Kerala Syllabus Class 8 Maths Model Question Paper Set 1

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 1 help manage time.

Class 8 Maths Kerala Syllabus Model Question Paper Set 1

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section – A
This section has 4 questions of score 1 each. Select the correct answer.

Question 1.
In a right angled triangle with its perpendicular sides 9 cm and 12 cm, the radius of its circumcircle is:
(a) 6 cm
(b) 7.5 cm
(c) 10 cm
(d) 15cm
Answer:
(b) 7.5 cm

Hypotenuse =15 cm
⇒ Radius = \(\frac{15}{2}\) = 7.5 cm

Question 2.
A histogram is drawn for …
(a) Discrete data
(b) Continuous data
(c) Alphabetical data
(d) Tally marks
Answer:
(b) continuous data

Question 3.
Four statements are given below Identify the true statement
(a) Equal triangles have equal area
(b) Equiangular triangles (Triangles having equal angles) are equal triangles
(c) Equality one angle and any two sides of two triangles make them equal
(d) One side and angle at the ends can fix a triangle

(a) (a) and (b) are true
(b) (a) and (c) are true
(c) (b) and (c) are true
(d) (a) and (d) are true
Answer:
(d) (a) and (d) are true
Equality means identical or having same shape and side

Question 4.
101² – 100²
(a) 1
(b) 201
(c) 100
(d) none of these
Answer:
101² – 100² = 101 + 100 = 201

Section – B
This section has 4 questions of score 2 each.

Question 5.
Find the sum of the angles marked in the star.

Answer:

In triangle BQD, sum of the marked angles is the outer angle at Q
In triangle PCE, the marked angle sum is the outer angle at P
All marked angles make the sum 180°

Question 6.
The age of Ammu’s mother is now four times of Ammu. After 20 years mothers age will be two times Ammu’s age
(a) If Ammu’s age is x then write the equation.
Answer:
Ammu’s age x, Mothers age 4x

(b) Find the age of Ammu and her mother.
Answer:
After 20 years Ammu’s age = x +
20,mothers age = 4x + 20
4x + 20 = 2(x + 20), 4x + 20
= 2x + 40, x = 10
Ammu’s age 10, Mother age 40

Question 7.
Write the following in the simplified form
\(\frac{12-11+10-9+8-7}{6-5+4-3+2-1}\)
Answer:
\(\frac{12-11+10-9+8-7}{6-5+4-3+2-1}=\frac{1+1+1}{1+1+1}=\frac{3}{3}\)
= 1

Question 8.
The area of a rectangle is 2100 sq.cm.
(a) Write length and breadth as multiple of x.
Answer:
Length = 7x
Breadth = 3x

(b) Find the length and breadth of the rectangle.
Answer:
Area = 2100 sq.cm
length × breadth = 2100
7x × 3x = 2100
21x² = 2100
x² = 100
x = 10
Length = 70 cm
Breadth = 30 cm

Section – C
This section has 4 questions of socre 3 each.

Question 9.
Draw a square with side length 4 cm and construct its circumcircle.
Answer:

Question 10.
Draw a rectangle of diagonal 5.5 cm with the angle between the diagonals 50°.
Answer:
Draw a diagonal 5.5 cm long. Draw its perpendicular bisector. Take angle 50° from this midpoint. Distance on either side of the mid point on the first diagonal is same as on the second diagonal. Join the points. We get the required rectangle.

Question 11.
[A] i. Five angles of a hexagon have measures 100°, 110°, 120°, 130°, 140°. What is the measure of the sixth angle?
ii. Squares are drawn on the sides of a regular hexagon.

Find x
Answer:
(i) Inner angle sum of the hexagon is 720°
Sixth angle is
= 720 – (100 + 110 + 120 + 130 + 140)
= 720 – 600
= 120°

(ii) One inner angle of regular hexagon is 120°
Two squares in a comer make the angles 90°.
Angle around the comer is 360°
x = 360 – (120 + 90 + 90) = 60

OR

[B] Draw 22 \(\frac{1}{2}\) cone using scale and compass.
Answer:
Draw a line and mark a point on it. Draw a perpendicular line through the point.
90° angle is formed. Draw the angle bisctor. 45° angle is formed. Draw the angle bisctor.
22 \(\frac{1}{2}\) angle is formed.

