Kerala Syllabus Class 9 Maths Model Question Paper Set 4 English Medium

Teachers recommend solving Maths Model Question Paper for Class 9 Kerala Syllabus Set 4 to improve time management during exams.

Kerala Syllabus Std 9 Maths Model Question Paper Set 4 English Medium

Time: 2½ Hours
Max Score: 80 Marks

Instructions:

• There is a ‘cool off’ time 15 minutes in addition to the writing time. Use this time to get familiar with questions and plan your answers.
• Read the instructions carefully before answering the questions.
• Keep in mind, the score and time while answering the questions. Give explanations wherever necessary.
• No need to simplify irrationals like √2, √3, π etc. using approximations unless you are asked to do so.

Answer any 3 Questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
The difference between the areas of the two squares is 56, and the sum of their perimeters is 56. Find the side length of each square.
Answer:
Let ‘x’ be the side of first square and ‘y’ be the side of second square.
x² – y² = 56 …………… (1)
4x + 4y = 56
x + y = \(\frac{56}{4}\) = 14 ……………… (2)
(1) → (x + y)(x – y) = 56
14 (x – y) = 56
x – y = 4 ………….. (3)
(2) + (3) → x = \(\frac{14 + 4}{2}\) = 9
Therefore, y = 5

Question 2.
The difference of two integers is 11 and \(\frac{1}{5}\)^th of their sum is 5. What are the numbers?
Answer:
Let x and y be the two integers and x > y then we have,
x – y = 11 …………… (1)
\(\frac{x+y}{5}\) = 5 ⇒ x + y = 25 ……………. (2)
Adding (1) and (2) we get, x = \(\frac{25 + 11}{2}\) = 18
Subtracting equation (2) from (1) we get, y = \(\frac{25 – 11}{2}\) = 7
Thus, the number are 18 and 7.

Question 3.
In triangle ABC, the line parallel to BC meets AB and AC at D and E respectively. AE = 4.5 cm,
\(\frac{A D}{D B}\) = \(\frac{2}{5}\). Then find EC.

Answer:
We know that \(\frac{A D}{D B}\) = \(\frac{A E}{E C}\)
\(\frac{2}{5}\) = \(\frac{4.5}{E C}\)
EC = 4.5 × \(\frac{5}{2}\) = 11.25 cm.

Question 4.
Find the area and perimeter of a rectangle with sides 3√8 cm and 7√8 cm.
Answer:
Perimeter = 3√8 + 7√8 = 10√8 cm
Area = 3√8 × 7√8
= 3 × √8 × 7 × √8
= 3 × 7 × √8 × √8
= 21 × 8
= 168 cm²

Answer any 4 Questions from 5 to 10. Each question carries 3 scores. (4 × 3 = 12)

Question 5.
In the figure ∠Q = 90°, QR = 5 cm, SR = 3 cm, QS is perpendicular to PR.
a) Find the length of QS.
b) Find the length of PS.

Answer:
a) ΔSQR is a right triangle. So,
QS² = QR² – SR²
= 5² – 3²
= 25 – 9
= 16
∴ QS = \(\sqrt{16}\) = 4 cm

b) PS × SR = QS²
PS × 3 = 4²
PS = \(\frac{16}{3}\) cm

Question 6.
In the figure ∠A = ∠P, ∠B = ∠Q, AB = 5 cm. BC = 4 cm, AC = 2 cm, PR = 6 cm.

a) What is the length of PQ?
b) What is the ratio of the perimeters of ΔABC and ΔPQR?
Answer:
a) PQ = 5 × 3 = 15 cm, QR = 12 cm
b) \(\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}\) = \(\frac{11}{6 + 15 + 12}\) = \(\frac{11}{33}\) = 1 : 3

