Students can access the CBSE Sample Papers for Class 10 Maths Basic with Solutions and marking scheme Set 9 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Basic Set 9 with Solutions
Maximum Marks: 80
Time Allowed: 3 hours
General Instructions:
1. This Question Paper has 5 Sections A, B, C, D and E.
2. Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
3. Section B has 5 Short Answers-I (SA-I) type questions carrying 2 marks each.
4. Section C has 6 Short Answers-IT (SA-II) type questions carrying 3 marks each
5. Section D has 4 Long Answers (LA) type questions carrying 5 marks each.
6. Section E has 3 source based/case based/passage based integrated units of assessment (4 marks each) with sub¬ parts of the values of 1,1 and 2 marks each respectively.
7. All questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Qs of 5 marks has been given provided. An internal choice has been provided in 2 marks questions of Section E.
8. Draw neat figures wherever required. Take π = 22/7, wherever required if not stated.
Section – A(20 Marks)
Question 1.
The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is
(A) 13
(B) 65
(C) 875
(D) 1,750
Solution:
(A) 13
Explanation: Required largest number = HCF of
(70 – 5) and (125 – 8) = HCF of 65 and 117 = 13.
Question 2.
The simplest form of √35 / √5 is:
(A) 1
(B) 2
(C) √5
(D) √7
Solution:
(D) √7
Explanation: We have
√35 / √5 = \(\sqrt{(5 \times 7)}\) / √5 = √7
Question 3.
Which of these is the polynomial whose zeroes are \(\frac{1}{3}\) and (\(\frac{-3}{4}\)).
(A) 12x2 + 5x – 3
(B) 12x2 – 5x – 3
(C) 12x2 + 13x + 3
(D) 12x2 – 13x – 3
Solution:
(A) 12x2 + 5x – 3
12x2 + 5x – 3 = 0 × 12
12x2 + 5x – 3 = 0
So, we have, P(x) = 12x2 + 5x – 3
Question 4.
The larger of the two supplementary angles exceeds the smaller by 20°. Find smaller angle.
(A) 80°
(B) 100°
(C) 90°
(D) 70°
Solution:
(A) 80°
Explanation: According to the question
x – y = 20° …(i)
x + y = 180° …(ii)
solving eq. (i) and (ii); we get
x = 100° and y = 80°
on solving eq. (i) and (ii), we get
∴ smaller angle = 80°
Question 5.
If the product of the roots of equation kx2 + 2x + 3k = 0 is equal to their sum then the value of k is:
(A) –\(\frac{1}{3}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{2}{3}\)
(D) –\(\frac{2}{3}\)
Solution:
(D) –\(\frac{2}{3}\)
Question 6.
If A(4, -2), B(7, -2) and C(7, 9) are the vertices of a ΔABC, then ΔABC is
(A) equilateral triangle
(B) isosceles triangle
(C) right angled triangle
(D) isosceles right angled triangle
Solution:
(C) right angled triangle
Explanation: A(4, –2), B(7, –2) and C(7, 9) are the vertices of a triangle.
Using distance formula,
Clearly, they are not equilateral or isosceles.
Also, AC2 = AB2 + BC2
Which mean it is following Pythagoras theorem.
∴ ΔABC is a right angled triangle.
Question 7.
In ΔABC, DE || BC. According to Basic proportionality theorem which of the following is true?
(A) \(\frac{A B}{B D}=\frac{A C}{E C}\)
(B) \(\frac{A D}{B D}=\frac{A E}{E C}\)
(C) \(\frac{A D}{A B}=\frac{A E}{A C}\)
(D) All of these
Solution:
(D) All of these
Explanation: Given that DE || BC
According to BPT,
From (i), (ii) and (iii), All options given in the question are true.
Question 8.
Shown below are two triangles such that length of two sides of each is known.
(Note: The figures are not to scale.)
Along with the given information, which of these is sufficient to conclude whether ΔKLM is similar to ΔPQR?
(i) ∠KLM = ∠PQR
(ii) Ratio of KM : PR
(A) only (i)
(B) only (ii)
(C) either (i) or (ii)
(D) the given information is enough to conclude that ΔKLM ~ ΔPQR as ratio of sides is known)
Solution:
(C) either (i) or (ii)
Explanation: In ∆KLM & ∆PQR
\(\frac{M L}{R Q}=\frac{3}{6}=\frac{1}{2}\) and \(\frac{K L}{P Q}=\frac{4}{8}=\frac{1}{2}\)
Therefore, if ∠KLM = ∠PQR, By SAS similarity rule triangles are similar.
