Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 7 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 7 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A-E.
- Section A has 20 MCQs carrying 1 mark each
- Section B has 5 questions carrying 02 marks each.
- Section C has 6 questions carrying 03 marks each.
- Section D has 4 questions carrying 05 marks each.
- Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
- All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E
- Draw neat figures wherever required. Take π =22/7 wherever required if not stated.
Section – A
(Section A consists of 20 questions of 1 mark each.)
Question 1.
The prime factorisation of 96 is: (1)
(a) 25 × 3
(b) 26
(c) 24 × 3
(d) 24 × 32
Answer:
(a) 22 × 3
Explanation: The prime factorisation of 96 is:
96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3
Question 2.
The zeroes of the poLynomial p(x) = 4x2 – 12x +9 is: (1)
(a) \(\frac{1}{2}\), 0
(b) 0,\(\frac{1}{2}\)
(c) \(\frac{5}{6}\),\(\frac{1}{2}\)
(d) \(\frac{3}{2}\),\(\frac{3}{2}\)
Answer:
(d) \(\frac{3}{2}\),\(\frac{3}{2}\)
Explanation: p(x) = 4x2 – 12x + 9 = (2x – 3)2 = 0
Thus, x= \(\frac{3}{2}\) and \(\frac{3}{2}\) are the two zeroes of p(x).
Question 3.
If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4. then the values of a and b is: (1)
(a) a = 3, b = 1
(b) a = 5, b = 2
(c) a = 1, b = 3
(d) a =4, b = 2
Answer:
(a) a = 3, b = 1
Explanation: As x = o, y = b is the solution of: x – y = 2 and x + y = 4
we have,
a – b = 2 and a + b = 4
⇒ a = 3 and b = 1
Question 4.
Discriminant of the quadratic equation 2x2 + 4x -7=0 is: (1)
(a) 28
(b) 72
(c) 36
(d) 48
Answer:
(b) 72
Explanation:
The discriminant of 2x2 + 4x- 7 = 0 is [(4)2 – 4(2)(-7)] = 16 + 56 = 72
Question 5.
The roots of quadratic equation x2 – 4x + 2 = 0 (1)
(a) 2 ± \(\sqrt{2}\)
(b) \(\sqrt{2}\),2
(c) 3,2
(d) 3 ± \(2 \sqrt{2}\)
Answer:
(a) 2 ± \(\sqrt{2}\)
Explanation:
Here, quadratic equation is: p(x) = x2 – 4x + 2 On comparing it with ax2 + fax + c = 0
Then, a = 1, b = – 4, c = 2
D = b2 – 4ac = (- 4)2 – 4 × 1 × 2
= 16 – 8 = 8
Then, roots, x = \(\frac{-b+\sqrt{D}}{2 a}\)
= \(\frac{-(-4)±\sqrt{8}}{2}\)
= \(\frac{4±2\sqrt{2}}{2}\)
= 2 ± \(\sqrt{2}\)
Hence, roots of the equation are 2 + \(\sqrt{2}\) and 2 – \(\sqrt{2}\).
Question 6.
If Sn = 5n2 + 3n, then its nth term is: (1)
(a) 5n – 1
(b) 10n2
(c) 10n – 2
(d) 8n – 3
Answer:
(c) 10n – 2
Explanation:
an = Sn – Sn-1
= 5n2 + 3n – [ 5 (n – 1)2 + 3(n – 1)]
= 5n2 + 3n – [ 5n2 + 5 – 10n + 3n – 3]
= 10n – 2
Question 7.
If the common difference of an A.P. is 5, then find a18 – a13. (1)
(a) 38
(b) 40
(c) 18
(d) 25
Answer:
(d) 25
Explanation:
a18 – a13 = a + 17d – a – 12d
= 5d = 5 × 5
= 25
Question 8.
In an A.P., a = -6 and d = 2. The sum of its first 20 terms is: (1)
(a) 270
(b) 200
(c) 195
(d) 260
Answer:
(d) 260
Explanation: Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
So, S20 = \(\frac{20}{2}\) [2(-6)+(20 – l)(2)]
= 10[-12 + 38]
= 10 × 26 = 260
Question 9.
