 cm and 16 cm and diagonal is 14 cm.
(vii) Parallelogram whose base is 10 cm and height is 14 cm.
(viii) Rhombus of diagonals to 10 and 24 cm.
(ix) Two sides of trapezium are 36 and 24 cm and its altitude is____.

Worksheet Solutions: Circles

True and False

Q1: If two arc of a circle are congruent. Then corresponding chord are unequal.
Ans: False (Corresponding chords are equal)

Q2:  Two perpendicular bisector of chord intersect at center of circle.
Ans: True ( each perpendicular bisector of chord passes through the center so center is common point for the two perpendicular bisectors of the chords)

Q3: The line joining the mid-point of a chord to centre perpendicular to chord. 
Ans: True (A line joining the mid-point of a chord to the centre of circle, perpendicular to the chord)

Q4: It is possible to draw two circles from three non-collinear points.
Ans: False (One and only one circle can be passed through three co-linear points in a plane.)

Answer the following Questions

Q1: If O is the center of circle of radius 5 cm OP perpendicular to AB and OQ perpendicular to CD, AB||CD, AB = 6cm and CD = 8 cm. Determine PQ.
Ans:

Q2: AB and CD are the two chords of the circle such that AB = 6 cm, CD = 12 cm and AB||CD, if the distance between AB and CD is 3 cm, find the radius of the circle.
Ans:

Q3: Prove that the line joining to the centre of circle to the mid-point of a chord, is perpendicular to the chord.
Ans:

O is the centre of the circle and AB is the Chord ,OM is the line segment intersecting at M , Mid point of AB

In ΔAMO and ΔBMO

AM = BM (M is the mid point of AB)

OA = OB (radius of circle)

OM = OM (Same side)

Hence

ΔAMO ≌ ΔBMO (By SSS)

∠AMO = ∠BMO ( BY C.P.C.T)

∠AMO +∠ BMO = 180 (linear pair angles)

∠AMO = 90

Q4: Given an arc of circle how you will find its centre and complete the circle.
Ans:

Construction : 

Step 1: Take 3 points P,Q,R on circumference of arc join P to Q and R to Q

Step 2: Draw Perpendicular bisector of line PQ and RQ these intersect at point O

Step 3: join O to P , O to Q and O to R

O is the centre of given arc where OQ and OR and OP are the radius of circle

Q5: Two equal chord AB and CD of circle with center O, when produced meet at a point E, proving that BE=DE and AE=CE
Ans:

Let OL and OM be two perpendiculars from centre O to chord AB and CD respectively so L and M are the mid points of Chord AB and CD

Since AB = CD (Given)

AB/2 = CD/2 or LB = MD

In ΔOLE and ΔOME

OL = OM (Equal Chords having equal distance from the centre)

<OLE = <OME (both 90 degrees)

OE = OE (common side)

So BY R.H.S

ΔOLE ≌ ΔOME

So LE = ME ( by C.P.C.T)

LE = LB+BE

ME = MD+DE

LB+BE = MD+DE

But LB = MD (proved above)

So BE = DE

Q6: Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. prove that QA=QB
Ans:

Let O and O’ be the centre of the circle where C(o, r) ≌ C(o’, r) circles are equal

So PQ is the common arc in both the circle

arc(PDQ) = arc(PCQ)

<QAP = <QBP (equal chords make equal angles on the circles)

in triangle ABQ <A = <B

so

Side AQ = Side BQ (Sides opposite to equal sides are equal)

True and False

Q1: If two arc of a circle are congruent. Then corresponding chord are unequal.

Q2:  Two perpendicular bisector of chord intersect at center of circle.

Q3: The line joining the mid-point of a chord to centre perpendicular to chord.

Q4: It is possible to draw two circles from three non-collinear points.

Answer the following Questions

 Q1: If O is the center of circle of radius 5 cm OP perpendicular to AB and OQ perpendicular to CD, AB||CD, AB = 6cm and CD = 8 cm. Determine PQ.

Q2: AB and CD are the two chord of the circle such that AB = 6 cm , CD = 12 cm and AB||CD, if the distance between AB and CD is 3 cm, find the radius of the circle.

Q3: Prove that the line joining to the centre of circle to the mid-point of a chord, is perpendicular to the chord.

Q4: Given an arc of circle how you will find its centre and complete the circle.

Q5: Two equal chord AB and CD of circle with center O, when produced meet at a point E, prove that BE=DE and AE=CE

Q6: Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA=QB.