Kerala Syllabus Class 8 Maths Model Question Paper Set 3

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Class 8 Maths Kerala Syllabus Model Question Paper Set 3

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.
The angle around the centre of a circle is divided into three equal parts, each 120° by drawing radii. The ends of radii are joined to make a triangle. The triangle so formed is
(a) Right triangle
(b) Isosceles triangle
(c) Equilateral triangle
(d) None of these
Answer:
(c) Equilateral triangle

Triangles formed around the centre of the circle are equal, one angle 120°Other two angles are 30°. So angles of the triangle will be 60°

Question 2.
From the following, which of the number will be the sum of the digits in the ones and tens place a square of a number which ends in 5
(a) 2
(b) 5
(c) 7
(d) 3
Answer:
(c) 7

Question 3.
Read the given statements.
Statement I: The diagonals of a rhombus bisect each other at 90°.
Statement II: The sides of a rhombus are equal in length
A) Statement I is true, Statement II is true
B) Statement I is false, Statement II is false
C) Statement I is true, Statement II is the reason for Statement I
D) Statement I is true, Statement II is not the reason for Statement I
Answer:
D) Statement I is true, Statement II is not the reason for Statement I
the reason for Statement I The diagonals of a rhombus do bisect each other at right angles. However, the reason given that the sides are equal does not directly cause the diagonals to bisect at 90°.

Question 4.
Area of the trapezium shown in the following figure is ___________.

(a) 4.5 cm²
(b) 6 cm²
(c) 5.5 cm²
(d) 6.5 cm²
Answer:
(c) 5.5 cm²

Area = \(\frac{1}{2}\) × 2 × (2+ 3.5) = 5.5 cm²

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

Question 5.
A polygon has 7 sides
(a) What is the sum of its inner angles Y.
(b) What is the sum of the outer angles?
Answer:
(a) Angle sum of a polygon of n sides is 180° × (n – 2)
Angle sum is 180 × 5 = 900°

(b) 360°

Question 6.
Sum of two numbers is 60. One number is four times the other. Find the number.
Answer:
When number and four times the number is added we get five times the number
Number is \(\frac{1}{5}\) part of 60.
Number is = 12

Question 7.
(a) What is 1 – 2 ?
Answer:
-(2 – 1) = -1

(b) Find 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
Answer:
-1 – 1 – 1 – 1 – 1 = -5

Question 8.
In a circle of radius 13 cm, a chord is at a distance of 5 cm from the centre. Calculate the length of the chord.
Answer:
Half of the chord = \(\sqrt{13^2-5^2}\) = 12 cm
Therefore, the full length of the chord = 24 cm.

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

Question 9.
If x = 0.333 … then,
(a) Write the fractional form of x ?
Answer:
\(\frac{3}{9}\)

(b) Write x² as a fraction
Answer:
\(\frac{1}{9}\)

(c) Write the decimal form of x² + x
Answer:
x² + x = \(\frac{1}{2}\) = 0.444 ……..

Question 10.
Using this equation x² – y² = (x + y)(x – y)
(a) What is 2² – 1²?
Answer:
(2 + 1) (2 – 1) = 3 x 1 = 3

(b) What is 9² – 10²?
Answer:
19

(c) Find 1² – 2² + 3² – 4² + ….. + 9² – 10² ?
Answer:
-3 – 7 – 19 – 11 – 15 – 19 = -(3 + 7 + 11 + 15 + 19)
= -55

Question 11.
[A] Perimeter of a rectangle is 50 cm. Length is 5 cm more than breadth
(a) What is length + breadth ?
(b) If x is the length then write the equation
(c) Find the sides
Answer:
(a) 25 cm

(b) x + x + 5 = 25
2x + 5 = 25, 2x = 20, x = 10

(c) Sides are 10 cm, 15 cm

OR

[B] A man’s age after 15 years will be two times his age before 15 years.
(a) If the present age is x then write the equation.
(b) Find the present age?
Answer:
(a) Age after 15 years is x + 15, age before 15 years is x – 15
2(x – 15) = x + 15
2x – 30 = x + 15
⇒ x = 45