Write the relationship between the coefficients, if the following pair of equations is inconsistent. ax + by + c = 0, a’x+b’y + d = 0. (1)
(a) \(\frac{a}{a^{\prime}} \neq \frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}\)
(b) \(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}} \neq \frac{c}{c^{\prime}}\)
(c) \(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}\)
(d) 0
Answer:
(b) \(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}} \neq \frac{c}{c^{\prime}}\)
Explanation:
the required relationship is:
\(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}} \neq \frac{c}{c^{\prime}}\)
Question 10.
In a ∆ABC, right-angled at B, if AB : AC = 1 : 2, then the value of \(\frac{2 tan A}{1+tan^{2} A}\) is: (1)
(a) \(\frac{1}{2}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{\sqrt{3}}{2}\)
(d) \(\frac{2}{\sqrt{3}}\)
Answer:
(c) \(\frac{\sqrt{3}}{2}\)
Explanation:
Question 11.
If a tower 6 metres high casts a shadow on the ground that is 2-^3 metres long, then the elevation of the sun is: (1)
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer:
(a) 60°
Explanation:
As per the given question:
Question 12.
The area of a circle, whose circumference is 22 cm, is: (1)
(a) 38 cm
(b) 36 cm2
(c) 38.5 cm2
(d) 40 cm2
Answer:
(c) 38.5 cm2
Explanation:
Let r be the radius of the circle.
Then,
2πr = 22 cm
⇒ r = \(\frac{22}{2 \times \frac{22}{7}}\)
⇒ r = \(\frac{7}{2}\) cm or 3.5 cm
Thus,
Area = πr2 = \(\frac{22}{7}\) × 3.5 × 3.5 cm2 = 38.5 cm2
Question 13.
If the ratio between the volumes of two spheres is 8 : 27, then the ratio between their surface areas is: (1)
(a) 2 : 3
(b) 1 : 2
(c) 25 : 16
(d) 4 : 9
Answer:
(d) 4 : 9
Explanation:
Let, the radius of 2 sphere be r and R
Question 14.
The class-mark of the class interval 10-25 is: (1)
(a) 17.5
(b) 16
(c) 14
(d) 18
Answer:
(a) 17.5
Explanation: The class mark of 10 – 25 is \(\frac{10 + 25}{2}\) = 17.5
Question 15.
One card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is either red or a queen. (1)
(a) \(\frac{5}{14}\)
(b) \(\frac{1}{14}\)
(c) \(\frac{5}{13}\)
(d) \(\frac{7}{13}\)
Answer:
(d) \(\frac{7}{13}\)
Explanation:
In a pack of 52 cards, there are 26 red cards and 2 black queens.
So, total possible outcomes = 26 + 2 = 28
P(a red or a queen) = \(\frac{28}{52}\) = \(\frac{7}{13}\)
Question 16.
How many face cards are there in a pack of 52 cards? (1)
(a) 12
(b) 10
(c) 14
(d) 16
Answer:
(a) 12
Explanation: Face cards are jacks, queen and kings
∴ Total face cards = 4 + 4 + 4=12
Question 17.
Determine the upper limit of the modal class of the following frequency distribution:
| Class | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
| Frequency | 13 | 10 | 15 | 8 | 11 |
(a) 16
(b) 19.5
(c) 18
(d) 17.5
Answer:
(d) 17.5
Explanation:
The given frequency distribution in the exclusive form is:
| Class | 0.5 – 5.5 | 5.5-11.5 | 11.5- 17.5 | 17.5 – 23.5 | 23.5 – 29.5 |
| Frequency | 13 | 10 | 15 | 8 | 11 |
Here, the modal class is 11.5 – 17.5
So, the upper limit of the modal class is 17.5.
Question 18.
The empirical relationship among the three measures of central tendency mean, mode and median. (1)
(a) 3 Median = Mode + 2 Mean
(b) Mode = 2 Median – Mean
(c) Mean = 3 Mode + 2 Median
(d) Median = Mode – Mean
Answer:
(a) 3 Median = Mode + 2 Mean
Explanation: The required relationship is: 3Median = Mode + 2Mean
DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option 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 (IQ are true and 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.
