Kerala Syllabus Class 9 Maths Model Question Paper Set 2 English Medium

Teachers recommend solving Maths Model Question Paper for Class 9 Kerala Syllabus Set 2 to improve time management during exams.

Kerala Syllabus Std 9 Maths Model Question Paper Set 2 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.
In square ABCD, AC = 12 cm. Find the length of a side of the square, correct to a cm.
Answer:
AC is the diagonal.
Length of the diagonal = 12 cm
side × √2 = 12
side = \(\frac{12}{\sqrt{2}}\) = \(\frac{12 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}\) = \(\frac{12 \times \sqrt{2}}{2}\) = 6 × √2 = 6 × 1.41 = 8.46 cm

Question 2.
The perimeter of a rectangle is 40 centimeters, and its breadth is 4 centimeters less than the length. Compute the length and breadth.
Answer:
Let, length = x
breadth = y
perimeter = 40 cm
⇒ 2(x + y) = 40
x + y = 20 ……………. (1)
breadth is 4 cm less than the length gives,
x – y = 4 …………. (2)
(1) + (2) → 2x = 24
x = 12
(1) → 12 + y = 20
y = 8
Thus, the length = 12 cm.
the breadth = 8 cm.

Question 3.
AB, CD, and EF are parallel lines. AC = x, CE = x + 1, BD = 12, DF = 16.

a) What is the relation between AC, CE, BD and DF?
b) Find x.
Answer:
a) \(\frac{A C}{C E}\) = \(\frac{B D}{D F}\)

b) \(\frac{x}{x+1}\) = \(\frac{12}{16}\)
\(\frac{x}{x+1}\) = \(\frac{3}{4}\)
4x = 3x + 3
x = 3

Question 4.
Compare the two triangles given below and find the lengths of QR and PR.

Answer:
\(\frac{A C}{P Q}\) = \(\frac{A B}{Q R}\) = \(\frac{B C}{P R}\)
\(\frac{6}{3}\) = \(\frac{8}{Q R}\) = \(\frac{4}{P R}\)
2 = \(\frac{8}{Q R}\) = \(\frac{4}{P R}\)
2 = \(\frac{8}{Q R}\) ⇒ QR = 4
2 = \(\frac{4}{Q R}\) ⇒ PR = 2

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

Question 5.
a) Find the sum and product of 2 + √3 and 2 – √3
b) Write a pair of irrational numbers whose sum is a rational number.
Answer:
a) (2 + √3)+ (2 – √3) = 4
(2 + √3)(2 – √3) = 4 – 2√3 + 2√3 – 3
= 1

b) (5 + √3), (4 – √3)

Question 6.
In the figure, the lines AB and CD intersect at point P. Prove that the length of PB is a third of the length of AP.

Answer:
ΔAPC and ΔBPD are similar.
So, \(\frac{A C}{B D}\) = \(\frac{A P}{P B}\)
\(\frac{6}{2}\) = \(\frac{A P}{P B}\)
3 = \(\frac{A P}{P B}\)
PB = \(\frac{A P}{3}\)
Thus PB is a third of the length of AP.

Question 7.
In the figure, the line DB is extended and the perpendicular to BC at C meets at E. Here, AB is parallel to CE and AD is parallel to CB.

a) Prove that the triangles ADB and CBE have the same angles.
b) Compute the length of CE.
Answer:
a) In ΔADB and ΔCBE
∠A = ∠C = 90°
AB || CE ⇒ ∠ABD = ∠CEB(Corresponding angles)
AD || CB ⇒ ∠ADB = ∠CBE (Corresponding angles)
Therefore, both the triangles have same angles.

b) Consider ΔADB and ΔCBE. Since all the angles are equal, they are similar triangles.
∴ \(\frac{A D}{C B}\) = \(\frac{A B}{C E}\) ⇒ \(\frac{3}{1}\) = \(\frac{7}{C E}\)
⇒ 3CE = 7
⇒ CE = \(\frac{7}{3}\) = 2\(\frac{1}{3}\) cm

