Students can access the CBSE Sample Papers for Class 9 Maths with Solutions and marking scheme Set 3 will help students understand the difficulty level of the exam.
CBSE Sample Papers for Class 9 Maths Set 3 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This question paper contains 38 questions. All questions are compulsory.
- The Question paper is divided into Five sections – Sections A, B, C, D and E.
- In Section A, question number 1 to 18 are multiple choice questions (MCQs) and question number 19 and 20 are Assertion-Reason based questions of 1 mark each.
- In Section B, questions number 21 to 25 are very short answer (VSA) type questions of 2 marks each.
- In Section C, questions number 26 to 31 are short answer (SA) type questions carrying 3 marks each.
- In Section D, questions number 32 to 35 are long answer (LA) type questions carrying 5 marks each.
- In Section E, question number 36 to 38 are case-based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks questions in each case study.
- There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C and 2 questions in Section D.
- Draw neat figures wherever required. Take π = \(\frac{22}{7}\) wherever required if not stated.
- Use of calculator is not allowed.
Section-A
Consists of Multiple Choice Type questions of 1 mark each.
Question 1.
A polynomial is expressed as x3 + bx2 + cx + d = 0. The same polynomial can be written in factor form as x + px + qx+ r = 0.
How is the constant term in the polynomial related to its factors p, q, and r?
(A) d = p + q + r
(B) d = (p + q) × r
(C) d = p × q × r
(D) d = pq + qr + pr
Answer:
(C) d = p × q × r
Question 2.
The number obtained on rationalizing the denominator of \(\frac{1}{\sqrt{7}-2}\) is
(A) \(\frac{\sqrt{7}+2}{3}\)
(B) \(\frac{\sqrt{7}-2}{3}\)
(C) \(\frac{\sqrt{7}+2}{5}\)
(D) \(\frac{\sqrt{7}+2}{45}\)
Answer:
(A) \(\frac{\sqrt{7}+2}{3}\)
Explanation:
\(\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}\)
Question 3.
Which of the following is not true?
(A) A line has one dimension.
(B) A plane has two dimensions.
(C) A circle can be drawn with any radius and at any point.
(D) Two distinct lines can pass through a point in the same direction.
Answer:
(D) Two distinct lines can pass through a point in the same direction.
Question 4.
If a + b + c = 0, then a3 + b3 + c3 is equal to
(A) 0
(B) abc
(C) 3abc
(D) 2abc
Answer:
(C) 3abc
Explanation:
a + b + c = 0
a + b = -c …..(i)
On cubing both sides
(a + b)3 = (-c)3
⇒ a3 + b3 + 3ab(a + b) = (-c)3
⇒ a3 + b3 + 3ab(-c) = (-c)3 [∵ By using equation (i)]
⇒ a3 + b3 – 3abc = (-c)3
⇒ a3 + b3 + c3 = 3abc
Question 5.
In the quadrilateral ABCD given below, ∠DAC = 90° and AB = AC – AD = DE = EB.
What is the value of ∠EAC?
(A) 15°
(B) 30°
(C) 45°
(D) 90°
Answer:
(A) 15°
Explanation:
In ∆ADE,
AD = DE (given)
So, ∠DAE = ∠DEA = a (angles opposite to equal sides are equal)
Now, ∠ADE + ∠DAE + ∠DEA = 180°
⇒ 30° + a + a = 180°
⇒ 2a = 180° – 30°
⇒ a = 75°
∴ ∠DAE = 75°
∠DAC = ∠DAE + ∠EAC
⇒ 90° = 75° + x
⇒ x = 90° – 75°
⇒ x = 15°
Hence ∠EAC = 15°
Question 6.
Given below is the map giving the position of four housing societies in a township connected by a circular road A.
Society 2 and 3 are connected by straight Road B, Society 4 and 2 are connected by straight Road C and Society 4 and 3 are connected by Road D. Point P denotes the position of a park. The park is equidistant to all four societies. Rubina claims that it is not possible to construct another circular road connecting all four societies. The length of Road B is equal to the length of Road D. Which of the following options can be true for the roads in the township?
(A) Road B bisects Road D.
