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S.chand Class 8 Maths Solution Chapter 7 Linear Equations Exercise 7 A

Exercise 7 A

Solve the following equations and check your answer:

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Question 1

(i) $10 p-(3 p-4)=4(p+1)+9$

(ii) $7+2(a+1)-3 a=5 a$

(iii) $4(x+3)-2(x-1)=3 x+3$

Sol :

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Question 2

(i) $\frac{x}{3}-5=8$

(ii) $\frac{t+8}{3}=t$

Sol :

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Question 3

(i) $4 y+\frac{y}{5}=21$

(ii) $\frac{2 m}{3}+\frac{3 m}{4}=17$

Sol :

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Question 4

$\frac{2}{3}\left(x+\frac{3}{5}\right)=\frac{7}{2}$

Sol :

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Question 5

$\frac{x+3}{7}-\frac{2 x-5}{3}=\frac{3 x-5}{5}-25$

Sol :

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Question 6

$\frac{3 x-2}{10}-\frac{x+3}{7}+\frac{4 x-7}{3}=x-1$

Sol :

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Question 7

$2.8 v=54+v$

Sol :

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Question 8

$0.26 x+0.09 x=8-0.45 x$

Sol :

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Question 9

$\frac{1}{6}(4 y+5)-\frac{2}{3}(2 y+7)=\frac{3}{2}$

Sol :

Solve the following equations :

Question 10

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$\frac{3}{x+1}=\frac{5}{2 x}$

Sol :

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Question 11

$\frac{6}{3 m+1}=\frac{9}{5 m-3}$

Sol :

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Question 12

$\frac{2 x+3}{5}=\frac{4 x+9}{11}$

Sol :

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Question 13

$\frac{5 x-7}{3 x}=2$

Sol :

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Question 14

$\frac{0.4 z-3}{1.5 z+9}=\frac{7}{5}$

Sol :

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Question 15

$\frac{x-2}{x-4}=\frac{x+4}{x-2}$

Sol :

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Question 16

$\frac{y-2}{y-5}=\frac{y+3}{y+5}$

Sol :

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Question 17

$\frac{2 y-4}{3 y+2}=-\frac{2}{3}\left(\right.$ Hint. Write $-\frac{2}{3}$ as $\left.\frac{-2}{3}\right)$

Sol :

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Question 18

$\frac{\frac{x}{4}-\frac{3}{5}}{\frac{4}{3}-7 x}=-\frac{3}{20}$

Sol :

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Question 19

$\frac{(2 x+3)-(5 x-7)}{6 x+11}=\frac{-8}{3}$

Sol :

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Question 20

$\frac{17(2-x)-5(x+12)}{1-7 x}=8$

Sol :

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Question 21

$\frac{(4+x)(5-x)}{(2+x)(7-x)}=1$

Sol :

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Question 22

$\frac{x^{2}-(x+1)(x+2)}{5 x+1}=6$

Sol :

Multiple Choice Questions (MCQs)

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Question 23

If $\frac{x+3}{2}-\frac{3 x+1}{4}=\frac{2(x-2)}{3}-2$, then the value of $x$ is

(a) $\frac{61}{11}$

(b) 5

(c) $-5$

(d) $-\frac{61}{11}$

Sol :

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Question 24

The solution of $\frac{2 x+3}{2 x-1}=\frac{3 x-1}{3 x+1}$ is

(a) $\frac{1}{8}$

(b) $-\frac{1}{8}$

(c) $\frac{8}{3}$

(d) $\frac{-8}{3}$

Sol :

High Order Thinking Skills (HOTS)

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Question 25

Solve and give the positive value of $x$ which satisfies the given equation :

$(2 x+4)^{2}-(x-5)^{2}=26 x$

Sol :

S.chand Class 8 Maths Solution Chapter 6 Factorisation of Algebraic Expressions Exercise 6 G

Exercise 6 G

Factorise:

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Question 1

$4 x^{2}+5 x+1$

Sol :

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Question 2

$2 x^{2}+11 x+14$

Sol :

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Question 3

$2 x^{2}+11 x+12$

Sol :

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Question 4

$3 x^{2}+13 x+4$

Sol :

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Question 5

$2 x^{2}-5 x-12$

Sol :

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Question 6

$4 x^{2}+8 x-5$

Sol :

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Question 7

$13 x^{2}+37 x-6$

Sol :

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Question 8

$40 n^{2}+n-6$

Sol :

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Question 9

$4 z^{2}-16 z+15$

Sol :

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Question 10

$6-9 c-27 c^{2}$

Sol :

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Question 11

$1-t-6 t^{2}$

Sol :

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Question 12

$2 a^{2}+7 a b-15 b^{2}$

Sol :

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Question 13

$4 x^{2}+24 x+20$

Sol :

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Question 14

$12 a^{2}+2 a-4$

Sol :

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Question 15

$12 m^{3}+6 m^{2}-6 m$

Sol :

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Question 16

$6 x^{5}-22 x^{3}-8 x$

Sol :

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Question 17

$6 z^{6}-21 z^{4}-12 z^{2}$

Sol :

Multiple Choice Questions (MCQs)

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Question 18

A rectangular parking lot has an area of $5 x^{2}+17 x+6$ square units. One of its sides is

(a) $(5 x+6)$

(b) $(x+2)$

(c) $(5 x+2)$

(d) $(5 x+1)$

Sol :

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Question 19

The complete factorisation of $-12 b^{5}-b^{3}+b$ is

(a) $-b\left(12 b^{4}+b^{2}-1\right)$

(b) $b(2 b-1)(2 b+1)\left(1+3 b^{2}\right)$

(c) $-b\left(4 b^{2}-1\right)\left(3 b^{2}+1\right)$

(d) $b(1-2 b)(1+2 b)\left(1+3 b^{2}\right)$

Sol :

High Order Thinking Skills (HOTS)

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Question 20

The binomial $x-3$ is not a factor of which of the following trinomials ?

(a) $2 x^{2}-x-15$

(b) $3 x^{2}-13 x+12$

(c) $2 x^{2}-8 x+6$

(d) $2 x^{2}-7 x-3$

Sol :

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18C

Exercise 18 C

Question 1

Name the shape of the cross- section in each solid?

(DIAGRAM TO BE ADDED)

Question 2

What cross- section would you get when you gibe a (a) Vertical cut (b) Horizontal cut to the following object?

(i) a basket ball

ii) a die

(iii) an unsharpened pencil

(iv) A 1litre carton of milk

(v) a joker's cap

(vi) A chocolate box(triangular

Question 3

Multiple choice question

A slice of cheese is cut from the cylinder shaped cheese as shown. What name can be given to the cross-section?

(a) Circle

(b) Square

(c) Rectangle

(d) Triangle

Question 4

Name a three-dimensional figure from which a cross-section of a circle can be made.

(a) Cube

(b) Cuboid

(c) Cone

(d) Pyramid

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18B

Exercise 18B

Question 1

Copy each net into your book and add the face that is needed to make the net of named solid . Shade the added face .

Prime:- Two parallel congruent Polygonal faced connected

Pyramid : - Polygon base and triangular faces meet at a common vertex.

(diagram to be added)

Question 2

Label each net as of :

(a) cuboid

(b) cube

(d) rectangular pyramid

(e) cone

(f) cylinder

(g) square pyramid

(h) triangular prism

Question 3

Draw the net of a cube. Put the number I to 6 on the faces so that the number on opposite faces add to 7. Can this be done in more than one. If so, give all ways.

