Students can access the CBSE Sample Papers for Class 10 Maths Basic with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Basic Set 5 with Solutions
Time: 3 Hours
Maximum Marks: 80
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-II (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-mark questions of Section E.
8. Draw neat figures wherever required, Take. x = 2, wherever required if not stated.
Section-A(20 Marks)
Question 1.
If a and b are co-prime numbers, then find the HCF (a, b).
(A) 2
(B) 1
(C) 0
(D) None of these
Solution:
(B) 1
Explanation: Since, 1 is the only common factor of co-prime numbers.
Thus, HCF (a, b) = 1
Question 2.
What will be the simplest form of \(\frac{\sqrt{45}+\sqrt{20}}{\sqrt{5}}\) ?
(A) 5√5
(B) 5
(C) 25
(D) √5
Solution:
(B) 5
Explanation: \(\frac{3 \sqrt{5}+2 \sqrt{5}}{\sqrt{5}}=\frac{5 \sqrt{5}}{\sqrt{5}}\) = 5
Question 3.
How many zero(s) does the polynomial 293x2 – 293x have?
(a) 0
(B) 1
(C) 2
(D) 3
Solution:
(C) 2
Explanation: As we have 293x2 – 293x which is a polynomial of degree 2 and the maximum number of roots of a polynomial is equal to its degree. So, given polynomial can have only two zero.
Question 4.
What about the number of solutions if the given system of equations represent coincident lines?
(A) One solution
(B) Infinite or many solutions
(C) No solution
(D) None of these
Solution:
(B) Infinite or many solutions
Explanation: As we know that, Each point on the Coincident line will be a solution for given pair of linear equations, So, infinite or many solution exists.
Question 5.
Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹ 2 coins are, respectively :
(A) 35 and 15
(B) 35 and 20
(C) 15 and 35
(D) 25 and 25
Solution:
(D) 25 and 25
Explanation: Let the number of ₹ 1 coins = x and the number of ₹ 2 coins = y
So, according to the question,
x + y = 50 ….(i)
x + 2y = 75 ….(ii)
Subtracting equation (i) from (ii) y = 25
Substituting value of y in (i) x = 25
So, y = 25 and x = 25.
Question 6.
The distance of the point P(3, -4) from the origin is
(A) 7 units
(B) 5 units
(C) 4 units
(D) 3 units
Solution:
(b) 5 units
Detailed Solution :
Here, origin O(0, 0)
Point P(3, – 4)
By using distance formula
OP = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
OP = \(\sqrt{(3-0)^2+(-4-0)^2}\)
= \(\sqrt{3^2+(-4)^2}\)
\(=\sqrt{9+16}=\sqrt{25}\) = 5 units.
Question 7.
In the following figure, ST || QR, point S divides PQ in the ratio 4: 5. If ST = 1.6 cm, what is the length of QR?
(Note: The figure is not to scale.)
(A) 0.71 cm
(B) 2 cm
(C) 36 cm
(D) (cannot be calculated from the given data.)
Solution:
(C) 36 cm
Explanation:
As given, ST is parallel to QR then in triangle PST and ΔPQR, ∠S = ∠Q and ∠T = ∠R (corresponding angle)
Therefore, ΔPST and ΔPQR are similar by (AA)
Now, \(\frac{P S}{P Q}=\frac{S T}{Q R}\) = (As we know ratio of corresponding sides of similar triangles are always equal)
Question 8.
In the given figure, if ∠A = 90°, ∠B = 90°, OB = 45cm, OA = 6 cm and AP = 4 cm, then find QB.
(A) 1 cm
(B) 2 cm
(C) 3 cm
(D) 4 cm
Solution:
(C) 3 cm
Solution:
(C) 3 cm
Question 9.