(B) Road B and Road D make an acute angle.
(C) Road B, Road C and Road D are of equal length.
(D) Road Band Road D subtend equal angles at society 1.
Answer:
(D) Road Band Road D subtend equal angles at society 1.
Explanation:
According to property of circles, Angles in the same segment of a circle are equal.
Question 7.
A zoo is in the shape of an isosceles trapezium. It is divided into three zones – Zone 1, Zone 2 and Zone 3.
To avoid the entry of animals in zones 2 and 3, a 1.8 km long wired fencing is installed. Which of the following calculations shows the area for animals?
(A) \(\sqrt{1.35 \times 0.65 \times 1.15}\) km2
(B) \(\sqrt{2.15 \times 0.35 \times 0.65 \times 1.15}\) km2
(D) \(\sqrt{3.15 \times 1.35 \times 1.65 \times 1.15}\) km2
(C) \(\sqrt{4.30 \times 1.35 \times 0.65 \times 1.15}\) km2
Answer:
(B) \(\sqrt{2.15 \times 0.35 \times 0.65 \times 1.15}\) km2
Explanation:
Here, a = 1.8 km, b = 1.5 km, c = 1 km
Question 8.
56x = 1252, then x = _______
(A) 2
(B) 3
(C) 1
(D) 4
Answer:
(C) 1
Explanation:
56x = 1252
⇒ 56x = 56
⇒ 6x = 6 (As base is same)
⇒ x = 1
Question 9.
Which of the following is true for the line with equation: 1.x + 0.y – 4 = 0?
(A) The distance of the line from the x-axis is 1.
(B) The distance of the line from the y-axis is 4.
(C) The distance of the line from the y-axis is -1.
(D) The distance of the line from the x-axis changes from 1 to -4.
Answer:
(B) The distance of the line from the y-axis is 4.
Explanation:
1x + 0y – 4 = 0
⇒ 1x + 0y = 4
⇒ x = 4
Thus, the line is at a distance of 4 from y-axis.
Question 10.
If \(49 x^2-(\sqrt{b})^2=\left(7 x+\frac{1}{2}\right)\left(7 x-\frac{1}{2}\right)\), then the value of b is
(A) 0
(B) \(\frac{1}{\sqrt{2}}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{2}\)
Answer:
(C) \(\frac{1}{4}\)
Explanation:
Question 11.
If 30x = y, then standard form of equation will be
(A) 30x – y + 0 = 0
(B) 30x + y + 0 = 0
(C) 30x – y – 0 = 0
(D) 30x – y = 0
Answer:
(A) 30x – y + 0 = 0
Explanation:
Standard form of a linear equation in two variables is ax + by + c = 0.
Question 12.
How much is the volume of a hemisphere if the radius of the base is 3.5 m?
(A) 87.53 m3
(B) 90.22 m3
(C) 75.34 m3
(D) 89.83 m3
Answer:
(D) 89.83 m3
Explanation:
Volume of the hemisphere = \(\frac{2}{3} \pi r^3\)
= \(\frac{2}{3} \times \frac{22}{7} \times(3.5)^3\)
= 89.83 m3
Question 13.
The area of an isosceles triangle having base of 2 cm and the length of one of the equal sides 4 cm, is
(A) \(\sqrt{15}\) cm2
(B) \(\sqrt{\frac{15}{2}}\) cm2
(C) 2\(\sqrt{15}\) cm2
(D) 4\(\sqrt{15}\) cm2
Answer:
(A) \(\sqrt{15}\) cm2
Explanation:
Question 14.
A parking lot for a city mall is shown below. The painted lines that separate the parking spaces are parallel.
Parking space number 378 is inclined at 60° to the horizon line. At what angle is parking space 380 inclined to the horizontal line?
(A) 120°
(B) 90°
(C) 60°
(D) 30°
Answer:
(C) 60°
Explanation:
Parking space 378 = 60°
∴ Parking space 380 = 60° (∵ corresponding angles are equal)
Question 15.
What is the type of solution of the equation 30x = y formed?