Question 7.
The product of two natural numbers is 300, and their sum is 35, then
a) What is the product of the two natural numbers that come just after each of these two natural numbers?
b) What is the product of the two natural numbers that come just before each of these two natural numbers?
Answer:
Let the numbers are x and y.
According to the question, xy = 300 and x + y = 35
a) (x + 1 )(y + 1) = xy + (x + y) + 1
= 300 + 35 + 1
= 336

b) (x – 1 )(y – 1) = xy – (x + y) + 1
= 300 – 35 + 1
= 266

Question 8.
The product of two odd numbers is 899, and the sum is 60. What is the product of the two odd numbers that come just after each of these two odd numbers?
Answer:
Let the odd numbers be x and y, then,
xy = 899 and x + y = 60
The odd number just after x is x + 2
The odd number just after y is y + 2
(x + 2)(y + 2) = xy + 2(x + y) + z²
= 899 + 2 × 60 + 4
= 1023

Question 9.
In the picture below, the midpoints of the sides of ΔABC are joined to make ΔPQR. Line AR is the median of ΔABC.

(i) Prove that this median bisects the top side of the small triangle.
(ii) Prove that the centroid of the large and small triangles are the same.
Answer:

(i) Since P, Q, R are the mid points of the sides APRQ is a parallelogram. So, its diagonals bisect each other.
i.e, PS = SQ
i.e, the median AR bisected PQ
(ii) Draw BQ and CP. As we did earlier, we can show that PM = MR and RN = NQ

ie, PN, QM, SR are the medians of triangle PQR, then G is the centroid of triangle PQR.
Already P, Q, R are the mid points of sides AB, AC and BC. Then AR, BQ, CP are medians. These are along SR, QM and PN. So, AR, BQ and CP also pass through the point G. This means G is the centroid of both triangles.

Question 10.
The area of a square is 3 cm²,
a) Find one of its lengths.
b) Find its perimeter.
c) Find the length of its diagonal.
Answer:
a) One side = √3 cm
b) perimeter = 4√3 cm
c) Length of its diagonal = \(\sqrt{(\sqrt{3})^2+(\sqrt{3})^2}\)
= √6 cm

Answer any 8 Questions from 11 to 21. Each question carries 4 scores. (8 × 4 = 32)

Question 11.
The cost of two tables and one chair is 9000 rupees, and the cost of one table and four chairs is 8000 rupees.
a) Form two equations based on the given facts.
b) What is the price of one chair? What is the price of one table?
Answer:
a) Let, t be the price of a table and c be the price of a chair.
From the question we have;
2t + c = 9000 ……………… (1)
t + 4c = 8000 …………… (2)

b) (2) × 2 → 2t + 8c = 16000 …………… (3)
(3) – (1) → 7c = 7000
c = 1000
(2) → t + 4 × 1000 = 8000
t = 8000 – 4000
= 4000
Thus, price of a chair = 1000
price of a table = 4000

Question 12.
In the figure, the vertical parallel lines divide line AB into three equal parts

a) AP: PQ: QR = _____:_______:________
b) Draw an equilateral triangle with a perimeter of 13 centimetres.
Answer:
a) AP: PQ: QR = 1: 1: 1

Question 13.
The sum of five times of a number and two times of another number gives 20. The sum of two times the first number and six times the second number gives 34. Find the numbers.
Answer:
Let m and n be the two numbers.
From the question we have,
5m + 2n = 20 ………….. (1)
2m + 6n = 34 …………… (2)
(1) × 2 → 10m + 4n = 40 … (3)
(2) × 5 → 10m + 30n = 170 … (4)
(4) – (3) → 26n = 130
n = 5
(1) → 5m + 2 × 5 = 20
5m = 20 – 10 = 10
m = 2
Thus 2 and 5 are the numbers.

