Bhagya

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Here, r = 6 cm and θ = 60°

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let radius of the circle be r.
∴ 2πr = 22

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Length of minute hand = radius of the circle (r) = 14 cm
∵ Angle swept by the minute hand in 60 minutes = 360°
∴ Angle swept by the minute hand in 360°
5 minutes = \(\frac{360^{\circ}}{60}\) × 5 = 30°
Now, area of the sector with r = 14 cm and θ = 30°
= \(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 14 × 14 cm²
= \(\frac{11 \times 14}{3} \mathrm{cm}^{2}=\frac{154}{3} \mathrm{cm}^{2}\)
Thus, the required area swept by the minute hand in 5 minutes = \(\frac{154}{3}\) cm²

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major segment.(Use π = 3.14)
Solution:
Length of the radius (r) = 10 cm
Sector angle (θ) = 90°

= [314 – 78.5] cm² = 235.5 cm²

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
Solution:
Here, radius(r) = 21 cm and θ = 60°
(i) Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × 21 cm = 2 × 22 × 3 cm = 132 cm

∴ Length of the arc APB

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and \(\sqrt{3}=\) = 1.73)
Solution:
Here, radius (r) = 15 cm and
Sector angle (θ) = 60°
∴ Area of the sector
\(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{60^{\circ}}{360^{\circ}} \times \frac{314}{100}\) × 15 × 15 cm²
= \(\frac{11775}{100}\) cm² = 117.75 cm²
Since ∠O = 60° and OA = OB = 15 cm
∴ AOB is an equilateral triangle.

⇒ AB = 15 cm and ∠A = 60°
Draw OM ⊥ AB,
In ∆AMO

Now area of the minor segment = (Area of minor sector) – (ar ∆AOB)
= (117.75 – 97.3125) cm² = 20.4375 cm²
Area of the major segment = [Area of the circle] – [Area of the minor segment]
= πr² – 20.4375 cm²
= [\(\frac{314}{100}\) × 15²] – 20.4375 cm²
= (706.5 – 20.4375) cm² = 686.0625 cm²

Question 7.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and \(\sqrt{3}=\) = 1.73 )
Solution:
Here θ = 120° and r = 12 cm

Draw OM ⊥ AB
In ∆AOB, ∠O = 120°
By angle sum property,
∠A + ∠B + ∠O = 180°
⇒ ∠A + ∠B = 180° – 120° = 60°
∵ OB = OA = 12 cm
⇒ ∠A = ∠B = 30°

= 36 × 1.73 cm² = 62.28 cm²
∴ Area of the minor segment = [Area of sector] – [Area of ∆AOB]
= [150.72 cm²] – [62.28 cm²] = 88.44 cm²

Question 8.
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10m long instead of 5 m. (Use π = 3.14)

Solution:
Here, length of the rope = 5 m
∴ Radius of the circular portion grazed by the horse(r) = 5 m
(i) Area of the circular portion grazed

(ii) When length of the rope is increased to 10 m, then, r = 10 m
Area of the new circular portion grazed
= \(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{90^{\circ}}{360^{\circ}} \times \frac{314}{100}\) × (10)² m²
= \(\frac{1}{4}\) × 314 m² = 78.5 m²
∴ Increase in the grazing area = (78.5 – 19.625) m² = 58.875 m²

Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution:
Diameter of the circle = 35 mm
∴ Radius (r) = \(\frac{35}{2}\) mm
(i) Circumference = 2πr
= 2 × \(\frac{22}{7} \times \frac{35}{2}\) mm = 22 × 5 mm = 110 mm
Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm
∴ Length of 5 pieces = 5 × 35 mm = 175 mm
∴ Total length of the silver wire = (110 + 175) mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

Question 10.
An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
Here, radius (r) = 45 cm
Since circle is divided into 8 equal parts,
∴ Sector angle corresponding to each part

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Here, radius (r) = 25 cm
Sector angle (θ) = 115°
∴ Total area cleaned at each sweep of the blades

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution:
Here, radius (r) = 16.5 km and sector angle (θ) = 80°
∴ Area of the sea surface over which the ships are warned

Question 13.
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use \(\sqrt{3}=\) = 1.7)

Solution:
Here, r = 28 cm
Since, the circle is divided into six equal sectors.

Now, area of 1 design = Area of segment APB = Area of sector ABO – Area of ∆AOB ………..(2)
In ∆AOB, ∠AOB = 60°, OA = OB = 28 cm
∴ ∠OAB = 60° and ∠OBA = 60°
⇒ ∆AOB is an equilateral triangle.
⇒ AB = AO = BO ⇒ AB = 28 cm
Draw OM ⊥ AB
∴ In right ∆AOM, we have

= 14 × 14 × 1.7 cm² = 333.2 cm² ……………(3)
Now, from (1), (2) and (3), we have
Area of segment APB
= 410.67 cm² – 333.2 cm² = 77.47 cm²
⇒ Area of 1 design = 77.47 cm²
∴ Area of the 6 equal designs = 6 × (77.47) cm² = 464.82 cm²
Hence, the cost of making the design at the rate of ₹ 0.35 per cm²
= ₹ 0.35 × 464.82 = ₹ 162.68.