(A) a unique solution
(B) only two solutions
(C) no solution
(D) infinitely many solutions
Answer:
(D) infinitely many solutions
Explanation:
Because for every value of x, there is a corresponding value of y and vice versa.
Question 16.
How much cloth material will be required to cover 4 small domes of Taj Mahal each of radius of 2 m?
(A) 99 m2
(B) 100.57 m2
(C) 98.23 m2
(D) 75 m2
Answer:
(B) 100.57 m2
Explanation:
Required cloth material = 4 × Curved surface area of hemispherical domes
= 4 × 2πr2
= 4 × 2 × \(\frac{22}{7}\) × 22
= 8 × \(\frac{22}{7}\) × 4
= 32 × \(\frac{22}{7}\)
= 100.57 m2
Question 17.
What do we call a triangle if the angles are in the ratio 5 : 3 : 7?
(A) Acute angled triangle
(B) Obtuse angled triangle
(C) Right-angled triangle
(D) Complete angled triangle
Answer:
(A) Acute angled triangle
Explanation:
Let the angles of triangle are 5x, 3x and 7x, then
5x + 3x + 7x = 180°
or, 15x = 180°
Thus, x = 12°
∴ Angles are 60°, 36°, 84°.
∵ Each angle is less than 90°.
∴The triangle is an acute-angled triangle.
It can also be a scalene triangle, since all angles are of different measure. Hence, corresponding sides are unequal.
Question 18.
Formula for the volume of hemisphere is
(A) \(\frac{1}{3} \pi r^3\)
(B) \(\frac{2}{3} \pi r^3\)
(C) 2πr3
(D) \(\frac{4}{3} \pi r^3\)
Answer:
(B) \(\frac{2}{3} \pi r^3\)
Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Question 19.
Assertion (A): Value of \((32)^{2 / 5}+(-7)^0\) is a rational number.
Reason (R): (a)0 ≠ 1; where a > 0 and a is a real number.
Answer:
(C) Assertion (A) is true, but Reason (R) is false.
Explanation:
In case of Assertion:
\((32)^{2 / 5}+(-7)^0=\left(2^5\right)^{2 / 5}+(-7)^0\)
= \(2^{5 \times \frac{2}{5}}+1\)
= 22 + 1
= 4 + 1
= 5
∴ Assertion is true.
In case of Reason:
(a)0 = 1
∴ Reason is false.
Question 20.
Assertion (A): The class interval needs to be continuous while drawing a Histogram.
Reason (R): Histogram is a rectangular diagram using frequency distributions which are joined to one another.
Answer:
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Explanation:
In case of Assertion (A): Histograms depicts continuous data which is taken in the form of class interval.
∴ Assertion is true, Reason is true as it is a definition of Histogram, but Reason is not correct explanation of Assertion.
Section-B
Consists of 5 questions of 2 marks each.
Question 21.
If a = 2 and b = 3, then find the value of:
(i) \(\left(a^b+b^a\right)^{-1}\)
(ii) \(\left(a^a+b^b\right)^{-1}\)
Answer:
Question 22.
The figure below consists of a square and an equilateral triangle connected together with a common side.
In the design, DF and IG are two iron rods perpendicular to BC. The measure of ∠BAC = 120°. The central triangle AFG is equilateral. What is the measure of ∠FDA?
Answer:
∆AFG is an equilateral triangle. (given)
∴ ∠AFG = 60°
DF ⊥ BC (given)
∴ ∠DFG = 90°
∠DFA = ∠DFG – ∠AFG
= 90° – 60°
= 30°
DF = DA
∴ ∠DFA = ∠DAF = 30° (Angles opposite to equal sides are equal)
Now, in ∆DAF
∠DFA + ∠DAF + ∠FDA = 180°
30° + 30° + ∠FDA = 180°
Hence, ∠FDA = 120°.
Question 23.
In a school camp, 40 students were divided into two groups to play a game. The table given below shows the scores of team A and team B.
Time(s) in Minutes | Cumulative Score of Team A | Cumulative Score of Team B |
0 – 5 | 14 | 20 |
5 – 10 | 35 | 27 |
10 – 15 | 30 | 31 |
15 – 20 | 35 | 31 |
20 – 25 | 44 | 37 |
25 – 30 | 52 | 50 |
Which team scored more points during last 5 minutes? Justify your answer.