Question 8.
In the figure A, B, C are the midpoints of the sides of triangle PQR,

a) If AC = 4 centimetres, find QR.
b) If the perimeter of the triangle ABC is 16 centimetres, find the perimeter of the triangle PQR.
Answer:
a) The length of the line -joining the midpoints of two sides of a triangle is half the length of the third side
∴ \(\frac{Q R}{2}\) = AC
⇒ QR = 2AC
= 2 × 4
= 8 cm

b) Perimeter of ΔABC = AB + BC + AC
but AB = \(\frac{1}{2}\)PR
AC = \(\frac{1}{2}\)QR
CB = \(\frac{1}{2}\)PQ
Now, Perimeter of ΔABC = AB + BC + AC = 16
\(\frac{1}{2}\)[PR + QR + PQ] = 16
PR + QR + PQ = 16 × 2 = 32 cm
i.e., perimeter of ΔPQR = 32 cm

Question 9.
Area of a square is 2 m².
a) What is the length of its side?
b) Calculate the perimeter of the square.
c) What is the length of its diagonal?
Answer:
Given, Area of the square is 2 m²
a) Length of one side of the square = \(\sqrt{\text { Area of the square }}\)
= √2m

b) Perimeter of the square = 4 × side
= 4 × √2
= 4√2 m

c) Length of diagonal = side × √2
= √2 × √2
= 2 m

Question 10.
Four years ago, the age of the father was three times that of his son. Eight years later, the age of the father will be twice that of his son. Find the present age of the father and the son.
Answer:
Let x be the present age of father and y be the present age of son.
4 years ago, father’s age was x – 4 and son’s age was y – 4.
According to the question,
(x – 4) = 3(y – 4)
x – 4 = 3y – 12
x – 3y = – 12 + 4
x – 3y = -8 ………… (1)
8 years later, father’s age becomes x + 8 and son’s age becomes y + 8.
According to the question:
(x + 8) = 2 (y + 8)
x + 8 = 2y + 16
x – 2y = 8 ……………. (2)
(2) – (1) → – 2y + 3y = 8 + 8
y = 16
Putting y = 16 in (2) we get,
x – 2 × 16 = 8
x = 8 + 32 = 40
∴ Present age of father = 40
Present age of son = 16

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

Question 11.
Calculate \(\sqrt{18}\) – \(\frac{1}{\sqrt{2}}\) = correct to a millimetre.
Answer:

Question 12.
The diagonals of a quadrilateral are perpendicular to each other. The midpoints of its sides are joined to form another quadrilateral. What is the speciality of this quadrilateral? What is the reason?
Answer:

Given: ABCD is a quadrilateral having AC perpendicular to BD. P, Q, R, S are midpoints of its sides. The line joining the midpoint of 2 sides of a triangle is parallel to 3rd side. ⇒ PQ || BD
∴ ∠Q = 90° (parallel lines property).
Similarly ∠P = ∠Q = ∠R = ∠S = 90° ; PQRS is a rectangle.

Question 13.
In the figure, a line is drawn inside an isosceles triangle parallel to the base. What is its length?

a) What is the length of such a parallel line 6 centimeters down from the top of the triangle?
b) Prove that the length of such a line varies proportionally as its downward distance.
Answer:
In the figure Δ ABC is isosceles triangle.

Draw a line AF perpendicular to the line PQ and BC.
Now, AE = 3 cm and AF = 9 cm
Draw a line parallel to BC and it meets AF at G.
Now consider Δ APQ and Δ ABC, their corresponding angles are equal to each other.
Therefore, sides opposite to the equal angles are proportional and the altitudes of the triangles are also proportional.
∴ \(\frac{A E}{A F}\) = \(\frac{P Q}{B C}\)
⇒ \(\frac{3}{9}\) = \(\frac{P Q}{6}\)
⇒ PQ = \(\frac{6 \times 3}{9}\)
= 2 cm

a) In Δ ARS and Δ ABC
Given that AG = 6 cm
∴ \(\frac{A G}{A F}\) = \(\frac{R S}{B C}\)
⇒ \(\frac{6}{9}\) = \(\frac{R S}{6}\)
⇒ RS = \(\frac{6 \times 6}{9}\) = 4 cm

b) Let xy be the line parallel to BC and distance between A and xy be cm.
∴ \(\frac{t}{A F}\) = \(\frac{x y}{6}\)
⇒ \(\frac{t}{9}\) = \(\frac{x y}{6}\)
⇒ \(\frac{t}{x y}\) = \(\frac{9}{6}\) = \(\frac{3}{2}\)
i.e., t and xy are proportional.