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

Solution:
(D) Here, radius = R
Angle of a sector (θ) = p
∴ Area of the sector

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why’
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and 8 for each
additional km.
(ii) The amount of air present In a cylinder when a vacuum pump removes 1/4 of the
air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs 150 for
the first metre and rises by 50 for each subsequent metre.
(iv) The amount of money in the account every year, when 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Let us consider, the first term
(T₁) Fare for the first 1 km = ₹ 15
Since, the taxi fare beyond the first 1 km is ₹ 8 for each additional km.
∴ Fare for 2 km = ₹ 15 + 1 × ₹ 8
T₂ = a + 8 [where a = 15]
Fare for 3 km = ₹ 15 + 2 × ₹ 8
⇒ T₃ = a + 16
Fare for 4 km= ₹ 15 + 3 × ₹ 8
⇒ T₄ = a + 24
Fare for 5 km = ₹ 15 + 4 × ₹ 8
⇒ T₅ = a + 32
Fare for n km = ₹ 15 + (n – 1)8
⇒ T_n = a + (n – 1)8
We see that above terms form an AP with common difference 8.

(ii) Let the amount of air present in the cylinder be x

The above terms are not in A.P.

(iii) Here, the cost of digging for first 1 metre = ₹ 150
The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200
The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250
The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300
∴ The terms are: 150, 200, 250, 300,…..
Since, 200 – 150 = 50 and 250 – 200 = 50
(200 – 150) (250 – 200) = 50
∴ The above terms form an AP with common difference 50.

(iv)

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given
as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 1/2
(y) a = -1.25, d = -0.25
Solution:
(i) ∵ T_n = a + (n – 1)d
∴ For a = 10 and d = 10, we have:
T₁ =10 + (1 – 1) × 10 = 10 + 0 = 10
T₂ = 10 + (2 – 1) × 10 = 10 + 10 – 20
T₃ = 10 + ( 3 – 1) × 10 = 10 + 20 = 30
T₄ = 10 + (4 – 1) × 10 = 10 + 30 = 40
Thus, the first four terms are:
10, 20, 30, 40

(ii) ∵ T_n = a + (n – 1)d
∴ For a = -2 and d = 0,we have:
T₁ = -2 + (1 – 1) × 0 = -2 + 0 = -2
T₂ = -2 + (2 – 1) × 0 = -2 + 0 = -2
T₃ = -2 + (3 – 1) × 0= -2 + 0 = -2
T₄ = -2 + (4 – 1) × 0 = -2 + 0 = -2
∴ Thus, the first four terms are: -2, -2, -2, -2

(iii) ∵ T_n = a + (n – 1)d
For a = 4 and d = -3, we have:
T₁ = 4 + (1 – 1) × (-3) = 4 + 0 = 4
T₂ = 4 + (2 – 1) × (-3) = 4 + (-3) = 1
T₃ = 4 + (3 – 1) × (-3) = 4 + (-6) = -2
T₄ = 4 + (4 – 1) × (-3) = 4 + (-9) = -5
Thus, the first four terms are:
4, 1, -2, -5

(iv)

(v) ∵ T_n = a + (n – 1)d
∴ For a = -1.25 and d = -0.25, we have
T₁ = -1.25 + (1 – 1) × (-0.25) = -1.25 + 0
= -1.25
T₂ = -1.25 + (2 – 1) × (-0.25) = -1.25 + (-0.25) = -1.50
T₃ = -1.25 + (3 – 1) × (-0.25) = -1.25 + (-0.50) = -1.75
T₄ = -1.25 + (4 – 1) × (-0.25) = -1.25 + (-0.75) = -2.0

Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3,…..
(ii) -5, -1, 3, 7,….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots .\)
(iv) 0.6, 1.7, 2.8, 3.9, …..
Solution:
(i) We have : 3, 1, -1, -3, …
⇒ T₁ ⇒ 3 = a = 3
T₂ = 1, T₃= -1, T₄= -3
∴ T₂ – T₁ = 1 – 3 = -2
T₄ = T₃ = -3 – (-1) = -3 + 1 = -2 ⇒ d = -2

(ii) We have : -5, -1, 3, 7,….
⇒ T₁ = -5 ⇒ a = -5, T₂ = -1, T₃ = 3, T₄ = 7
∴ T₂ – T₁ = -1 – (-5) = -1 + 5 = 4
and T₄ – T₃ = 7 – 3 = 4 ⇒ d = 4
Thus a = -5, d = 4

(iii)

Question 4.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16,
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2, 2, ….
(v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots\)
(vi) 0.2, 0.22, 0.222, 0.2222, ….
(vii) 0, -4, -8, -12,
(ix) 1, 3, 9, 27, …..
(x) o, 2a, 3a, 4a,…
(xi) a, a₂, a₃, a₄,….
(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)
(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots .\)
(xiv) 1², 3², 5², 7², ….
(xv) 1², 5², 7², 73, …..
Solution:
(i)

∴ The given numbers do not form an AP

(ii)

(iii) We have : -1.2, -3.2, -5.2, -7.2,
∴ T₁ = -1.2, T₂ = -3.2, T₃ = -5.2, T₄ = -7.2
T₂ – T₁ = -3.2 + 1.2 = -2
T₃ – T₂ = -5.2 + 3.2 = -2
T₄ – T₃ = -7.2 + 5.2 = -2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = -2 d = -2
∴ The given numbers form an AP such that d = -2.
Now, T₅ = T₄ + (-2) = -7.2 + (-2) = -9.2,
T₆ = T₅ + (-2) = -9.2 + (-2) = -11.2 and
T₇ = T₆ + (-2) = -11.2 + (-2) = -13.2
Thus, d = -2 and T₅ = -9.2, T₅ = -11.2 and T₆ = -13.2

(iv) We have : -10, -6, -2, 2,
T₁ = -10, T₂ = -6, T₃ = -2, T₄ = 2
T₂ – T₁ = -6 + 10 = 4
T₃ – T₂ = -2 + 6 = 4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4 ⇒ d = 4
∴ The given numbers form an AP Now, T₅ = T₄ + 4 = 2 + 4 = 6,
T₆ = T₅ + 4 = 6 + 4 = 10,
T₇ = T₆ + 4 = 10 + 4 = 14
Thus, d = 4 and T₅ = 6, T₆ = 10, T₇ = 14

(v) We have :
3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….

