## Kerala Syllabus 8th Standard Textbooks Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 8th Standard Textbooks Solutions Guide Pdf Free Download all Subjects in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Class 8 Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 8th Standard Textbooks Solutions Pdf Part 1 and Part 2.

Students can download SCERT Kerala Textbooks for Class 8 English Malayalam Medium

## Kerala State Syllabus 8th Standard Textbooks Solutions

We hope the given Kerala Syllabus 8th Standard Textbooks Solutions Guide Pdf Free Download all Subjects in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 8th Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

## Kerala Syllabus 8th Standard Hindi Solutions Guide

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## Kerala State Syllabus 8th Standard Hindi Textbooks Solutions

Kerala Syllabus 8th Standard Hindi Guide

Kerala State Syllabus 8th Standard Hindi Textbooks Solutions Part 1

इकाई 1

इकाई 2

इकाई 3

Kerala State Syllabus 8th Standard Hindi Textbooks Solutions Part 2

इकाई 4

इकाई 5

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## Kerala Syllabus 8th Standard English Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 8th Standard English Solutions Guide Pdf Free Download of Textbook Questions and Answers, Chapter Wise Notes, English Textbook Activity Answers, Chapters Summary in Malayalam, English Study Material, Kerala Reader English Book Answers, English Teachers Hand Book are part of Kerala Syllabus 8th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 8th Standard English Textbooks Solutions Pdf of Kerala Class 8 Part 1 and 2.

## Kerala State Syllabus 8th Standard English Textbooks Solutions

Kerala Syllabus 8th Standard English Guide

Unit 1 Hues and Views

Unit 2 Wings and Wheels

Unit 3 Seeds and Deeds

Unit 4 Flowers and Showers

Unit 5 Share and Care

Learn Grammar concepts easily by using our English Grammar Notes website as it is free and easy to download and also we have provided you with the connected rules.

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## Kerala Syllabus 8th Standard Social Science Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 8th Standard Social Science Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 8th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 8th Standard Social Science Textbooks Solutions Pdf of Kerala Class 8 Part 1 and 2.

## Kerala State Syllabus 8th Standard Social Science Textbooks Solutions

Kerala Syllabus 8th Standard Social Science Guide

Kerala State Syllabus 8th Standard Social Science Textbooks Solutions Part 1

• Chapter 1 Early Human Life
• Chapter 2 The River Valley Civilizations
• Chapter 3 In Search of Earth’s Secrets
• Chapter 4 Our Government
• Chapter 5 Ancient Tamilakam
• Chapter 7 Economic Thought

Kerala State Syllabus 8th Standard Social Science Textbooks Solutions Part 2

• Chapter 8 Towards The Gangetic Plain
• Chapter 9 From Magadha To Thaneswar
• Chapter 10 Blanket of the Earth
• Chapter 11 Economic Planning In India
• Chapter 12 Water on Earth
• Chapter 13 Social Groups and Social Control

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## Kerala Syllabus 8th Standard Basic Science Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 8th Standard Basic Science Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 8th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 8th Standard Basic Science Textbooks Solutions Pdf of Kerala Class 8 Part 1 and 2.

## Kerala State Syllabus 8th Standard Basic Science Textbooks Solutions in English Medium

Kerala Syllabus 8th Standard Basic Science Guide

Kerala State Syllabus 8th Standard Basic Science Textbooks Solutions Part 1

Kerala State Syllabus 8th Standard Basic Science Textbooks Solutions Part 2

## Kerala State Syllabus 8th Standard Basic Science Textbooks Solutions in Malayalam Medium

Kerala Syllabus 8th Standard Basic Science Guide Malayalam Medium

Kerala State Syllabus 8th Standard Basic Science Textbooks Solutions Part 1 Malayalam Medium

Kerala State Syllabus 8th Standard Basic Science Textbooks Solutions Part 2 Malayalam Medium

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## Kerala Syllabus 8th Standard Maths Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 8th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 8th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 8th Standard Maths Textbooks Solutions Pdf of Kerala Class 8 Part 1 and 2.

## Kerala State Syllabus 8th Standard Maths Textbooks Solutions in English Medium

Kerala Syllabus 8th Standard Maths Guide

Kerala State Syllabus 8th Standard Maths Textbooks Solutions Part 1

Kerala State Syllabus 8th Standard Maths Textbooks Solutions Part 2

## Kerala State Syllabus 8th Standard Maths Textbooks Solutions in Malayalam Medium

Kerala Syllabus 8th Standard Maths Guide Malayalam Medium

Kerala State Syllabus 8th Standard Maths Textbooks Solutions Part 1 Malayalam Medium

Kerala State Syllabus 8th Standard Maths Textbooks Solutions Part 2 Malayalam Medium

We hope the given Kerala Syllabus 8th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus 8th Standard Maths Textbooks Answers Guide Pdf of Kerala Class 8 Part 1 and 2, drop a comment below and we will get back to you at the earliest.

## Kerala State Syllabus 8th Standard Maths Solutions Chapter 1 Equal Triangles

### Equal Triangles Text Book Questions and Answers

Textbook Page No. 11

Question 1.
In each pair of triangles below, find all pair of matching angles and write them down.

Solution:
(i) ∠A = ∠R (The angles opposite to 5cm sides)
∠B = ∠P (The angles opposite to the sides of length 4 cm)
∠C = ∠Q (The angles opposite to the sides of length 6 cm)

(ii) ∠L = ∠Y (The angles opposite to the sides of length 10cm)
∠M = ∠Z (The angles opposite to the side of length 4 cm)
∠N = ∠X (The angles opposite to the side of length 8cm)

Question 2.
In the triangles below AB = QR, BC = RP, CA = PQ

Compute ∠C of ∆ABC and all angle of ∆PQR.
Solution: C = 80° (Use the property that the sum of three angles of a triangle is 180°)
AB = QR
∴ ∠C = ∠P
∠C = 80°
∴ ∠P = 80°

BC = RP
∴ ∠A = ∠Q
∠A = 40°
∴ ∠Q = 40°

CA = PQ
∴ ∠B = ∠R
∠B = 60°
∴ ∠R = 60°
(The angle opposite to equal sides are equal)

Question 3.
In the triangle below.
AB = QR BC = PQ CA = RP

Compute the remaining angles of both the triangles.
Solution:
AB = QR ∴ ∠C = ∠P
BC = PQ ∴ ∠A = ∠R
CA = RP ∴ ∠B= ∠Q
∠A = 60° ∴ ∠R = 6o°
∠Q = 70° ∴ ∠B = 70°
∠A = 60, ∠B = 70° then ∠C = 180 – (60° + 70°)
∴ ∠C = 50° ∴ ∠P = 50°

Question 4.

Are the angles of ∆ABC and ∆ABDequal in the figure above? Why?
Solution:
The side AB is common to both the triangles in the figure.
The side of ∆ABC are equal to the sides of ∆ABD. So the angles of ∆ABC are equal to the sides of ∆ABC.

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Question 5.
In the quadrilateral ABCD shown below, AB = AD, BC = CD

Compute all the angles of the quadrilateral?
Solution:
AB = AD, BC = CD
AC is the common side
The sides of the triangles ABC and ADC are equal. So their angles are also equal. AB = DC
∴ ∠ACD = ∠ACB = 50° (Angles opposite to equal sides of a triangle are equal)
BC = CD
∴ ∠BAC = ∠DAC = 30° (Angles opposite to equal sides are equal in a triangle )
∴ ∠D = ∠B = 100°

Textbook Page No. 15

Question 1.
In each pair of triangles below find the pairs of matching angles and write them down.

