Author name: Prasanna

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
Two circles with centre O, and A having radius 5 cm and 3 cm respectively.
The distance between the centres i.e. OA = 4 cm CD is the common chord. OA is perpendicular to CD and bisects CD.

OA ⊥ CD and AC = AD
In ∆OAC, OC = 5 cm, OA = 4 cm.
By Pythagoras theorem
AC² = OC² – OA²
⇒ AC² = 9
⇒ AC = 3 cm
Hence CD passes through the centre of a smaller circle and equal to the diameter of the circle.
CD = 2AC = 2 × 3 = 6 cm

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are corresponding segments of the other chord.
Solution:
Given: AB and CD are two equal chords of the circle which intersect each other at P.

To prove that: Segment ACB ≅ Segment CBD.
Proof: We have given,
chord AB = chord CD
∴ \(\widetilde{\mathrm{AB}} \cong \widetilde{\mathrm{CD}}\)
We know that, the segment made between congruent arcs and equal chords are congruent.
∴ Segment ACB ≅ Segment CBD.

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Given: AB and CD are two equal chords of a circle whose centre is O. Chord AB and CD intersect each other at E.

To prove that: ∠1 = ∠2
Construction: Draw OM ⊥ AB and ON ⊥ CD, and join OE.
Proof: In ∆OME and ∆ONE
OM = ON (Equal chords are equidistance from the centre)
∠OME = ∠ONE (Each 90°)
OE = OE (Common)
So, by R-H-S congruency condition
∆OME ≅ ∆ONE
∠1 = ∠2 (By CPCT)

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D. Prove that AB = CD (see Fig. 10.25)

Solution:
Let OM be perpendicular from O on line l.
We know that the perpendicular from the centre of a circle to a chord; bisect the chord. Since BC is a chord of the smaller circle and OM ⊥ BC.
∴ BM = CM ……(i)
Again, AD is a chord of the larger circle and OM ⊥ AD.
∴ AM = DM ……(ii)
Subtracting equation (i) from (ii), we get
AM – BM = DM – CM
or, AB = CD.

Question 5.
Three girls Reshma, Salma, and Mandeep are playing a game by standing on & circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Resma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Let Reshma, Salma and Mandip are standing on the point R, S, and M respectively where O is the centre of the circle.

RS = SM = 6m
OR = 5m
∆RSM is Issosclus Triangle
∴ SP ⊥ RM and also P is mid-point of RM
RP = PM
ar (∆ORS) = \(\frac {1}{2}\) × OS × PR
= \(\frac {1}{2}\) × 5 × x …….(i)
Draw OT ⊥ RS here OR = OS (Radii of circle)
So, ∆ORS is Issosceles Triangle
∴ RT = TS = 3 cm
In right ∆OTS,
OS = 5 cm, TS = 3 cm
Using Pythagoras theorem,
OT² = OS² – TS²
⇒ OT² = (5)² – (3)²
⇒ OT² = 16
⇒ OT = √16 = 4 cm
Again, ar (∆ORS) = \(\frac {1}{2}\) × RS × OT
= \(\frac {1}{2}\) × 6 × 4 ……(ii)
From equation (i) and (ii)
\(\frac {1}{2}\) × 5 × x = \(\frac {1}{2}\) × 6 × 4
or \(\frac {5x}{2}\) = 12
x = \(\frac{12 \times 2}{5}=\frac{24}{5}\)
We have to calculate
RM = 2RP = 2 × \(\frac{24}{5}\) = \(\frac{48}{5}\)
∴ RM = 9.6 m
Distance between Reshma and Mandip is equal to 9.6 m.

Question 6.
A circular park of a radius of 20 m is situated in a colony. Three boys Ankur, Syed, and David are sitting at an equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Solution:
Let three boys Ankur, Syed and David are sitting on the point A, B, and C respectively.
O is the centre of the circle.
According to question AB = BC = CA = x (Let)