Question 14.
a) Draw a line and mark five points on it which are 3 units apart.
b) Mark the points -2, -1, 0, 1, 2 on a number line.
c) Mark the point √2 on a line.
Answer:

Question 15.
Draw a triangle of perimeter 11 centimeters and sides in the ratio 2 : 3 : 3.
Answer:

Question 16.
In ΔPQR, ∠Q = 90°, PQ = QR

ABCQ is a square.
a) Find the measures of ∠P and ∠R.
b) Which are the two similar triangles in the picture?
c) If CR = 1 cm, find the length of PA.
d) What is the area of ABCQ?
Answer:
a) Given that PQ and QR are equal.
Angles opposite to the equal sides are equal to each other.
∴ ∠P = ∠R – 45°

b) Δ PAB and Δ RCB.

c) Since Δ PAB and Δ RCB are similar, ratio of their corresponding side are equal.
\(\frac{P A}{C R}\) = \(\frac{A B}{B C}\)
\(\frac{P A}{1}\) = 1 [∵ AB = BC]
PA = 1 cm

d) Let x be the length of each side of the square.
we have Δ PQR and Δ BCR are similar
⇒ \(\frac{P Q}{B C}\) = \(\frac{Q R}{C R}\)
⇒ \(\frac{1+x}{x}\) = \(\frac{1+x}{1}\)
⇒ x = 1
∴ Area of ABQC = 1sq cm

Question 17.
a) Which among the following products are natural numbers?
(\(\sqrt{10}\) × √2, \(\sqrt{12}\) × √3, \(\sqrt{10}\) × \(\frac{1}{\sqrt{2}}\), \(\sqrt{10}\) × \(\frac{1}{\sqrt{3}}\))
b) The area of a rectangle is \(\sqrt{128}\) square metres and its length is 4 metres. Find its breadth.
Answer:
a) \(\sqrt{12}\) × √3
b) Given,
Area of a rectangle = \(\sqrt{128}\) m²
length = 4 m
We know, Area of a rectangle = length × breadth
∴ breadth = \(\frac{\text { Area of rectangle }}{\text { length }}\) = \(\frac{\sqrt{128}}{4}\) = \(\frac{8 \sqrt{2}}{4}\) = 2√2 m

Question 18.
In triangle ABC, P and Q are midpoints of AC and BC respectively. X and Y are midpoints of AM and BM respectively.

a) AB = 10 centimetres. Find the lengths of PQ and XY.
b) Prove that AX = XM = MQ.
Answer:
a)
PQ = \(\frac{1}{2}\)AB
= \(\frac{1}{2}\) × 10
= 5 cm
XY = \(\frac{1}{2}\)AB
= \(\frac{1}{2}\) × 10
= 5 cm

b) AQ is the median of the Δ ABC
∴ AM : MQ = 2 : 1
Or MQ = \(\frac{A M}{2}\) …………. (1)
Since X is the midpoint of AM
we can write, AX = XM = \(\frac{A M}{2}\) ………….. (2)
from (1) & (2)
AX = XM = MQ

Question 19.
Draw a triangle with perimeter 13 centimetres whose sides are in the ratio 2 : 3 : 4.
Answer:

Question 20.
In the right triangle ABC, P and Q are the midpoints of AB and BC respectively. O is the centre of the circle.

a) What is the measure of angle APO?
b) Prove that triangles APO and CQO are similar.
Answer:
a) ∠APO = ∠OQC = 90°
b) ∠APO = ∠OQC = 90°
Let ∠A = x
∴ ∠C = 90° – x°
Since ∠A = x°, ∠AOP = 90° – x°
Since ∠C = 90° – x°, ∠QOC = x°
i. e., ∠A – ∠QOC = x°
∠C = ∠AOP = 90 – x°
All corresponding angles of these two triangles are equal
∴ ΔAPQ and ΔCQO are similar.