And if ratio KM:PR is also 1:2 then by SSS similarity rule both triangles are similar.
Therefore, any one of the cases given above is sufficient to conclude that both triangles are similar.
Question 9.
If PQ = 28 cm, then find the perimeter of ΔPLM.
(A) 50 cm
(B) 56 cm
(C) 60 cm
(D) 64 cm
Solution:
(B) 56 cm
Explanation:
∵ PQ = PT
PL + LQ = PM + MT
PL + LN = PM + MN …(i)
(LQ = LN, MT = MN)
(Tangents to a circle from a common point
Perimeter (ΔPLM) = PL + LM + PM
= PL + LN + MN + PM
= 2(PL + LN) [From eq.(i)]
= 2(PL + LQ)
= 2 × 28 = 56 cm
Question 10.
In the figure below, what is the length of AB?
(Note: The figure is not to scale.)
(A) 45√3 m
(B) \(\frac{45}{\sqrt{3}}\) m
(C) 45(√3 – 1) m
(D) 45(√3 + 1) m
Solution:
(C) 45(√3 – 1) m
Explanation:
AC = 45 √3 m
AB = AC – BC
= 45 √3 – 45
= 45(√3 − 1) m
Question 11.
If cos A = \(\frac{4}{5}\), then the value of tan A is
(A) \(\frac{3}{5}\)
(B) \(\frac{3}{4}\)
(C) \(\frac{4}{3}\)
(D) \(\frac{1}{8}\)
Solution:
(B) \(\frac{3}{4}\)
Question 12.
The value of (sec A + tan A) (1 – sin A) is
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
Solution:
(D) cos A
Explanation:
(sec A + tan A) (1 – sin A)
Question 13.
In the figure below, RT = 1 cm and OQ = 3 cm.
What is the area of the shaded region?
(A) (12.5 π – 12) cm2
(B) (6.25 π – 12) cm2
(C) (12.5 π -15) cm2
(D) (6.25 π – 15) cm2
Solution:
(B) (6.25 π – 12) cm2
Explanation:
Given that, RT = 1 cm and OQ = 3 cm
OS is the radius,
OS = OR + RT = OR + 1 …(i)
In ∆SOR,
According to Pythagoras theorem,
OS2 = RS2 + OR2
⇒ (OR + 1)2 = (3)2 + (OR)2 [RS = OQ]
⇒ OR2 + 1 + 2OR = 9 + OR2
⇒ 2OR = 8
⇒ OR = \(\frac{8}{2}\) = 4 cm
Hence, Area of rectangle ORSQ = Length × breadth = 3 × 4 = 12 cm2
Area of shaded region = Area of quadrant OPTRO – Area of rectangle QRSQ
= \(\frac{\theta}{360^{\circ}}\) × πr2 – length × breadth
= \(\frac{90}{360}\) × π × 5 × 5 – 12
= 6.25 π – 12 cm2
Question 14.
Find the area of the circle inscribed in a square of side 12 cm.
(A) 20π cm2
(B) 25π cm2
(C) 30π cm2
(D) 36π cm2
Solution:
(D) 36π cm2
Explanation:
Side of square = 12 cm
Diameter of square = side of square = 12 cm
Radius of square = r = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Area of circle = πr2 = π × 6 × 6 = 36π cm2
Question 15.
The probability that a non-leap year selected at random will contain 53 Sundays is
(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{3}{7}\)
(D) \(\frac{5}{7}\)
Solution:
(A) \(\frac{1}{7}\)
Explanation: Number of days in non-leap year = 365
Number of weeks = \(\frac{365}{7}\) = 52\(\frac{1}{7}\) = 52 weeks
Number of days left = 1
For example, it may be any of 7 days which from Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday; so, T(E) = 7
F(E) = 1 (Sunday)
P(F) = \(\frac{F(E)}{T(E)}=\frac{1}{7}\)
Question 16.
Find the median of the data, using an empirical relation when it is given that Mode = 12.4 and Mean = 10.5.