One has been done for you :

(diagram to be added)

Question 4

Name the different plane shapes needed to draw the net of :

(a) a cube

(b) a triangular prism

(d) a cylinder

(diagram to be added)

Question 5

Which of these nets will not fold up to give a cube?

[Hint. Make the cardboard cutouts of the following nets and fold them to sce whether a cube is formed or not]

(diagram to be added)

Question 6

Only one of these nets could form a pyramid. Which one ?

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18A

Exercise 18 A

Question 1

Leonhard Euler (1707-83) was a famous mathematician who discovered a rule about solids.-He observed the number of faces, vertices and edges of many solids before discovering this rule.

(a) Let us see if you can derive the Euler's rule $(F+V-E=2$ ) by completing the following table.

(Diagram to be added)

Question 2

For each solid, count the number of faces, vertices and edges. Check the Euler's rule for each one.

(Diagram to be added)

(a) (Diagram to be added)

Sol: F= 7

V=10

E=15

F+V-F=2

17-15=2

(b) (Diagram to be added)

Sol: $F=6$

$V=8$

$E=12$

$F+V-E$

$6+6-12$

$14-12$ = 2

Sol:

$\begin{aligned} F &=7 \\ V &=7 \\ E &=12 \\ 14-12 \\ &=2 \end{aligned}$

(d) (Diagram to be added)

Sol: $F=9$

$V=9$

$E=16$

$18-16$

$=2$

Question 3

Fill in the blanks. Name a solid that has :

(i) 4 faces

(ii) 4 vertices

(iii) 2 edges

(iv) 6 vertices

(v) No edges.

(vi) 6 faces.

(Diagram to be added)

Multiple choice question

Question 4

Which of these solids has the maximum number of vertices ?

(a) Cons

(b) Cylinder

(c) Cuboid

(d) Pyramid

Question 5

If $\mathrm{F}=6$ and $\mathrm{V}=4$, then the value of $\mathrm{E}$ using Euler's formula is

(a) 10

(b) 8

(d) 4

SChand Composite Mathematics Class 7 Chapter 17 Symmetry Exercise 17A

Exercise 17 A

Question 1

Examine the following shapes and tell which of these have a line or lines of symmetry and which do not have any line of symmetry?

(Diagram to be added)

Question 2

Draw the lines of symmetry in each of the following figures :

Question 3

Draw the line of symmetry in each of the following letters of the English alphabet.

Question 4

Name the seven letters of the English alphabet which have no line of symmetry.

Question 5

Each of the shapes shown Below has two lines of symmetry. Using squared paper copy each these shapes and then draw the two lines of symmetry.

(IMAGE TO BE ADDED)

Question 6

A number of half- shapes with the line of symmetry indicated by dotted line are shown below. Draw the complete symmetrical shape using squared paper.

(IMAGE TO BE ADDED)

Question 7

Use the two lines of symmetry to copy and complete each of these shapes. [Hint. Use just une line of symmetry at a time.]

(IMAGE TO BE ADDED)

Question 8

How many lines of symmetry do the following have :

(i) a scalene triangle

(ii) a square

(iii) an equilateral triangle

(iv) a rhombus

(v) a parallelogram

(vi) a right triangle with equal legs

(vii) a semi-circle

(viii) an angle with Equal arms.

Question 9

Copy the word MATHS onto a grid, as shown. (a) Reflect it about $A B$. (b) Reflect it about CD.

(IMAGE TO BE ADDED)

Question 10

Answer true or false :

(i) A parallelogram has two lines of symmetry.

(ii) The letter $Z$ has one line of symmetry.

(iii) A right angled triangle can have at most one line of symmetry.

(iv) A triangle with more than one line of symmetry is an equilateral triangle.

(v) A pentagon which has more than one line of symmetry must be regular.

Question 11

Multiple Choice Questions (MCQs)

The number of letters in the word SNAIL that have line symmetry is

(a) 0

(b) 1

(d) 3

Question 12

Which of the following does not have a line symmetry ?

(IMAGE TO BE ADDED)

Question 13

Which of the following road signs has a line symmetry ?

(IMAGE TO BE ADDED)

Question 14

High Order Thinking Skills (HOTS)

What do you call a rectangle that has four lines of symmetry ?

SChand Composite Mathematics Class 7 Chapter 16 Chance and Probability Exercise 16B

Exercise 16 B

Question 1

Refer to the shape chart and fill in the blanks in the following table giving the probability of touching each figure. One is done for you.

(Diagram to be added)

$\begin{array}{|c|c|c|c|c|c|}\hline \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } & \text { (vi) } \\\hline \frac{15}{32} & \frac{8}{32} \text { ar } \frac{1}{4} & \frac{4}{32} \text { or } \frac{1}{8}& \frac{3}{32} & \frac{1}{32} & \frac{1}{32} \\\hline\end{array}$

Question 2

A box contains 4 packs of chocolates. Shruti takes out a pack without looking at the packs. What is the chance (probability) that she picks:

(a) Perk

(b) Kitkat

2 Kitkat

1 Perk

1 Dairy milk

Sol: Probability= $\frac{Number of successful outcomes }{Number of possible outcomes}$

$\frac{1}{4}$

Question 3

A jar has one blue and 9 green marbles in it. What is the probability of drawing

(a) a green marble ?

(b) a blue marble?

(diagram to be added)

Sol: (a) $p=\frac{9}{10}$

(b) $p=\frac{1}{10}$

Question 4

The six faces of a cube are numbered from 1 to 6 . The cube is rolled. What is the chance that it will land with :

(a) a 4 up

(b) a 2 up?

Sol: (a) $p=\frac{1}{6}$

(b) $p=\frac{1}{6}$

Question 5

Five girls put their names in a box. They are anjali , Haze, Ayushree, Ishita and Rashi. Rashi draws a name from the box. What is the chance that she draws her own name?

Ans: $p=\frac{1}{5}$

Question 6

An ordinary pack of $5 z$ cards is well shuffled. The top card is then turned over. What is the probability that :

(a) the top card will be a red card.

(b) the top card is a number card.

[Clue : How many red cards are there? How many number cards are there ?]

Sol: (a) Number of Outcomes = 26

$P=\frac{26}{52}$ or $\frac{1}{2}$

(b) Total Number card $=9 \times 4=36$

$P=\frac{36}{52}$ or $\frac{9}{13}$

Question 7

A gift pack of chips contains 2 cheese and onion, 2 plain salted, 3 Masala munch and 1 Pudina. Tanya takes out a pack of chips without looking at the packs. What is the chance that she picks :

(a) plain salted

(b) Masala

(d) Pudina

Sol: (a) $p=\frac{2}{8}$ or $\frac{1}{4}$

(b) $p=\frac{3}{8}$

(d) $p=\frac{1}{8}$

Question 8

An ordinary die is rolled. What is the probability that the number of dots on its upper face is

(a) 3

(b) less than 3

(d) 7 ?

Sol: (a) $p=\frac{1}{6}$

(b) $\frac{2}{6}$ or $\frac{1}{3}$

(d) $\frac{0}{6}$

Question 9

A jar contains 3 white, 4 blue, 5 red and 2 green marbles. If a marble is drawn at random from the jar, what is the probability that the marble is :

(a) white

(b) red

(e) not white

(f) not red?