In the given figure, AT is tangent to the circle with centre ‘O’ such that OT = 4cm and ∠OTA = 30°. Then AT is equal to
(A) 4cm
(B) 2 cm
(C) 2√3cm
(D) 4√3 cm
Solution:
(C) 2√3cm
Explanation: Join OA. OA is radius and AT is tangent they both meet at point A.
∠OAT = 90°,
(Tangent of a circle is a line that touches the circle exactly at one point. The point at which tangent touches the circle is known as ‘point of contact’. The radius of the circle and tangent are perpendicular to each other at the point of contact).
Given that, OT = 4 cm
Now, \(\frac{A T}{4}=\frac{\text { base }}{\text { hypotenuse }}\) = cos 30°
⇒ AT = 4 × \(\frac{\sqrt{3}}{2}\) = 2√3 cm.
Question 10.
If sec θ + tan θ = p, then tan θ is
(A) \(\frac{p^2+1}{2 p}\)
(B) \(\frac{p^2-1}{2 p}\)
(C) \(\frac{p^2-1}{p^2+1}\)
(D) \(\frac{p^2+1}{p^2-1}\)
Solution:
(B) \(\frac{p^2-1}{2 p}\)
Explanation: sec θ + tan θ = p is the given equation.
Since, 1 + tan2 θ = sec2 θ
Or, sec θ = \(\sqrt{1+\tan ^2 \theta}\)
Put this value of sec θ in equation (i), we get
\(\sqrt{1+\tan ^2 \theta}\) + tan θ = p
Or, \(\sqrt{1+\tan ^2 \theta}\) = p – tan θ
On squaring both sides, we get
1 + tan2 θ = p2 + tan2 θ – 2p tan θ
Or, 1 = p2 – 2p(tan θ)
Or, 1 – p2 = -2p(tan θ)
Or, tan θ = \(\frac{p^2-1}{2 p}\)
Question 11.
The value of 9 sec2 A – 9 tan2 A is
(A) 1
(B) 9
(C) 8
(D) 0
Solution:
(B) 9
Explanation:
9sec2 A – 9tan2 A = 9(sec2 A – tan2 A)
= 9(1) = 9
Question 12.
A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is:
(A) 60°
(B) 45°
(C) 30°
(D) 90°
Solution:
(A) 60°
Question 13.
If the volume of a cube is 8 cm3, then the total surface area is
(A) 24 cm2
(B) 384 cm2
(C) 48 cm2
(D) 40 cm2
Solution:
(A) 24 cm2
Explanation: Volume of cube = 8 cm3
a3 = 8
a3 = 23
a = 2 cm
total surface area of cube = 6a2
= 6(2)2
= 6(4)
= 24 cm2
Question 14.
The area of a sector for a circle of radius 1.4 cm and central angle 90° is:
(A) 154 cm2
(B) 154 cm2
(C) 3.08 cm2
(D) 308 cm2
Solution:
(A) 154 cm2
Explanation: The area of the sector
\(\begin{aligned}
& =\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 1.4 \times 1.4
\end{aligned}\) = 1.54 cm2
Question 15.
A dice is rolled twice. The probability that 5 will not come up either time is
(A) \(\frac{11}{36}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{13}{26}\)
(D) \(\frac{25}{36}\)
Solution:
(D) \(\frac{25}{36}\)
Explanation: All possible events are written below:
(1 1) (1 2) (1 3) (1 4) (1 5) (1 6)
(2 1) (2 2) (2 3) (2 4) (2 5) (2 6)
(3 1) (3 2) (3 3) (3 4) (3 5) (3 6)
(4 1) (4 2) (4 3) (4 4) (4 5) (4 6)
(5 1) (5 2) (5 3) (5 4) (5 5) (5 6)
(6 1) (6 2) (6 3) (6 4) (6 5) (6 6)
Total events = 36
Out of the events in which 5 will not come up either time are (1, 1) (1, 2) (1, 3) (1, 4) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 6).
No. of required events = 25
Required probability = \(\frac{25}{36}\)
Question 16.
