Practicing with Class 8 Maths Question Paper Pdf Kerala Syllabus and First Term Question Paper 2022-23 will help students prepare effectively for their upcoming exams.
Class 8 Maths First Term Question Paper 2022-23 Kerala Syllabus
Time : 1½ Hours
Score : 40
Answer any 4 questions from 1 to 6. Each question carries 2 scores. (4 × 2 = 8)
Question 1.
In the figure, AB = QR, BC = PR, AC = PQ, ∠A = 50°, ∠B = 60°
(a) What is the measure of ∠C?
Answer:
∠C = 180 – (50 + 60) = 70°
(b) What is the measure of ∠P ?
Answer:
Since ∆ABC and ∆QRP are equal triangles, the angles opposite to the equal sides will be equal.
So ∠P= ∠C = 70°
Question 2.
2 added to three times a number gives 32. What is the number?
Answer:
32 – 2 = 30
∴ Number = \(\frac{30}{3}\) = 10
Question 3.
In a polygon with all angles equal, one inner angle is 144°.
(a) What is the measure of one outer angle?
Answer:
One outer angle = 180° – 144° = 36°
(b) How many sides does it have?
Answer:
No. of sides = \(\frac{360}{36}\) = 10
Question 4.
In the figure, AB = BC and ∠A = 40°
(a) What is the measure of ∠C ?
Answer:
Since AB = BC
∠A = ∠C = 40°
(b) What is the measure of ∠B?
Answer:
∴ ∠B = 180 – (40 + 40) = 100°
Question 5.
If 6 (x – 2) = 3 (x + 1), find ‘x’
Answer:
6(x – 2) = 3(x + 1)
∴ 6x – 12 = 3x + 3
3x = 15
x = \(\frac{15}{3}\) = 5
Question 6.
In triangle ABC, ∠A = 50°. ∠B = 70°
(a) What is the measure of ∠ACB?
Answer:
∠ACB = 180 – (50 + 70)
= 180°- 120°
= 60°
(b) What is the measure of ∠ACD?
Answer:
∠ACD = 180 – 60
= 120°
Answer any 4 Questions from 7 to 12. Each question carries 3 scores. (4 × 3 = 12)
Question 7.
(a) If x is an odd number, what is the next odd number?
Answer:
x + 2
(b) Sum of two consecutive odd numbers is 72. What are the numbers?
Answer:
x + x + 2 = 72
∴ 2x + 2 = 72
2x = 70
∴ x = \(\frac{70}{2}\) = 35
∴ numbers = 35, 37
Question 8.
(a) Draw a line 8.5 centimetres long.
(b) Draw its perpendicular bisector.
(c) Draw an equilateral triangle of side 4.25 centimetres long.
Answer:
Draw a line AB = 8.5 cm. Draw its perpendicular bisector.
Then AP= PB 4.25 cm. Now, draw an equilateral triangle with PB as length of the side.
Then ∆PBC is an equilateral triangle of side 4.25 cm.
Question 9.
The inner angle of a regular polygon is double that of its outer angle.
(a) Find the measure of one outer angle?
Answer:
Let outer angle Inner angle = 2x
∴ x + 2x = 180
3x = 180
x = \(\frac{180}{3}\) = 60°
One outer angle = 60°
(b) Find the measure of one inner angle?
Answer:
One inner angle = 180 – 60 = 120°
Question 10.
A hundred rupee note was changed to ten ru-pee notes and twenty rupee notes, eight notes in all.
(a) If we take the number of twenty rupee notes as ‘x’, what is the number of ten rupee notes?
Answer:
If number of 20 rupee notes = x, then
number of 10 rupee notes = 8 – x.
(b) Find the number of 10 rupee notes and 20 rupee notes?
Answer:
Total amount = 100
10(8 – x) + 20 x = 100
80 – 10x + 20x = 100
∴ 80 + 10x = 100
10x = 100 – 80 = 20
x = \(\frac{20}{10}\) = 2
20 rupee notes = 2
∴ 10 rupee notes = 8 – 2 = 6
Question 11.