Solution:
(i) BC = PR (If two sides of a triangle and the angle made by them are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal.)
∴ ∠B = ∠R
∴ ∠C = ∠P
∴ ∠A = ∠Q (The opposite angles of equal sides of two triangles are also equal)

(ii) MN = XY (If two sides of triangle and the angle made by then are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal)
∴ ∠L = ∠Z
∠M = ∠Y
∠N = ∠X (If two sides of a triangle are equal, the angles opposite to these sides are also equal)

Question 2.
In the figure below, AC and BE are parallel lines.

(i) Are the lengths of BC and DE equal. Why?
(ii) Are BC and DE parallel? Why?
Solution:
(i) Given AC and BE are parallel lines.
∴ ∠CAB = ∠EBD
When we consider the triangles ∆CAB, ∆EBD (Corresponding angles)
BC = DE (The two sides of ∆ CAB and the angle made by them are equal to the two sides of ∆ EBD and the angle made by them. So the thirif side of triangle are also equal.)

(ii) Yes, they are parallel.
∠ABC = ∠BDE (The angles opposite to the equal sides of equal triangles are equal) But they are corresponding angles. BC and DE are parallel.

Question 3.
Is ABCD in the figure, a parallelo¬gram? Why?

Solution:
AC = BD
AB is the common side.
The angles between the sides AC, AB and BD, AB are equal.
The opposite sides of quadrilateral ∆CBD are equal. The angles opposite to equal sides of triangles ∆ABC and ∆ABD are equal. So the opposite angles in quadrilateral ACBD are also equal.
∴ ACBD is a parallelogram.

Question 4.
In the figure below, M is the midpoint of the line AB. Compute the other two angles of ∆ABC

Solution:
AM = BM (Given M is the mid point of AB)
CM = CM (common)
∠AMC = 90° = ∠BMC
∴ The two sides in ∆AMC and ∆BMC and the angle made by them are equal.
So the third side and other angles are equal.
∠A = 50° ∴ ∠B = 50°
∠ACM = 40° ∴ ∠BCM = 40°
∴ ∠C = 80°

Question 5.
In the figure below, the lines AB and CD are parallel and M is the mid point of AB.

(i) Compute the angle of ∆AMD, ∆MBC and ∆DCM?
Solution:
Given AB = 12 cm and M is the mid-point of AB.
∴ AM = MB = 6 cm
AM = CD
AB||CD ∴ AM||CD
∴ AMCD is a parallelogram.
∴ ∠AMD = ∠CDM (Alternate interior angles)
∠ADM = ∠CMD (Alternate interior angles)
∠A = ∠DCM = 40° = ∠CMB
∴ ∠MCB = 80° [180 – (60 + 40)]
(i) The angles of ∆AMD, ∆MBC and ∆DCM are 40°, 60° and 80° respectively.
(ii) Both of them are parallelograms.

Textbook Page No. 21

Question 1.
In each pair of triangles below, find matching pairs of sides and write their names.

Solution:
(i) BC = PQ
AC = PR (The two angles and the side in between them in ∆ ABC are equal to the two angles and the side in between them in ∆PQR. So the third angles of the triangles ∠C and ∠R are also equal. Also BC and PQ, opposite to the 50° angle are also equal. The sides AC and PR opposite to the 70° angle are also equal.

(ii) ∠N = 70°
∠Z= 80°
MN = XZ
∠M = ∠Z(Sides opposite to equal angles are also equal)

Question 2.
In the figure, AP and BQ equal and parallel are lines drawn at the ends of the line AB. The point of inter section of PQ and AB is marked as M.

(i) Are the sides of ∆AMP equal to the sides of ∆BMQ? Why?
(ii) What is special about the position of M on AB.
(iii) Draw a line 5.5 cm long. Using a set square, locate the midpoint of this line.
Solution:
(i) Yes, they are equal
∠P = ∠Q
∠A = ∠B (alternate angles formed by cutting the parallel lines AP and QB by PQ and AB.)
AP = QB
∴ The third angle of ∆APM and ∆BMQ and opposite sides of equal angles are equal.
(ii) AM = BM. So M is the midpoint of AB.
(iii) Draw a line segment of length 5.5 cm. Draw perpendiculars of equal lengths upward at one end of the line and downwards at the other end. Join the ends. This line divides the first one equally.

Question 3.
In the figure, ABCDE is a pentagon with all sides of the same length and all angles of the same size. The sides AB and AE extended, meet the side CD extended at Px and Q.

(i) Are the sides of ∆BPC equal to the sides of ∆EQD? Why?
(ii) Are the sides of AP and AQ of ∆ APQ equal? Why?
Solution:
(i) Yes, this are equal the sides and angles of a pentagon are equal.
∴ BC = DE
∠PBC = ∠PCB (Exterior angles of a regular pentagon)
∠QDE = ∠QED (Exterior angles of a regular pentagon)
∆QDE = ∆QED (If two angles and side of one triangle are equal are equal to two angles and corresponding side of the other triangle then their sides are equal.
BP = EQ and PC = DQ

(ii) AB = AE sides of regular pentagon.
BP = EQ
∴ AP = AQ [AB + BP = AC + EQ]

Question 4.
In ∆ABC and ∆PQR shown below.
AB = QR BC = RP CA = PQ

(i) Are CD and PS equal? Why?
(ii) What is the relation between the areas of ∆ABC and ∆PQR?
Solution:
(i) AB = QR
BC = RP
∠A = ∠Q
∴ ∆ABC and ∆QRP are equal triangles. Given all sides of ∆ABC are equal to the sides of ∆QRP
∴ CD and PS are equal. (Opposite sides of equal angles

(ii) AB = QR and CD = PS
⇒ 1/2 AB × CD = 1/2QR × PS
∴ The areas of ∆ABC and ∆PQR are equal.

Question 5.
In the quadrilateral ABCD shown below the sides AB and CD are parallel. M is midpoint of the side BC. The lines DM and AB extended meet at N.

(i) Are the areas of ∆DCM and ∆BMN equal? Why?
(ii) What is the relation between the areas of the quadrilateral ABCD and the triangle ADN.
Solution:
(i) M is the midpoint of the line BC.
∴ CM = MB; BN || DC
∴ ∠DCM = ∠NBM (Alternate angles)
∠DMC = ∠NMB (Vertically opposite angles)
∴ ∆DCM and ∆BMN are equal triangles. So their areas are equal.

(ii) The areas of ∆DCM and ∆BMN are equal and quadrilateral AB, MD common
∴ The area of the quadrilateral
ABCD and the area of ∆ADN are equal.

Question 6.
Are the two diagonals of a rectangle equal. Why?
Solution:
ABCD is a rectangle.
Consider the ∆ABD and ∆ABC
AB = AB, common AD = BC (opposite sides of rectangle); ∠A = ∠B =90°

Both sides of ∆ABD and ∆ABC and the angle formed by them are equal. So the third sides BD and AC are equal. So the diagonals of the rectangle are equal.

Textbook Page No. 26, 27

Question 1.
Some are equal isosceles triangles are drawn below, In each, one angle is given. Find the other angles.