So, ∆ABC is an equilateral triangle
OA = OB = OC = Radius of die circle = 20 meter
Join A, O and extand to D.
As ∆ABC is an equilateral triangle
\(\frac{\mathrm{OA}}{\mathrm{OD}}=\frac{2}{1}\)
or \(\frac{\mathrm{20}}{\mathrm{OD}}=\frac{2}{1}\) (∵ OA = 20 m)
OD = 10 meter
In right angle triangle ACD
(AD)² = (CD)² + (AD)²
⇒ (x)2 = \(\left(\frac{x}{2}\right)^{2}\) + (30)² (∵ AD ⊥ BC and D is the mid point)
⇒ x² = \(\frac{x^{2}}{4}\) + 900
⇒ x² – \(\frac{x^{2}}{4}\) = 900
⇒ \(\frac{4 x^{2}-x^{2}}{4}\) = 900
⇒ \(\frac{3 x^{2}}{4}\) = 900
⇒ x² = \(\frac{900 \times 4}{3}\)
⇒ x² = 1200
⇒ x = √1200
⇒ x = 20√3 meteres
Therefore, the length of the string of each phone is 20√3 meters.

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration InText Questions

NCERT In-text Question Page No. 206
Question 1.
Measure and write the length of the four sides of the top of your study table.
AB = …………… cm BC = ………….. cm
CD = ………….. cm DA = …………… cm

Now, the sum of the lengths of the four sides
= AB + BC + CD + DA
= ……….. cm + ……….. cm + ………….. cm + ……………. cm
= …………… cm
What is the perimeter?
Answer:
AB = 140 cm, BC = 80 cm,
AB = 140 cm, DA = 80 cm
∴ AB + BC + CD + DA
= 140 cm + 80 cm + 140 cm + 80 cm
= 140 cm
The perimeter 440 cm.
Note :This is an activity so the measures of AB, BC, CD and DA will be different for different table tops.

Question 2.
Measure and write the lengths of the four sides of a page of your notebook. The sum of the lengths of the four sides
= AB + BC + CD + DA
= ………….. cm + ………………. cm + ……………. cm + ……………. cm
= ……………… cm
What is the perimeter of the page?
Answer:
Please try yourself.

Question 3.
Meera went to a park 150 m long and 80 m wide. She took one complete round
on its boundary. What is the distance covered by her?
Answer:
The park is in the shape of rectangle.
Since she takes one complete round which is equal to perimeter of the rectangular park.
Perimeter of rectangular park
= 2 (length + breadth)
= 2 ( 150+ 80)
= 2 × 230
= 460 m
Therefore, the distance covered by her is 460m.

Question 4.
Find the perimeter of the following figures:
(a) Perimeter = AB + BC + CD + DA
= …………… + …………… + ……………. + ……………….+
= ………………

(b) Perimeter = AB + BC + CD + DA
= ……………. + …………. + ………….. + …………… +
= ……………..

(c) Perimeter = AB + BC + CD + DE + EF + FG + GH + HI+ IJ + JK+ KL + LA
= ……………… + ……………… + …………… + …………… +
……………… + …………….. + …………….. + ……………. +
……………. + …………….. + ……………. + …………….. +

(d) Perimeter=AB + BC + CD + DE + EF + FA
= ……………. + ……………. + ……………… + …………….. +
……………… + ……………. +
= ………………

Answer:
(a) AB = 40 cm, BC =10 cm,
AD = 10 cm and DC = 40 cm
∴ Perimeter = AB + BC + AD + DC
= 40 cm + 10 cm + 10 cm + 40 cm
= 100 cm

(b) AB = 5 cm, BC = 5 cm,
AD = 5 cm and AD = 5 cm
∴ Perimeter = AB + BC + CD + AD
= 5 cm + 5 cm + 5 cm + 5 cm = 20 cm

(c) Here, perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm = 28 cm

(d) Here, perimeter = AB + BC + CD + DE + EF + FA
= 100 m + 120 m + 90 m + 45 m + 60 m + 80 m = 495m

NCERT In-text Question Page No. 208
Question 1.
Find the perimeter of the following rectangle :

Answer:
Do yourself.

NCERT In-text Question Page No. 211
Question 1.
Find various objects from your surroundings which have regular shapes and find their perimeters.
Answer:
It is a project work. You can find objects having regular shapes such as a ‘name plae’ a road signal’, etc. Measure their sides and find perimeters.

NCERT In-text Question Page No. 215
Question 1.
Draw any circle on a graph sheet. Count the squares and use them to estimate the area of the circular region.
Answer:
It is an ‘activity’. Please do it yourself.