Statement A (Assertion): If the circumference of a circle is 176 cm, then its radius is 28 cm.
Statement R (Reason): Circumference = 2π × radius. (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Explanation:
Circumference (C) = 176 cm
2πr= 176
⇒ 2 × \(\frac{22}{7}\) × r = 176
⇒ \(\frac{44}{7}\) × r = 176
⇒ r = 176 × \(\frac{7}{44}\) = 28 cm
∴ The radius of the circle is 28 cm.
Question 20.
Statement A (Assertion): If the value of mode and mean is 60 and 66 respectively, then the value of median is 64.
Statement R (Reason): Median = mode + 2 mean (1)
Answer:
(c) Assertion (A) is true but reason(R) is false.
Explanation: Median = -(Mode + 2 mean) = (60 + 2 × 66) = 64
Section – B
(Section B consists of 5 questions of 2 marks each)
Question 21.
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, find the coordinates of P and Q.
OR
Find the third vertex of a triangle, if two of its vertices are at (-3,1) and (0, -2) and the centroid is at the origin. (2)
Answer:
Letthe coordinates of P and Q be (0, y) and (x,0) respectively.
∴ Mid-point of PQ : (\(\frac{0+x}{2}\), \(\frac{y+0}{2}\))
i.e (\(\frac{x}{2}\),\(\frac{y}{2}\))
Equating it with (2, -5), we have,
\(\frac{x}{2}\) = 2; \(\frac{y}{2}\) = -5
⇒ x = 4, y = -10
Thus, the coordinates of P and Q are (0, -10) and (4, 0) respectively.
OR
Given that two points are (- 3, 1) and (0, -2) and its centroid is (0, 0).
Let the third vertex be (x, y). Then,
(\(\frac{x-3+0}{3}\),\(\frac{y+1-2}{3}\)) = (0,0)
⇒ \(\frac{x-3}{3}\) = 0 and \(\frac{y-1}{3}\) = 0
⇒ x = 3 and y = 1
Thus, the third vertex is (3,1).
Question 22.
Explain why 3 × 5 × 7 × 9 × 11 + 11 is a composite number.
OR
If n = 23 × 34 × 54 × 7, where n is a natural number, then find the number of consecutive zeros in n. (2)
Answer:
11(3 × 5 × 7 × 9 + 1) = 11 (945 + 1)
= 11 × 946 = 11 × 2 × 11 × 43
Since, it has more than 2 factors.
Therefore, it is a composite number.
OR
According to questions,
n = 23 × 34 × 54 × 7
= 23 × 34 × 53 × 5 × 7
= (2 × 5)3 × 34 × 5 × 7
= (10)3 × 34 × 5 × 7
= 34 × 5 × 7 × 1000
Thus, in the given natural number ‘ri there are 3 zeros.
Question 23.
A tangent from point A, which is placed 5 cm away from the circle’s centre, has a length of 4 cm. Find the circle’s radius. (2)
Answer:
We know that the tangent at any point of a circle is ⊥ to the radius through the point of contact.
∴ ∠OPA = 90°
∴ OA2 = OP2 = AP2
(By Pythagoras theorem)
⇒ (5)2 = (OP)2 + (4)2
⇒ 25 = (OP)2 + 16
⇒ OP2 = 9
⇒ OP = 3 cm
(OP = – 3 cm is moved)
Question 24.
Find the missing frequency for the given data if mean of distribution is 52. (2)
| Wages in Rs | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| No. of Workers | 5 | 3 | 4 | f | 2 | 6 | 13 |
Answer:
Given data is
Find the missing frequency for the given data is mean of distribution is 52.
| Wages in Rs | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| No. of Workers | 5 | 3 | 4 | f | 2 | 6 | 13 |
Given data is
| Wages | fi | xi | fixi |
| 10-20 | 5 | 15 | 75 |
| 20-30 | 3 | 25 | 75 |
| 30-40 | 4 | 35 | 140 |
| 40-50 | f | 45 | 45 f |
| 50-60 | 2 | 55 | 110 |
| 60-70 | 6 | 65 | 390 |
| 70-80 | 13 | 75 | 975 |
| Total | 33 + f | 1765 + 45f |
Then, Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)
\(52=\frac{1765+45 f}{33+f}\)
⇒ 7f = 1765 – 1716 = 49
⇒ f = 7
Hence, missing frequency is 7.