(A) 11.47
(B) 11.35
(C) 11.23
(D) 11.13
Solution:
(D) 11.13
Explanation:
Question 17.
Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1. What is the ratio of their volumes.
(A) 3:1
(B) 1:3
(C) 2:3
(D) 3:4
Solution:
(A) 3:1
Explanation:
Let h1 and h2 be height and r1, r2 be radii of two cones, then ratio of their volumes
Hence, ratio of their volumes is 3:1.
Question 18.
If the probability of winning a game is 0.07, what is the probability of losing it?
(A) 0.83
(B) 0.87
(C) 0.93
(D) 0.97
Solution:
(C) 0.93
Explanation:
P(winning the game) = 0.07
P(losing the game) = 1 – 0.07 = 0.93
Directions: In the following questions, A Statement of Assertion (A) is followed by a statement of Reason (R).
Mark the correct choice as.
Question 19.
Assertion (A): The points (0, -3) lies on the y-axis.
Reason (R): The of the point (0, -3) is zero.
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Solution:
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
Explanation: The x-coordinate of the point (0, –3) is zero therefore it lies on the y-axis.
Therefore, Both A and R are true and R is the correct explanation for A.
Question 20.
Assertion (A): 6n ends with the digit zero, where n is a natural number.
Reason (R): Any number ends with digit zero, if its prime factor is of the form 2m × 5n, where m, n are natural numbers.
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Solution:
(D) (A) is false but (R) is true.
Explanation: 6n = 2n × 3n, to get the last digit zero
we need 2m × 5n.
Section – B(10 Marks)
Question 21.
Solve: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3.
Solution:
0.2x + 0.3 y = 1.3 …(i)
0.4x + 0.5y = 2.3 …(ii)
Multiply equation (i) with (ii), we get
0.4x + 0.6y = 2.6
0.4x + 0.5y = 2.3
On subtracting,
0.1y = 0.3
y = 3.
Put value of y in equation (i),
0.2x + 0.3(3) = 1.3
0.2x + 0.9 = 1.3
0.2x = 0.4
x = 2
Question 22.
In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔABD ~ ΔCBE
(ii) ΔPDC ~ ΔBEC
Solution:
(i) In ΔABD and ΔCBE
∠ADB = ∠CEB = 90°
∠ABD = ∠CBE (Common angle)
⇒ ∆ABD ~ ∆CBE (AA criterion)
(ii) In ΔPDC and ∆BEC
∠PDC = ∠BEC = 90°
∠PCD = ∠BCE (Common angle)
⇒ ΔPDC ~ ΔBEC (AA criterion)
Question 23.
In given figure, if AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠OTA = 30°, then find the length of AT (in cm).
Solution:
cos θ = \(\frac{\text { Base }}{\text { Hypotenuse }}\)
\(\frac{A T}{O T}\) = cos 30°
∴ AT = OT cos 30°
Or, AT = 4 × \(\frac{\sqrt{3}}{2}\)
= 2 √3 cm
OR
In the given figure PQ is chord of length 6 cm of the circle of radius 6 cm. TP and TQ are tangents to the circle at points P and Q respectively. Find ∠PTQ.
Solution:
Here, PQ = 6 cm,
OP = OQ = 6 cm
∴ PQ = OP = OQ
∴ ∠POQ = 60°
(angle of equilateral D)
∠P = ∠Q = 90° (radius ^ tangent)
∴ ∠PTQ + 90° + 90° + 60° = 360° (angle sum property)
∠PTQ = 120°
Question 24.
Find the value of:
sin 30°. cos 60° + cos 30°. sin 60°
Is it equal to sin 90° or cos 90°?
Solution:
sin 30° cos 60° + cos 30° sin 60°
It is equal to sin 90° = 1 but not equal to cos 90° as cos 90° = 0.
Question 25.
In fig., sectors of two concentric circles of radii 7 cm and 3.5 cm are given. Find the area of shaded region. [Use π = 22/7]
Solution:
Area of larger sector OCDO = \(\frac{\theta}{360^{\circ}}\)π(7)2
Area of smaller sector OABO = \(\frac{\theta}{360^{\circ}}\)π(\(\frac{7}{2}\))2
Area of required shaded region ACDB = Area of larger sector OCDO – Area of smaller sector OABO
= 9.6 cm2
OR
Find the number of metallic circular discs with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Section – C(18 Marks)
Question 26.