Question 14.
AD and BE are the medians of triangle ABC. They intersect at G.

a) Find AG: GD
b) Find BG if GE = 3 centimetres.
c) If the area of triangle ABC is 60 square centimetres, find the area of triangle ABD.
Answer:
a) 2 : 1

b) BG:GE = 2 : 1 ⇒ BG is double that of GE
⇒ BG = 2 × GE
= 2 × 3
= 6 cm

c) As AD is the median* it divides the triangle into two equal parts. So,
Area of ΔABD = \(\frac{1}{2}\) × area of ΔABC
= \(\frac{1}{2}\) × 60
= 30 sq. cm

Question 15.
In the figure, the length of the rectangle is the √2 + 1 metres, and its breadth is √2 – 1 metres

a) Find its area.
b) The area of a rectangle is 1 square metre and its length is 2 + √3 metres. Find its breadth correct to a centimetre. (√3 ≈ 1.732)
Answer:
a) Area = (√2 + 1)(√2 – 1)
= (√2)² – 1²
= 2 – 1
= 1 sq.cm

b) Area = length × breadth
1 = (2 + √3) × breadth
breadth = \(\frac{1}{(2+\sqrt{3})}\)
= \(\frac{(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}\)
= \(\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^3}\)
= \(\frac{2-\sqrt{3}}{4-3}\)
= 2 – √3
= 2 – 1.73
= 0.27 cm

Question 16.
a) Write the natural number equal to \(\frac{\sqrt{12}}{\sqrt{3}}\)
b) Draw a number line and mark √5 in this line.
Answer:

Question 17.
Draw a triangle of angles same as those of the triangle shown and sides scaled by 1\(\frac{1}{2}\).

Answer:
6 × 1.5 = 9
7 × 1.5 = 10.5

ΔAPQ is the required triangle.

Question 18.
Four vertices of a square are on the circle.

The length of the diagonal of the square is 4√2 centimetres, then
a) What is the radius of the circle?
b) Find the area of the circle.
c) If the radius of the circle is 4√2 centimetres, what is the area of the square?
Answer:
a) 2√2 cm
b) Area = πr² = π × (2√2)² = 8π cm²
c) Area = πr² = π × (4√2)² = 32π cm²

Question 19.
The product of two natural numbers just after each of two natural numbers is 2021, and the product of the natural numbers just before each is 2001. What is the product of numbers?
Answer:
Let the numbers be x and y
According to the question
(x + 1)(y + 1) = 2021
xy + (x + y) + 1 = 2021 ………………. (1)
(x – 1)(y – 1) = 2001
xy – (x + y) + 1 =2001 ………….. (2)
(1) + (2) → 2xy + 2 = 4022
2xy = 4020
xy = \(\frac{4020}{2}\) = 2010

Question 20.
The perimeter of a rectangle is 40 centimetres and its area is 70 square centimetres. Find the area of the rectangle with each side 3 centimetres shorter.
Answer:
Perimeter = 40 cm
Area = 70 square cm.
Let x be the length and y be the breadth,
2(x + y) = 40 ⇒ x + y = 20
xy = 70
Now, Area of the rectangle with each side 3 centimetres shorter
= (x – 3)(y – 3)
= xy – 3x – 3y + 9
= xy – 3(x + y) + 9
= 70 – 3 × 20 + 9
= 19 square cm.

Question 21.
(i) Take four consecutive natural numbers or their negatives as a, b, c, d and calculate a – b – c + d.
(ii) Explain using algebra why it is zero for all such numbers.
(iii) What do we get if we calculate a + b – c – d instead of a – b – c + d?
(iv) What about a – b + c – d?
Answer:
i) Let, a = 1, b = 2, c = 3, d = 4
a – b – c + d = 1 – 2 – 3 + 4
= – 1 – 3 + 4
= – 4 + 4
= 0

ii) Let a = x, b = x + 1, c = x + 2, d = x + 3
a – b – c + d = x – (x + 1) – (x + 2) + x + 3
= x – x – 1 – x – 2 + x + 3
= 0

iii) a + b – c – d = 1 + 2 – 3 – 4
= 3 – 3 – 4
= 0 – 4
= -4

iv) a – b + c- d = 1 – 2 + 3 – 4
= – 1 + 3 – 4
= 2 – 4 = -2

Answer any 6 Questions from 22 to 29. Each question carries 5 scores. (6 × 5 = 30)