Question 25.
If a number x is chosen at random from the numbers -2, -1, 0,1, 2, then what is the probability that x2 < 2 ? (2)
Answer:
Total outcomes = 5
Favourable outcomes (i.e., x2 < 2) are 4, 1, 0, 1, 4 i.e. 3
∴ P(x2 < 2) = \(\frac{3}{5}\)
Hence, the required probability is \(\frac{3}{5}\)
Section – C
(Section C consists of 6 questions of 3 marks each.)
Question 26.
If p is a prime number, then prove that \(\sqrt{p}\) is irrational. (3)
Answer:
Let us assume, to the contrary that \(\sqrt{p}\) is rationaL
So, we canfind co-prime integers a’ and ‘b’ (b* 0),
such that \(\sqrt{p}\) = \(\frac{a}{b}\)
⇒ \(\sqrt{p}\) b = a
⇒ pb2 = a2 …(i)
⇒ a2 is divisible by p
⇒ a is divisible by p.
So, we can write a = pc for some integer c.
Therefore, a2 = p2c2 (squaring both side)
⇒ pb2 = p2c2 (from (i))
⇒ b2 = pc2
⇒ b2 is divisible by p
⇒ b is divisible by p
⇒ p divides both a and b.
⇒ ‘a’ and ‘b’ have at least p’ as a common factor
But this contradicts the fact that ‘a and ‘b’ are coprime.
This contradiction arises because we have assumed that \(\sqrt{p}\) is rational.
Therefore, \(\sqrt{p}\) is irrational.
Question 27.
Find the zeros of the polynomial 2x2 – (1 + 2\(\sqrt{2}\))x + \(\sqrt{2}\). (3)
Answer:
p(x) = 2x2 – (1 + 2\(\sqrt{2}\))x + \(\sqrt{2}\) = 0
= 2x2 – x + 2\(\sqrt{2}\)x + \(\sqrt{2}\) = 0
= x(2x – 1) – \(\sqrt{2}\) (2x – 1) = 0
= (2x – 1) (x – \(\sqrt{2}\)) = 0
Either, 2x – 1 = 0 or x – \(\sqrt{2}\) = 0
⇒ x = \(\frac{1}{2}\) or x = \(\sqrt{2}\)
Thus, zeros of the given polynomial are \(\frac{1}{2}\) and \(\sqrt{2}\).
Question 28.
Solve for x:\(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \cdot x \neq 0,1,2\) (3)
Answer:
Here,
\(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \cdot x \neq 0,1,2\)
\(\frac{(x-1)+2(x-2)}{(x-2)(x-1)}\) = \(\frac{6}{x}\)
\(\frac{x-1+2x-4}{(x^{2}-2x-x+2)}\) = \(\frac{6}{x}\)
⇒ (3x – 5)x = 6(x2 – 3x + 2)
⇒ 3x2 – 5x = 6x2 – 18x + 12
⇒ 3x2 – 13x + 12 = 0
⇒ 3x2 – 9x – 4x + 12 = 0
⇒ 3x(x – 3) – 4(x – 3) = 0
⇒ (3x – 4) (x – 3) = 0
⇒ x = \(\frac{4}{3}\),3
Hence, the values of x are \(\frac{4}{3}\) and 3.
Question 29.
In the figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with the chord PQ, then determine ∠POQ.