Find the greatest number of six digits exactly divisible by 18, 24 and 36.
Solution:
LCM of 18, 24 and 36
18 = 2 × 32
24 = 23 × 3
36 = 22 × 32
LCM (18, 24, 36) = 23 × 32 ⇒ 72
The largest 6 digit number is 999999
∴ The required number = 9,99,999 – 63 = 9,99,936.
So, 9,99,936 is the required largest number. Since digit number divisible by 18, 24 and 36.
Question 27.
Find the sum of the integers between 100 and 200 that are divisible by 6.
Solution:
The series as per question is 102, 108, 114, …….., 198 which is an A.P.
Given, a = 102, d = 6 and l = 198
Then 198 = 102 + (n – 1)6
Question 28.
Solve for x: x2 + 5x – (a2 + a – 6) = 0.
Solution:
Given,
Thus, x = a – 2 or x = –(a + 3)
OR
Divide 27 into two parts such that the sum of their reciprocals is \(\frac{3}{20}\).
Solution:
Let two parts be x and 27 – x.
⇒ x2 – 27x + 180 = 0
⇒ (x – 15) (x – 12) = 0
⇒ x = 12 or 15.
∴ The two parts are 12 and 15.
Question 29.
(A) In the given figure, AOB is a diameter of the circle with centre O and AC is a tangent to the circle at A. If ∠BOC = 130°, then find ∠ACO.
Solution:
∠OAC = 90° (as radius ^ tangent)
∠BOC = ∠OAC + ∠ACO
(Exterior angle property)
Or, 130° = 90° + ∠ACO
Or, ∠ACO = 130° – 90° = 40°
(B) In figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.
Solution:
∠TOQ = 180° – 70° = 110° [angle of supplementary]
Question 30.
If 1 + sin2 θ = 3 sin θ cos θ, prove that tan θ = 1 or \(\frac{1}{2}\).
Solution:
Given, 1 + sin2 θ = 3 sin θ cos θ
On dividing by sin2 θ on both sides, we get
\(\frac{1}{\sin ^2 \theta}\) + 1 = 3 cot θ
[∵ cot o = \(\frac{\cos \theta}{\sin \theta}\)]
⇒ cosec2 θ + 1 = 3 cot θ
⇒ 1 + cot2 θ + 1 = 3 cot θ
⇒ cot2 θ – 3 cot θ + 2 = 0
⇒ cot2 θ – 2 cot θ – cot θ + 2 = 0
⇒ cot θ(cot θ – 2) – 1(cot θ – 2) = 0
⇒ (cot θ – 2)(cot θ – 1) = 0
If cot θ = 1 or \(\frac{1}{2}\)
Then, tan θ = 1 or \(\frac{1}{2}\)
OR
If cos θ + sin θ = √2 cos θ, show that cos θ – sin θ = √2sin θ.
Solution:
Given, cos θ + sin θ = √2 cos θ
Or, Squaring both side
cos2 θ + sin2 θ + 2 sin θ cos θ = 2 cos2 θ
⇒ 2 sin θ.cos θ = cos2 θ – sin2 θ
⇒ 2 sin θ. cos θ = (cos θ + sin θ) (cos θ – sin θ)
√2 sin θ = cos θ − sin θ
Hence proved.
Question 31.
Two different dice are thrown together. Find the probability of:
(i) getting a number greater than 3 on each dice.
(ii) getting a total of 6 or 7 of the numbers on two dice.
Solution:
(i) Favourable outcomes are (4, 4) (4, 5) (4, 6) (5, 4), (5, 5) (5, 6) (6, 4) (6, 5) (6, 6)
∴ No. of favourable outcomes = 9
P(a number > 3 on each dice) = \(\frac{9}{36}\) or \(\frac{1}{4}\)
(ii) Favourable outcomes are (1, 5) (2, 4) (3, 3) (4, 2) (5, 1), (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
∴ No. of favourable outcomes = 11
P(a total of 6 or 7) = \(\frac{11}{36}\)
Section – D(20 Marks)
Question 32.
Swati can row her boat at a speed of 5 km/hr in still water. If it takes her 1 hour more to row the boat 5.25 km upstream than to return downstream, find the speed of the stream.
Solution:
Let the speed of the stream be x km/hr.