Question 21.
The difference between two numbers is 6 and the difference between its squares is 48.
a) Form the equations indicating above statement.
b) What is the sum of the numbers?
c) What are the numbers?
Answer:
Let two numbers be x and y.
a) According to the question,
x – y = 6 …………… (1)
x² – y² = 48 …………. (2)

b) x² – y² = 48
(x + y)(x – y) = 48
(x + y) × 6 = 48
x + y = \(\frac{48}{6}\) = 8 ……….. (3)

c) Adding (1) and (2) we get,
2x = 14
x = \(\frac{14}{2}\) = 7
from(1) ⇒ 7 – y = 6y = 1
Thus 7 and 1 are the numbers.

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

Question 22.
The area of a square is 8 cm².
a) What is its side length?
b) Draw this square.
c) What is the length of its diagonal?
Answer:
a) (side)² = area
= 8
side = √8 = \(\sqrt{4 \times 2}\) = 2√2 cm

b)

c) Length of the diagonal = side × √2 = 2√2 × √2 = 2 × 2 = 4 cm

Question 23.
The total number of cars and autorickshaws in a garage is 23. The total number of wheels of these vehicles is 81.
a) If the number of cars is taken as x, then what is the total number of wheels of cars?
b) Find the number of cars and autorickshaws.
Answer:
Given, Total number of cars and autorickshaws in a garage = 23
Total number of wheels of these vehicles = 81
a) Let number of cars = x
One car has four wheels
∴ Total number of wheels of cars = 4x

b) Number of autorickshaws = 23 – x
We have, 4 x + 3 (23 – x) = 81
4 x + 69 – 3x = 81 x = 81 – 69 = 12
Number of cars = x = 12
Number of autorickshaws = 23 – x = 23 – 12 = 11

Question 24.
In the figure, PQ and BC are parallel. AB = 8 centimetres and PB = 2 centimetres.

a) FindAP: AB.
b) In triangle XYZ, XY = 8 centimetres.
∠X = 50°, ∠Y= 40°. Draw the triangle having angles same to that of XYZ and with sides 3/4^th of the sides of XYZ.
Answer:
a) We know, AB = AP + PB
AP = AB – PB = 8 – 2 = 6 cm
Thus,
\(\frac{A P}{A B}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)
Therefore, AP : AB = 3 : 4

b) The length of the new side = 8 × \(\frac{3}{4}\) = 6 cm

Thus, ΔXY’Z’ is the required triangle having angles same to that of XYZ and with sides the sides \(\frac{3}{4}\)^th of XYZ.

Question 25.
In triangle ABC. ∠ACB = 90° and CD is perpendicular to AB.

a) If ∠A= 50°, find ∠ACD.
b) Prove that angles of triangles ACD and BCD are equal.
c) Prove that if the perpendicular from the right-angled vertex of a right triangle divides the
opposite side into parts of lengths a and b and if the length of the perpendicular is h, then h² = ab
Answer:
a) Consider the right triangle ACD
∠ACD = 90° – ∠A
∠ACD = 90° – 50°
∠ACD = 40°
So, ∠ACD = 40°

b) In Δ ACD,
∠CDA = 90°
∠ACD = 40° (proved above)
In Δ BCD,
∠CDB = 90°, ∠B = 180° – 90° – 50° = 40°
Therefore,
∠BCD = 90°- ∠B = 90° – 40° = 50°
Thus, ∠ACD = ∠BCD = 40° , ∠ACD = ∠BCD = 90° and ∠ACD = ∠BCD = 50°
Hence angles of Δ ACD and Δ BCD are equal.

c)
Here Δ ADC and Δ CDB are similar triangles.
Therefore, \(\frac{C D}{B D}\) = \(\frac{A D}{C D}\)
\(\frac{\mathrm{h}}{\mathrm{~b}}\) = \(\frac{a}{h}\)
Thus, h² = ab
Hence proved.