OR
O is point inside a triangle ABC The bisectors of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA at point D, E and F respectively, show that AD x BE x CF = DB x EC x FA. (3)
Answer:
Here, ∠QPR = 50°
and ∠OPR = 90°
[∵ Radius is perpendicular to tangent]
∴ ∠OPQ + ∠QPR = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
Now, OP = OQ [Radii of same circle]
∴ ∠OPQ = ∠OQP = 40°
In ∆OPQ,
∠POQ + ∠OPQ + ∠OQP = 180°
⇒ ∠POQ + 40° + 40° = 180°
⇒ ∠POQ = 180° – 80° = 100°
OR
Given : A AABC, in which O is a point inside it. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CD at points D, E and F respectively.
=> DB × EC × FA = AD × BF × CF
or AD × BE × CF = DB × EC × FA
Hence, proved.
Question 30.
The diagram shows a round about at the junction of four roads (of equal width).
The central park is in the form of a circle with centre O and radius 14 m.
The curbs BC, DE, FG and HA are in the form of arcs that lie on a circle with centre O and radius 21 m. The angles subtended by these curbs at O are 60°, 45°, 45°, 90°.
(A) Find the total lengths of the curbs; 1 1/2
(B) Find the area of the circular road surrounded the central park. 1 1/2
Answer:
(B) Area of the circular road surrounding the central park
= [π(21)2 – π(14)2] m2
= \(\frac{22}{7}\)× (21 + 14) (21 – 14) m2
= 770 m\(\frac{22}{7}\).
Question 31.
Rnd the median of the following data:
| Marks (out of 90) | No. of Students |
| 0-10 | 2 |
| 10-20 | 2 |
| 20-30 | 4 |
| 30-40 | 6 |
| 40-50 | 6 |
| 50-60 | 5 |
| 60-70 | 2 |
| 70-80 | 4 |
| 80-90 | 4 |
| Total | 35 |
OR
Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. Find the probability that both will visit the shop:
(A) the same day
(B) different days
(C) consecutive days. (3 )
Answer:
Then, \(\frac{N}{2}\) = \(\frac{35}{2}\) = 17.5
Then, median class is 40 – 50
l = 40 f = 6, c.f. = 14, h = 10
Then Me = l + \(\frac{N/2 – cf}{f}\) × h
= 40 + \(\frac{17.5 – 14}{6}\) × 10
= 40 + \(\frac{3.5}{6}\) × 10 = 40 + 5.83
= 45.83
Thus, median of the given data is 45.83
OR
Two customer can visit the shop in 6 × 6 ways = 36 ways.
Total no. of events = 36
(A) Two customer can visit the shop on the same day in one of the following ways, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
∴ P(of visiting on same day) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
(B) ∴ P(of visiting on different day) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
(C) Two customers can visit the shop on 2 consecutive days in the following ways : (Mon, Tue), (Tue, Wed), (Wed, Thurs), (Thurs, Fri), (Fri, Sat) = 5 ways ,
∴ P (of visiting on consecutive days) = \(\frac{5}{36}\)
Section D
(Section D consists of 4 questions of 5 marks each.)
Question 32.
State and prove basic proportionality theorem.
OR
D and E are points on the sides CA and CB respectively of a AABC, right-angled at C. Prove that: AE2 + BD2 = AB2 + DE2 (5)
Answer:
Statement “If a side is parallel to one side of a triangle and it intersects the other two sides in 2 distinct points, then it divides the other 2 sides in same proportion.”
Proof:
Given: A ∆ABC, a line DE parallel to BC intersect AB and D and AC at E.
To prove: \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
Construction : Draw EF ⊥ AB and DG ⊥ AC Join BE and CE
Proof: Since, EF ⊥ AB
EF is the height of triangle ADE and DBE.
Area of ∆ADE = \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × AD × EF
Area of ∆DBE = \(\frac{1}{2}\) × DB × EF
But ∆DBE and ∆DCE are on the same base DE and between the same parallel straight lines BC and DE.
Area of ∆DBE = Area of ∆DCE …(iii)
From, (i), (ii) and (iii), we have
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
Hence proved.
OR
In the figure, ACB is a right angled triangle, right-angled at C. D and E are points on sides CA and BC respectively.
We Join DE, BD and AE.