(d) green

Sol; (a) $p=\frac{3}{14}$

(b) $p=\frac{5}{14}$

(d) $p=\frac{2}{14}$ or $\frac{1}{7}$

(e) $p=\frac{14-3}{14}=\frac{11}{14}$

Question 10

Draw a probability scale. Mark each of these outcomes on your scale.

Give reasons for your answers.

(a) The school bus will break down tomorrow.

(b) The next baby to be born will be a boy.

(d) A heavy stone will float when it is dropped in the sea.

(e) The winner of the women's Olympic 100 meter final will be aged under 35 years.

(Diagram to be added)

Question 11

- Ashita has 20 movies in her video collection and 5 of the movies feature her favourite actor. If she randomly choses a movie, what is the probability that she will choose one featuring her favourite actor ?

(a) $\frac{1}{5}$

(b) $\frac{1}{4}$

(d) $\frac{3}{5}$

Question 12

A card is drawn from a pack of 100 cards numbered 1 to 100 . Find the probability of drawing a square number.

(a) $\frac{1}{10}$

(b) $\frac{9}{10}$

(d) $\frac{2}{5}$

Question 13

During a class survey, it was found out that cheese pizza is the favourite snack for 30 out 40 students. Which per cent is closest to the probability that a student's favourite snack is cheese pizza ?

(a) $50 \%$

(b) $60 \%$

(c) $75 \%$

(d) $80 \%$

Question 14

High order thinking skills ( Hots )

You roll a die. Write the probability of rolling a prime number as a decimal.

SChand Composite Mathematics Class 7 Chapter 16 Chance and Probability Exercise 16A

Exercise 16 A

Question 1

In the following events, match correctly to indicate whether the outcomes are possible, certain, or impossible.

(1) You will throw a 7 with a normal die.

(2) Your friend will call you tonight.

(3) A cat will give birth to puppies next year.

(4) Thursday will be the day after Wednesday next week.

(5) You will have cornflakes, samosas and toasts in your breakfast today.

(6) It will be winter in Australia when it is summer in India.

(7) You will be able to see a live dinosaur in the city zoo.

(8) You will be awarded a prize for your good performance.

Ans:

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline b & a & b & c & a & c & b & a \\\hline\end{array}$

Question 2

Categorize each outcome as likely or unlikely.

(1) You will not get reservation in the train due to heavy Dusshra rush.

(2) Your friend will go to moon next month.

(3) Someone in your class will be absent next week.

(4) It will snow in mussorie in January.

(5) There will be floods in Delhi in March next year.

(6) You will become an army officer when you grow up.

$\begin{aligned}&\text { Ans. }\\&\begin{array}{|c|c|c|c|c|c|}\hline 1 & 2 & 3 & 4 & 5 & 6 \\\hline \mathbf{a} & \mathrm{b} & \mathrm{a} & \mathrm{a} & \mathrm{b} & \mathrm{a} \\\hline\end{array}\end{aligned}$

Question 3

Choose the likelihood which matches the outcome of each event:

(1) The school football team will win the local toumament.

(2) Mr. Shah will cut the grass on his lawn when it is snowing.

(3) While going to school, you will pass by a white car.

(4) The mountaineer will be hurt if he falls off the mountain.

(5) The number on the top face of an ordinary die will be an even number.

(6) The next baby to be born will be a female.

(7) The number on the top face of an ordinary die when rolled will be less than 7 .

(8) The Titanic will float back up to the top of the ocean.

(9) You will live for 100 years.

(10) A dog will have kittens.

(11) Monday will be the day after Tuesday next week.

(12) You will have a birthday next year.

(a) Impossible

(b) Unlikely

(d) Likely

$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|}\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\hline \mathrm{d} & \mathrm{b} & \mathrm{d} & \mathrm{d} & \mathrm{c} & \mathrm{c} &\mathrm{e} & \mathrm{a} & \mathrm{b} & \mathrm{a} & \mathrm{a} & \mathrm{e} \\\hline\end{array}$

Question 4

Multiple Choice Questions (MCQs)

The probability of being chosen for a team is $9 \%$. This event can be described as:

(a) likely

(b) certain

(d) unlikely

Question 5

It will be a Thursday in one of the next 7 days.

The event is

(a) likely

(b) unlikely

(c) certain

(d) having an even chance.

Question 6

High Order Thinking Skills (HOTS)

A person with $\mathrm{O}, \mathrm{A}, \mathrm{B}$ or $\mathrm{AB}$ blood group can safely donate blood to a person with $\mathrm{AB}$ positive blood group.

Describe the event "a randomly chosen person could donate blood to a person with $\mathrm{AB}$ positive blood group" as certain or likely or unlikely or impossible.- Certain

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15D

Exercise 15 D

Question 1

During the festive season, the sales of the various gadgets is as under :

$\begin{array}{|c|c|c|c|c|}\hline \text { Mixer } & \text { Microwave } & \text { Toaster } & \text { DVD Player } & \text { I Pod} \\\hline 250 & 180 & 200 & 260 & 50 \\\hline\end{array}$

Represent the above information by a bar graph.

(Diagram to be added)

Question 2

The marks obtained by a student in different subjects are:

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & \text { English } & \text { Maths } & \text { Science } & \text { Hindi }&\text { Social Science } \\\hline \text { Marks } & 70 & 95 & 80 & 60 & 75 \\\hline\end{array}$

(Diagram to be added)

Question 3

The approximate population (in lac) based on 2011 census of 8 Indian States:

$\begin{array}{|c|c|}\hline \text { State } & \text { Population (in lac) } \\\hline \text { Andhra Pradesh } & 850 \\\hline \text { Bihar } & 1050 \\\hline \text { Chhatisgarh } & 260 \\\hline \text { Gujarat } & 600 \\\hline \text { Madhya Pradesh } & 730 \\\hline \text { Maharashtra } & 1130 \\\hline \text { Uttar Pradesh } & 2000 \\\hline \text { West Bengal } & 920 \\\hline\end{array}$

(Diagram to be added)

Question 4

Use the bar graph below to answer the question.

(Diagram to be added)

1. How many students scored above 75 ?

(ii) What score was made by the most students ?

(iii) If the pass score was 70 , how many did not pass ?

(iv) Form a frequency table with the help of the graph and find the mean score.

(Diagram to be added)

Question 5

The marks in a spelling test are shown in this table.

(a) Draw a bar graph to represent this set of data.

(b) What mark did most pupils score ?

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline 10 & 6 & 7 & 9 & 5 & 7 & 9 & 3 \\\hline 3 & 9 & 8 & 7 & 6 & 10 & 7 & 6 \\\hline 9 & 5 & 10 & 3 & 8 & 8 & 5 & 9 \\\hline 7 & 8 & 6 & 5 & 9 & 6 & 7 & 5 \\\hline 4 . & 9 & 8 & 7 & 6 & 8 & 10 & 7 \\\hline\end{array}$

(Diagram to be added)

Question 6

$\begin{aligned}&\text { Draw a double bar graph for the given information : }\\&\begin{array}{|l|c|c|c|c|c|c|}\hline \text { Month , } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } & \text {May } & \text { June } \\\hline \text { Actual Rainfall } & 3 \mathrm{~cm} & 5 \mathrm{~cm} & 6 \mathrm{~cm} & 7\mathrm{~cm} & 8 \mathrm{~cm} & 6 \mathrm{~cm} \\\text { Predicted Rainfall } & 2 \mathrm{~cm} & 3 \mathrm{~cm} & 3 \mathrm{~cm} & 6\mathrm{~cm} & 10 \mathrm{~cm} & 9 \mathrm{~cm} \\\hline\end{array}\end{aligned}$

(Diagram to be added)

Question 7

The table shows Arun and Sanjay's test scores. The tests were marked out of 20 .