(vi) We have : 0.2, 0.22, 0.222, 0.2222,….

∴ The given numbers do not form an AP

(vii) We have : 0, – 4, – 8, – 12,
∴ T₁ = 0, T₂ = -4, T₃ = -8, T₄ = -12 T₂ – T₁ = -4 – 0 = -4
T₃ – T₂ = – 8 + 4 = – 4
T₄ – T₃ = -12 + 8 = -4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = -4 ⇒ d = -4
∴ The given numbers form an AP
Now, T₅ = T₄ + (- 4) = -12 + (- 4) = -16
T₆ = T₅ + (-4) = -16 + (-4)= -20
T₇ = T₆ + (- 4) = -20 + (- 4) = -24
Thus, d = – 4 and T₅ = -16, T₆ = -20,
T₇ = -24.

(viii)

(ix) We have, 1, 3, 9, 27,….

(x) We have : a, 2a, 3a, 4a,

(xi)

(xii)

(xiii)

(xiv)
We have : 1², 3₂, 5₂, 7₂,….

(xv) We have : 1₂ , 5₂, 7₂, 7₂,….

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
We have ar(∆ABC) = 64 cm²
ar(∆DEF) = 121 cm² and EF = 15.4 cm[Given]
∵ ∆ABC ~ ∆DEF
∴ \(\frac { { ar }(\Delta ABC) }{ { ar }(\Delta DEF) } =\left( \frac { BC }{ EF } \right) ^{ 2 }\)
[Ratios of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides]

Thus, BC = 11.2 cm

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
In trapezium ABCD, AB || DC. Diagonals AC and BD intersect at O.
In ∆AOB and ∆COD,

∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
∴ Using AA criterion of similarity, we have
∆AOB ~ ∆COD

i. e., ar(∆AOB) : ar(∆COD) = 4 : 1

Question 3.
In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac { { ar }(\Delta ABC) }{ { ar }(\Delta DBC) } =\frac { AO }{ DO } \)

Solution:
We have, ∆ABC and ∆DBC are on the same base BC. Also BC and AD intersect at O.
Let us draw AE⊥BC and DF⊥BC.
In ∆AOE and ∆DOF,
∠AEO = ∠DFO = 90° ……….. (1)
Also, ∠AOE = ∠DOF …………… (2)
[Vertically Opposite Angles]

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
We have ∆ABC and ∆DEF, such that ∆ABC ~ ∆DEF and ar(∆ABC) = ar(∆DEF).
Since the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.


i.e., the corresponding sides of ∆ABC and ∆DEF are equal.
⇒ ∆ABC ≅ ∆DEF [By SSS congruency]

Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:
We have a ∆ABC in which D, E and F are mid points of AB, BC and CA respectively. D, E and F are joined to form ∆DEF.

Now, in ∆ABC, D and F are the mid-points of sides AB and AC.
∴ \(\frac{A D}{D B}=\frac{A F}{F C}\) = 1
∴ By the converse of the basic proportion¬ality theorem, we have,
DF||BC ⇒ DF||BE
Similarly; EF||AB ⇒ EF||BD
Since, DF||BE and DB||EF
∴ Quadrilateral BEFD is a parallelogram.
⇒ FE = BD = \(\frac{1}{2}\)AB …………. (1)
Similarly, quadrilateral ECFD is a parallelogram.
⇒ DF = EC = \(\frac{1}{2}\)BC ………… (2)
and DE = FC = \(\frac{1}{2}\)AC ………….. (3)
Now, in ∆ABC and ∆DEF

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
We have two triangles ABC and DEF such that ∆ABC ~ ∆DEF

AM and DN are medians corresponding to BC and EF respectively.
∵ ∆ABC ~ ∆DEF
∴ The ratio of their areas is equal to the square of the ratio of their corresponding sides.

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
We have a square ABCD, whose diagonal is AC. Equilateral ∆BQC is described on the side BC and another equilateral ∆APC is described on the diagonal AC.

∵ All equilateral triangles are similar.
∴ ∆APC ~ ∆BQC
∴ The ratio of their areas is equal to the square of the ratio of their corresponding sides.

Tick the correct answer and justify:
Question 8.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution:
(C) : We have an equilateral ∆ABC and D is the mid point of BC. DE is drawn such that BDE is also an equilateral traingle. Since all equilateral triangles are similar,

∴ ∆ABC ~ ∆BDE
⇒ The ratio of their areas is equal to the square of the ratio of their corresponding sides.
∴ \(\frac { { ar }(\Delta ABC) }{ { ar }(\Delta BDE) } =\left( \frac { AB }{ BD } \right) ^{ 2 }\) …………. (1)
∵ AB = AC = BCfsides of equilateral AABC]
and BD = \(\frac{1}{2}\)BC [∵ D is the mid point of BC]
⇒ BC = 2BD = AB …………… (2) [∵ AB = BC]
From (1) and (2), we have