The mode of the numbers 2, 3, 3, 4, 5, 4, 4, 5, 3, 4, 2, 6, and 7 is:
(A) 2
(B) 3
(C) 4
(D) 5
Solution:
(C) 4
Explanation: Mode = 4 (occurring most number of times)
Question 17.
A cylinder, a cone, and a hemisphere have same base and same height. The ratio of their volumes is ……
(A) 1:3:3
(B) 3:3:1
(C) 3:1:2
(D) 2:1:3
Solution:
(C) 3:1:2
Explanation: Ratio of volumes = Volume of cylinder
∴ Volume of cone: Volume of hemisphere
\(\begin{aligned}
& =\pi r^2 r: \frac{1}{3} \pi r^2 r: \frac{2}{3} \pi r^3 \quad[r=h] \\
& =1: \frac{1}{3}: \frac{2}{3}
\end{aligned}\) (each term is divide by πr3
= 3:1:2
Question 18.
Find the class marks of the classes 10 – 25 and 35 – 55.
(A) 17.5, 17.5
(B) 17.5, 45
(C) 18.5, 50
(D) 10, 55
Solution:
(B) 17.5, 45
Explanation: Class mark of 10-25 = \(\frac{10+25}{2}\) = \(\frac{35}{2}\) = 17.5
and, Class mark of 35-55 = \(\frac{35+55}{2}\) = \(\frac{90}{2}\) = 45
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): If the distance between the point (4, p) and (1, 0) is 5, then the value of p is 4.
Reason (R): The point which divides the line segment joining the points (7, -6) and (3, 4) in ratio 1: 2 internally lies in the fourth quadrant.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but Reason (R) does not give 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.
Solution:
(B) Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
Explanation: In case of assertion: Distance between two points (x1, y1) and (x1, y2) is given as d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
where, (x1, y1) = (4, p)
(x1, y2) = (1, 0)
And d = 5
Put the values, we have
53 =(1 – 4)3 + (0 – p)3
25 = (-3)3 + (-p)3
25 – 9 = p3
16 = p3
p = ± 4, As given p = 4 is correct.
∴ Assertion is true.
In case of reason: Let (x, y) be the point
So, the required point (x, y) = (\(\frac{17}{3},-\frac{8}{3}\) lies in IVth quadrant.
∴ Reason is correct.
Hence, Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
Question 20.
Assertion (A): LCM of 200 and 240 is 1200.
Reason (R): HCF of two co-prime numbers is 1.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but Reason (R) does not give 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.
Solution:
(B) Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
Explanation: Prime factorisation of 200 = 23 × 52 and 240 = 24 × 3 × 51
LCM (200, 240) = 24 × 3 × 52 = 1200
So, assertion is true.
It is true that HCF of co-prime numbers is 1.
Therefore, Both A and R are true but R is not the correct explanation of A.
Section-B(10 Marks)
Question 21.
For what value of ‘K’, the system of equations kx + 3y = 1, 12x + ky = 2 have no solution.
Solution:
The given equations can be written as kx + 3y – 1 = 0 and 12x + ky – 2 = 0
Here, a1 = k, b1 = 3, c1 = -1 and a2 = 12, b2 = k, c2 = -2
The equation for no solution
and k ≠ 6
Or, k = ± 6, Hence k = -6
Question 22.
In the given figure, PQ and PR are tangents to the circle with centre O such that ∠QPR = 60°, then find ∠OQR.
Solution:
∠QPR = ∠60°(Given)
∠QOR + ∠QPR = 180° (supplementary angles)
∴ ∠QOR = 180° – 60° = 120°
From ΔOQR,
∠OQR + ∠ORQ + ∠QOR = 180
(Angle sum property of triangle)
∠OQR + ∠OQR + 120 = 180
(∠OQR = ∠ORQ, Opposite angles of equal sides )
∠OQR + ∠OQR = 180 – 120 2∠OQR = 60, ∠OQR = \(\frac{60}{2}\) = 30°
OR
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ
Solution:
Let ∠OPQ be θ.