Answer:
Team B scored more than Team A.
As during the last 5 minutes, the score of team B is 13 (= 50 – 37) and the score of Team A is 8 (= 52 – 44) in the last five minutes.
Question 24.
Expand by using identity (2x – y + z)2.
OR
Expand: \(\left(\frac{1}{3} x-\frac{2}{3} y\right)^3\)
Answer:
By using the identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(2x + (-y) + z)2 = (2x)2 + (-y)2 + z2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x) = 4x2 + y2 + z2 – 4xy – 2yz + 4xz
OR
Question 25.
If x = 3 – 2√2, find the value of \(\sqrt{x}+\frac{1}{\sqrt{x}}\).
OR
Taking √2 = 1.414 and π = 3.141, evaluate \(\frac{1}{\sqrt{2}}\) + π upto three places of decimal.
Answer:
Section-C
Consists of 6 questions of 3 marks each.
Question 26.
If x = -√2 – 1, then find the value of \(\left(x-\frac{1}{x}\right)^3\).
Answer:
Question 27.
Raghvan claims that the magnitude of the angle ABC is greater than the magnitude of the area of the right triangle PQR.
Is his claim correct? Why?
Answer:
No, Raghvan’s claim is not correct.
As, the measure of an angle cannot be compared to the area of a triangle.
Question 28.
In the fig. given below, an exterior angle of a triangle is 130° and the two interior opposite angles are equal. Find each of these angles.
OR
In the figure given below ∠APC = 100° and ∠BPD = 146°. Find ∠CPD?
Answer:
In a ∆ABC
x + x = 130° (exterior angle is equal to the sum of interior opposite angles)
⇒ 2x = 135°
⇒ x = 65°
∴ ∠A = ∠B = 65°
∠ACB + ∠ACD = 180° (Linear pair)
⇒ ∠ACB + 130° = 180°
⇒ ∠ACB = 50°
Hence, ∠ABC = 65°, ∠BAC = 65° and ∠ACB = 50°
OR
According to figure
∠BPC + ∠CPD = 146°
and ∠APD + ∠CPD = 100°
Let ∠CPD = x
∴ ∠BPC = 146° – x
and ∠APD = 100° – x
∠APD + ∠CPD + ∠BPC = 180° (straight line angles)
⇒ 100° – x + x + 146° – x = 180°
⇒ 246° – x = 180°
⇒ x = 66°
Hence, ∠CPD = 66°
Question 29.
The following table gives the pocket money (in ₹) given to children per day by their parents:
Pocket Money | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Number of Children | 12 | 23 | 35 | 20 | 10 |
Represent the data in the form of a histogram.
OR
Consider the marks out of 100, obtained by 50 students of a class in a test, given as below.
Marks | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
Number of Students | 15 | 10 | 10 | 11 | 4 |
Draw a frequency polygon representing the data.
Answer:
Marks | Number of Students |
0 – 10 | 12 |
10 – 20 | 23 |
20 – 30 | 35 |
30 – 40 | 20 |
40 – 50 | 10 |
OR
Marks | Number of Students |
0 – 20 | 15 |
20 – 40 | 10 |
40 – 60 | 10 |
60 – 80 | 11 |
80 – 100 | 4 |
Frequency polygon graph is ABCDEFG.
Question 30.
In a rectangle ABCD, E is a point on AB such that \(\frac{2}{3}\). If AB = 6 km and AD = 3 km, then find DE.
Answer:
According to the question,
AE = \(\frac{2}{3}\)AB
= \(\frac{2}{3}\) × 6
= 4 km
In right ∆ADE, right-angled at A,
DE2 = AD2 + AE2
⇒ DE2 = (3)2 + (4)2
⇒ DE2 = 9 + 16
⇒ DE2 = 25
⇒ DE = 5 km
Question 31.