⇒ ar(∆ABC) : ar(∆BDE) = 4 : 1

Question 9.
Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution:
(D) : We have two similar triangles such that the ratio of their corresponding sides is 4 : 9.
∴ The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
∴ \(\frac { { ar }(\Delta -I) }{ { ar }(\Delta -II) } =\left( \frac { 4 }{ 9 } \right) ^{ 2 }=\frac { 16 }{ 81 } \)
⇒ ar(∆-I): ar(∆-II) = 16 : 81

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solution:
(i) we have p(x) = x² – 2x – 8
= x² + 2x + – 4x – 8
= x(x + 2) – 4(x + 2)
= (x – 4)(x + 2)
For p(x) = 0, we must have (x – 4)(x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = -2
∴ The zeroes of x² – 2x – 8 are 4 and -2
Now, sum of zeroes


Thus, the relationship between the zeroes and the coefficients in the polynomial x² – 2x – 8 is verified.

(ii) We have p(s) = 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s(2s – 1) – l(2s – 1)
= (2s – 1)(2s – 1)

Thus, the relationship between the zeroes and coefficients in the polynomial 4s² – 4s + 1 is verified.

(iii) We have p(x) = 6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
For p(x) = 0, we have,

Thus, the relationship between the zeroes and coefficients in the polynomial 6x² – 3 – 7x is verified.

(iv) We have, p(u) = 4u² + 8u = 4u(u + 2)
For p(u) = 0,
we have
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = -2
∴ The zeroes of 4u² + 8u are 0 and – 2.
Now, 4u² + 8u can be written as 4u² + 8u + 0.

Thus, the relationship between the zeroes and coefficients in the polynomial 4u² – 4s + 1 is verified.

(v) We have, p(t) = t² – 15 = (t)² – \((\sqrt{15})^{2}\)

Thus, the relationship between zeroes and the coefficients in the polynomial t² – 15 is verified.

(vi) We have, p(x) = 3x² – x – 4
= 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1)(3x – 4)
For p(x) = 0 we have,
Either (x + 1) = 0 ⇒ x = -1

Thus, the relationship between the zeroes and coefficients in the polynomial 3x² – x – 4 is verified

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), -1
(ii) \(\sqrt{2}, \frac{1}{3}\)
(iii) 0, \(\sqrt{5}\)
(iv) 1, 1
(v) \(-\frac{1}{4}, \frac{1}{4}\)
(vi) 4,1
Solution:
(i) Since, sum of zeroes, \((\alpha+\beta)=\frac{1}{4}\)
Product of zeroes, αβ = -1
∴ The required quadratic polynomial is

have samezeroes, therefore (4x² – x – 4) is the required quadratic polynomial.

(ii) Since, sum of zeroes, \((\alpha+\beta)=\sqrt{2}\)
product of zeroes, αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial is

(iii) Since, sum of zeroes, (α + β) = 0
Product of zeroes, αβ = \(\sqrt{5}\)
∴ The required quadratic polynomial is

(iv) Since, sum of zeroes, (α + β) = 1
Product of zeroes, αβ = 1
∴ The required quadratic polynomial is

(v) Since, sum of zeroes, (α + β) = \(-\frac{1}{4}\)
Product of zeroes, αβ = \(\)[\frac{1}{4}/latex]
∴ The required quadratic polynomial is

same zeroes, therefore, the required quadratic polynomial is (4x² + x + 1)

(vi) Since, sum of zeroes, (α + β) = 4 and
product of zeroes, αβ = 1
∴ The required quadratic polynomial is
x² – (α + β)x + αβ = x² – 4x + 1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.6 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6

Question 1.
In the figure, PS is the bisector of ∠QPR of ∆PQR. Prove that \(\frac{Q S}{S R}=\frac{P Q}{P R}\).

Solution:
We have, ∆PQR in which PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS
Let us draw RT || PS to meet QP produced at T, such that
∠1 = ∠RPS [Alternate angles]
Also, ∠3 = ∠QPS [Corresponding angles]

But ∠RPS = ∠QPS [Given]
∴ ∠1 = ∠3
∴ PT = PR
Now, in ∆QRT, PS || RT [By construction]
∴ Using the Basic Proportionality Theorem, we have

Question 2.
In the figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that
(i) DM² = DN.MC
(ii) DN² = DM.AN

Solution:

Question 3.
In the figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² – AB² + BC² + 2BC.BD

Solution:
∆ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
∵ In ∆ADB, ∠D = 90°
∴ Using Pythgoras Theorem, we have
AB² = AD² + DB² ….. (1)
In right ∆ADC, ∠D = 90°
∴ Using Pythagoras Theorem, we have
AC² = AD² + DC²
= AD² + [BD + BC]²
= AD² + [BD² + BC² + 2BD.BC]
⇒ AC² = [AD² + DB²] + BC² + 2BC – BD
⇒ AC² = AB² + BC² + 2BC – BD [From (1)] Thus, AC² = AB² + BC² + 2 BC.BD

Question 4.
In the figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC.BD.

Solution:
We have ∆ABC in which ∠ABC < 90°
and AD ⊥ BC
In right ∆ADB, ∠D = 90°
Using Pythagoras Theorem, we have
AB² = AD² + BD² …… (1)
Also in right ∆ADC, ∠D = 90°
Using Pythagoras Theorem, we have AC² = AD² + DC²
= AD² + [BC – BD]² = AD² + [BC² + BD² – 2BC.BD]
= [AD² + BD²] + BC² – 2BC.BD = AB² + BC² – 2BC.BD [From (1)]
Thus, AC² = AB² + BC² – 2BC.BD, which is the required relation.