$\begin{array}{|l|c|c|c|c|c|c|c|}\hline & \text { English } & \text { Maths } & \text { Science } & \text { History } & \text {Geography } & \text { Hindi } & \text { Art } \\\hline \text { Arun } & 20 & 7 & 12 & 14 & 16 & 18 & 11 \\\text { Sanjay } & 17 & 13 & 18 & 9 & 19 & 10 & 8 \\\hline\end{array}$

(a) Draw a dual bar chart to compare their results.

(b) If anyone scored less than 10 theN had to do the test again. Which tests did (i) Sanjay (ii) Arun have to do again ?

(Diagram to be added)

Question 8

Draw a double-bar graph to compare the data in the given table :

Enrollment in Computer Courses

$\begin{aligned}&\begin{array}{|l|c|c|c|c|}\hline \text { Year/Course } & \text { Basic } & \text { Logo } & \begin{array}{c}\text { Computer } \\\text { Science }\end{array} & \begin{array}{c}\text { Using } \\\text { Software }\end{array} \\\hline \text { Year One } & 130 & 40 & 10 & 20 \\\text { Year Two } & 260 r & 110 & 30 & 90 \\\hline\end{array}\\&\text { Which scale will you prefer to use ? }\end{aligned}$

(Diagram to be added)

Question 9

The performance of a student in 1st term and 2 nd term is given. Draw a double bar graph choosing appropriate scale and answer the following :

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Subject } & \text { English } & \text { Hindi } & \text { Maths } & \text { Science } &\text { S. Science } \\\hline \text { Ist term (M.M. 100) } & 67 & 72 & 88 & 81 & 73 \\\text { 2nd term (M.M. 100) } & 70^{\circ} & 65 & 95 & 85 & 75 \\\hline\end{array}$

(a) In which subject has the child improved his Performance the most, ?

(b) In which subject is the improvement the least?

(Diagram to be added)

Question 9

The following graph shows the points scored in five games of basketball. Study the graphs and answer the questions that follow.

(i) How many games did our team win ?

(ii) What is the highest score ? Who scored and in which game ?

(iii) By how many points did our team win the third game ?

(iv) What is the opponent's team highest score ?

(Diagram to be added)

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15C

Exercise 15 C

Question 1

Determine the mode for each of the following sets of data :

(a) 5,3,5,2,3,5,1,7

Ans: Mode 5

(b) 1,7,13,19,25

Ans: No mode

Ans: 4 and 6

(d) 51,68,70,76,76,55,70,73,68,70,81

Ans: 70

(e) 8,4,8,4,8,4,8,4,8,4,8,4

Ans: 4 and 8

(f) 29,29,10,10,3,3,14,10,2,2

Ans: 10

(g) 26,32,26,21,83,26,83,85

Ans: 26 and 83

Question 2

Find the mode of the following distributions:

(a)

$\begin{array}{|l|l|l|l|l|l|l|}\hline \text { Size } & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { fre } & 4 & 6 & 2 & 5 & 3 & 1 \\\hline\end{array}$

Ans: Mode :3

(b)

$\begin{array}{|l|l|l|l|l|l|}\hline \begin{array}{l}\text { Size of } \\\text { shoes }\end{array} & 6 & 7 & 8 & 9 & 10 \\\hline \begin{array}{l}\text { No of } \\\text { person }\end{array} & 12 & 20 & 40 & 15 & 7 \\\hline\end{array}$

Ans: Mode : 8

Question 3

Determine (a) the mode (b) the median , and (c) the mean for each of the following sets of data:

(i) $1,16,11,6,11,8,14,11,21$

Sol: mode = 11

1,6,8,11,11,14,16,21

Median= 11

Mean $\bar{x}=\frac{Sum of observation}{No of observation}$

$=\frac{1+6+8+11+11+11+14+16+21}{9}$= $\frac{99}{9}=11$

(ii) $3,10,10,12,14,16,16,18,18,25$

Mode = 10,16 and 18

Sol: 3,10,10,12,14,16,18,18,25

Median = $\frac{14+16}{2}$

=$\frac{30}{2}=15$

Mean $\bar{x}=\frac{3+10+10+12+14+16+16+18+18+25}{10}$

=$\frac{142}{10}$ = 14.2

(iii) $23,2,42,6,36,11,29,9,15$

Mode = No mode

2,6,9,11,15,23,29,36,42

Median = 15

Mean

$\begin{aligned} \bar{x} &=\frac{2+6,+9+11+15+23+29+36+42}{9} \\ &=\frac{173}{9}=19 \frac{2}{9} \end{aligned}$

(iv) $8,8,8,8,18,18,18,18$

Sol: Mode = 8 and 18

8,8,8,8,18,18,18,18

Median = $\frac{8+18}{2}$

$=\frac{26}{2}=13$

$\operatorname{mean} \vec{x} =\frac{8,8,8,8,18,18,18,18}{8}$

$=\frac{104}{8}=13$

Question 4

Find the mode, and the mean for the set of data below.

Number of fiction books read by 28 eight grade pupils:

$4,6,1,7,3,9,5,7,8,5,4,6,10,6,9,5,6,6,8,6,8,5,10,7,2,5,3,7$.

Sol: Mode = 6

Mean = $=\frac{4+6+1+7+3+9+5+7+8+5+4+6+10+6+9+5+6+6+8+6+8+5+10+7+2+5+3+7}{28}$

$=\frac{168}{28}$ = 6

Question 5

Find the mode and the mean of the following set showing number of hours of operating life of 25 flashlight batteries:

$20,21,19,22,18,23,25,22,23,20,23,20,22,21,24,21,22,23,19,21,22,22,24,26,22 .$

Sol: Mode = 22

Mean = $\frac{545}{25}=21.8$

Question 6

Tell whether the statement is true or false:

(i) The mode is always one of the numbers in a data.- True

(ii) The mean is one of the numbers in a data - False

(iii) The median is always one of the numbers in a data- False

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15B

Exercise 15 B

Question 1

The median of a set of numbers is the middle number when all the numbers are arranged in order of size, i.e., in descending or ascending order.

Find the median of the following.

(1) 2,3,5,7,9 = 5

(2) 4,8,12,16,20,24, 28,32

=16,20

median $=\frac{16+20}{2}=\frac{36}{2}=18$

(3) $60,33,63,61,44,48,51$

Sol: $33,44,48,51,60,61,63$

(4) $13,22,25,8,11,19,17,31,16,10$

Sol: $8,10,11,13,16,17,19,22,25,31$

median $=\frac{16+17}{2}=\frac{33}{2}=16.5$

Question 5

15 students secured the following marks in a test statistic . find the median marks

35,2,13,17,20,30,19,29,11,10,29,23,18,25,17

Sol: Arrange in Ascending order.