∴ ∠TPQ =∠OPT – ∠OPQ
∴ ∠TPQ = (90°- 0)
(∠OPT = 90 , The radius of the circle and tangent are perpendicular to each other at the point of contact)
Since TP = TQ (Tangents)
∴ ∠TQP = (90° – θ) (Opposite angles of equal sides)
Now, ∠TPQ + ∠TQP + ∠PTQ = 180°
⇒ 90° – θ + 90°- θ + ∠PTQ = 180°
⇒ ∠PTQ = 180° -180° + 2θ
⇒ ∠PTQ = 2θ
⇒ ∠PTQ = 2∠OPQ. Hence proved.
Question 23.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\)
Solution:
Question 24.
Prove that: \(\sqrt{\frac{1-\cos A}{1+\cos A}}\) cosec A – cot A
Solution:
Question 25.
Find the perimeter of a quadrant of a circle of radius 14 cm.
Solution:
Perimeter of a quadrant = 2(Radius) + Pad\(\frac{\theta}{360}\) × 2πr
= 2(14) + = \(\frac{90}{360}\) x2x \(\frac{22}{7}\) x14 [For quadrant, 0 = 90°]
= 28 + \(\frac{90}{360}\) × 88
= 28 + 22
= 50 cm
OR
Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.
Solution:
Let radius of first circle = R = 24 cm and radius of second circle = r = 7 cm
And let radius of third circle = A cm
Area of first circle = πR2 = π(24)2 = 576 π cm2
Area of second circle = πr” = π(7)2 = 49 π cm2
Area of third circle = Area of first circle + Area of second circle
πA2 = 576 π + 49 π
Or, A = 625
Or, A=25cm
Therefore, radius of third circle = 25 cm.
Section -C(18 Marks)
Question 26.
Prove that √2 is an irrational number.
Solution:
Let us assume √2 be a rational number and its simplest form be a/b, a and b are co-prime positive integers and b ≠ 0.
So, √2 = a/b
⇒ a2 = 2b2
Thus a2 is a multiple of 2
⇒ a is a multiple of 2
Let a = 2 m for some integer m
b2 = 2m2
Thus, b2 is a multiple of 2
⇒ b is a multiple of 2
Hence, 2 is a common factor of a and b.
This contradicts the fact that a and b are co-primes
Hence, √2 is an irrational number.
Question 27.
The fifth term of an A.P is 20 and the sum of its seventh and eleventh terms is 64. Find the common difference.
Solution:
Let the first term be a and common difference be d.
Then, a +4d = 20 (Given T5 = 20) …(i)
and a + 6d + a + 10d = 64
a + 8d = 32
(Given T17 + T11 = 64) …(ii)
Solving equations (i) and (ii), we get d = 3
Hence, common difference, d = 3
Question 28.
Find the quadratic polynomial, the sum and product of zeroes are -3 and 2 respectively. Hence find the zeroes.
Solution:
Sum of zeroes = -3 and Product of zeroes = 2
The quadratic polynomial will be x2 – (Sum of zeroes)
x + Product of zeroes
x2 – (-3z) + 2 = 0
x2 + 3x + 2 = 0
x2 + 2x + x + 2 = 0
x(x + 2) +1 (x + 2) = 0
(x + 2)(x + 1) = 0
x = -2 and -1
OR
Find the zeroes of the quadratic polynomial x2 – 2√2x and verify the relationship between the zeroes and the coefficients.
Solution:
We know that, p(x) = x2 – 2√2x
⇒ x(x – 2√2) = 0
⇒ x = 0 & 2√2
Sum of zeroes = 0 + 2√2 = 2√2
Product of zeroes = (0) × ( 2√2, ) = 0
Sum of zeroes = -b/a = 2√2
Product of zeroes = c/a = 0
Question 29.
