Kerala Syllabus Class 8 Maths Model Question Paper Set 4

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Class 8 Maths Kerala Syllabus Model Question Paper Set 4

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.
Which of the following is a quadrilateral in which all the angles and sides are equal?
(a) Rhombus
(b) Rectangle
(c) Parallelogram
(d) Square
Answer:
(d) A square
All angles are 90 °.

Question 2.
The maximum temperature for seven consecutive days in a city is 26 ° C and the minimum temperature is 21 ° C. Temperature difference is,
(a) 6°
(b) 5°
(c) 10°
(d) 9°
Answer:
(b) 5°
26 – 21 = 5

Question 3.
Two statements are given below
p₁ : Whatever be the number of sides of a polygon the outer angle sum is 360°
p₂ inner angle sum increases by 180° if the number of sides increases by 1.
(a) p₁ and p₂ are true
(b) P₁ is true p₂ is false
(c) p₁ is false p₂ is true
(d) p₁ and p₂ are false.
Answer:
(a) p₁ and p₂ are true

Question 4.
In triangle ABC, AB: BC = 3:4, BC: CA =8:9 then AB: BC: CA =
(a) 3 : 5 : 8
(b) 8 : 12 : 15
(c) 6 : 8 : 9
(d) 1 : 4 : 9
Answer:
(b) 8 : 12 : 15

The actual lengths of AB and BC are 3 centimetres and 4 centimetres multiplied by the same number. Denoting this number as x, we have AB = 3x cm BC = 4x cm
Again, the actual lengths of BC and CA are 8 centimetres and 9 centimetres multiplied by another number. Denoting this number as y, we have BC = 8y cm CA = 9y cm
Now, 4x = 8y ⇒ y = \(\frac{1}{2}\)x
So, CA = 9 × \(\frac{1}{2}\)x = \(\frac{9}{2}\)x
AB : BC : CA = 3x: 4x: \(\frac{9}{2}\)x = 6: 8: 9

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

Question 5.
If x = \(\frac{1}{2}\) find x + x². Write this in the decimal form.
Answer:

Question 6.
A chord that is 8 cm long lies 3 cm away from the centre of a circle. Calculate the diameter of the circle.
Answer:
The perpendicular drawn from the centre to the chord bisects the chord.

OB² = 3² + 4² = 25
OB =5 cm
Therefore the diameter = 2 × 5 = 10 cm

Question 7.
In the diagram AB = AD, BC = DC

(a) If ∠BAC = 20° then what is ∠A ?
(b) Name pairs of equal triangles in the figure
Answer:
(a) Triangle BAC and triangle DAC are equal ∠BAC = ∠DAC = 20°
∠A = ∠BAC + ∠DAC = 20° + 20° = 40°

(b) Triangle AOB and triangle AOD are equal Triangle BOC and triangle DOC are equal Triangle ABC and triangle ADC are equal Triangle ABD and triangle CBD are equal

Question 8.
23² = 529
(i) What is 24²?
(ii) What is 26²?
Answer:
23² = 529
(i) 24² = (23 + 1)²
= 23² + 2 × 23 × 1 + 1²

= 529 + 46 + 1 = 576
(ii) 26² = (23 + 3)²
= 23² + 2 × 23 × 3 + 3²
= 529 + 138 + 9 = 676

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

Question 9.
The measure of an outer angle of a regular polygon is 2x, and the measure of an inner angle is 4x.
(a) Use the relationship between inner and outer angles to find x.
(b) Find the measure of one inner and outer angle.
(c) Find the number of sides in the polygon and the type of polygon.
Answer:
(a) Sum of the inner angle sum at a comer is 180° 6x = 180, x = 30
(b) Angles are 60°, 120°
(c) Regular hexagon

Question 10.
Sum of the digits of a two digit number is 10. On reversing the digits it is found that the number decreases by 36. Find the number.
Answer:
Digit in ten’s place is x. Digit in one’s place is 10 – x
Number is 10x + 10 – x
When the digits are reversed number becomes 10(10 – x) + x
It is given that 10(10 – x) + x = 10x + 10 – x – 36
100 – 10x + x = 9x – 26, 100 – 9x = 9x – 26, 126 = 18x,
x = \(\frac{126}{18}\) = 7
Number is 73

Question 11.
[A] In the following figure, four equal rectangles joined to form a square.