$10,11,13,17,17,18,19,20,23,25,28,29,29,30,35$

Question 6

The median of the number 85, 86,78,89 and 64 is

Sol: $64,78,85,86,89$

Question 7

The marks scored by 10 students are 5,9,8,7,2,3,4,9, 6 and 8 . The median marks are '

Sol: $2,3,4,5,6,7,8,8,9,9$

Median $=\frac{6+7}{2}=\frac{13}{2}=6.5$

Question 8

If the median of 46, 64,88, 40 , x ,76,35,91,56,32 and 91 is 58 . find the value of x

Sol: 32,35,40,46,56 , 64,7, 88, 91 ,91,

x= 58 answer

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15A

Exercise 15 A

Question 1

Find the mean of :

(a) First 5 natural number

Sol: $\bar{x}=\frac{1+2+3+4+5}{5}=\frac{15}{5}=3$

(b) First 7 whole numbers

Sol: $\bar{x}=\frac{0+1+2+3+4+5+6}{7}=\frac{21}{7}=3$

Sol: $\bar{x}=\frac{2+3+5+7}{4}=\frac{17}{4}=4.25$

(d) Rs7, Rs 19 , Rs 31, Rs 43, Rs 70

Sol: $\bar{x}=\frac{7+19+31+43+70}{5}=\frac{170}{5}=Rs 34$

Question 2

Seema obtained the following scores (Out of 100) $80,85,90,71,60,100$ . What is her mean score?

Sol: $\bar{x}=\frac{80+85+90+71+60+100}{6}=\frac{486}{6}=81$

Question 3

Shaleen's last six batting scores were 138, 144, 155 ,142, 167,172. What was his mean score?

Sol: $\bar{x}=\frac{138+144+155+142+167+172}{6}$

$\bar{x}=\frac{918}{6} =\bar{x}=153$

Question 4

Madhu worked $2 \frac{1}{2}$ hours on Monday, $3 \frac{1}{4}$ hrs. on Tuesday, and $2 \frac{3}{4}$ hrs. on Wednesday. What is the mean number of hours she worked on these three days?

Sol: Total number of hours worked = $2 \frac{1}{2}+3 \frac{1}{4}+2 \frac{3}{4}$

$=\frac{5}{2}+\frac{13}{4}+\frac{11}{4}$

$=\frac{10+13+11}{4}=\frac{34}{4}$

Mean $\bar{x}=\frac{\text { Sun of observation}}{\text { Number of observation }}$

$=\frac{\frac{34}{4}}{3}=\frac{34}{12}=\frac{17}{6}=2 \frac{5}{6}$ hours

Question 5

Ayushree sat for six tests and ananya sat for seven tests. Their percentage scores were:

$\begin{array}{|l|c|c|c|c|c|c|c|}\hline \text { Ayushree } & 68 & 75 & 70 & 45 & 57 & 77 & \\\hline \text { Ananya } & 52 & 87 & 64 & 531 & 74 & 81 & 86 \\\hline\end{array}$

Sol: Mean of ayushree = $\frac{68+75+70+45+57+77}{6}$

$=\frac{392}{6}=65.33$

Mean of ananya = $\frac{52+87+64+53+74+81+86}{7}$

$=\frac{497}{7}=71$

Ananya > Ayushree

Question 6

Find the mean of first ten odd natural numbers

Sol: $=\frac{1+3+5+7+9+11+13+15+17+19}{10}$

$=\frac{100}{10}=10$

Question 7

If the mean of 16, 14, x ,23 ,20 is 18 . find the value of x

Sol: $18=\frac{16+14+x+23+20}{5}$

$90=x+73 \Rightarrow x=90-73$

$x=17$ Ans

Question 8

If the mean of x, x+ 2, x+4 ,x+6 , x+8 is 24, find x

Sol: $24=\frac{x+x+2+x+4+x+6+x+8}{5}$

$120=5 x+20 \Rightarrow 120-20=5 x$

$\Rightarrow 100=5 x$

$\Rightarrow x=\frac{100}{5} \Rightarrow x=20$ answer

Question 9

Nisha secured 73,86,78 and 75 marks in four tests. What is the least number of marks she can secure in her next test, if she has to have a mean score of 80 marks in five test?

Sol: Let the number =x

$80=\frac{73+86+78+75+x}{5}$

$400=312+x=x=400-312$

$x=88$ marks

Question 10

The mean of 6 numbers is 24. If one number is excluded, the mean of remaining numbers becomes 22 find the excluded number .

Sol: Mean = $\frac{Sum of observation}{No. of Observation }$

$24=\frac{5.0}{6} \Rightarrow 5.0 \cdot=144$

22 = $\frac{Sum of obs of 5 No}{No. of obs}$

$22=\frac{5.0}{5}$

$110=5.0$

Excluded No. = 144-110= 34 answer

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14F

Exercise 14 F

Question 1

Find the area of each circle. In question 1 take $\pi=3 \frac{1}{7}$ or $\frac{22}{7}$ and in question 2 , take $\pi$ 3.14

(i) Area $=\pi r^{2}$

Sol:: $\frac{22}{7} \times 14^{2}$ $\frac{22}{7} \times 14 \times 14=616 \mathrm{~cm}^{2}$

(ii) Diameter = 56cm

Sol: Radius = 28cm

$\because$ Area $=\pi r^{2}=\frac{22}{7} \times 28 \times 28=2464 \mathrm{~cm}^{2}$

(iii) Diameter $=2 \frac{4}{5} \mathrm{~cm}=\frac{14}{5} \mathrm{~cm}$

Radius $=\frac{D}{2}=\frac{\frac{14}{5}}{2}=\frac{14}{5 \times 2}=\frac{14}{10} \mathrm{cm}$

Area $=\pi r^{2}=\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}$

$\frac{154}{25}$ or $6 \frac{4}{25} \mathrm{~m}^{2}$

(iv) $R=1 \frac{3}{4} \mathrm{~cm}=\frac{7}{4} \mathrm{~cm}$

Sol: Area = $\pi r^{2}=\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}=\frac{77}{8}=9, \frac{5}{8} \mathrm{cm}^{2}$

Question 2

(i) radius $=10 \mathrm{~cm}$ Take $=\pi=3 \cdot 14$

Sol: Area $=\pi r^{2}$

$=3.14 \times 10 \times 10=314\operatorname{cm}^{2}$

(ii) $D=12 \mathrm{~mm}$

$R=6 \mathrm{~mm}$

Sol: Area $=\pi r^{2}=3.14 \times 6 \times 6=113.04 \mathrm{mm}^{2}$

(iii) diameter $=30 \mathrm{~km}$

$R=15 \mathrm{~km}$

Sol: Area $=\pi r^{2}=3.14 \times 15 \times 15=706.5 \mathrm{~km}^{2}$

(iv) $R=0.5 \mathrm{~m}$

Sol: Area = $\pi r^{2}=3.14 \times 0.5 \times 0.5=0.785 \mathrm{~m}^{2}$

Question 3

Taking $\pi=\frac{22}{7}$, find the radii of the circles whose areas are

(i) $\frac{22}{7} \mathrm{~cm}^{2}$

Sol: Area $=\frac{22}{7} \mathrm{~cm}^{2}$

$\pi r^{2}=\frac{22}{7} \Rightarrow r^{2}=\frac{22}{7} \times \frac{7}{22} \quad \Rightarrow r^{2}=1$

r=1cm answer

(ii) Area $=154 \mathrm{~cm}^{2}$

Sol: $\pi r^{2}=154 \Rightarrow r^{2}=\frac{154}{22} \times 7 \Rightarrow r^{2}=7 \times 7$

r= 7cm answer

(iii) Area $=314 \frac{2}{7} \mathrm{~mm}^{2}$ $[\because 314 \times 7=2193+2=2200]$