In given fig., two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q.
Solution:
Since PT = TC ….(i)
and QT = TC ….(ii) [tangents of circle from external point]
So, PT = QT [from eq (i) and eq (ii)]
Now, PQ = PT + TQ
⇒ PQ = PT + PT
⇒ PQ = 2PT
⇒ 1/2 PQ = PT = TQ
Question 30.
Prove that \(\frac{1+\tan ^2 A}{1+\cot ^2 A}\) = sec2 A – 1
Solution:
OR
From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60° and 45°. Find the distance between two cars.
Solution:
Hence, the distance between two cars = 189.28 m.
Question 31.
From all the two digit numbers a number is chosen at random. Find the probability that the chosen number is a multiple of 7.
Solution:
All possible outcomes are 10, 11, 12 ….. 98, 99.
No. of all possible outcomes = 90.
All favourable outcomes are 14, 21, 28 …. 98
No. of favourable outcomes = 13
∴ P(getting a number multiple of 7) = 13/90
Section -D(20 Marks)
Question 32.
For what values of a and b does the following pair of linear equations have infinite number of solutions?
2x + 3y = Z and a(x + y)—b(x-y) = 3a + b – 2.
Solution:
For equation
2x + 3y – 7 =0
a1 = 2, b1 = 3 and c1 = -7
and a(x + y)-b (x – y) = 3a + b – 2
ax + ay – bx + by = 3a + b – 2
(a – b) x + (a + b) y- (3a + b – 2) = 0
a2 = a – b, b2 = a + b and c2 = -(3a + b – 2)
For infinitely many solutions
OR
(a) Solve for x:\(\frac{x}{x-1}+\frac{x-1}{x}\) whole \(\frac{1}{4}\), x ≠ 0,1.
(b) Is 1 a root for the quadratic equation, \(\frac{1}{x-1}=\frac{2}{x-2}\) +4?
Solution:
\(\frac{2 x^2-2 x+1}{x^2-x}=\frac{17}{4}\)
4(2x2 – 2x + 1) = 17(x2 – x)
8x2 – 8x + 4 = 17x2 – 17x
9x2 – 9x – 4 =0
9x2 – (12-3)x – 4 = 0
9x2 – 12x + 3x – 4 = 0
3x(3x – 4) + 1(3x – 4) = 0
(3x – 4)(3x + 1) = 0
3x – 4 = 0, 3x + 1 = 0
3x = 4, 3x = -1
x = 4/3 x = -1/3
(b) No. because \(\frac{1}{x-1}\) is not defined for x = 1
Question 33.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Given: A circle with centre O is inscribed in a quadrilateral ABCD.
To prove : ∠AOB + ∠COD = 180°
Proof: In ΔAEO and ΔAFO,
OE = OF [radii of circle]
∠OEA = ∠OFA = 90° [radius is ⊥ to tangent]
The point of contact is perpendicular to the tangent.
OA = OA [common side]
ΔAEO ≅ ΔAFO [by RHS]
∠7 = ∠8 (By cpct) …(i)
Similarly, ∠1 = ∠2 …(ii)
∠3 = ∠4 …(iii)
∠5 = ∠6 …(iv)
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° [angle around a point]
2∠1 + 2 ∠8 + 2 ∠4 + 2 ∠5 = 360°
∠1 + ∠8 + ∠4 + ∠5 = 180°
(∠1 + ∠8) + (∠4 + ∠5) = 180°
∠AOB + ∠COD = 180° (Hence Proved)
Question 34.
A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find the volume of the solid. (Use π = \(\frac{22}{7}\))
Solution:
OR
Water is flowing at the rate of 15 km/h through a cylindrical pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time the level of water in pond rise by 21 cm?
Solution:
Question 35.
The following table gives the monthly consumption of electricity of 100 families:
Find the median of the above data.