(b) Present age 45years

Question 12.
[A] A triangle has sides 20 cm, 15 cm, and 25 cm. What type of triangle is it? Calculate the radius of its circumcircle.
Answer:
15² + 20² = 225 + 400 = 625 = 25²
Here we get a right angled triangle.
For a right triangle the circumcentre is the midpoint of the hypotenuse.
So the circumradius R = half the hypotenuse 25
R = \(\frac{25}{2}\) = 12.5 cm

OR

[B] In a trapezium ABCD, the parallel sides are 8 cm, 6 cm. The angles in the largest side is 60°, 75°.
(a) Draw a rough figure
Answer:

(b) Draw trapezium ABCD in actual measures.
Answer:

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

Question 13.
Draw a 7.5 centimeter long line and construct a right triangle whose perpendicular sides are of 3.75 centimetre long.
Answer:
Draw a 7.5 cm long line.
Draw the perpendicular bisector of this line. Complete the right triangle by drawing the half length of the line on the perpendicular line.

Question 14.
Look at the pattern given below:
1 + 3 = 4 = 2²
l + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
(a) Complete the next line.
(b) Find the sum of the first 10 odd numbers.
(c) What is the 20th odd number.
(d) What is —1 — 3 — 5 …….. 39
Answer:
(a) 1 + 3 + 5 + 7 + 9 = 25 = 5²
(b) 10² = 100
(c) 39
(d) -(1 + 3 + 5 + ………. + 39) = -400

Question 15.
[A] A rubber farmer records the number of rubber sheets obtained on each day of a month. Using thi data, prepare a frequency table and answer the following questions:

i. What is the maximum number of sheets obtained in a single day?
ii. How many days did he get 16 sheets?
iii. How many days did he get the minimum number of sheets?
iv. How many days did he get less than 15 sheets?
v. What is the total number of sheets obtained in the whole month?
Answer:

Number of sheets	Number of days
9	3
10	4
11	3
12	3
13	5
14	6
16	6

i. 16
ii. 6 days
iii. 3 days
iv. 24 days
v. 85 sheets

OR

[B] Find the area of the parallelogram in the figure.

Answer:

Area of ∆ABF = \(\frac{1}{2}\) × 6 × 3 = 9 cm
AB = \(\sqrt{6^2+3^2}\)
= \(\sqrt{36+9}=\sqrt{45}\)
\(\frac{1}{2}\) × AB × h
\(\frac{1}{2}\) × \(\sqrt{45}\) × h = 9
h = \(\frac{9 \times 2}{\sqrt{45}}=\frac{18}{\sqrt{45}}\)
Area of ABCD = \(\sqrt{45} \times \frac{18}{\sqrt{45}}\) = 18 cm
[or Area of Parallelogram ABCD = Area of the rectangle = 6 × 3 = 18]

Question 16.
[A] Draw a parallelogram of sides 6 cm, 4cm and area 18 cm².
Answer:
First draw a horizontal line of length 6 cm as the bottom side of the parallelogram. Since area of the parallelogram is 18 sq.cm, its distance to. the opposite side (height) must be 3 cm.
So, draw a perpendicular of height 3 cm from one end of the line. Draw the line through its top end, parallel to the first line, then the top side of the parallelogram is a part of this line.
Mark 4 cm from left end of the bottom line on the top side and complete the parallelogram.

OR

[B] Sum of innerangle of a polygon with ‘n’ sides is (n – 2) 180°. The sum of outer angles will be always 360°.
(i) What is the sum of innerangles of a polygon with 20 sides?
(ii) What is the sum of outer angles of a polygon with 100 sides. .
(iii) Which is the polygon having sum of inner angles and outer angles equal.
(iv) Can 1080° be the sum of inner angles of a polygon?
Answer:
i) (20 – 2) 180 = 18 × 180° = 3240°
ii) 360°
iii) Rectangle
iv) Yes, because 1080° is a multiple of 180