Practicing with Class 8 Maths Question Paper Pdf Kerala Syllabus and Annual Exam Question Paper 2023-24 will help students prepare effectively for their upcoming exams.
Class 8 Maths Annual Exam Question Paper 2023-24 Kerala Syllabus
Time : 1½ Hours
Score : 40
Answer any 4 Questions from 1 to 5. Each question carries 2 scores. (4 × 2 = 8)
Question 1.
In the figure, AC = BC = 5 centimetres. ∠C = 60°.
(a) AB = ____________
(b) ∠A = ____________
Answer:
(i) Here, A ABC is an equilateral triangle.
∴ AB = 5 cm
(ii) ∠A = 60°
Question 2.
Five is subtracted from five times of a num-ber gives 100. What is the number?
Answer:
5 is subtracted, so adding 5 to loo we get 105.
5 is multiplied, so dividing 105 by 5 we get \(\frac{105}{5}\) =21.
Thus, 21 is the required number.
Question 3.
In the figure, MNOPQ is a regular petagon.
(a) ∠OPQ = ____________.
(b) ∠ONR = ____________.
Answer:
a) Sum of all the inner angles of a pentagon = (5 – 2) × 180° = 3 × 180° = 540°.
As it is a regular pentagon all the inner angles are equal.
Thus, ∠OPQ = \(\frac{540}{5}\) = 108°
b) Sum of all the inner angles of a pentagon = 360°.
Thus, ∠ONR = \(\frac{360}{5}\) = 72°
Question 4.
The three angles in a triangle are in the same ratio.
(a) Which type of a triangle is this? (Equilateral triangle, Isosceles triangle, Right triangle)
(b) Find one angle of this triangle.
Answer:
a) Equilateral triangle.
b) Sum of all the inner angles of an equilateral triangle 180°.
Thus, one angle = \(\frac{180^{\circ}}{3}\) = 60°.
Question 5.
(a) 5 – (-3) = ………..
(b) 10 + (-10) = ………
Answer:
a) 5 – (-3) = 5 + 3 = 8
b) 10 + (-10) = 10 – 10 = 0
Answer any 4 Questions from 6 to 11. Each question carries 3 scores. (4 × 3 = 12)
Question 6.
a² – b² = (a + b)(a – b)
(a) 146² – 145² = ____________
(b) Using the above identity, find 31 × 29
Answer:
a) 146² – 145²
= (146 + 145) (146 – 145)
= 291 × 1
= 291
b) 31 × 29
= (30 + 1) (30 – 1)
= 30² – 1²
= 900 – 1
= 899.
Question 7.
A person deposited 20000 rupees in a bank which pay 10% interest compounded annually. How much would he get back after two years?
Answer:
P = 20000 rupees
r = 10%
n = 2 years
Amount get after 2 years
= p(1 + \(\frac{r}{100}\))n
= 20000 (1 + \(\frac{10}{100}\))2
= 20000 × \(\frac{110}{100} \times \frac{110}{100}\)
= 24200 rupees
Question 8.
In the figure, PQRS is a quadrilateral. QS = 10 centimetres, PA= 4 centimetres, BR = 5 centimetres.
(a) What is the sum of heights of the triangle PQS and the triangle RQS?
Answer:
a) Height of triangle PQS = 4 cm.
Height of triangle RQS = 5 cm.
Thus, sum of the heights = 5 + 4
= 9 cm.
(b) Find the area of the quadrilateral PQRS.
Answer:
Area = \(\frac{1}{2}\) × QS × (4 + 5)
= \(\frac{1}{2}\) × 10 × 9
= 5 × 9
= 45 cm².
Question 9.
(a) What is (-1)²?
Answer:
(-1)² = (-1) × (-1) = 1
(b) y = x² + 5, if x = -1, find y.
Answer:
y = x² + 5
Put x = -1
y = (-1)² + 5
= 1 + 5
= 6
Question 10.
The scores obtained by some children in a mathematics quiz competition are tabled below.
| Score | Number of children |
| 0 – 5 | 2 |
| 5 – 10 | 6 |
| 10 – 15 | 11 |
| 15 – 20 | 5 |
| 20 – 25 | 1 |
(a) What is the total number of children participated in the quiz competition?
Answer:
Total number of children
= 2 + 6 + 11 + 5 + 1
= 25.
(b) In which class does the score of the first placed child belongs?
Answer:
The first placed child should have the highest score. So, it belongs in the class 20-25.