Solution:
Monthly Consumption | Number of families(f) | Cumulative frequency (C.f) |
130-140 | 5 | 5 |
140-150 | 9 | 14 |
150-160 | 17 | 31 |
160-170 | 28 | 59 |
170-180 | 24 | 83 |
180-190 | 10 | 93 |
190-200 | 7 | 100 |
Section – E(12 Marks)
Question 36.
Jaspal Singh takes a loan from a bank for his car. Jaspal Singh repays his total loan of ₹ 118000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month.
Answer the following questions:
(i) If the given problem is based on A.P, then what is the first term and common difference?
(ii) Find the amount to be paid by him in 25th instalment.
(iii) Find the total amount to be paid by him in 25th and 30th instalment.
Solution:
(i) The number involved in this case form an A.P in which first term (a) = 1000 and common difference (d) = 100.
(ii) The amount paid by him in 25th instalment is:
T25 = a + 2Ad [Tn = a + (n – 1)d]
= 1000 + 24 × 100 = 1000 + 2400
= ₹ 3400.
(iii) Total amount paid by him in 25th and 30th instalment
T25+ T30
= [a + (25 – 1)d] + [a + (30 – 1)d]
= [a + 24d] + [a + 29d]
= 2a + 53d
= 2 × 1000 + 53 × 100
= 2000 + 5300
= 7300
OR
The difference amount paid by him in 26th and 28th instalment.
Solution:
The amount paid by him in 26th instalment,
T26 = a + 25d
= 1000 + 25 × 100
= 1000 + 2500
= ₹ 3500
The amount paid by him in 28th instalment,
T28 = a + 27d
= 1000 + 27 × 100
= 1000 + 2700
= ₹ 3700
∴ The difference amount paid by him in 26th and 28th instalment is:
= ₹ (3700 – 3500)
= ₹ 200
Question 37.
An electrician has to repair an electric fault on the pole of height 5 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work (see figure).
(i) What is the length of BD?
(ii) What should be the length of ladder, when inclined at an angle of 60° to the horizontal?
(iii) How far from the foot of the pole should he place the foot of the ladder?
Solution:
(i) From figure, the electrician is required to reach at the point B on the pole AD.
OR
If the horizontal angle is changed to 30°, then what should be the length of the ladder?
Solution:
In ΔBDC,
∴ sin 30° = \(\frac{B D}{B C}\)
⇒ \(\frac{1}{2}\) = \(\frac{3.7}{B C}[latex]
⇒ BC = 3.7 × 2 = 7.4 m
Question 38.
In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs [latex]1 / 4^{\text {th }}\) the distance AD on the 2th line and posts a green (G) flag. Preet runs \(1 / 5^{\text {th }}\) distance AD on the eighth line and posts a red (R) flag.
Answer the following questions:
(i) Find the position of green flag.
(ii) Find the position of red flag.
(iii) What is the distance between both the flags?
Solution:
(i) From the given figure, abscissa = 2 and ordinate = \(\frac{1}{4}^{\text {th }}\) of 100
= \(\frac{1}{4}\) × 100 = 25
So, required point = G(2, 25)
(ii) From the given figure, abscissa = 8 and ordinate = \(\frac{1}{5}^{\text {th }}\) of 100
= \(\frac{1}{5}\) × 100 = 20
So, required point = R (8, 20)
(iii) Position of Green flag = (2, 25)
Position of Red flag = (8, 20)
Distance between both the flags
\(\begin{aligned}
\left|\sqrt{(8-2)^2+(20-25)^2}\right| & =\left|\sqrt{6^2+(-5)^2}\right| \\
& =\sqrt{36+25} \\
& =\sqrt{61} \text { units }
\end{aligned}\)
OR
If Rashmi has to post a blue flag exactly half way between the line segment joining the flags, where should she post her flag?
Solution:
Position of blue flag
Mid-point of line segment joining the green and red flags
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\) = (5, 22.5)