Two right angled triangles are given in the figure. Lengths of two sides are given.
Answer:
(a) Find the length of third side of each triangle.
Answer:
AC = \(\sqrt{8^2+6^2}\) = 10 cm
QR = \(\sqrt{10^2-6^2}=\sqrt{64}\) = 8 cm
(b) Write equal angles of triangle ABC and triangle PQR?
Answer:
∆ABC and ∆PQR are equal
∠B = ∠Q, ∠C = ∠R, ∠A = ∠P
Question 12.
Consider a polygon having 42 sides.
(a) What is the sum of the inner angles of this polygon?
Answer:
Sum of inner angles = (n – 2) 180°
= (42 – 2) 180°
= 7200°
(b) What is the sum of the inner angles of a polygon with one side more?
Answer:
Sum of inner angles of a polygon with One side more = 7200 + 180 = 7380°
Answer any 5 Questions from 13 to 18. Each question carries 4 scores. (5 × 4 = 20)
Question 13.
In figure, ∠AOB = 71° OC is the bisector of ∠AOB.
(a) What is the measure of ∠AOC?
Answer:
∠AOC = \(\frac{71}{2}\) = 35\(\frac{1}{2}\)
(b) Draw triangle PQR with PQ = 6 centimetres,
∠P = 35\(\frac{1}{2}\) , ∠Q = 60°
Answer:
Draw PQ = 6 cm, ∠P = 71° and ∠Q = 60°.
Draw bisector of ∠P.
Now in ∆PQR, ∠P = \(\frac{71}{2}\) = 35.5°
Question 14.
Length of a rectangle is 3 metres more than two times the breadth. Perimeter of the rectangle is 36 metres.
(a) Length + Breadth = ____________
Answer:
\(\frac{36}{2}\) = 18 cm
(b) Find the length and breadth of the rectangle.
Answer:
If breadth = x
Length = 2x + 3
2x + 3 + x = 18
3x + 3 = 18
3x = 18 – 3 = 15
x = \(\frac{15}{3}\) = 5 cm
∴ Length = (2 × 5) + 3 = 13 cm
Question 15.
In the figure, a regular hexagon, a regular pentagon and a square put together.
(a) Write the measures ∠AQR, ∠AQB and ∠PQB.
Answer:
∠AQR = 90°, ∠AQBO = 108°, ∠PQB = 120°
(b) Find the measure of ∠PQR
Answer:
∠PQR = 360 – (90 + 108 + 120) = 42°
Question 16.
In the figure PQRS is a quadrilateral. PR is one of its diagonals.
(a) Find the measures of ∠PSR and ∠QPR
Answer:
∠PSR – 180 – (30 + 40) = – 110°
∠QPR = 180 – (30 + 110) = o
(b) Find the perimeter of quadrilateral PQRS.
Answer:
Perimeter of PQRS = (6 + 7) × 2 = 26 cm
(c) Write the most suitable name for the quadrilateral PQRS.
Answer:
Parallelogram
Question 17.
ABCDEF is a polygon
(a) What is the suitable name for this polygon?
Answer:
Hexagon
(b) How many diagonals can be drawn from any one of the vertices?
Answer:
3
(c) What is the sum of inner angles of this polygon?
Answer:
180 × 4 = 720°
(d) What is the sum of inner angles of a 7 sided polygon?
Answer:
720 + 180 = 900°
Question 18.
Look at the pattern.
1 + 2 + 3 = 6 = 3 × 2
2 + 3 + 4 = 9 = 3 × 3
3 + 4 + 5 = 12 = 3 × 4
……………………………
……………………………
……………………………
……………………………
(a) Write the next two lines of this pattern.
Answer:
4 + 5 + 6 = 15 = 3 × 5
5 + 6 + 7 = 18 = 3 × 6
(b) 99 + 100 + 101 = 3 × ____________
Answer:
99 + 100 + 101 = 3 × 100
(c) If the middle number of three consecutive natural numbers is ‘x’, what is their sum? .
Answer:
3x