Sol: $\pi \pi^{2}=\frac{2200}{7}$

$r^{2}=\frac{2200}{7} \times \frac{7}{22} \Rightarrow r^{2}=100$

$\mathrm{r}=10 \mathrm{~mm}$ Answer

Question 4

Taking $\pi=3.14$, find the diameters of the circles whose areas are

(i) $2826 \mathrm{~m}^{2}$

(ii) $7.065 \mathrm{~mm}^{2}$

(iii) $254.34 \mathrm{~cm}^{2}$

Sol: (i) Area $=2826 \mathrm{~m}^{2}$

$\pi r^{2}=2826 \Rightarrow r^{2}=\frac{2326}{3.14} \quad \Rightarrow r^{2}=900$

Taking square root r= 30 m Answer D= 60m answer

Sol:(ii) Area = 7.065 $\mathrm{mm}^{2}$

=$\pi r^{2}=7.065$

=$\pi^{2}=\frac{7.065}{3.14}$

$=r^{2}=2.25$

$r=1.5 \mathrm{~mm}$

$D=3.0 \mathrm{~mm}$

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14E

Exercise 14 E

Question 15

The circumference of the front wheel of a car is 10 dm and that of the hind wheel in 16dm. How may revolutions more will the first wheel make than the second is covering a distance of 96 km?

Sol: Circumference of front wheel =10dm

Circumference of back wheel = 16 dm

Distance to be cover= 96 km or 9,60,000 dm

More no . of revolution by front wheel = Rev. by front wheel - Rev by back wheel

$\left(\frac{960,000}{10}\right)-\left(\frac{960,000}{16}\right)$

$=96000-60,000$

=36000

Question 16

There is a circular pond and foot- path runs along its boundary. A man walks around it, exactly once, keeping close to the edge. If his step is 55cm long and he takes exactly 400 steps to go round the pond . what is the diameter of the pond ?

Sol: Circumference of the pond = $55 \times 400$ $=22,000 \mathrm{~cm}$ or 220 m

$\left[\begin{array}{c}C=2 \pi r \\ \text { or } \\ \pi d\end{array}\right]$

$\pi d=220$

$d=\frac{220 \times 7}{22} \quad \Rightarrow d=70 \mathrm{cm}$

Question 17

A piece of wire is bent in the shape of an equilateral triangle of each side 6.6cm. It is rebent to form a circular ring. what is the diameter of the ring?

[Hint. Perimeter of triangle $=$ Circumference of circular ring, i.e., $\pi \times d=3 \times 6.6]$

(image to be added )

Sol: Circumference of circular ring= Perimeter of triangle

$\pi d=3 \times 6.6$

$d=\frac{3 \times 6.6}{22} \times 7 \Rightarrow d=6.3$ cm

Question 18

A wire is wound to form a circle of radius 56 cm. If the same wire is converted into a square. Find the side of this square.

Sol: Perimeter of square = Circumference of circle

$4 \times$ side $=2 \times \frac{22}{7} \times 56$

Side of the square = $2 \times \frac{22}{7} \times \frac{56}{4}=88 \mathrm{~cm}$

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14D

Exercise 14 D

Question 9

A rectangular field is 24 m long and 15 m wide. How many triangular flower beds each of base 3m and altitude 4m can be laid in this field?

Sol: No of triangular flower = $\frac{Area of rectangular field}{Area of triangular flower}$

$=\frac{l \times B}{\frac{1}{2} \times \text { Base } \times \text { Height }}$

$=\frac{24 \times 15}{\frac{1}{2} \times 3 \times 4}=60$

Question 10

A right angled triangle has the largest side as 13 cm and one of the sides containing the right angle as 12cm. It's area in $\mathrm{Cm}^{2}$ is :

(IMAGE TO BE ADDED)

Sol:

$\begin{aligned} H^{2} &=B^{2}+p^{2} \\ \Rightarrow P^{2} &=H^{2}-B^{2} \\ &=13^{2}-12^{2}=169-144=25 \end{aligned}$

$\Rightarrow p^{2}=25^{\circ}$

$\because \quad p=5 \mathrm{~cm}$

Area of Triangle = $\frac{1}{2} \times$ Base $\times$ height $=\frac{1}{2} \times 12 \times 5$

= 30 $\mathrm{Cm}^{2}$

Question 11

The area of a right angled triangle is 40 times its base. what is its height ?

Sol: Area = $40 \times B$

Area of Triangle = $\frac{1}{2} \times$ Base $\times$ Height

Height $=\frac{2 \times \text { Area }}{\text { Base }}=\frac{2 \times 40 \times B}{B}$

= 80 cm

Question 12

In two triangles, the ratio of their areas is 4:3 and that of their height is 3:4 .find the ratio of their bases.

Sol: Ratio of areas of triangles $A_{1}: A_{2}$ is $4 x: 3 u$

Ratio of their height $H_{1}: H_{2}$ is $3 y: 4 y$

now area of triangle = $\frac{1}{2} \times$ Base $\times$ Height

Base $B_{1}=\frac{2 \times \text { Area of } \Delta\left(A_{1}\right)}{\text { Height }\left(H_{1}\right)}$

Basc B $2=\frac{2 \times \text { Area of } \triangle\left(A_{2}\right)}{\text { Height }\left(\mathrm{H}_{2}\right)}$

$\frac{B_{1}}{B_{2}}=\frac{2 \times \frac{A_{1}}{H_{1}}}{2 \times \frac{A_{2}}{H_{2}}}$

$\frac{4 x}{3 y} \times \frac{4 y}{3x}$

$\frac{B_{1}}{B_{2}}=\frac{16}{9}$

$B_{1}: B_{2}=16: 9$

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14C

Exercise 14 C

Question 1

Find the area of each parallelogram.

$\begin{array}{|l|l|l|l|l|l|}\hline & \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } \\\hline \text { Base } & 8 \mathrm{~cm} & 12 \mathrm{~mm} & 6.5 \mathrm{~m} & 1 \mathrm{~m}5 \cdot \mathrm{cm} & 4.2 \mathrm{dm} \\\hline \text { Height } & 3 \mathrm{~cm} & 8.7 \mathrm{~cm} & 4.8 \mathrm{~m} & 45\mathrm{~cm} & 25 \mathrm{~cm} \\\hline\end{array}$

(i) Area = $=B \times H$

$=8 \times 3$

$=24 \mathrm{~cm}^{2}$

(ii) Base $212 \mathrm{~mm}$

$1 c m=10 \mathrm{~mm}$

B= 1.2cm

Area = $=B \times H$

$=1.2 \times 3.7=10.44\mathrm{~cm}2$

(iii)

$\begin{aligned} A &=B \times H \\ &=6.5 \times 4.8 \\ &=31.20 \\ & m^{2} \end{aligned}$

(iv) $1 m=100 \mathrm{~cm}$

$B=105 \mathrm{~cm}$

Area $=105 \times 45^{-}$

$=4725 \mathrm{~cm}^{2}$ $=0.4725 \mathrm{~m}^{2}$

Question 2

(i) Area of ||gm ABCD = $48 \mathrm{~cm}^{2}$ DE = 6cm AB =?

(DIAGRAM TO BE ADDED)

Sol: Area $=B \times 4$

$\begin{aligned}&48=D E \times A B \\&A B=\frac{43}{6} \\&A B=8 \mathrm{~cm}\end{aligned}$

(ii)

Area of $\| \mathrm{gm}$ PQRS $=252 \mathrm{~cm}^{2}$ PQ = 9CM RT= ?