Evaluate (√2 + √3)2 + (√5 – √2)2
Answer:
( 2 + √3)2 + (√5 – √2)2
= (√2)2 + (√3)2 + 2 × √2 × √3 + (√5)2 + (√2)2 – 2 × √5 × √2
= 2 + 3 + 2√6 + 5 + 2 – 2√10
= 12 + 2√6 – 2√10
= 2(6 + √6 – √10)
Section-D
Consists of 4 questions of 5 marks each.
Question 32.
Prove that (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3 = 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a).
OR
Prove that x3 + y3 + z3 – 3xyz = \(\frac{1}{2}\)(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Answer:
Let x = a2 – b2, y = b2 – c2, z = c2 – a2
Now, x + y + z = a2 – b2 + b2 – c2 + c2 – a2 = 0
∴ x + y + z = 0
x3 + y3 + z3 = 3xyz
i.e., (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3 = 3(a2 – b2)(b2 – c2)(c2 – a2)
= 3(a + b)(a – b)(b + c)(b – c)(c + a)(c – a)
= 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a)
Hence proved.
OR
RHS = \(\frac{1}{2}\)(x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]
= \(\frac{1}{2}\)(x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx]
= \(\frac{1}{2}\)(x + y + z)[2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= \(\frac{1}{2}\)(x + y + z) . 2[x2 + y2 + z2 – xy – yz – zx]
= x3 + y3 + z3 – 3xyz [∵ Using identity x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)]
= LHS
Hence proved.
Question 33.
(a) A hemispherical tank has inner radius of 2.8 m. Find its capacity in liters.
(b) Find the volume of a right circular cone with radius 6 cm and height 7 cm.
Answer:
(a) Radius of hemispherical tank, r = 2.8 m
Capacity of hemispherical tank = \(\frac{2}{3} \pi r^3\)
= \(\frac{2}{3} \times \frac{22}{7} \times(2.8)^3\)
= \(\frac{2}{3} \times \frac{22}{7}\) × 2.8 × 2.8 × 2.8
= 45.997 m3
Now, 1m3 = 1000 liters
∴ Capacity of hemispherical tank in liters = 45997 liters
(b) Volume of right circular cone = \(\frac{1}{3} \pi r^2 h\)
= \(\frac{1}{3} \times \frac{22}{7} \times(6)^2 \times 7\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 36 × 7
= 264 cm3
Question 34.
In figure, AB = AC, CH = CB and HK || BC. If ∠CAX = 137°, then find ∠CHK.
OR
In the given figure, ABCD and BPQ are straight lines. If BP = BC and DQ is parallel to CP. prove that:
(i) CP = CD
(ii) DP bisects ∠CDQ
Answer:
∠XAK + ∠KAH = 180° (Linear pair)
∠KAH = 180° – 137° = 43° (∵ ∠CAX = ∠XAK = 137°, Given)
AB = AC (Given)
∴ ∠ABC = ∠ACB (Angles opp. to equal sides)
∠ABC + ∠ACB = 137° (exterior angle)
∠ABC = ∠ACB = \(\frac{137}{2}\) = 68.5°
CH = CB (Given)
∠CBA = ∠CHB = 68.5°
∴ ∠HCB = 180° – 137° = 43°
∠CHK = ∠HCB = 43° (Alternate angles)
OR
(i) BP = BC (Given)
∴ ∠BCP = ∠BPC = y (Angle opposite to equal sides)
∴ ∠1 = y (Corresponding angles)
4x = 2y (exterior angle)
∠2 = x (Alternate angles)
y = x + ∠3 (∵ 4x = 2y, \(\frac{4x}{2}\) = y)
⇒ 2x = x + ∠3
⇒ x = ∠3
CP = CD (side opposite to equal angles)
(ii) Also, we have ∠2 = x, x = ∠3
∴ DP bisects ∠CDQ
Hence Proved.
Question 35.
The figure below consists of a square and an equilateral triangle connected together with a common side.
What is the measure of ‘x’?