Solution:
(i) 30°, 75°, 75°
(ii) 40°, 70°, 70°
(iii) 20°, 8o°, 8o°
(iv) 100°, 40°, 40°

Question 2.
One angle of an isosceles triangle is 90°. What are the other two angles?
Solution:
The other two angles are equal. So they are 45°, 45°

Question 3.
One angle of an isosceles triangle is 6o°. What are the other two angles.
Solution:
The other two angles are equal. So they are 60°, 60°

Question 4.
In the figure below, O is the centre of the circle and A, B are points on the circle.
Compute ∠A and ∠B?

Solution:
OA = OB (radius of circle)
∆AOB is a isosceles triangle; ∴ ∠A = ∠B
∠O = 60°
∴ ∠A = ∠B = 60°

Question 5.
In the figure below, O is the centre of the circle and A, B, C are points on the circle.

What are the angles of ∆ABC?
Solution:
∆AOB, ∆AOC, ∆BOC are isosceles triangles. Each triangles are with angles 120°, 30° and 30°.
∴ ∠A = ∠B = ∠C = 30° + 30° = 60°

Text Book Page No. 29

Question 1.
Draw a line of 6.5 centimetres long and draw its perpendicular bisector.
Solution:
Draw a line segment AB of length 6.5 c.m with A and B as centres draw arcs on both sides of AB with equal radii. The radius of each of these arcs must be more the half the length of AB. Let these arcs cut each other at points C and D. Join CD which cuts AB at M
Then AM = BM. Also ∠AMC = 90°
Hence, the line segment CD is the perpendicular bisector of AB as it bisects AB at M and is also perpendicular to AB.

Question 2.
Draw a square, each side 3.75 centimetres long?
Solution:
Draw AB = 3. 75 cm at. A Construct ∠PAB = 90° from AP, cut AD = 3.75 cm
Taking D as centre, draw an arc of radius 3.75 cm and taking B as centre, draw one more are of radius 3.75 cm.
Let the two arcs intersect at point C. Join BC and DC.
Then ABCD is the required square.

Question 3.
Draw an angle of 750 and draw its bisector?
Solution:
Draw a line segment AB of any suitable length with A as centre. Draw an arc of any size to cut AB at D. With D as centre. Draw another arcs of some size to cut the previous arc at C.
Now ∠CAD = 60°. Draw ∠EAB = 90° and bisect ∠EAC.
∴∠PAC = 150 ∠DAC + ∠CAP = 60 + 15 = 75°
∴ ∠BAP = 75°
Then bisect
∠BAP AQ to the bisectors of ∠PAB

Question 4.
Draw a circle of radius 2.25 centimetres.
Solution:
Draw a line of length 4.5 cm. Draw its perpendicular bisector it meet at point ‘O’.
‘O’ is the centre of the circle and radius = 2.25 cm. Then complete the circle.

Question 5.
Draw ∆ ABC, with AB = 6 cm,
∠A = 22$$\frac{1}{2}$$°, ∠B = 67$$\frac{1}{2}$$°
Solution:
Draw the line AB in 6 cm length. Draw angle A at 45° and draw its bisector. Draw angle 135° at B and draw its bisector. Mark the point as C where bi-sectors meet.

Question 6.
Draw a triangle and perpendicular bisectors of all three sides. Do all these three bisectors intersect at the same point?
Solution:
Draw a triangle ABC. By using compass mark the arcs on both sides from each ends.
Draw the same for all sides. They intersect at same point

Question 7.
Draw a triangle and the bisectors of the three angles. Do all three bisectors intersect at the same point.
Solution:
Draw a triangle PQR and by using com-pass draw the bisectors of angles. All three bisectors meet at the same point.

Question 8.
Prove that if both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram.
Solution:
When the diagonal of a quadrilateral with equal opposite sides is drawn, we get two equal triangles. The angels opposite to the diagonal in the triangles are equal. That is the opposite sides and angles in the quadrilateral are equal. So the quadrilateral is a parallelogram.

Question 9.
In the figure, ABCD is parallelogram and AP = CQ

Prove that PBQD is a parallelogram
Solution:
DC = AB ………. (1)
QC = AP ……. (2) (as ABCD is a parallelogram)
(1) – (2) ⇒ DC – QC = AB – AP; ∴ DQ = PB
When, ∆ APD, ∆ CQB are considered.
AP = QC
∠A = ∠C, The two sides and angle formed by them in these triangles are equal. So the third sides PD and BQ are equal.
∴ Two pairs of opposite sides in the quadrilateral PBQD are equal. So PBQD is a parallelogram.

Question 10.
Prove that if all sides of a parallelogram are equal, them each diagonal is the perpendicular bisector of the other.
Solution:

The diagonal BD divides the parallelogram into two isosceles triangles. [The angles opposite to the equal sides in an isosceles triangles are equal.] So the diagonal DB bisect ∠D and ∠B.
Similarly the diagonal AC bisect A and ∠C. 4x +4y = 360° ⇒ x + y = 90°
The four triangles formed by intersec¬ting the diagonals are equal triangles. Each one 90 angle. So each diagonal is the perpendicular bisector of the other. In ∆AMD ∠AMD = 180 – (x – y) = 180 – 90 = 90° ⇒ BD ⊥ AC

Question 11.
In the figure below O is the centre of the circle and AB is the diameter. C is the point on the circle.

(i) Compute ∠CAB
(ii) Draw another figure like this with a different number for the size of ∠COB. Calculate ∠CAB
Solution:
(i) ∠BOC = 50°
∴ ∠COA = 180° – 50°= 130° (straight angle)
∴ AOC is an isosceles triangle.
∴ ∠A = ∠C = $$\frac{180-130}{2}$$ = 25°

(ii) ∠O = 70°
∴ ∠COA = 180 – 30 = 150°
OA = OC
∴ ∠CAB = ∠ACO =$$\frac{180-150}{2}=\frac{30}{2}$$ = 15°

Question 12.
In the figure below, O is the centre of the circle and AB is a diameter. C is a point on the circle.

(i) Compute ∠ACB
(ii) Draw another figure like this, changing the size of ∠COB and calculate ∠ACB.
Solution:
(i) ∠BOC = 50°
∠AOC = 130°
∆ AOC and ∆ BOC are isosceles triangles.
∠OAC = ∠OCA = $$\frac{180-130}{2}$$ = 25°
∠OBC = ∠OCR = $$\frac{180-50}{2}$$ = 65°
∠ACB = ∠OCA+ ∠OCB
25° + 65° = 90°

(ii) ∠AOC = 180 – 80 = 100°
∠OAC + ∠OCA = 180 – 100 = 8o°
∴ ∠OAC = ∠OCA = 80 ÷ 2 = 40°
∆ OBC
∠OBC + ∠OCB = 180 – 80 = 100°
∠OBC =∠OCB = 100 ÷ 2 = 50°
∠ACB = ∠OCA + ∠OCB
40° + 50° = 90°

Question 13.
How many different isosceles triangles be drawn with one angle 50° and any one side 7 centimetres.
Solution:
An isosceles triangle can be drawn with one angle 50° as angles either 50°, 50°, 8o° or 50°, 65°, 65°. In both the cases, 7 cm can be taken as equal sides or can be without 7 cm one as side. So there can be 4 ways of drawing diagram.

Question 14.
Draw ∆ ABC with AB = 7 cm, ∠A = 67$$\frac{1}{2}$$, ∠B = 15° without using protector.
Solution:
Draw AB with length 7 cm. Extend both the sides. Draw the perpendicular from A. Draw the bisector through the left 90° angle among the 90° angles obtained. Draw an angle as 90° + 45° = 135°, Draw its bisector.
Now ∠A = 67$$\frac{1}{2}$$. Draw an angle 60° in B to construct an equilateral triangle. Draw the bisector of its bisector. Then ∠B = 15°. We get ∆ ABC.