Question 5.
In the figure, AD is a median of triangle ABC and AM ⊥ BC. Prove that


Solution:

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
We have a parallelogram ABCD.

AC and BD are the diagonals of parallelogram ABCD.
∵ Diagonals of a parallelogram bisect each other.
∴ O is the mid-point of AC and BD.
Now, in ∆ABC, BO is a median.

Question 7.
In the figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC – ∆DPB
(ii) AP.PB = CP.DP

Solution:
We have two chords AB and CD of a circle. AB and CD intersect at P.
(i) In ∆APC and ∆DPB,
∴ ∠APC = ∠DPB ….. (1)
[Vertically opp. angles]
∠CAP = ∠BDP …… (2)
[Angles in the same segment]
From (1) and (2) and using AA similarity, we have
∆APC ~ ∆DPB

(ii) Since, ∆APC ~ ∆DPB [As proved above]
∴ Their corresponding sides are proportional,

⇒ AP.BP = CP.DP, which is the required relation.

Question 8.
In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ∆PAC – ∆PDB
(ii) PA.PB = PC.PD

Solution:
We have two chords AB and CD when produced meet outside the circle at P.
(i) Since in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle,

Question 9.
In the figure, D is a point on side BC of ∆ABC such that \(\frac{B D}{C D}=\frac{A B}{A C}\). Prove that AD is the bisector of ∠BAC.

Solution:
Let us produce BA to E such that AE = AC
Join EC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds ?

Solution:
Let us find the length of the string that Nazima has out.
In right ∆OAB, OB² = OA² + AB²
∴ OB² = (2.4)² + (1.8)²
⇒ OB² = 5.76 + 3.24 = 9.00
\(\Rightarrow \quad O B=\sqrt{9.00}=3 \mathrm{m}\)
i.e., Length of string she has out = 3 m

Since, the string is pulled in at the rate of 5 cm/sec,
∴ Length of the string pulled in 12 seconds

In the ∆PBC, let PB be the required horizontal distance of fly.
Since, PB² = PC² – BC² [By Pythagoras theorem]
∴ PB² = (2.4)² – (1.8)² = 5.76 – 3.24 = 2.52

Thus, the horizontal distance of the fly from Nazima after 12 seconds
= (1.59 + 1.2) m (approximately)
= 2.79 m (approximately)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Mode:
Here, the highest frequency is 23.
The frequency 23 corresponds to the class interval 35 – 45.
∴ The modal class is 35 – 45
Now, class size, h = 10
Lower limit, l = 35
Frequency of the modal class (f₁) = 23
Frequency of the class preceding the modal class (f₀) = 21
Frequency of the class succeeding the modal class (f₂) = 14
∴ Mode = l + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × h

Mean: Let assumed mean, a = 40
Class size, h = 10
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-40}{10}\)

∴ Required mean = 35.38 years.
Interpretation:
The maximum number of patients admitted in the hospital are of age 36.8 years while the average age of patients is 35.37 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Solution:
Here, the highest frequency = 61.
∵ The frequency 61 corresponds to class 60 – 80
∴ The modal class is 60 – 80
∴ We have l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38

Thus, the modal lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
Mode:
∵ The maximum number of families i.e., 40 have their total monthly expenditure is in interval 1500 – 2000.
∴ The modal class is 1500 – 2000 and l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33

Thus, the required modal monthly expenditure of the families is ₹ 1847.83.
Mean: Let assumed mean, a = 3250
and class size, h = 500
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-3250}{500}\)
∴ We have the following table;

∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σf_iu_i] = 3250 + 500 × \(\left[\frac{-235}{200}\right]\)
= 3250 – \(\frac{1175}{2}\) = 3250 – 587.50 = 2662.5
Thus, the mean monthly expenditure = ₹ 2662.50

Question 4.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
Mode : Since greatest frequency i.e., 10 corresponds to class 30 – 35.
∴ Modal class = 30 – 35 and h = 5, l = 30, f₁ = 10, f₀ = 9, f₂ = 3
∴ Mode = l + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × h
= 30 + \(\left[\frac{10-9}{20-9-3}\right]\) × 5
= 30 + \(\left[\frac{10-9}{20-9-3}\right]\) × 5 = 30 + 0.625 = 30.6 (approx)
Mean: Let the assumed mean, a = 37.5 and class size, h = 5
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-37.5}{5}\)
∴ We have the following table:


Interpretation:
The maximum teacher-student ratio is 30.6 while average teacher-student ratio is 29.2

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data
Solution:
∵ The class 4000 – 5000 has the highest frequency i.e., 18
∴ Modal class = 4000 – 5000.
Also, h = 1000, l = 4000, f₁ = 18, f₀ = 4, f₂ = 9

Thus, the required mode is 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:
∵ The class 40 – 50 has the maximum frequency i.e., 20
∴ Modal class = 40 – 50
∴ f₁ = 20, f₀ = 12, f₁ = 11 and l = 40. Also, h = 10

Thus, the required mode is 44.7 cars.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following:
(i) sin60° cos30° + sin 30° cos 60°
(ii) 2tan² 45° + cos²30° – sin²60°

Solution:
(i) We have




= \(\frac{\frac{1}{12}\times67}{\frac{4}{4}}=\frac{67}{12}\)

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\)
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0

(iii) sin 2A = 2sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution:

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) 0°< A + B ≤ 90°; A > B, find A and B.
Solution:
We have,
tan 60° = \(\sqrt{3}\), tan 30° = \(\frac{1}{\sqrt{3}}\) ……………. (1)
Also, tan(A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) …………… (2)
From (1) and (2), we get
A + B = 60° …………… (3)
and A – B = 30° ………….. (4)
On adding (3) and (4), we get
2A = 90° ⇒ A =45°
On subtracting (4) from (3), we get
2B = 30° ⇒ B = 15°

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin6 increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False:
Let us take A = 30° and B = 60°, then
L.H.S = sin (30° + 60°) = sin 90° = 1

∴ L.H.S. ≠ R.H.S.
(ii) True:
Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.
(iii) False:
Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.
(iv) False:
Let us take θ = 30°
sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
⇒ sin 30° ≠ cos 30°
(v) True:
We have, cot 0° = not defined

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.