Question 26.
The product of the larger of two numbers increased by one and the smaller decreased by one is 540. The product of the larger decreased by one and the smaller increased by one is 560.
i) What is the product of the numbers themselves?
ii) What is their difference?
iii) What are the numbers?
Answer:
Let x be the larger number and y be the smaller number. •
i) Given,
(x + 1 )(y – 1) = 540
(x – 1)(y + 1) = 560
Rewriting the equation we get,
(x + 1) (y – 1) = xy – x + y – 1 = 540 ……………… (1)
(x – 1) (y + 1) = xy + x – y – 1 = 560 ……………(2)
Adding (1) and (2) we get,
2xy – 2 = 1100
2xy = 1102
xy = 551 …………….(3)

ii) Subtracting (1) from (2) we get,
xy + x – y – 1 – (xy – x + y – 1) = 20
2x – 2y = 20 x – y = 10

iii) Let us express x in terms of y
x = y + 10
Substituting this in (3) we get,
y (y + 10) = 551
y² + 10y = 551
When y = 19
19 × (19 + 10) = 19 × 29 = 551
Thus, x = y + 10 = 19 + 10 = 29
Therefore, two numbers are x = 29 and y = 19.

Question 27.
a) If AB = √2 cm, BC = √8 cm and AC = \(\sqrt{18}\) cm, prove that AB + BC = AC.
b) Find the area and perimeter of the rectangle in the figure.

Answer:
a) Given, AB = √2 cm , BC = √8 = 2√2 cm and AC = \(\sqrt{18}\) = 3√2 cm
Thus,
AB + BC = √2 + 2√2 = 3√2 = AC
Thus,
AB + BC = AC
Hence proved.

b) Area of rectangle = AB × AD = √2 × 2√2 = 2 × 2 = 4 cm²
Perimeter of rectangle = 2(AB + AD) = 2(√2 + 2√2)= 2 × 3√2 = 6√2 cm

Question 28.
a) Raju bought 5 notebooks and 4 pens for rupees 148. The price of 4 notebooks and 5 pens is 140 rupees. What is the price of a book? What is the price of a pen?
b) The price of 4 pen and 3 pencil is 49. The price of 9 pen and 3 pencil is 99. What is the price of a pen? What is the price of a pencil?
Answer:
a) Let x be the price of the notebook and y be the price of the pen
Given, 5x + 4y = 148 ………….. (i)
4x + 5y = 140 ……………… (ii)
(i) × 4 = 20x + 16y = 592 ………………. (iii)
(ii) × 5 = 20x + 25y = 700 ………………… (iv)
Subtracting (iii) from (iv) we get,
9y = 108
y = \(\frac{108}{9}\) = 12
Substitute y value in (ii),
4x + 5(12) = 140
4x = 140 – 60 = 80
x = \(\frac{80}{4}\) = 20
Thus, the price of the book = Rs. 20
Price of a pen = Rs. 12

b) Let p be the price of a pen and q be the price of a pencil.
Given, 4p + 3q = 49 ……………. (i)
9p + 3q = 99 …………….. (ii)
(ii) – (i) →
(9p + 3q) – (4p + 3q) = 50
5p = 50
p = 10
Substitute the value p in (i) we get,
4(10) + 3q = 49
3q = 9
q = 3
Thus, the Price of pen = Rs. 10
Price of a pencil = Rs. 3

Question 29.
∠B = 90°, AC = \(\sqrt{208}\) cm. Area of the triangle is 48 sq. cm. Then,

a) What is the value of x² + y²?
b) What is the value of xy?
c) Find x and y
Answer:
(a) Consider the right-angled triangle ABC where,
AC² = AB² + BC²
(\(\sqrt{208}\))² = x² + y²
208 = x² + y²
Thus,
x² + y² = 208

(b) Area of Δ ABC = \(\frac{1}{2}\) × AB × BC
48 = \(\frac{1}{2}\) × x × y
96 = x × y
Thus, xy = 96

(c) We know that,
(x + y)² = x² + y² + 2xy
Substituting x² + y² = 208 and xy = 96 we get,
(x + y)² = 208 + 2 x 96
(x + y)² = 208 + 192
(x + y)² = 400
x + y = \(\sqrt{400}\) = 20
x and y are those numbers whose sum is 20 and product is 96.
Thus, x and y are 12 cm and 8 cm respectively.