Question 1.

O is the centre of the circle in the diagram. If AB = BC,
(a) Then prove that ∠AOB = ∠BOC
(b) If OA = AB = BC, then find the values of ∠AOB and ∠BOC?
(c) Find out how many equilate¬ral triangles can be drawn in a circle with length of its side is radius.
Solution:
(a) OA = OB = OC, AB = BC
∆ OAB and ∆ OBC are equal triangles.
∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC.
(b) If OA = AB then ∆ OAB is an equilateral triangle.
If OB = BC, ∆ OBC is equilateral triangle.
∴ ∠AOB = ∠BOC = 60°
(c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.

Question 2.

If AB = AD, BC = CD in the diagram, then prove that ∠ABC = ∠ADC
Solution:
The three sides of triangles ∆ ABC are equal. The angles opposite to the sides are also equal.
AB = AD, BC = DC, AC = AC
AC is the common side. So the angles opposite to this side ∠ADC and ∠ABC are also equal,

Question 3.
Draw a rhombus with sides and a diagonal as 5 cm.
Solution:
Draw a line of length 5 cm. Draw equilateral triangles on both ends of the line with length 5 cm and line as one side.

Question 4.

In the figure, AB = DE, BC = EF and AC = DF. Can ∠BPD = ∠C? Prove it?
Solution:
The sides of ∆ DEF and ∆ ABC are equal. The angles opposite to equal sides are equal.
∠E = ∠B, ∠F = ∠C, ∠D = ∠A
But ∠C = 180 – (∠B + ∠A)
∠P = 180 – (∠B + ∠D)
∴ ∠D = ∠A
∴ ∠C = ∠P = 180 – (∠B + ∠A)
∴ ∠C = ∠P

Question 5.

In the figure, AB = PQ, AC = PR, BC = QR. PQ is parallel to AB.
(a) Then show that BC is parallel to QR.
(b) Also show that PR is parallel to AC
Solution:

AB = PQ, AC = PR, BC = QR
∴ The angles of ∆ ABC and ∆ PQR are equal.
∠A = ∠P, ∠B =∠Q, ∠C = ∠R
(a) AB||PQ, ∴ ∠B = ∠PMN (corresponding angle);
∴ ∠PMN = ∠Q
∴ MN||QR
∴ BC||QR

(b) BC||QR
∠R = ∠MNP = ∠C
∴ NP||AC, PR||AC

Question 6.
Diagonals of three parallelograms with equal areas are given. Draw the parallelograms.
(i) length of diagonal 7 cm
(ii) length of diagonal 6 cm
(iii) length of diagonal 5 cm.
Solution:
(i) Draw a line of length 7 cm. Draw triangles of sides 7 cm, 6 cm, 5 cm at both the ends of the line to get a parallelogram by joining both the triangles.
(ii) Draw a line of length 6 cm and follow the above method.
(iii) Draw a line of length 5 cm and follow the above method.

Question 7.

O is the centre of circle and AC, BD are diameters in the figure. Prove that AB = CD
Solution:
Consider ∆ODC and ∆OAB.
OD = OC = OA = OB (radi ∠AOB = ∠DOC; Two triangles are equal So the third sides of the triangles AB and CD are equal.)

Question 8.

ABCD in the figure is a parallelogram. P, Q, R and S are the mid points of the sides of the parallelogram. The prove that PQ = RS, and QR = PS.
Solution:
Consider the triangles ∆APS and ∆CRQ
AP = CR, (half of the equal lines AB and CD)
AS = CQ (half of the lines with equal lengths AD and BC.)
∠A = ∠C (opposite angles of the parallelogram are equal)
When two sides and the angle made by them, in a triangle are equal then the third sides are also equal.
∴ QR = PS
Similarly if ∆ DSR and ∆ BQP are considered, PQ = RS is obtained.

Question 9.

P is the midpoint of the sides AB and DF in the figure.
(a) Prove that BD = AF
(b) Is EF parallel to BC? Why?
(c) If D is the midpoint of BC, A is the midpoint of EF and Q is the mid point of DE then can Q be the midpoint of AC? Why?
Solution:
(a) Consider ∆APF and ∆DBP
FP = DP and AP = PB
∠APF = ∠DPB. The sides and the angle made by them in the triangles are equal. So the third sides BD and AF are equal.

(b) FP = PD. So the angles opposite to them are also equal.
∠FAP and ∠DBP are equal.
∴ FA||BD andBC|| EF.

(c) Consider BD = AF, in Question (a)
∴ BC = EF, Consider ∆ AEQ and ∆ DCQ
AE = DC, QE = DQ
∠AEQ = ∠CDQ
Two sides of a triangle and the angle made by them are equal. So the third sides are also equal, ie AQ = QC.
∴ So Q is the midpoint of AC

Question 10.
Draw a parallelogram if one of its diagonal is 8 cm length and one side is 6 cm. and the angle formed by the side and the diagonal is 40.
Solution:
Draw a diagonal of length 8 cm. Draw a line of 6 cm with 40 angle at its one end. Draw the same in its opposite direction.

Question 11.
One angle of an isosceles triangle is 80. Find the other possible angles of the triangle.
Solution:
8o°, 8o°, 20°
8o°, 50°, 50°

Question 12.

O is the centre of the circle in the diagram. Radiaus is 3 cm and ∠AOB = 60°. Find the perimeter of ∆ AOB?
Solution:
∆ OAB is an isosceles triangle.
∴ ∠A =∠B
∠O = 60
∴ ∠OAB is an equilateral triangle so perimeter = 3 + 3 + 3 = 9 cm.

Question 13.

OM is perpendicular to AB in the diagram. Prove that M is the mid point of AB.
Solution:
OA = OB
∴ ∆ OAB is an isosceles triangle.
∴ ∠A = ∠B
When ∆ OMA and ∆ OMB are considered, OM is the common side
∠AMO = ∠BMO = 90 ∠AOM = ∠BOM
One side of the triangle and angles at the ends of sides are equal. So the other two sides are also equal.
∴ AM = MB
∴ M is the midpoint of AB.

Question 14.

In the figure ∠ABC = ∠ADC and AB = AD. Prove that A BCD is an isosceles triangle?
Solution:
∴ ∆ ABD is an isosceles triangle.
It is given that ∠ABC = ∠ADC
∴ ∠CBD = ∠CDB
∴ CD = CB
∴ ∆ BCD is an isosceles triangle.

Question 15.
In ∆ ABC, AB = AC = 10 cm. M is the midpoint of BC. If BC = 12 cm, Find AM? Also find the area of ∆ ABC?
Solution:

Question 16.
Show that, the triangle obtained by joining the mid points of the sides of an isosceles triangle is also an isosceles triangle.
Can we get an equilateral triangle by joining the mid points of the sides of an equilateral triangle?
Solution:

∆ ABC is an isosceles triangle. P, Q, R are the mid points of the sides of the triangle. Consider ∆ PBR and ∆ QRC.
PB = QC
BR = CR
∠B = ∠C
Two sides and the angle made by them are equal. The third sides PR and QR also equal.
∴ ∆ PQR is an isosceles triangle. Similarly the triangle obtained by joining the mid points of the sides of an equilateral triangle is an equilateral triangle.

Question 17.