Question 22.
In the figure, AB = 2 centimetres, AC = 4 centimetres and BC = 5 centimetres. The length of OP, OQ and OR are twice the length of OA, OB and OC respectively.

a) What are the lengths of sides of triangle PQR?
b) AB = 4 centimetres, ∠A = 40°, ∠B = 60°. Draw the triangle ABC, and draw a triangle whose sides are twice the sides of triangle ABC.
Answer:
a) sides of ΔPQR are twice that of ΔABC. Thus,
PQ = 2 × AB = 2 × 2 = 4 cm
QR = 2 × BC = 2 × 5 = 10 cm
RP = 2 × CA = 2 × 4 = 8 cm

Question 23.
We know,
\(\sqrt{12}\) = \(\sqrt{4 \times 3}\) = √4 × √3 = 2√3
\(\sqrt{18}\) = \(\sqrt{9 \times 2}\) = √9 × √2 = 3√2
a) Write \(\sqrt{32}\) and \(\sqrt{50}\) as shown above.
b) Find \(\sqrt{50}\) + \(\sqrt{32}\)
c) Find \(\sqrt{50}\) – \(\sqrt{32}\)
Answer:

Question 24.
(a) Draw a line of length 13 centimetres and divide it in the ratio 2 : 3 : 4
(b) Draw a triangle of perimeter 13 centimetres and sides in the ratio 2 : 3 : 4
Answer:

Question 25.
In the figure, the radius of a larger circle is 2 times the radius of a smaller one.

a) If PQ = 5 centimetres, find AB.
b) Draw a triangle of sides 4 centimetres, 5 centimetres and 6 centimetres and draw another triangle with sides scaled by two.
Answer:
a) AB = 2 × PQ = 2 × 5 = 10 cm

Question 26.
Find the product of the following pairs of numbers. Which are the pairs whose product is a natural number?
a) 3, \(\sqrt{12}\)
b) \(\sqrt{0.3}\), \(\sqrt{1.2}\)
c) √5, √7
d) \(\sqrt{0.5}\), \(\sqrt{8}\)
e) \(\sqrt{7 \frac{1}{2}}\), \(\sqrt{3 \frac{1}{3}}\)
Answer:

Question 27.
The measures given in the quadrilateral are in centimetres. Draw a quadrilateral with same angle and sides scaled by 11/2 times.

Answer:
6 × 1.5 = 9
5 × 1.5 = 7.5
4 × 1.5 = 6

Question 28.
In the multiplication table we’ve made, draw a square of nine numbers, instead of four, and mark the numbers at the four corners

i) What is the difference of the diagonal sums?
ii) Explain using algebra why we get the difference as 4 in all such squares. What about a square of sixteen numbers?
Answer:
i) When adding the diagonal pair, we get
6 + 20 = 26
10 + 12 = 22
Difference = 4

ii) We can algebraically express the numbers in the square as follows.

n	n + 2	n + 4
n + 3	n + 6	n + 9
n + 6	n + 10	n + 14

By adding numbers in the four corners diagonally, we get
n + (n + 14) = 2n + 14
n + 4 + (n + 6) = 2n + 10
Difference = 4

iii)

6	8	10	12
9	12	15	18
12	16	20	24
15	20	25	30

By adding the numbers of corner, we get
6 + 30 =36
15 + 12 = 27
Difference = 9

Question 29.
There are two squares. One side of the larger square is 5 centimetres more than the smaller. Difference between their areas is 55 square centimetres. Find the sides of each square.
Answer:
Let, side length of the larger square = x
side length of the smaller square = y
We have, x – y = 5 ………….. (1)
x² – y² = 55
(x + y)(x – y) = 55
(x + y)5 = 55
x + y = 11 ……………. (2)
(1) + (2) → 2x = 16
x = 8
(2) → 8 + y = 11
y = 3
Side of the larger square = 8 cm.
Side of the smaller square = 3 cm.