Area of the inner square is \(\frac{1}{4}\) part of the area of the outer square. Find the ratio between larger and smaller sides of the rectangle.
Answer:
Let l be the larger side (length) and b be the smaller side(breadth) of the rectangle.
Side of the inner square = l – b
Area of the inner square = (l – b)²
Side of the outer square = l + b
Area of the outer square = (l + b)²
Area of the inner square is \(\frac{1}{4}\) part of the area of the outer square.
\(\frac{(l-b)^2}{(l+b)^2}=\frac{1}{4}\)
\(\frac{l-b}{l+b}=\frac{1}{2}\)
2 (l – b) = l + b
⇒ 2l – l = b + 2b
⇒ l = 3b
⇒ \(\frac{l}{b}=\frac{3}{1}\)
So, the ratio between larger and smaller sides of the rectangle is 3:1

OR

[B] The figure shown below is draw using 2 equal rhombuses. Draw the figure as per the given measures.

Answer:
Draw AB 4 cm first as in the fig. then take 60° at B and 120° at A, on each side of AB there complete the fig:

Question 12.
[A] Look at the pattern given below. These are the decimal forms of fractions:
\(\frac{1}{11}\) = 0.0909 …….
\(\frac{2}{11}\) = 0.1818……..
\(\frac{3}{11}\) = 0.2727…….
……………………………….
(a) Write the next line.
Answer:
\(\frac{4}{11}\) = 0.3636…….

(b) Write the fraction form of 0.6363…
Answer:
\(\frac{7}{11}\)

(c) Write the decimal form of 0.8181… + 0.7272…
Answer:
\(\frac{9}{11}+\frac{8}{11}=\frac{17}{11}\)
= 1 + \(\frac{6}{11}\)
= 1.5454…….

OR

[B] The angles of a triangle are in the ratio 2:3:5. How much is each angle?
Answer:
180 × \(\frac{2}{10}\) = 36°
180 × \(\frac{3}{10}\) = 54°
180 × \(\frac{5}{10}\) = 90°

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

Question 13.
Draw an isosceles triangle whose length of its one side is 9.5 centimetres, and the height to this side is 4.75 centimetre.
Answer:
Draw a line of length 9.5 cm. draw the perpendicular bisector of this line.
Mark the half length of the line on the top of its perpendicular bisector.
From this point join the both ends of the line and form a triangle.

Question 14.
Draw a rhombus of diagonals 6 cm and 8 cm.
Answer:
Draw 6 cm long line and its perpendicular bisector. Mark two points upper and lower 4 cm from the midpoint. Join these points with the ends of the first diagonal. We get rhombus.

Question 15.
[A] Electricity consumption of 30 families in a locality is given in the table below. Draw a histogram.

Electricity Consumption	Number of families
100 – 200	5
200 – 300	7
300 – 400	8
400 – 500	6
500 – 600	4

Answer:

OR

[B] In the given figure a rhombus is formed by joining the midpoints of the sides of the rectangle. AB =6 centimetres. BC = 4 centimetres.

(a) What is the area of the rectangles?
Answer:
6 × 4 = 24 cm²

(b) Find the area of the rhombus
Answer:
\(\frac{1}{2}\) × 24 = 12cm²

(c) Find the area of rhombus with diagonals 12 centimetres and 8 centrimetres
Answer:
\(\frac{1}{2}\) × 12 × 8 = 48 cm²

Question 16.
[A] Read the mathematical concept given below and answer any four questions that follow. Each question carries 1 score. (4 × 1 = 4)
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
……………….
……………….
The numbers which can be expressed as the sum of consecutive natural numbers starting from 1 are known as trianglular numbers.
1² = 1
2² = 4
3² = 9
………………..
………………..
The squares of natural numbers are known as square numbers.
That is,
1, 3, 6, 10, ……… Are trianglular numbers.
1, 4, 9, ………… Are square numbers.
(a) Which is the fouth triangular number?
(b) How many first natural numbers are added to get 5th triangular number?
(c) Which is the 6th square number?
(d) Find the two digit number which is both triangular and square number
(e) What is the difference between the 6th square number and the 6th triangular number?
(f) Write a square number between 70 and 90
Answer:
(a) 10
(b) 5
(c) 36
(d) 36
(e) 15
(f) 81

OR

[B] 3² – 1² = 4 × 2
4² – 2² = 4 × 3
5² – 3² = 4 × 4
6² – 4² = 4 × 5
(i) Write the next twolines of this pattern.
(ii) What is the principle behind this pattern.
(iii) Explain it using algebra.
Answer:
i) 7² – 5² = 4 × 6
8² – 6² = 4 × 7

ii) The difference in the squares of alternate natural numbers always be a multiple of 4

iii) (n + 2)² – n² = 4(n + 1)