Solution:
(i) In ∆ABC and ∆PQR,
We have : ∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
∴ The corresponding angles are equal.
∴ Using the AAA similarity rule,
∆ABC ~ ∆PQR

(ii) In ∆ABC and ∆QRP,

i.e., Using the SSS similarity, ∆ABC ~ ∆QRP

(iii) In ∆LMP and ∆DEF,

∴ Triangles are not similar,

(iv) In ∆MNL and ∆QPR,
\(\frac{M L}{Q R}=\frac{M N}{Q P}\) [\(\frac{1}{2}\) each]
and ∠NML = ∠PQR
∴ Using SAS similarity, we have
∆MNL ~ ∆QPR.

(v) In ∆ABC and ∆FDE,
\(\frac{A B}{D F}=\frac{B C}{E F}\) [\(\frac{1}{2}\) each]
Now, angle between DF and EF is 80°. But angle between AB and BC is unknown.
∴ Triangles are not similar.

(vi) In ∆DEF and ∆PQR,
∠D = ∠P = 70°
[∵ ∠P = 180° – (80° + 30°) = 180° – 110° = 70°]
∠E = ∠Q = 80°
∠F = ∠R = 30° [∵∠F = 180°]
∴ Using the AAA similarity rule,
∆DFE ~ ∆PRQ.

Question 2.
In the figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
We have, ∠BOC = 125° and ∠CDO = 70°
since, ∠DOC + ∠BOC = 180° [Linear pair]
⇒ ∠DOC = 180° – 125° = 55° ……………… (1)
In ∆DOC,
Using the angle sum property for ∆ODC, we get
∠DOC + ∠ODC + ∠DCO = 180°
⇒ 55° + 70° + ∠DCO = 180°
⇒ ∠DCO = 180° – 55° – 70° = 55°
Again,
∠DOC = ∠BOA ……………. (2) [vertically opposite angles]
and ∠OCD = ∠OAB = 55° ………….. (3)
[corresponding angles of similar triangles]
Thus, from (1), (2) and (3)
∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{O A}{O C}=\frac{O B}{O D}\).
Solution:
We have a trapezium ABCD in which AB || DC. The diagonals AC and BD intersect at O.

In ∆OAB and ∆OCD,
∠OBA = ∠ODC (Alternate angles)
and ∠OAB = ∠OCD(Alternate angles)
Using AA similarity rule, ∆OAB ~ ∆OCD
So, \(\frac{O B}{O D}=\frac{O A}{O C}\) (Ratios ot corresponding sides of the similar triangles)
⇒ \(\frac{O A}{O C}=\frac{O B}{O D}\)

Question 4.
In the figure, \(\frac{Q R}{Q S}=\frac{Q T}{P R}\) and ∠1 = ∠2. show that ∆PQS ~ ∆TQR.

Solution:
In ∆PQR
∵ ∠1 = ∠2 [Given]
∴ PR = QP ……………… (1)
[∵ In a ∆ sides opposite to equal angles are equal]

and ∠SQP = ∠RQT = ∠1
Now, using SAS similarity rule,
∆PQS ~ ∆TQR.

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
In ∆PQR,
T is a point on QR and S is a point on PR such that ∠RTS = ∠P.

Now in ∆RPQ and ∆RTS,
∠RPQ = ∠RTS [Given]
∠PRQ = ∠TRS [Common]
Using AA similarity, we have ∆RPQ ~ ∆RTS.

Question 6.
In the figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
Solution:
We have,
∆ABE ≅ ∆ACD
Their corresponding parts are equal, i.e.,
AB = AC, AE = AD


and ∠DAE = ∠BAC (common)
∴ Using the SAS similarity, we have ∆ADE ~ ∆ABC.

Question 7.
In the figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC

Solution:
We have a ∆ABC in which altitude AD and CE intersect each other at P.
⇒ ∠D = ∠E = 90° …………. (1)
(i) In ∆AEP and ∆CDP,
∠AEP = ∠CDP [From (1)]
∠EPA = ∠DPC [Vertically opp. angles]
∴ Using AA similarity, we get
∆AEP ~ ∆CDP

(ii) In ∆ABD and ∆CBE,
∠ADB = ∠CEB [From (1)]
Also, ∠ABD = ∠CBE [Common]
Using A A similarity, we have
∆ABD ~ ∆CBE

(iii) In ∆AEP and ∆ADB,
∵ ∠AEP = ∠ADB [From (1)]
Also, ∠EAP = ∠DAB [Common]
∴ Using AA similarity, we have
∆AEP ~ ∆ADB

(iv) In ∆PDC and ∆BEC,
∵ ∠PDC = ∠BEC [From (1)]
Also, ∠DCP = ∠ECB [Common]
∴ Using AA similarity, we have
∆PDC ~ ∆BEC

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
We have a parallelogram ABCD in which AD is produced to E and BE is joined such that BE intersect CD at F.