In the figure ∠B = ∠C = 90. If AB = CE and BE = CD, then find angles of ∆AED?
Solution:
If ∆ ABE and ∆ ECD are considered, AB = EC, BF = DC and ∠B = ∠C. Two sides of the triangle and the angle made by them are equal. The third sides AE and DE are also equal. ∆ADE is an isosceles triangle.
∆BAE = ∆DEC
∠BEA = ∠EDC (Angles opposite to the equal sides are also equal)
∠BAE + ∠BEA = 90
∴ ∠BEA + ∠+ BEA = 90
∴ ∠AED = 90°
∴ ∆AED is an isosceles triangle.
∴ ∠EAD = ∠EDA = 45°

Question 18.
Construct the following triangles by using only scale and compass.
(a) In ∆ ABC, AB = 6 cm, ∠A = 45°, ∠B = 75°
(b) In ∆ PQR, PQ = 7 cm,
∠P = 52$$\frac{1}{2}$$° , ∠ Q = 82$$\frac{1}{2}$$° 2
Solution:
(a) Draw AB = 6 cm
Draw AP making angle 45° with AB.
Draw BQ making angle 75° with AB.
Let AP and BQ intersect at C.
∴ ABC is the required triangle.

## Kerala State Syllabus 9th Standard Maths Solutions Chapter 5 Circles

### Kerala Syllabus 9th Standard Maths Circles Text Book Questions and Answers

Textbook Page No. 68

Question 1.
Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.

AE = AE (Common side)
BC = BD (Radii of the same circle)
ΔABC = ΔABD (Three sides are equal)
In equal triangles, angles opposite to equal sides are equal.
So, ∠CAE = ∠DAE
∠CAE = ∠DAE
AE = AE (Common side)
ΔAEC = ΔAED (Two sides and the angle between them)
In equal triangles, sides opposite to equal angles are equal.
So, CE = DE (∠CAE = ∠DAE)
CE = DE ………(1)
In equal triangles, angles opposite to equal sides are equal. So, ∠AEC = ∠AED
∠AEC + ∠AED = 180° (Linear pair)
∠AEC = ∠AED = 90° ………(2)
From equation (1) and (2)
The line joining the centres of the circles is the perpendicular bisector of the chord.

Question 2.
The picture on the right shows two circles centred on the same point and a line intersecting them. Prove that the parts of the line between the circles on either side are equal.

OE bisects the chords perpendicularly AD and BC
BE = CE
AE = DE
AE – BE = DE – CE
AB = CD

Question 3.
The figure shows two chords drawn on either sides of a diameter: What is the length of the other chord?

PO is the perpendicular bisector of AB, OQ is the perpendicular bisector of AC.
∠OAQ = ∠OAP = 30° (Given)
∠OQA = ∠OPA = 90° (Right angles)
∴ ∠AOQ = ∠AOP (Third angle also equal)
AO = AO (Common side)
ΔOQA = ΔOPA
In equal triangles, sides opposite to equal angles are equal. So AP = AQ.
AP = ½AB (Perpendicular from the centre of a circle to a chord bisects the chord)
AQ = ½AC
Since AP = AQ
½AB = ½AC
AB = AC
So length of AC = 3 cm

Question 4.
A chord and the diameter through one of its ends are drawn in a circle. A chord of the same inclination is drawn on the other side of the diameter.

Prove that the chords are of the same length.

AB is the diameter
AC and AD are the chords
In triangle OAC,
∠OAC = ∠OCA
∠OAC = ∠OAD; ∠OCA = ∠ODA
∠AOC = ∠AOD; AO = AO
Triangles are equal.
Sides opposite to equal angles are also equal.

Which among Mex, RCH2X, R2CHX , and R3CX is most reactive towards sn2 reaction​.

Question 5.
The figure shows two chords drawn on either sides of a diameter. How much is the angle the other chord makes with the diameter?

AD is the diameter and O is the centre of the circle.
∠OAB = 40°
Consider ΔOAB and ΔOAC
AB = AC = 3cm
OC = OB (Radius of the circle)
OA = OA (Common side)
Three sides ΔOAB of are equal to three sides of ΔOAC
In equal triangles, angle opposite to equal sides are equal.
∴ ∠OAB = ∠OAC
∴ ∠OAC = 40°

Question 6.
Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.
AB, AC are the chords of same length AD is the diameter of the circle.

When we consider ΔAOB, ΔAOC
AB = AC (Given)
OA = OA (Common side)
Three sides ΔOAB of are equal to three sides of ΔOAC.
In equal triangles, angle opposite to equal sides are equal.
∠BAO = ∠CAO
∴ the diameter AD bisects ∠A.

Question 7.
Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.

Prove that this polygon is a regular octagon.
The diameters are parallel to the sides

AD = BD (Perpendicular from the centre of a circle to a chord bisects the chord)
DC = DC (Common side)
Two sides and the angle between them of ΔADC are equal to two sides and the angle between them of ΔBDC.
So ΔADC & ΔBDC are equal.
∴ AC = BC
In the same way we can see that other sides of the octagon are also equal.
So it is a regular octagon.

Textbook Page No. 72

Question 1.
Prove that chords of the same length in a circle are at the same distance from the centre.

AB, CD are the chords of same length.
AB = CD
AP = ½AB (Perpendicular from the centre of a circle to a chord bisects the chord)
Similarly CQ = ½CD
AP = CQ [Since AB = CD]
Consider the right angled triangle ΔAOP and ΔCOQ
OP² = OA² – AP²
OP² = OB² – CQ² [Since OA = OB, AP = CQ]
OP² = OQ²
∴ OP = OQ
So, the chords of the same length in a circle are at the same distance from the centre.

Question 2.
Two chords intersect at a point on a circle and the diameter through this point bisects the angle between the chords. Prove that the chords have the same length.
OA = OC (radius of the same circle)
OB = OB (common side)

∠OBA = ∠OBC (given)
∠BAO = ∠BCO [Base angle of isosceless triangle ΔOCB & ΔOCA]
∠AOB = ∠BOC;
∴ ΔAOB = ΔBOC
So the sides AB and BC opposite to equal angles are also equal.

Question 3.
In the picture on the right, the angles between the radii and the chords are equal. Prove that the chords are of the same length.

Perpendiculars are drawn from the centre of the circle to the chords.

Consider ΔAOM, ΔCON;
OM = ON
∠AMO = ∠CNO = 90°
∠A = ∠C (given)
ΔAOM ≅ ΔCON (A.A.S)
In equal triangles, angle opposite to equal sides are equal.
AM = CN
½AB = ½CD
∴ AB = CD

Textbook Page No. 73

Question 1.
In a circle, a chord I cm away from the centre is 6 cm long. What is the length of a chord 2 cm away from the centre?

$$\sqrt{3^{2} + 1^{2}} = \sqrt{9 + 1} = \sqrt{10}$$
AB = $$\sqrt{\sqrt{10}^{2} – 2^{2}} = \sqrt{10 – 4} = \sqrt{6}$$
Length of the chord $$\sqrt 6 + \sqrt 6 = 2\sqrt 6$$

Question 2.
In a circle of radius 5cm, two parallel chords of lengths 6cm and 8cm are drawn on either side of a diameter. What is the distance between them? If parallel chords of these lengths are drawn on the same side of a diameter, what would be the distance between them?