Question 2.
Trace shapes of leaves, flower petals and other such objects on the graph paper and find their areas.
Answer:
It is an ‘activity’. Please do it yourself.

NCERT In-text Question Page No. 217
Question 1.
Find the area of the floor of your classroom.
Answer:
This is an activity. Please do it yourself.

Question 2.
Find the area of any one door in your house.
Answer:
This is an activity. Please do it yourself.

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
There are three different pairs of circle.

It is clear that in Fig. I, there is no any common point in pair of circle O and O’.
In Fig. II, there is only one point A is common point in pair of circle P and P’.
In Fig. III, there are two points X and Y are common points in pair of circle Q and Q’.
The maximum number of common points are two.

Question 2.
Suppose you are give a circle. Give a construction to find its centre.
Solution:
To find the centre of the circle we take following steps:

Step (i) – Take three points A, B and Con die circle.
Step (ii) – Join AB and AC.
Step (iii) – Draw perpendicular bisector of AB and BC.
Step (iv) – Both perpendicular bisector intersect each other at point O.
Step (v) – Point O is centre of the required circle.
Remark: We know that centre lie on perpendicular bisector of chord. Therefore centre must be lie on perpendicular bisector of AB. Again centre also must be lie on perpendicular bisector of BC. Both cases are possible only when centc lie on their point of intersection.

Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
Given: Two circles O and O’ intersect each other at A and B. AB is a common chord for both circles.

To prove that: OO’ is perpendicular bisector of AB.
Construction: Draw line segments OA, OB, O’A and O’B.
Proof: In ∆OAO’ and ∆OBO’ we have
OA = OB (Radii of same circle)
O’A = O’B (Radii of same circle)
and OO’ = OO’ (Common)
So, by S-S-S congruency condition
∆OAO’ ≅ ∆OBO’
∴ ∠AOO’ = ∠BOO’ (By C.P.C.T)
or ∠AOM = ∠BOM …..(i)
Now, In ∆AOM and ∆BOM
OA = OB (Radii of samecircle)
∴ ∠AOM = ∠BOM (From (i))
OM = OM (Common)
So, by S-A-S congruency condition
∆AOM ≅ ∆BOM
∴ AM = BM (By CPCT)
and ∠AMO = ∠BMO (By CPCT)
But, ∠AMO + ∠BMO = 180° (Liner pair)
2∠AMO = 180° (∵ ∠AMO = ∠BMO prove above)
or, ∠AMO = 90°
Hence, OO’ lie on the perpendicular bisector of AB.

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.3

Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Answer:
(a) Area of rectangle
= length × breadth
= 3 cm × 4 cm = 12 cm²

(b) Area of rectangle
= length × breadth
= 12 m × 21 m = 252 m²

(c) Area of rectangle
= length × breadth
= 2 km × 3 km = 6 km²

(d) Area of rectangle
= lenght × breadth
= 2m ( lm = 100cm)
= 200 cm × 70 cm = 14000sq.cm

Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Answer:
(a) Area of square = side × side
= 10 cm × 10 cm = 100 cm²

(b) Area of square = side × side
= 14 cm × 14 cm = 196 cm²

(c) Area of square = side × side
= 5 m × 5 m = 25 m²

Question 3.
The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Answer:
(a) Area of rectangle
= length × breadth
= 9 m × 6 m = 54 m²

(b) Area of rectangle
= length × breadth
= 3 m × 17 m = 51 m²

(c) Area of rectangle
= length × breadth
= 4 m × 14 m = 56 m²
Thus, the rectangle (c) has largest area, and rectangle (b) has smallest area.

Question 4.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Answer:
Length of rectangle
= 50 m and Area of rectangle = 300 m² Since, Area of rectangle
= length × breadth
Therefore, breadth
\(\frac{\text { Area of rectangle }}{\text { Length }}\) = \(\frac{300}{50}\) = 6m
Thus, the breadth of the garden is 6 m.

Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
Answer:
Length of land
= 500 m and Breadth of land = 200 m
Area of land = length × breadth
= 500 m × 200 m = 1,00,000 m²
Cost of tilling 100 sq. m of land = 8
∴ Cost of tilling 1,00,000 sq. m of land
\(\frac{8 \times 100000}{100}\) = 8000

Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Answer:
Length of table = 2 m
Breadth of table = 1 m 50 cm = 1.50 m
Area of table = length × breadth
= 2 m × 1.50 m = 3 m²

Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Answer:
Length of room = 4 m
Breadth of room = 3 m 50 cm = 3.50 m
Area of carpet = length × breadth
= 4 × 3.50= 14 m²

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Answer:
Length of floor
= 5 m and breadth of floor = 4 m
Area of floor = length × breadth
= 5m × 4m = 20m²
Now, Side of square carpet = 3 m
Area of square carpet = side × side
= 3 × 3 = 9 m²
Area of floor that is not carpeted = 20m² – 9m² = 11m²

Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Answer:
Side of square bed = 1 m
Area of square bed = side × side
= 1m × 1m = 1m²
∴ Area of 5 square beds = 1 × 5 = 5 m²
Now, Length of land = 5 m
Breadth of land = 4 m
∴ Area of land = length × breadth
= 5 m × 4 m = 20 m²
Area of remaining part
= Area of land – Area of 5 flower beds = 20 m² – 5 m² = 15 m²

Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Answer:

Area of HKLM = 3 × 3 = 9 cm²
Area of IJGH = 1 × 2 = 2 cm²
Area of FEDG = 3 × 3 = 9 cm²
Area of ABCD = 2 × 4 = 8 cm²
Total area of the figure
= 9 + 2 +9 + 8
= 28 cm²

Area of ABCD = 3 × 1 = 3 cm²
Area of BDEF = 3 × 1 = 3 cm²
Area of FGHI = 3 × 1 = 3 cm²
Total area of the figure = 3 + 3 + 3 = 9 cm²

Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Answer:
(a) Area of rectangle ABCD
= 2 × 10 = 20 cm²
Area of rectangle DEFG
= 10 × 2 = 20 cm²
Total area of the figure
= 20 + 20 = 40 cm²

(b) There are 5 squares each of side 7 cm.
Area of one square
= 7 × 7 = 49 cm²
Area of 5 square
= 49 × 5 = 245 cm2

(c) Area of rectangle ABCD
= 5 × 1 = 5 cm²
Area of rectangle EFGH
= 4 × 1 = 4 cm²
Total area of the figure
= 5 + 4 cm²

Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Answer:
(a) Area of region
= 100 cm × 144 cm = 14400 cm²
Area of one tile
= 5 cm × 12 cm = 60 cm²
Number of tiles = \(\frac{\text { Area of region }}{\text { Area of one tile }}\) = \(\frac{14400}{60}\) = 240
Thus, 240 tiles are required.

(b) Area of region
= 70 cm × 36 cm = 2520 cm²
Area of one tile
= 5 cm × 12 cm = 60 cm²
Number of tiles
= \(\frac{\text { Area of region }}{\text { Area of one tile }}\) = \(\frac{2520}{60}\) = 42
Thus, 42 tiles are required.

NCERT Solutions for Class 7 English

An Alien Hand NCERT Solutions for Class 7 English An Alien Hand Chapter 10

An Alien Hand NCERT Text Book Questions and Answers

An Alien Hand Exercises Question and Answer 

Discuss the following topics in groups.

Question 1.
If you had to live in a home like Tilloo s, what parts of life would you find most difficult? What compensations might there be?
(Encourage the students to discuss their answers in groups.)
Answer:
If I had to live in a home like Tilloo’s, I would miss the oudoors and the trees and plants. I would definitely miss the clean, fresh air very often as I might find it difficult to breathe in a claustrophobic environment. Living closer to the core of a planet would mean that it would be uncomfortably hot there.The only compensation that I can see in Tilloo’s life is that he is living with his family that loves him a lot.

Question 2.
What, if anything, might drive mankind to make their homes underground?
Answer:
A nuclear attack, or an alien attack can make human beings live under the ground. This resonates with the World War II when the Jews had to find homes under the ground to remain protected and hidden from the Nazis.

Question 3.
Do you think there is life on other planets? Can you guess what kind of people there may be on them? In what ways are they likely to be different from us?
Answer:
Yes, there is a huge possibility that there is life on other planets because the universe is unimaginably vast. The Earth is but merely one of the planets in one of the galaxies. And there are billions and billions of galaxies in the universe.I think the people on other planet may look like us, but may be far more advanced in terms of technology.

They may have a completely different culture. In fact, they may even eat different food than us. The way they use energy in their lives may be significantly different as well.