Now, in ∆ABE and ∆CFB
∠BAE = ∠FCB [Opp. angles of a || gm are always equal]
∠AEB = ∠CBF [∵ Parallel sides are intersected by the transversal BE]
Now, using AA similarity, we have ∆ABE ~ ∆CFB.

Question 9.

Prove that:
(i) ∆ABC ~ ∆AMP
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\)
Solution:
We have ∆ABC, right angled at B and ∆AMP, right angled at M.
∴ ∠B = ∠M = 90°
(i) In ∆ABC and ∆AMP,
∠ABC = ∠AMP [From (1)]
and ∠BAC = ∠MAP [Common]
∴ Using AA similarity, we get
∆ABC ~ ∆AMP

(ii) ∵ ∆ABC ~ ∆AMP [As proved above]
∴ Their corresponding sides are proportional.
⇒ \(\frac{C A}{P A}=\frac{B C}{M P}\)

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that 0 and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:
CD AC
(i) \(\frac{C D}{G H}=\frac{A C}{F G}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
Solution:
We have, two similar ∆ABC and ∆FEG such that CD and GH are the bisectors of ∠ACB and ∠FGE respectively.

(i) In ∆ACD and ∆FGH,
Since ∆ABC ~ ∆FEG
∠A = ∠F …………….. (1)
and ∠ACB = ∠FGE ⇒ \(\frac{1}{2}\) ∠ACB = \(\frac{1}{2}\) ∠FGE
⇒ ∠ACD = ∠FGH ……………. (2)
From (1) and (2),
∆ACD ~ ∆FGH [AA similarity]
∴ Their corresponding sides are proportional,
∴ \(\frac{C D}{G H}=\frac{A C}{F G}\)

(ii) In ∆DCB and ∆HGE,
Since ∆ABC ~ ∆FEG
⇒ ∠B = ∠E …………….. (1)
and ∠ACB = ∠FGE
∴ \(\frac{1}{2}\)∠ACB = \(\frac{1}{2}\)∠FGE
⇒ ∠DCB = ∠HGE ……………. (2)
From (1) and (2),
∆DCB ~ ∆HGE [AA similarity]

(iii) From (i) part, we get
∆ACD ~ ∆FGH
⇒ ∆DCA ~ ∆HGF

Question 11.
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD⊥BCand EF⊥AC, prove that ∆ABD ~ ∆ECF.

Solution:
We have an isosceles ∆ABC in which
AB = AC.
⇒ Angles opposite to them are equal
∠ACB = ∠ABC ……………. (1)
In ∆ABD and ∆ECF,
∠ECF = ∠ABD [from (1)]
and AD⊥BC and EF⊥AC
⇒ ∠ADB = ∠EFC = 90°
∴ ∆ABD ~ ∆ECF [AA similarity]

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see figure). Show that ∆ABC ~ ∆PQR.

Solution:
We have ∆ABC and ∆PQR in which AD and PM are medians corresponding to sides BC and QR respectively such that
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\)

∴ Using SSS similarity, we have ∆ABD ~ ∆PQM
∴ Their corresponding angles are equal.
⇒ ∠ABD = ∠PQM ⇒ ∠ABC = ∠PQR
Now, in ∆ABC and ∆PQR, \(\frac{A B}{P Q}=\frac{B C}{Q R}\) …………. (1)
Also, ∠ABC = ∠PQR …………… (2)
From (1) and (2),
∆ABC ~ ∆PQR. (SAS similarity)

Question 13.
O is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB CD.
Solution:
We have a ∆ABC and a point D on its side BC such that ∠ADC = ∠BAC.

In ∆BAC and ∆ADC,
∵ ∠BAC = ∠ADC [Given]
and ∠BCA = ∠DCA [Common]
Using AA similarity, we have
∆BAC ~ ∆ADC.
∴ Their corresponding sides are proportional.
⇒ \(\frac{C A}{C D}=\frac{C B}{C A}\)
⇒ CA × CA = CB × CD
⇒ CA² = CB × CD

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:
Given : ∆ABC and ∆PQR in which AD and PM are medians.
Also, \(\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}\) …………….. (1)
To Prove: ∆ABC ~ ∆PQR
Construction: Produce AD to E and PM to N such that AD = DE and PM = MN. Join BE, CE, QN and RN.

Proof: Quadrilaterals ABEC and PQNR are parallelograms, since their diagonals bisect each other at point D and M respectively.
⇒ BE = AC and QN = PR

⇒ ∆ABE ~ ∆PQN ⇒ ∠1 = ∠3 …………. (4)
Similarly, we can prove
⇒ ∆ACE ~ ∆PRN ⇒ ∠2 = ∠4 …………….. (5)
From (4) and (5)
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P ………….. (6)
Now, in ∆ABC and ∆PQR, we have
\(\frac{A B}{P Q}=\frac{A C}{P R}\) [From (1)]
and ∠A = ∠P [From (6)]
∴ ∆ABC ~ ∆PQR

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB = 6 m be the pole and BC = 4 m be its shadow (in right ∆ABC), whereas DE and EE denote the tower and its shadow respectively.
∵ EF = Length of the shadow of the tower = 28 m
and DE = h = Height of the tower

In ∆ABC and ∆DEF, we have ∠B = ∠E = 90°
∠C = ∠F [∵ Angular elevation of the sun at the same time is equal]
∴ Using AA similarity, we have
∆ABC ~ ∆DEF
∴ Their sides are proportional i.e., \(\frac{A B}{D E}=\frac{B C}{E F}\)
⇒ \(\frac{6}{h}=\frac{4}{28}\) ⇒ h = \(\frac{6 \times 28}{4}\) = 42
Thus, the required height of the tower is 42 m.