OM = $$\sqrt{5^{2} – 3^{2}}$$
= $$\sqrt{25 – 9}$$
=$$\sqrt {16}$$ = 4 cm
ON = $$\sqrt{5^{2} – 4^{2}}$$
= $$\sqrt{25 – 16}$$
=$$\sqrt 9$$ = 3 cm
The distance between the chords = 4 + 3 = 7cm
If the chords are on same side = 4 – 3 = 1cm

Question 3.
The bottom side of the quadrilateral in the picture is a diameter of the circle and the top side is a chord parallel to it. Calculate the area of the quadrilateral.

AB = $$\sqrt{2.5^{2} – 1.5^{2}}$$
= $$\sqrt{6.25 – 2.25}$$
=$$\sqrt 4$$ = 2 cm
The distance between the parallel sides = 2 cm
Area = $$\frac{1}{2}$$ × 2 × (5 + 3) = 8cm²

Question 4.
In a circle, two parallel chords of lengths 4 and 6 centimetres are 5 centimetres apart. What is the radius of the circle?

MN = 5
ON = x
OM = 5 – x
x² + 3² = (5 – x)² + 2²
x² + 9 = 25 – 10x + x² + 4
9 = 25 – 10x + 4
10x = 25 + 4 – 9
10x = 20
x = 20/10 = 2
Radius = $$\sqrt{2^{2} + 3^{2}}$$ = $$\sqrt {4 + 9}$$ = $$\sqrt {13}$$cm

Textbook Page No. 78

Question 1.
Draw three triangles with lengths of two sides 4 cm and 5 cm and angle between them 60°, 90° and 120°. Draw the circumcircle of each . (Note how the position of the circumcentre changes).

In this triangle all the angles are less than 90°. The circum centre ‘O’ is inside the triangle.

In the triangle with one angle is 90°. The circum centre is the midpoint of the hypotenuse.

In this triangle with an angle greater than 90°. The circumcentre ‘O’ is outside the triangle.

Question 2.
The equal sides of an isosceles triangle are 8 cm long and the radius of its circumcircle is 5 cm. Calculate the length of its third side.
ΔABC is an isosceles triangle The bisector of ∠A bisects BC
OM = x; BM = $$\sqrt{5^{2} – x^{2}}$$
When we consider ΔAMB,
AB² = AM² + BM²

82 = (5 + r)² + $$(\sqrt {5^{2} – x^{2}})^{2}$$
64 = 25 + 10x + x² + 25 – x²
64 = 10x + 50
14 = 10x
x = 14/10
= 1.4
BM = $$(\sqrt {5^{2} – 1.4^{2}}$$
BC = $$2(\sqrt {5^{2} – 1.4^{2}} = 2\sqrt {23.04} = \sqrt {92.16}$$
= 9.6 cm

Question 3.
Find the relation between the length of a side and the circumradius of an equilateral triangle.

In ΔABC,
AB = BC + AC (sides of an equilateral triangle)
∠DAO = 30°, ∠ADO = 90°
∴ ∠AOD = 60°
By using the properties of angles 30°, 60°, 90°.
If OA = r
OD = $$\frac{r}{2}$$
AD = $$\frac{\sqrt 3}{2}$$r
AB = 2 × $$\frac{\sqrt 3}{2}$$r = $$\sqrt 3$$r
One side of an equilateral triangle is $$\sqrt 3$$ times its circumradius.

### Kerala Syllabus 9th Standard Maths Circles Exam Oriented Text Book Questions and Answers

Question 1.
Draw a circle which passes through the points A, B and radius 5 cm.

The 5 cm line drawn from A meet the perpendicular bisector of AB at point O. The a circle drawn with O as centre and OA as radius will pass through A and B.

Question 2.
The distance between the points A and B is 3 cm. Find out the radius of the smallest circle which passes through these points? What is AB about this circle?
1.5 cm. Diameter.

Question 3.
Draw circles which passes through the points A and B and radius 3 cm, 4 cm and 5 cm.
Centres of the circle lie on the same straight line.

Question 4.
Two circles in the diagram have same radius. Prove that AC = BD.

OP is drawn perpendicular to the chord. OP bisect CD and AB.

Question 5.
A and B are the centres of two circles in the diagram. Circles meet at points O and P. MN || AB. Then prove that MN = 2 × AB.

Draw perpendicular lines AX, BY
∴ XM = XO similarly YN = YO
MN = MX + XY + YN
= OX + XY + OY = XY + XY
= 2XY = 2AB

Question 6.
AB is he diameter of the circle with centre C. PQ || AB, AB = 50cm, PQ = 14cm. Find BQ.

Draw CM perpendicular PQ.
PQ = 14cm,
∴ MQ = 7 cm, CN = 7;
CB = 25 cm
CQ = 25 cm
NB = 25 – 7 = 18
CM = $$\sqrt{25^{2} – 7^{2}} = \sqrt{625 – 49}$$
=$$\sqrt{576}$$ = 24
∴ NQ = 24
BQ = $$\sqrt{NQ^{2} – NB^{2}} = \sqrt{24^{2} + 18^{2}}$$
= $$\sqrt {576 + 524} = \sqrt {900}$$ = 30 cm

Question 7.
AB, AC are two chords of a circle and the bisector of ∠BAC is a diameter of the circle. Prove that AB = AC.
OA = OA (common side)
∠OPA = ∠OQA = 90°
(OP ⊥ AB, OQ ⊥ AC) ∠PAO
= ∠QAO (AE bisector)
∴ ΔOAP = ΔOAQ
OP = OQ therefore AB = AC (Equal chords of a circle are equidistant from the centre).

Question 8.
In the question above instead of assuming ∠OAB = ∠OCD assuming that AB = CD and then prove that ∠OAB = ∠OCD.

∠OAB = ∠OCD (Given)
∠P = ∠Q = 90°
∴ (OP ⊥ AB, OQ ⊥ CD ) ∴ ΔOAP ≅ ΔOCQ;
∴ OP = OQ, Therefore AB = CD
[Equal chords of a circle are equidistant from die centre]

Question 9.
What is the distance from the centre of a circle of a circle of radius 5 cm to a chord of length 8 cm.

Distance from the centre = cp = $$\sqrt{5^{2} – 4^{2}}$$
= $$\sqrt{25 – 16} = \sqrt{9}$$ = 3cm

## Kerala State Syllabus 8th Standard Maths Solutions Chapter 2 Equations

### Equations Text Book Questions and Answers

Textbook Page No. 34

Question 1.
“Six more marks and I would’ve got full hundred marks in the maths test.” Rajan was sad. How much mark did he actually get? the num¬ber.
Solution:
The marks Rajan got is six less from 100
∴ 100 – 6 = 94

Question 2.
Mother gave Rs.6o to Lissy for buying hooks. She gave back the 13 rupees left. For how much money did she buy books?
Solution:
Money Lissy got from mother = Rs. 6o
Amount she returned = Rs. 13
Amount Lissy used to buy books = 60 – 13 = Rs. 47

Question 3.
Gopalan bought a bunch of bananas. 7 of them were rotten which he threw away. Now there are 46. How many bananas were there in the bench?
Solution:
No. of rotten bananas = 7
No. of bananas left =46
∴ Total number of bananas
= 46 + 7 = 53

Question 4.
Vimala spent 163 rupees shopping and now she has 217 rupees. How much money did she have at first?
Solution:
Amount spent by Vimala for shopping = Rs. 163
Amount left with her after shopping Rs. 217
Total amount = 163 + 217 = Rs. 380

Question 5.
264 added to a number makes it 452. What is the number?
Solution:
Number = 452 – 264 = 188

Question 6.
198 subtracted from a number makes it 163. What is the number.
Solution:
Number = 198 + 163 = 361