In right triangle ACE, we have:
AE2 = AC2 + EC2 …(i)
In right triangle BCD, we have:
BD2 = BC2 + DC2 …(ii)
Adding (i) and (ii), we get:
AE2 + BD2 = (AC2 + EC2) + (BC2 + DC2)
= (BC2 + CA2) + (CE2 + CD2)
= AB2 + DE2.
Hence proved.
Question 33.
Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot A at the rate of 2 for 3 bananas and the second lot B at the rate of ₹ 1 per banana, and got a total collection of f 400. If he had sold the first lot A at the rate of f 1 per banana, and the second lot B at the rate of 4 for 5 bananas, his total collection would have been ₹ 460.
Determine the total number of bananas he had. (5)
Answer:
Let lot A contains ‘x’ bananas; and lot B contains ‘y’ bananas. Then, according to the question,
\(\frac{2}{3}\)x + y = 400 ; x + \(\frac{4}{5}\) y = 460
⇒ 2x + 3y = 1200 ; 5x + 4y = 2300
⇒ 10x + 15y = 6000 …(i)
and 10x + 8y = 4600 …(ii)
Subtracting (ii) from (i), we get
⇒ 7 y = 1400
⇒ y = 200.
From (i), we get
10x + 15 × 300 = 6000
⇒ 10x = 1500
⇒ x = 300
Thus, lot A contains 300 bananas and lot B contains 200 bananas.
Vijay had 500 bananas in all.
Question 34.
Two tangents are drawn from a point P to a circle with a centre of O. Prove that ∆APB is equilateral, if OP = diameter of the circle. (5)
Answer:
Join OP.
Suppose OP meets the circle at Q. Join AQ.
We have
i.e., OP = diameter
∴ OQ + PQ = diameter
∴ OQ + PQ = diameter
PQ = Diameter – radius [••• OQ = r]
∴ PQ = radius
Thus, OQ = PQ = radius
Thus, OP is the hypotenuse of right triangle
OAP and Q is the mid-point of OP
∴ OA = AQ = OQ
[v mid-point of hypotenuse of a right triangle is equidistant from the vertices]
⇒ AOAQ is equilateral
⇒ ∠AOQ = 60°
So, ∠APO = 30°
∴ ∠APB = 2∠APO = 60°
Also PA = PB
⇒ ∠PAB = ∠PBA
But ∠APB = 60°
∴ ∠PAB = ∠PBA = 60°
Hence, ∆APB is equilateral triangle.
Question 35.
Trigonometric ratios sin A, sec A, and tan A should be expressed in the form of cotA.
OR
The angles of depression of the top and bottom of building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distances between the building and the tower. (5)
Answer:
For sin A,
Bu using identity, cosec2 A – cot2 A = 1 .
OR
Let PQ be the tower of height h m and AB be the building of height 50 m.
From right triangle BDQ,
\(\frac{DQ}{BD}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
⇒ BD = \(\sqrt{3}\) DQ …..(i)
Also, from right triangle APQ,
\(\frac{PQ}{AP}\) = tan 60° = \(\sqrt{3}\)
or \(\frac{PQ}{BD}\) = \(\sqrt{3}\)
or BD = \(\sqrt{3}\) …..(ii)
From (i) and (ii), we have
DQ = \(\frac{PQ}{3}\)
Further,
PQ = PD + DQ
SECTION – E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)
Question 36.
Selvi is setting up a water purifier system in her house which includes setting up an overhead tank in the shape of a right circular cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid.
The underground water tank (sump) is a sturdy single moulded piece built to with stand underground pressure and is available in the storage capacity of 2000 L
These, along with hassle-free installation and minimum maintenance needs make it the ideal water storage solution.
Dimensions (sump):
1.57 m × 1.44 m × 95 cm.
Dimensions (overhead tank):
Radius is 60 cm and Height is 95 cm
Water flow conditions at the required overload capacity should be checked for critical pressure drop to ensure that valves are adequately sized.
On the basis of the above information, answer the following questions:
(A) Find the ratio of the capacity of the sump to the capacity of the overhead tank. (1)
(B) If overhead tank need to be painted to save it from corrosion, how much area need to be painted? (1)
(C) If water is filled in the overhead tank at the rate of 20 litre per minute, the tank will be completely filled in how much time?