Speed of the boat upstream = (5 – x) km/hr.
Sped of the boat downstream = (5 + x) km/hr.
Time taken for going 5.25 km upstream = \(\frac{5.25}{5-x}\) hours.
Time taken for going 5.25 km downstream = \(\frac{5.25}{5+x}\) hours.
Obviously, time taken for going 5.25 km upstream is more than the time taken for going 5.25 km, downstream.
It is given that the time taken for going 5.25 km. upstream is 1 hour more than the time taken for going 5.25 downstream.
⇒ 21x = 50 – 2x
⇒ 2x2 + 21x – 50 = 0
⇒ 2x2 + 25x – 4x – 50 = 0
⇒ x(2x + 25) – 2 (2x + 25) = 0
⇒ (2x + 25)(x – 2) = 0
⇒ x – 2 = 0, 2x + 25 = 0
⇒ x = 2 [∵ x ≠ −\(\frac{25}{2}\) as x > 0]
Hence, the speed of the stream is 2 km/hr.
OR
One fourth of a herd of camels was seen in the forest, Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution:
Let the total number of camels be x. Then,
Number of camels seen in the forest = \(\frac{x}{4}\)
Number of camels gone to mountains = 2√x
Number of camels on the bank of river = 15
Total number of camels = \(\frac{x}{4}\) + 2√x + 15
By hypothesis, we have
\(\frac{x}{4}\) + 2√x + 15 = x
⇒ 3x – 8√x – 60 = 0
⇒ 3y2 – 8y – 60 = 0, where x = y2
⇒ 3y2 – 18y + 10y – 60 = 0
⇒ 3y (y – 6) + 10 (y – 6) = 0
⇒ (3y + 10)(y – 6) = 0
⇒ y = 6 or, y = –\(\frac{10}{3}\)
Now, y = –\(\frac{10}{3}\)
⇒ x = (–\(\frac{10}{3}\))2 = \(\frac{100}{9}\) [∵ x = y2]
But, the number of camels cannot be a fraction,
∴ y = 6
⇒ x = 62 = 36
Hence, the number of camels = 36. [∵ x = y2]
Question 33.
(A) Two right triangles ABC and DEC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.
(B) Are two triangles having equal corresponding sides, similar?
Solution:
(A) To prove: AP × PC = BP × DP.
Proof: In DBAP and DCDP,
∠A = ∠D (Given)
∠BPA = ∠DPC (vertically opposite angles)
∴ ΔBAP ~ ΔCDP (By AA similarity)
Then \(\frac{A P}{D P}=\frac{B P}{P C}\)
Or, AP × PC = BP × DP.
Hence proved.
(B) Yes, Two triangles having equal corresponding sides are congruent and all congruent Ds have equal angles, hence they are similar too.
Question 34.
Water is flowing at the rate of 5 km/hour through a pipe of diameter 14 cm into a rectangular tank of dimensions 50 m × 44 m. Find the time in which the level of water in the tank will rise by 7 cm.
Solution:
Speed of flow of water = 5 km/hr = 5000 m/hr
Length of cylinder for water flowing in one hour = 5000 m
OR
A copper wire, 6 mm in diameter, is wound about a cylinder whose length is 24 cm and diameter 20 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 9.85 g per cm3.
Solution:
Diameter of wire = 6 mm
Radius of wire = 3 mm = \(\frac{10}{3}\) cm = 0.3 cm
Length of cylinder = Height of cylinder = 24 cm
Diameter of cylinder = 20 cm
Radius of cylinder= \(\frac{20}{2}\) = 10 cm
Length of one turn = circumference of base of cylinder
= 2πr = 2 × 3.14 × 10
= 62.8 cm
Number of turns = \(\frac{\text { Length of cylinder }}{\text { Diameter of wire }}\) = \(\frac{24}{0.6}\)
= 40 turns
Total length of wire in 40 turns
= 62.8 × 40
= 2512 cm
Volume of wire = πr2h
= 3.14 × 0.3 × 0.3 × 2512
= 709.89 cm3
Mass = Volume × density
= 709.89 × 10
= 7098.9 g or 7099 g (approx)
Question 35.
Height of students in class X are given in the following frequency distribution:
Find the mode of the given data. Also If mean of the given data is 161.17 cm, find the median.