(c) What is the number of children who scored more than 10?
(25,23, 17, 11)
Answer:
The number of children more than or equal 10 = 11 + 5 + 1 = 17.
Question 11.
In the figure, ABCD is a rectangle.
AB = 7 centimetres, AD = 4 centimetres, AP = CQ = 2 centimetres.
(a) What is the perpendicular distance between the line PB and QD?
Answer:
The perpendicular distance between PB and QD = DA = 4 cm.
(b) What is the length of PB?
Answer:
PB =AB – AP = 7 – 2 = 5 cm.
(c) Find the area of the parallelogram PBQD.
Answer:
The area of the parallelogram
PBQD = 5 × 4 = 20 cm².
Answer any 5 Questions from 12 to 18. Each question carries 4 scores. (5 × 4 = 2)
Question 12.
Draw a trapezium PQRS of given measures.
Answer:
First draw PQ = 7 cm. At P draw 70° angle and mark 4 cm on it,
PS = 4 cm.
Draw SR parallel to PQ. Draw at Q.
Question 13.
In the figure, ABCD is a quadrilateral. ∠D = 80°.
(a) ∠A + ∠B + ∠C = ………………
(b) If the measures of ∠A, ∠B and ∠C
Answer:
a) ∠A + ∠B + ∠C = 360° – 80° = 280°.
b) ∠A: ∠B: ∠C = 1 : 1 : 2.
So, we can take ∠A = x, ∠B = x, ∠C = ∠t
We have, x + x + 2x = 280°
4x = 280°
x = \(\frac{280}{4}\) = 70
Therefore, ∠A = 70°, ∠B = 70°
∠C = 2 × 70 = 140°
Question 14.
The runs scored by a cricket player in 40 one day matches are tabled below. Draw a histrogram.
| Runs | Number of matches |
| 0 – 20 | 5 |
| 20 – 40 | 7 |
| 40 – 60 | 6 |
| 60 – 80 | 12 |
| 80 – 100 | 4 |
| 100 – 120 | 6 |
Answer:
Question 15.
Complete the table.
Answer:
Question 16.
The area of a rhombus is 12 square centimetres and length of one of its diagonal is 6 centimetres.
a) What is the length of seocnd diagonal of this rhombus?
Answer:
Area = 12 cm²
\(\frac{1}{2}\) × d1 × d2 = 12
\(\frac{1}{2}\) × 6 × d2 = 12
3 × d2 = 12
d2 = \(\frac{12}{3}\) = 4
Therefore, the length of the second diagonal =4 cm.
b) Draw the rhombus using these measures.
Answer:
Question 17.
In the figure, ABCD is an isosceles trapezium, AB = 13 centimetres, CD = 7 centimeters, AD = 5 centimetres.
a) What is AE?
Answer:
AE + EP + PB = 13 cm.
As ABCD is an isosceles trapezium,
AE = PB and EP = 7 cm.
Thus, AE + EP + AE = 13.
2AE + 7 = 13.
2AE = 13 – 7 = 6.
AE = \(\frac{6}{2}\) = 3 cm.
b) DE = ____________
Answer:
∆ADE is a right triangle.
So, using Pythagoras theorem,
DE = \(\sqrt{A D^2-A E^2}\)
= \(\sqrt{5^2-3^2}\)
= \(\sqrt{25-9}\)
= \(\sqrt{16}\)
= 4 cm
c) Find the area of the isosceles trapezium ABCD.
Answer:
Area = \(\frac{1}{2}\) × DE(AB + CD)
= \(\frac{1}{2}\) × 4(13 + 7)
= 2 × 20
= 40 cm²
Question 18.
Read the understand the mathematical concept in the pattern given below. Write the answers to the following questions.
3² – 1² = 9 – 1 = 4 × 2
4² – 2² = 16 – 4 = 4 × 3
5² – 3² = 25 – 9 = 4 × 4
6² – 4² = 36 – 16 = 4 × 6
(a) Write the next line in the pattern.
(b) 9² – 7² = ____________ × 8
(c) 15² – ____________ = 4 × 14
(d) (x + 1)² – (x – 1)² = ____________
Answer:
a) 7² – 5² = 49 – 25 = 4 × 6
b) 9² – 7² – 81 -49 = 4 × 8
c) 15² – 13² = 4 × 14
d) (x + 1) – (x -1)²
= (x + 1 + x – 1) (x + 1 – x + 1)
= 2x × 2
= 4x.