OR
If the amount of water in the sump, at an instant, is 1500 litres , then find the water level in the sump at that instant? (2)
Answer:
(A)
(B) CSA. of cylindrical tank = 2πrH
= 2 × 3.14 × 60 × 95
= 35,796 cm2
= 3.5796 m2
= 3.6 m2
(C) Volume of water in cylindrical tank = πr2h
= 3.14 × 60 × 60 × 95
= 1073880 cm3
Now, 1l = 1000 cm3
∴ Volume of tank = 1073.88l
20l tank is filled in 1 minute
∴ 1073.88l tank is filled in \(\frac{1073.88 l}{20}\)
= 53.69
= 54 minutes
Question 37.
Rishu ¡s riding in a hot air bafloon. After reaching a point P, he spots a car parked at B on the ground at an angle of depression of 30°. The balloon rises further by 50 metres and now he spots the same car at an angle of depression of 45° and a lorry parked at B’ at an angle of depression of 30°. (Use \(\sqrt{3}\) = 1.73)
The measurement of Rishu facing vertically is the height. Distance is defined as the measurement of car/lorry from a point in a horizontal direction. If an imaginary line is drawn from the observation point to the top edge of the car/lorry, a triangle is formed by the vertical, horizontal and imaginary line.
On the basis of the above information, answer the following questions:
(A) If the height of the balloon at point P is ‘h’m and distance AB is ‘x’ m, then find the relation between ‘x’ and ‘h’. (1)
(B) Find the relation between the height of the balloon at point P’ and distance AB. (1)
(C) Find the height of the balloon at point P, and the distance AB on the ground.
OR
Find the distance B’B on the ground. (2)
Answer:
(C) On solving equation obtained in (A) and (B), we get
\(\sqrt{3}\)h = h + 50
⇒ h(\(\sqrt{3}\) – 1) = 50
⇒ h = \(\frac{50}{0.732}\) = 68.25
In ∆APB.
tan 30° = \(\frac{AP}{AB}\)
⇒ AB = \(\frac{AP}{tan 30°}\) = \(\frac{68.25}{1/\sqrt{3}}\)
= 68.25 × 1.732
= 118 m
OR
In ∆AP’B’
tan 30° = \(\frac{AP’}{AB’}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{68.25+50}{AB’}\)
⇒ AB’ = 118.25 × 1.732 = 204.809
BB’ = AB’ – AB
= 204.809 – 118
= 86.80 = 87 m
Question 38.
To conduct sport day activities in the rectangular school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each along DB.
100 flower pots have been placed at a distance of 1 m from each other along DA as shown in the figure below.
Radha runs \(\frac{1}{4}\)th of the distance DA on 2nd line and post a green flag at X . Preeti runs \(\frac{1}{5}\)th of the distance DA on other line post a red flag at Y.
On the basis of the above information, answer the following questions:
(A) Treating DB as x-axis and DA as y-axis, then find the position of green flag. (1)
(B) Treating DB as x-axis and DA as y-axis, the find the position of red flag. (1)
(C) Find the distance (in complete metres) between the two flags.
OR
Find the perimeter (in complete metres) of the triangular region OXY. (2)
Answer:
(A) Radha’s distance on x-axis is 2 and on y-axis, she is at \(\frac{1}{4}\) × 100 = 25°
Green flag coordinates are (2,25)
(B) X-coordinate = 8
Y-coordinate = \(\frac{1}{5}\) × 100 = 20
Coordinates of red flag (8,20).
(C) coordinates of green flag is (2,25)
∴ coordinates of red flag is (8,20)
∴ By distance formula
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{\left(8-2\right)^2+\left(20-25\right)^2}\)
= \(\sqrt{36 + 25}\)
= \(\sqrt{61}\)
= 7.8 cm
= 8 cm (approx)
OR
Perimeter = OX + OY + YY
= 25.07 + 21.54 + 7.81
= 54.42
= 55 m