Textbook Page No. 35

Question 1.
In a company the manager’s salary in five times that of a peon. The manager gets Rs, 40,000 a month. How much does a peon get a month?
Solution:
Salary of the manager = Rs.40,000.
$$\frac{1}{5}$$ th salary of the manager in the salary of the peon.
∴ Salary of the peon = $$\frac{40000}{5}$$
= 8000

Question 2.
The travellers of a picnic split equally the 5200 rupees spent. Each gave 1300 rupees. How many travellers were there?
Solution:
Total amount spent = Rs. 5200
Share of one = Rs.1300
∴ Members in the group = $$\frac{5200}{1300}$$ = 4

Question 3.
A number multiplied by 12 gives 756. What is the number?
Solution:
Number = $$\frac{756}{12}$$ = 63

Question 4.
A number divided by 21 gives 756. What is the number
Solution:
Number 756 × 21 = 15,876

Textbook Page No. 37

Question 1.
Anita and her friends bought pens. For five pens bought toge¬ther, they got a discount of three rupees and it cost them 32 rupees. Had they bought the pens separately, how much would each have to spend?
Solution:
Cost for 5 pens = Rs. 32
discount = Rs.3
Real cost = 32 + 3 = 35
∴ Real cost of 1 pen = $$\frac{35}{5}$$ = 7

Question 2.
The perimeter of a rectangle is 25 metres and one of its side is 5m. How many metres is the other side?
Solution:
Perimeter of the rectangle = 25m
One side = 5m
Double of the sum of length and breadth is perimeter = $$\frac{25}{2}$$ – 5 = 7.5

Question 3.
In each of the problems below, the result of doing some operations on a number is given. Find the number?
(i) Three added to double is 101
(ii) Two added to triple is 101
(iii) Three substracted from double is 101
(iv) Two substracted from triple is 101
Solution:

Question 4.
Half a number added to the number gived III. What is the number?
Solution:
1$$\frac{1}{2}$$ of the numbers = 111
ie $$\frac{3}{2}$$ of the number = 111
number = 111 × $$\frac{2}{3}$$ = 74

Question 5.
A piece of folk maths. A child asked a flock of birds, ‘How many are you”? A bird replied
“We and us again
with half of us
And half of that
with one more.
would make hundred” How many birds were there?
Solution:
We and us ⇒ double (2 times)

Textbook Page No. 41

Question 1.
The perimeter of a rectangle is 80 metre and its length is one metre more than twice the breadth. What are its length and breadth?
Solution:
If x be the breadth, Then length
= 2x + 1
2 (x + 2x + 1) = 80
2(3x + 1) = 80
6x + 2 = 80
6x = 80 – 2
x = $$\frac{80-2}{6}$$ = 13.
Length = 2 × 13 + 1 = 27 metre

Question 2.
From a point on a line another line is to be drawn such that the angle on one side is 50° more than the angle on the other side. How much is the smaller angle?
Solution:
Let one angle is x, the second angle = x + 50
Sum of two angles on a line = 180°
ie x + x + 50 = 180°
2x + 50 = 180°, 2x = 180 – 50
x = $$\frac{180-50}{2}$$ = 65
The angles are 65°, 115°

Question 3.
The price of a book is 4 rupees more than the price of a pen. The price of a pencil is 2 rupees less than the price of the pen. The total price of 5 books, 2 pens and 3 pencils is 74 rupees. What is the price of each?
Solution:
Let the cost of a pen = x
cost of a book = x + 4
Cost of a pencil = x – 2
ie.5(x + 4) + 2x + 3(x – 2) = 74
5x + 20 + 2x + 3x – 6 = 74
10 x + 14 = 74, 10x = 74 – 14
x = $$\frac{74-14}{10}$$ = 6
Cost of pen = Rs. 6
Cost of book = Rs. 10
Cost of pencil = Rs. 4

Question 4.
(i) The sum of three consecutive natural numbers is 36. What are the numbers
(ii) The sum of three consecutive even numbes is 36. What are the numbers?
(iii) Can the sum of three consecutive odd numbers be 36. Why?
(iv) The sum of three consecutive odd numbers is What are the numbers?
(v) The sum of three consecutive natural numbers is 33. What are the numbers?
Solution:
x + x + 1 + x + 2 = 36
3x + 3 = 36
x = $$\frac{36-3}{3}$$ = 11
∴ Numbers are 11, 12, 13
x – 1 + x + x + 1 = 36
3x = 36
x = 12
Or
∴ Numbers are 11, 12, 13

(ii) (x – 2) + x + (x + 2) = 36
3x = 36
x = 12
∴ The numbers are 10, 12, 14

(iii) The sum of three odd numbers is odd.
∴ It is not possible to get 36 as the sum of three odd numbers.

(iv) (x – 2) + x + (x + 2) = 33
3x = 33
x = 11
∴ Numbers are 9, 11, 13

(v) (x – 1)+ x + (x + 1) = 33
3x = 33
x = 11
The numbers are 10, 11, 12

Question 5.
(i) In a calender, a square of four numbers is marked. The sum of the numbers is 80. What are the numbers?
(ii) A square of nine numbers is marked in a calendar. The sum of all there numbers is 90. What are the numbers?
Solution:
(i) Let,
x x + 1
x + 7 x + 8
four numbers in the square of 4 numbers.
Sum of the numbers = x + (x + 1) + (x + 7) + (x + 8) = 80
4x + 16 = 80, 4x = 80 – 16

Textbook Page No. 44

Question 1.
Ticket rate for the science exhibition is rupees 10 for a child and 25 rupees for the adult. Rs. 740 was got from 50 persons. How many children among them?
Solution:
Let the numbers of children be x
Then the number of adult = 50 – x
∴ 10 x + 25 (50 – x) = 740
10 x + 1250 – 25 x = 740
1250 – 15 x = 740
15 x = 1250 – 740
x = $$\frac{1250-740}{15}$$ = 34
Number of adults = 50 – 34 = 16

Question 2.
A class has the same numbers of girls and boys. Only 8 boys were absent on a particular day and then the number of girls was double the number of boys and girls?
Solution:
Let the number of boys =number of girls = x
2 (x – 8) = x
2x – 16 = x, 2x – x = 16
x – 16 = 0
∴ x = 16
∴ The number of boys = No. of girls = 16

Question 3.
Ajayan is ten years older than Vijayan. Next year Ajayan’s age would be double that of Vijayan. What are their ages now?
Solution:
Let the age of Vijayan = x
Age of Ajayan = x + 10
Age of Vijayan after an year = x + 1
Age of Ajayan after an year = x + 11
2 (x + 1) = x + 11
2x + 2 = x + 11
2x – x = 11 – 2
x = 9
∴ Age of Vijayan = 9
Age of Ajayan = 19

Question 4.
Five times a number is equal to three times the sum of the number and 4. What is the number?
Solution:
Let the number = x
Five times number = 5x
4 more than the number = x + 4
Three times of it = 3(x + 4)
ie. 5 x = 3 (x + 4)
5 x = 3x + 12
5 x – 3 x = 12
2 x = 12
x = 6

Question 5.
In a co-operative society, the number of men is thrice the number of women 29 women and 16 men more joined the society and now the number of men is double the number of women. How many women were there in the society at first?
Solution:
Let the number of women = x
No of men = 3 x
No. of women when 29 more were joined = x + 29
No. of men when 16 more were joined = 3x + 16
3 x + 16 = 2 (x + 29)
3x + 16 = 2x + 58
3x – 2x = 58 – 16
x = 42
No. of women = 42;
No. of men = 3 × 42 = 126