Sol:

$\begin{aligned} \text { Area } &=P Q \times RT \\ \Rightarrow 252 &=9 \times R T \\ \Rightarrow R T &=\frac{252}{9} \Rightarrow R T=28 \mathrm{~cm} \end{aligned}$

Question 3

The side of a rhombs is 7.2cm and its altitude is 5cm. Find its area.

[Hint. Since a rhombus is a parallelogram with all its sides equal, the formula for area of a ||gm is applicable to it also]

Sol: Area = Base $ \times $ Height (altitude)

$=7.2 \times 5$

$=36 \mathrm{~cm}^{2}$

Question 4

The adjacent sides of a parallelogram are 36 cm and 27cm in length . if the perpendicular distance between the shorter sides is 12 cm, find the distacne between the longer sides.

(IMAGE TO BE ADDED)

Sol: Given AE = 12CM

Area of parallelogram ABCD = BASE $\times$ height

$=D C \times A E$

$=27 \times 12 \mathrm{~cm}^{2}$

As we know Area of ||gm ABCD = $ BC \times AF $ = $DC \times AE $

$\Rightarrow A F=\frac{27 \times 12}{36} \Rightarrow A F=9 \mathrm{~cm}$,

Question 5

The area of a parallelogram and a square are the same. If the perimeter of the square is $160 \mathrm{~m}$ and the height of the parallelogram is $20 \mathrm{~m}$, find the length of the corresponding base of the parallelogram.

Sol: Perimeter of square = 160 m

$4 \times$ side $=160 \Rightarrow$ side $=\frac{160}{4}$ = Side = 40 m

Area of square = side $^{2}$ $=40^{2}=1600 \mathrm{~m}^{2}$

Area of parallelogram = Base $ \times $ Height $=1600$

$\Rightarrow$ Base $=\frac{1600}{20}$

Base $=80 \mathrm{~m}$

Question 6

The area of a rhombus is $42 \mathrm{~m}^{2}$. If its perimeter is $24 \mathrm{~m}$, find its altitude.

Sol: Perimeter = 24 m

$4 \times $ side $=24 \Rightarrow$ side $=6 \mathrm{~m}$

Area = Base $ \times $ Height= 42

Height = $\frac{42}{6}$

H= 7m answer

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14B

Exercise 14 B

Question 1

Find the area of the shaded portion in each case.

(i) (Image to be added)

Sol: Area of shaded portion = Area of outer square -Area of Inner square

$=8^{2}-4^{2}$

$=64-16$

$=48 m^{2}$

(ii) (Image to be added)

Sol: : Area of shaded portion = Area of outer square -Area of Inner square

$=(24 \times 20)-(22 \times 18)$

$=480-396$

$=84 \mathrm{~m}^{2}$

Question 2

A photograph of sides 35 cm by 22 cm is mounted into a frame of external dimension 45 cm by 30 cm . find the area of the border surrounding the photograph .

Sol: Area of border = External area of frame - Area of photograph

$=(45 \times 30)-(35 \times 22)$

$=1350-770$

$=580 \mathrm{~cm}^{2}$

Question 3

A Verandah 1.25 m wide is constructed all along the outside of a room 5.5m long and 4m wide. find the cost of cementing the floor of this verandah at the rate of Rs 15 per sqm.

Sol: (IMAGE TO BE ADDED)

To find the cost cementing the floor of verandah we have to find the area.

Outer width = $\begin{aligned} & 5.5+1.25+1.21 \\=& 8 m \end{aligned}$

Outer length = $4+1.25+1.25=6.5 \mathrm{~m}$

Area of verandah = Area of outer - Area of inner

$=(8 \times 6.5)-(5.5 \times 4)$

$=52-22$

$=30 \mathrm{~m}^{2}$

Cost of cementing floor = $15 \times 30$

= Rs 450 Answer

Question 4

A sheet of paper measuring 30 cm by 20 cm. A strip 4cm wide is cut from it all around . Find the area of remaining sheet and also the area of the strip cut out .

Sol: Width after cutting 4cm= 30-4+4

$=30-8=22 \mathrm{~cm}$

Length = 20 - 4+4= $20-8=12 \mathrm{~cm}$

Area of remaining sheet = L$ \times $ B

$=22 \times 12=264 \mathrm{~cm}^{2}$

Area of strip = $(30 \times 20)-(264)$

$600-264=336 \mathrm{~cm}^{2}$

Question 5

Find the area of the crossroads at right angles to each other through the center of the field .

Sol: (IMAGE TO BE ADDED)

Area of cross roads

=Area of ABCD+ EFGH -IJKL

$75 \times 2+62 \times 2-2 \times 2$

$=150+124-4$

$274-4=270 \mathrm{~m}^{2}$ Answer

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14A

Exercise 14 A

Question 1

Copy and complete the following table:

$\begin{array}{c|c|c|c|}\hline \text { Length } & \text { Breadth } & \text { Area } & \text { Perimeter } \\\hline \text { (P) } 5 \mathrm{~m} & (3 \mathrm{~m}) & 15 \mathrm{~m}^{2} & 116 \mathrm{~m})\\\text { (ii) } 5 \mathrm{~cm} & (7 \mathrm{~m}) & \left(35 \mathrm{~m}^{2}\right) & 24\mathrm{~cm} \\\text { (iii) }(15 \mathrm{~cm}) & 9.5 \mathrm{~cm} & 142.5 \mathrm{~cm}^{2} & 49\mathrm{~cm} \\\text { (iv) }(16 \mathrm{~cm}) & 30 \mathrm{~cm} & 480 \mathrm{~cm}^{2} & (92\mathrm{~cm}) \\\hline\end{array}$

Question 2

Find (a) the area (b) the perimeter of the square PQRS if.

(i) $P Q=Q R=20 \mathrm{~cm} \quad$ Area $=P Q \times Q R=20 \times 20=400 \mathrm{~cm}^{2}$

Perimeter $=4 \times$ side $=4 \times 20=80 \mathrm{~cm}$

(ii) $P Q=Q R=1.2 \mathrm{~m} \quad$ Area $=P Q \times Q R=1.2 \times 1.2=1.14 \mathrm{~m}^{2}$

Perimeter $=4 \times$ side $=4 \times 1.2=4.8 \mathrm{~m}$

Question 3

Find the perimeter of the following squares:

(i) Area = $=64 \mathrm{~cm}^{2}$

Sol:

$\begin{aligned} S^{2} &=64 \\ \text { Side } &=8 \mathrm{~cm} \end{aligned}$

Perimeter $=4 \times$ side

$=4 \times 8$

$=32 \mathrm{~cm}$

(ii) area $=196 \mathrm{~cm}^{2}$

Sol: $s^{2}=196$

Taking square root

Side = 14 cm

Perimeter = $4 \times$ side

$=4 \times 14$

$=56 \text { Cm}$

(iii) Area = $2.25 \mathrm{m}^{2}$

Sol: $s^{2}=$ 2.25

Taking square root side = 1.5m

P= $4 \times 1.5$

=6m

Question 4

(i) The perimeter of the square is 20cm . find its area .

Sol: Side of square = $\frac{1}{4} \times $ perimeter of square

$=\frac{1}{4} \times 20=5 \mathrm{~cm}$

Area of square = Side $\times$ side

$=5 \times 5=25$ $\operatorname{cm}^{2}$

(ii) The perimeter of a square field in 3km. Find its area in hectares.