Question 12.
[A] (a) Write the following as powers of 10: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}\)
Answer:
\(\frac{1}{2}\)

(b) Write their decimal forms.
Answer:
\(\frac{1}{2}\) = 0.5, \(\frac{1}{4}\) = 0.25, \(\frac{1}{8}\) = 0.125

(c) Calculate: 1 – \(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}\)
Answer:
1 – \(\frac{1}{2}=\frac{1}{2}, \frac{1}{4}-\frac{1}{8}=\frac{1}{8}\)
\(\frac{1}{2}+\frac{1}{8}=\frac{4}{8}+\frac{1}{8}=\frac{5}{8}\) = 0.625

OR

[B] Four numbers forming a square in a calendar are given below:

(a) Find the difference of its diagonal products.
(b) Is it the same for all squares of four numbers in a calendar? Explain using algebra.
Answer:
(a) (5 × 11) – (4 × 12) = 7
(b) Yes

x	x + 1
x + 7	x + 8

Difference of diagonal product
= ((x + 1)(x + 7)) – (x(x + 8))
= x² + 8x + 7 – (x² + 8x)
= 7

Section – D
This section has 4 questions of score 4 each.

Question 13.
Draw a right triangle with two angles 45° and length of its hypotneous as 4.75 centimetres.
Answer:
Draw a line of length 9.5 cm. Draw its perpendicular bisector.
The part of the line on one side of the perpendicular bisector will be 4.75 cm.
The angle between the line and the perpendicular bisector is 90 0 Draw the angle bisector.
Draw a perpendicular from one end of the line to the angle bisector. It forms a right triangle.

Question 14.
(i) Draw an equilateral triangle with side length 4 cm and construct its circumcircle.
Answer:

(ii) The angles of a triangle are in the ratio 1:2:3. If the longest side is 18 cm, what is the radius of the circumcircle?
Answer:
It is a right-angled triangle.
The radius of the circumcircle is 9 cm, which is half of the hypotenuse.

Question 15.
[A] (i) In the figure AB = CF, EF = BD, ∠DBC = ∠AFE

(a) Are the triangles AFE and triangle BDC equal ? How do you know this?
Answer:
AB = CF
∴ AB + BF = CF + BF
ie., AF = CB,
BD = EF,
∠DBC = ∠AFE
Triangles are equal

(b) What are the equal sides and equal angles of these triangles?
Answer:
CD = AE, EF = BD, ∠DBC = ∠AFE

(ii) In the figure AB = AC, E is the mid point of AB, F is the mid point of AC .
(a) What are the equal triangles in the figure ?
Answer:
AB = AC ⇒ ∠B = ∠C
BE = CF, BC is common
Triangle BEC and triangle CFB are equal

(b) If BF = 12 cm then what is CE ?
Answer:
Sides opposite to equal angles are equal. CE = 12 cm

OR

[B] ABCDEF is a polygon

(a) What is the suitable name for this polygon?
(b) How many diagonals can be drawn from any one of the vertices?
(c) What is the sum of inner angles of this polygon
(d) What is the sum of inner anlgles of a 7 sided polygon?
Answer:
(a) Hexagon
(b) 3 w
(c) 180 × 4 = 720°
(d) 720 + 180 = 900°

Question 16.
[A] The scores of 30 students in a test are given below.
21, 32, 5, 12, 36, 24, 31, 15, 27, 34
44, 39, 37, 25, 7, 20, 23, 12, 9, 40
46, 42, 28, 16, 47, 27, 37, 18, 49, 26

(a) Prepare a frequency table as shown below.
Answer:

(b) How many students scored more than 30?
Answer:
7 + 6 = 13 students

OR

[B] In figure, ∠AOB = 71° OC is the bisector of ∠AOB

(a) What is the measure of ∠AOC ?
Answer:
∠AOC = \(\frac{71}{2}\) = 35\(\frac{1^0}{2}\)

(b) Draw triangle PQR with PQ = 6 centimetres,
∠P = 35\(\frac{1^0}{2}\), ∠Q = 60°
Answer:
Draw OQ = 6 cm, ∠P = 71° and
∠Q = 60°. Draw bisector of ∠P.
Now in ∆PQR, ∠P = \(\frac{71}{2}\) = 35.5°