Question 1.
“If you give rupees 5 to me both of us have equal amounts with us”. Ajay told to Vineeth. Then Vineeth told to Ajay “ If you give rupees 5 to me I will have 5 times more money than yours. Find out the amount both of them have?
Solution:
Let Ajay has Rs.x with him and when Vineeth gives Rs. 5, both of them have x + 5 rupees with them. If Vineeth gets RS.5 he has 5 times more than Ajay.
x + 5 + 5 = 5x
4 x = 10
x = $$\frac{10}{4}=\frac{5}{2}$$ = 2.5 Rupees
Amount with Ajay = Rs. 2.50
Amount with Vineeth = Rs.12.50

Question 2.
The perimeter of a triangle is 49 cm. One side is 7 cm more than the second side and 5 cm less than the third side. Find out the lengths of three sides?
Solution:
Let the second side = xcm
First side = x + 7 cm
Third side = x + 12 cm
Perimeter = x + x + 7 + x + 12 = 49
= 3x = 49 – 19 = 30
3x = 20, x = 10 cm
The sides are = 10 cm, 17 cm, 22 cm

Question 3.
The sum of two numbers is 9, 8 times the smaller number is 2 more than 6 times the bigger number. Write an equation to find the numbers?
Solution:
Let the smaller number = x
Bigger number = 9 – x
Equation is, 8 x = 6 (9 – x) + 2

Question 4.
Anitha and friends bought pens. When they bought a packet of 5 pens they got a discount of rupees 2. They paid rupees 18. Find the cost of each one buys separately?
Solution:
Total cost in including discount = Rs. 20
Cost of 5 pens = Rs.20
Cost of 1 pen = $$\frac{20}{5}$$ = 4

Question 5.
The cost of a chair and a table is Rs. 1500. The cost of table is 4 times the cost of chair. Find the expense of each?
Solution:
Let the cost of chair = x
Cost of table = 4x
x + 4 x = 1500
5x = 1500
x = 300
cost of chair = Rs. 300
Cost of table = Rs. 1200

Question 6.
There were 25 questions in the examination written by Jafar. Each correct answer gets 2 marks. There is a loss of mark for each wrong answer. Jafar answered all the questions. He got 35 marks. Find^ut the number of correct answers?
Solution:
Let the number of correct answers = x
Total number of questions = 25
No. of wrong answers = 25 – x
Marks for correct answers = 2x
marks losses for wrong answers = 25
ie. 2x – (25 – x) = 25
2x – 25 + x = 35
3x = 60
x = $$\frac{60}{3}$$ = 20, ∴ number of correct answers = 20.

## Kerala Syllabus 8th Standard Hindi Solutions Unit 5 Chapter 3 वह सुबह कभी तो आएगी

You can Download वह सुबह कभी तो आएगी Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 5 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 8th Standard Hindi Solutions Unit 5 Chapter 3 वह सुबह कभी तो आएगी

साहिर लुधियानवी साहिर लुधियानवी का जन्म 8 मार्च 1921 में लुधियाना में हुआ। आप प्रसिद्ध शायर तथा गीतकार थे। आपकी शिक्षा लुधियाना के खालसा हाइस्कूल में हुई। लाहौर तथा मुंबई आपकी कर्मभूमि रही। तल्खियाँ’ आपका पहला कविता संग्रह है। ‘आज़ादी की राह पर’ नामक फ़िल्म के लिए आपने पहली बार गीत लिखे। 25 अक्तूबर 1980 को दिल का दौरा पड़ने से आपका निधन हो गया।

### वह सुबह कभी तो आएगी Summary in Hindi

‘वह सुबह कभी तो आएगी’ में गीतकार एक अच्छी सुबह की प्रतीक्षा करते हैं। गीतकार कहते हैं, कभी वह सुबह आएगी तो इन बीती हुई काली सदियों के सिर से रात का आँचल नीचे की ओर जाएगा। तब दुःख का काला बादल पिघल जाएगा। दुःख के बदले सुख का सागर उमडेगा। आकाश खुशी से नाचेगा और धरती मधुर गीत गाएगी।

यहाँ गीतकार स्वतंत्र भारत का सपना देखते हैं।
गीतकार कहते हैं – कभी वह सुबह आएगी तो ये भूख और बेकारी के दिन बीत जाएँगे। संपत्ति और ठेकेदारी के दिन टूटेंगे। उनके स्थान पर एक अनोखी दुनिया की बुनियाद उठाई जाएगी।

संसार के सारे परिश्रमी लोग खेतों से उस दिन का वेतन लेकर आयेंगे। गरीबी के अंधेरे बिलों से वे बाहर आएँगे। तब वे मूल्यहीन या विवश नहीं होते। तब दुनिया शांति और खुशहाली से सजाएँगे।

### वह सुबह कभी तो आएगी Summary in Malayalam and Translation

” ആ പ്രഭാതം എന്നെങ്കിലും വരികയാണെങ്കിൽ’ എന്ന കവിതയിൽ കവി ഒരു നല്ല പ്രഭാതത്തിന്റെ വരവും കാത്തിരിക്കുകയാണ്.

– കവി പറയുന്നു, “ആ നല്ല പ്രഭാതം എന്നെങ്കിലും വരികയാണെങ്കിൽ കഴിഞ്ഞു. പോയ ആ കറുത്ത നാളുകളുടെ ശിരസ്സിൽ നിന്നും വസ്ത്രാഗം താഴേയ്ക്കുവീഴും. അന്ന് ദുഃഖത്തിന്റെ കാർമേഘം അലിഞ്ഞ് ഇല്ലാതാകും. ദുഃഖത്തിന്റെ സ്ഥാനത്ത് സന്തോ ഷത്തിന്റെ സമുദ്രം നിറഞ്ഞൊഴുകും. ആകാശം സന്തോഷംകൊണ്ട് നൃത്തം ചെയ്യും. ഭൂമി മധുരഗീതം ആലപിക്കും.

ഇവിടെ കവി സ്വതന്ത്രഭാരതമാണ് സ്വപ്നം കാണുന്നത്. – കവി പറയുന്നു, എന്നെങ്കിലും ആ പ്രഭാതം വന്നാൽ വിശപ്പിന്റെയും തൊഴിലില്ലാ യ്മയുടെയും ഈ ദിനങ്ങൾ കടന്നുപോകും. സമ്പത്തിന്റെയും ഏകാധിപത്യത്തിന്റെയും കാലം അവസാനിക്കും. അതിന്റെ സ്ഥാനത്ത് സുന്ദരമായ ഒരു പുതുയുഗത്തിന്റെ അടി ത്തറ ഉയർന്നുവരും.

ലോകത്തിലെ എല്ലാ അധ്വാനികളും ജോലികഴിഞ്ഞ് കൂലിയുമായി വരുന്ന ഒരു കാലം ഉണ്ടാകും. ദാരിദ്ര്യത്തിന്റെ ഇരുട്ടു നിറഞ്ഞ മാളത്തിൽ നിന്നും അവർ പുറത്തു വരും. അന്ന് അവർ വിവശരോ വിലയില്ലാത്തവരോ ആയിരിക്കുകയില്ല. അങ്ങനെ ലോകം ശാന്തിയും സമൃദ്ധിയും കൊണ്ട് അലങ്കരിക്കപ്പെടും.

वह सुबह कभी तो आएगी शब्दार्थ Word meanings