Sol: Side = $\frac{1}{4} \times $ perimeter

$=\frac{1}{4} \times 8=2 \mathrm{~km}$ or 2000m

Area of square = $side \times side$

$=2,000 \times 2000=40,00,000 \mathrm{~m}^{2}$

1 hectare = 10000 $m^{2}$

$1 m^{2}=\frac{1}{10000}$ Hectare

$\frac{4000000}{10000}$= 400 hectare

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13C

Exercise 13 C

Question 1

(a) What will be the other angles of a right angles isosceles triangle?

Sol: $x+x+90=180^{\circ} \Rightarrow 2x=90 \Rightarrow x=45^{\circ} \quad 45^{\circ}$

(b) Can you draw an obtuse angles isosceles triangle ?

Ans: Yes

Question 2

(a) The vertical angle of an isosceles triangle is 110 degree. What must be the size of its base angles?

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Sol: $\Rightarrow \quad x+x+110^{\circ}=180^{\circ}$

$\Rightarrow \quad 2 x=180^{\circ}-110^{\circ} \Rightarrow \quad 2 x=70^{\circ}$

x = $35^{\circ}$ Ans

(b) What is the size of each exterior angle of an equilateral triangle ?

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Sol: $180^{\circ}-60^{\circ}=x$

$x=120^{\circ}$

Question 3

$\triangle A B C$ is isosceles with AB= AC, if $\angle A=80^{\circ}$ what is the measure of angle b?

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Sol: Let $\angle B=\angle C=x$

$\Rightarrow x+x+80^{\circ}=180^{\circ}$

$\Rightarrow 2 x=180^{\circ}-80^{\circ}$

$\Rightarrow 2 x=100^{\circ}$

$\Rightarrow x=50^{\circ}$

$\angle B=\angle L=50^{\circ}$

Question 4

In $\triangle B B C, \angle A=\angle B=50^{\circ}$. Name the pair of sides which are equal .

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Side AC = BC

Question 5

In the figure given below AN =AC, $\angle B A C=52^{\circ}$ $\angle A C K=84^{\circ}$ and BCK is a straight line . prove that NB =NC

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Sol: $\triangle A B C$

$\angle A+\angle B=84^{\circ}$

$52^{\circ}+\angle B=84^{\circ}$

$\begin{aligned} \angle B &=84^{\circ}-52^{\circ} \\ \Rightarrow \angle B &=32^{\circ} \end{aligned}$

$\triangle A N C$

$\Rightarrow \quad 52^{\circ}+x+x=180^{\circ}$

$\Rightarrow \quad 2 x=180^{\circ}-52^{\circ}$ $2x=123^{\circ}$

$\Rightarrow x=\frac{128}{2} \quad \Rightarrow$

x = $64^{\circ}$

∵ BCK Straight line

$\angle B C N+\angle A C N+\angle A C K=180^{\circ}$

$\angle B C N+64^{\circ}+84^{\circ}=180^{\circ}$

$\angle B C N=180^{\circ}-148$

$\angle B C N=32^{\circ}$

$\because \angle B C N=\angle B$

BN = NC Hence proved

Question 6

In the figure AB = AC. Prove that BD = BC

Sol: $\Rightarrow x+x+40^{\circ}=180$

$\Rightarrow 2 x=180^{\circ}-40^{\circ}$

$\Rightarrow 2 x=140^{\circ}$

$\Rightarrow x=\frac{140^{\circ}}{2}$

$\Rightarrow x=70^{\circ}$

By ext. angle prop

$\Rightarrow 30^{\circ}+40^{\circ}=y$

$\Rightarrow y=70^{\circ}$

$\because x=y=70^{\circ}$

$\therefore B D=B C$ Hence proved

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13B

Exercise 13 B

Question 1

AB and CD bisect each other at K. Prove that AC = BD .

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Sol: AK= BK

$\angle A K C=\angle B K D$ (Vertically opposite angle )

CK = DK

By SAS $\triangle A K C, \cong \triangle B K D$

C.P.C.T $\quad A C=B D$ Hence proved

Question 2

The sides BA and CA have been produced such that BA = AD and CA = AE prove that DE||BC

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Sol: $A B=A D$

$\angle B A C=\angle E A D$ (Vertically opposite angle )

by $S A S \quad \triangle A B C \cong \triangle A E D$

by C.P.CT $\angle B=\angle D \quad \because B C \| D E$ Hence proved .

Question 3

$O A=O B, O C=O D$, $\angle A O B=\angle C O D$ prove that AC = BD

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Sol:

$\begin{aligned} O A &=O B \\ \angle A O B &=\angle C O D \\ \angle A O C+\angle C O B=\angle C O B+\angle B O D \\ \angle A O C &=\angle B O D \\ O C &=O D \end{aligned}$

By SAS $\triangle A O C \cong \triangle B O D$

by C.P.CT $A C=B D$ hence proved

Question 4

If AB and MN bisect each other at O and AM and BN are $\perp$ on xy. Prove that $\triangle S$, OAM and OBN are congruent and hence prove that AM= BN

(IMAGE TO BE ADDED)

Sol: $\angle A M O=\angle B N O=90^{\circ}$

$O B=O A$

$O M=O N$

by RHS $\triangle A M O \cong \triangle B N O$

by $C \cdot P \cdot c \cdot T \quad A M=B N$

Question 5

$\angle XYZ$ is bisected by YP . L is any point on YP and MLN is Perpendicular to YP. Prove that LM = LN

(image to be added)

Sol: $\angle YLM=\angle YL N=90^{\circ}$

YL = YL (Common)

$\angle M YL=\angle N YL$ (Bisect by YP )

by ASA $\triangle MY L \cong \triangle N Y L$

By C.P.C.T LM = LN Hence proved

Question 6

In the fig, PM =PN , PM $\perp A B$ AND $P N \perp A C$ Prove $\triangle A M P \cong \triangle A N P$

Sol: (image to be added)

$\angle A N P=\angle A M P=90^{\circ}$

$A P=A P \quad($ Comimon $)$

$M P=N P \quad$ (given)

By R.H.S $\triangle A M P \cong \triangle A N P$ Hence proved

Question 7

In the figure , AD = BC and AD ||BC. Prove that AB =DC

Sol: (IMAGE TO BE ADDED)

AD=BC

$\angle C A D=\angle A C B$ [Alt. angles]

AC = AC

By ASA $\triangle A C D \cong \triangle A B C$

by C.P.CT. $A B=D C$ Hence proved

Question 8

In the given figure, triangles ABC and DCB are right angles at A and D respectively and AC = DB . Prove that $\triangle A B C \cong \triangle D C B$

(IMAGE TO BE ADDED)

Sol: $\angle B A C=\angle B D C=90^{\circ}$

$B C=B C=$ Common

$A C=D B$

By RHS $\triangle A B C \cong \triangle D C B$ Hence proved

Exercise 7 A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Exercise 6 G

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Exercise 18 C

Exercise 18B

Exercise 18 A

Exercise 17 A

Exercise 16 B

Exercise 16 A

Exercise 15 D

Exercise 15 C

Exercise 15 B

Exercise 15 A

Exercise 14 F

Exercise 14 E

Exercise 14 D

Exercise 14 C

Exercise 14 B

Exercise 14 A

Exercise 13 C

Exercise 13 B

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