Answer:
In ΔABC
∠ABC = 90° + 60° = 150°
AB = BC (given)
∴ ∠BAC = ∠BCA = a (angles opposite to equal sides are equal)
Now, ∠BAC + ∠ABC + ∠ACB = 180°
⇒ a + a + 150° = 180°
⇒ 2a = 180° – 150°
⇒ a = 15°
∠BAD = 60° (Angle of an equilateral triangle)
⇒ ∠BAD = a + x (from fig)
⇒ 60° = 15° + x
⇒ x = 45°
Section-E
Cased-Based Subjective Questions.
Question 36.
A forest ranger keeps track of bears in his area. He plotted their location on a graph. The origin represents the ranger’s control room’s location. To access and maintain equipment, Road x and Road y have been laid and paved inside the forest. They pass through the control room.
One unit on the graph paper represents 1 km.
(i) Bear 467 has been injured. The forest rescue team starts from the control room and decides to use the paved road as much as possible. Which road should they take?
(ii) A tiger is at (11, 4). How far from it is the nearest bear?
OR
In the forest, rain shelters are at an interval of 2 km along paved roads. A forest ranger is travelling on Road x. He crosses a rain shelter located at (3, 0). What is likely to be the location of the next shelter?
(iii) How far is Bear 425 from Road x?
Answer:
(i) Forest rescue team should take Road y, as Bear 467 is at vertical position from control room (Origin), so forest rescue team will take vertical road y.
(ii) According to graph Bear 389 is at (11, 6), and Tiger is at (11, 4),
Here X-coordinates are same that is 11.
But difference between y-coordinates = (6 – 4) = 2 km
So Tiger is 2 km far from it’s nearest Bear.
OR
Rain shelter crossed by Forest Ranger = (3, 0)
Another Rain shelter is at 2 km along paved roads, So it will either be at (5, 0) or (1, 0)
(iii) According to graph Bear 425 is 13 km far from Road x.
Question 37.
Two brothers Ashish and Amit wanted to start a business together. They decided to share their amount depending upon the variable expenditure. The amount of two partners is given by the expression 12x2 + 11x – 15, which is the product of their individual share factors.
On the basis of above information answer the following questions.
(i) Which method of factorisation will be used to find individual share of Ashish and Amit?
(ii) Find the total expenditure of Ashish and Amit when x = ₹ 100?
OR
Find individual share factor of Ashish and Amit in terms of x?
(iii) Find value of x if their shares are equal?
Answer:
(i) Factorisation method used for calculating share of Ashish and Amit is Splitting the Middle Term.
(ii) Total expenditure = 12x2 + 11x – 15
Put x = 100
Total expenditure = 12 × (100)2 + 11 × 100 – 15
= 120000 + 1100 – 15
= ₹ 121,085
OR
Total Amount = 12x2 + 11x – 15
By splitting the middle term, we get
12x2 + 11x – 15 = 12x2 + 20x – 9x – 15
= 4x(3x + 5) – 3(3x + 5)
= (4x – 3)(3x + 5)
Share of Ashish and Amit are either (3x + 5) and (4x – 3) or (4x – 3) and (3x + 5) respectively.
(iii) According to question if their share are equal.
∴ 4x – 3 = 3x + 5
⇒ 4x – 3x = 5 + 3
⇒ x = 8
Question 38.
A plane mirror is mirror with a flat reflective surface. An incident ray is ray of light that strikes a surface. The reflected ray corresponding to a given incident ray, is the ray that represents the light reflected by the surface.
In figure, m and n are two plane mirror perpendicular to each other.
(i) What is the property of parallel lines?
(ii) Find value of ∠BOA?
OR
Find AO if BO = 3 cm, AB = 5 cm?
(iii) Incident ray CA and ray BD are parallel or perpendicular to each other?
Answer:
(i) Parallel lines never meet.
(ii) m and n are two plane mirrors perpendicular to each other.
So, AO is perpendicular to BO.
∴ ∠BOA = 90°
OR
m and n are two plane mirror perpendicular to each other.
So, AO perpendicular to BO.
Thus, triangle AOB is right-angled triangle.
OA2 + OB2 = AB2
⇒ OA2 + 32 = 52
⇒ OA2 + 9 = 25
⇒ OA2 = 25 – 9
⇒ OA2 = 16
⇒ OA = 4 cm
(iii) Incident ray CA is parallel to ray BD.