Solution:
Height of students | Frequency |
150-155 | 15 |
155 – 160 | 8 |
160 – 165 | 20 |
165 – 170 | 12 |
170 – 175 | 5 |
Median class = 160 – 165
= 160 + 3 = 163
As, mode = 163 and mean = 161.17
Therefore, According to empirical formula between mean, median and mode
3 Median = 2 Mean + mode
= 2(161.17) + 163
= 322.34 + 163
= 485.34
Median = \(\frac{485.34}{3}\) = 161 78.
Section – E(12 Marks)
Question 36.
The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream
(i) If speed of the stream be x km/h, then find the speed of the motorboat in upstream.
(ii) Write the relation between speed, distance and time?
(iii) Find the quadratic equation for the speed of the current?
OR
What is the speed of current?
Solution:
(i) We have, speed of the stream be x km/h
Speed of a motor boat is 20 km/h
So, the speed of motorboat in upstream will be (20 – x) km/h.
The speed will be less in upstream journey.
(ii) speed = \(\frac{\text { distance }}{\text { time }}\)
Speed is the distance travelled per unit time.
(iii) Let speed of the stream be x km/h
For covering the distance of 15 km the boat took one hour more for upstream than downstream.
We have,
⇒ \(\frac{15}{(20-x)}-\frac{15}{(20+x)}\) = 1
On simplifying we get,
x2 + 30x − 400 = 0
OR
On solving the quadratic equation x2 + 30x − 400 = 0
We get, x = 10 and –40(rejected)
So the speed of the current is 10 km/h.
Question 37.
Read the following text and answer the following questions on the basis of the same:
Using of mobile screen for long hours makes a child lazy, affect eyesight and give headache. Those who are addicted to playing PUBG can easily get stressed out or face anxiety issues in public due to lack of social interaction. To raise social awareness about it effects of playing PUBG, a school decided to start “Ban PUBG” campaign. Students are asked to prepare campaign board in the shape of rectangle.
(i) If we interchange both axes, find the coordinates of C.
(ii) Find the mid point of point A and point C.
(iii)Find the point of intersection of both axes.
OR
If Cost of 1 cm2 of board is? 10, find the cost of board. Given, 1 unit = 1 cm.
Solution:
(i) If both axes are interchanged, then abscissa and ordinate also interchange their places. Therefore, new coordinates of point C are (6, 8).
(ii) Midpoint of AC = \(\left(\frac{8+2}{2}, \frac{6+2}{2}\right)\)
(iii) Origin.
OR
Difference between abscissa of points A and B will be the Length = 8 – 2 = 6 units
Difference between abscissa of points A and D will be the Breadth = 6 – 2 = 4 units
Area of the board = L × B = 6 × 4 = 24 sq. units
Cost of 24 sq. units = ₹10 × 24 = ₹240.
Question 38.
Lakshaman Jhula is located 5 kilometers north-east of the city of Rishikesh in the Indian state of Uttarakhand. The bridge connects the villages of Tapovan to Jonk. Tapovan is in Tehri Garhwal district, on the west bank of the river, while Jonk is in Pauri Garhwal district, on the east bank. Lakshman Jhula is a pedestrian bridge also used by motorbikes. It is a landmark of Rishikesh. A group of Class X students visited Rishikesh in Uttarakhand on a trip. They observed from a point (P) on a river bridge that the angles of depression of opposite banks of the river are 60° and 30° respectively. The height of the bridge is about 18 meters from the river. Based on the above information answer the following questions.
(i) Find the distance PA.
(ii) Find the distance PB
(iii)Find the width AB of the river.
OR
Find the height BQ if the angle of the elevation from P to Q be 30°.
Solution:
Angle PAC = 60° (Alternate interior angle)
Angle PBC = 30° (Alternate interior angle)
(i) In Triangle PAC,
sin 60° = \(\frac{P C}{P A}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{18}{P A}\)
⇒ PA = 12√3 m
⇒ CB = 18 3 m
Width AB = AC + CB
= 6√3 + 18√3
= 24√3 m
OR
RB = PC = 18 m and
PR = CB = 18√3 m
tan 30° = \(\frac{Q R}{P R}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{Q R}{18 \sqrt{3}}\)
⇒ QR = 18 m
QB = QR + RB
= 18 + 18 = 36 m
Hence, height BQ is 36 m.