Question 16.
If MD and PM are medians of triangles ABC and PQR, respectively where, ∆ABC ~ ∆PQR, prove that \(\frac{A B}{P Q}=\frac{A D}{P M}\).
Solution:
We have ∆ABC ~ ∆PQR such that AD and PM are the medians corresponding to the sides BC and QR respectively.

∵ ∆ABC ~ ∆PQR
And the corresponding sides of similar triangles are proportional.
∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}\) ……………. (1)
∵ Corresponding angles are also equal in two similar triangles.
∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R …………… (2)
Since, AD and PM are medians.
∴ BC = 2BD and QR = 2QM
∴ From (1) \(\frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{B D}{Q M}\) ………… (3)
And ∠B = ∠Q ⇒ ∠ABD = ∠PQM
∴ From (3) and (4), we have
∆ABD ~ ∆PQM (SAS similarity)
∴ Their corresponding sides are proportional.
⇒ \(\frac{A B}{P Q}=\frac{A D}{P M}\)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
In right angle ∆ABC, we have
AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = 24² + 7² = 576 + 49 = 625 = 25²
⇒ AC = 25 cm

Question 2.
In the figure, find tan P – cotR?

Solution:
In right angle ∆PQR
Using the Pythagoras theorem, we get
QR² = PR² – PQ²
⇒ QR² = 13² – 12² = (13 – 12)(13 + 12) = 1 × 25 = 25
∴ QR = \(\sqrt{25}\) = 5 cm
Now, tanP = \(\frac{Q R}{P Q}=\frac{5}{12}\) , cotR = \(\frac{Q R}{P Q}=\frac{5}{12}\)
∴ tanP – cotR = \(\frac{5}{12}-\frac{5}{12}\) = 0.

Question 3.
If sinA = \(\frac{3}{4}\), calculate cosA and tanA.
Solution:
Let us consider, the right angle ∆ABC, we have
Perpendicular = BC and Hypotenuse = AC

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
In the right angle triangle ABC, we have 15 cot A = 8

Question 5.
Given sec θ = \(\frac { 13 }{ 12 } \), calculate all other trigonometric ratios.
Solution:

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:

From equations (i) and (ii) we get:
\(\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}\)
⇒ ∆CDA ~ ∆EFB [By SSS similarity]
⇒ ∠A = ∠B Hence Proved

Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate:
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)
(ii) cot²θ
Solution:

Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:
In right angled ∆ABC, ∠B = 90°
For ∠A, we have:
Base = AB and perpendicular = BC. Also, Hypotenuse = AC

Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
In right ∆ABC, ∠B = 90°
For ∠A, we have
Base = AB, Perpendicular = BC,
Hypotenuse = AC

Question 10.
In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.
Solution:
In right ∆PQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm
Let QR = x cm ⇒ PR = (25 – x) cm
∴ By Pythagoras theorem, we have
PR² = QR² + PQ²
⇒ (25 – x)² = x² + 5²
⇒ 625 – 50x + x² = x² + 25
⇒ – 50x = – 600
⇒ x = \(\frac{-600}{-50}\) = 12 i.e., QR = 12 cm
⇒ RP = 25 – 12 = 13 cm

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) secA = \(\frac{12}{5}\) for some value of angle A.
(iii) cosA is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sinθ = \(\frac{4}{3}\) for some angle θ.
Solution:
(i) False
∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.
(ii) True
∵ cos A is always less than 1.
∴ \(\frac{1}{\cos A}\) i.e., sec A will always be greater than 1.
(iii) False
∵ ‘cosine A’ is abbreviated as ‘cosA’
(iv) False
∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.
(v) False
∵ \(\frac{4}{3}\) is greater than 1 and sinθ cannot be greater than 1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:
We have, r₁ = 19 cm and r₂ = 9 cm
∴ Circumference of circle – I = 2πr₁ = 2π(19) cm
and circumference of circle – II = 2πr₂ = 2π(9) cm
Sum of the circumferences of circle-I and circle-II
= 2π(19) cm + 2π(9) cm
= 2π(19 + 9) cm
= 2π(28) cm
Let R be the radius of the circle – III.
∴ Circumference of circle – III = 2πR
According to the condition, 2πR = 2π(28)
⇒ R = \(\frac{2 \pi(28)}{2 \pi}\) = 28 cm
Thus, the radius of the new circle = 28 cm.

Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
We have,
Radius of circle – I, r₁ = 8 cm
Radius of circle – II, r₂ = 6 cm
∴ Area of circle – I = πr₁² = π(8)² cm²
Area of circle-II = πr₂² = π(6)² cm²
Let the radius of the circle – III be R cm.
∴ Area of circle-III = πR²
Now, according to the condition,
πr₁² + πr₂² = πR²
⇒ π(8)² + π(6)² = πR²
⇒ π(8² + 6²) = πR²
⇒ 8² + 6² = R²
⇒ 64 + 36 = R²
⇒ 100 = R²
⇒ 10² = R² ⇒R = 10
Thus, the radius of the new circle = 10 cm.

Question 3.
The given figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:
Diameter of the innermost (Gold scoring) region = 21 cm
Radius of the innermost (Gold scoring) region = \(\frac{21}{2}\) = 10.5 cm
∴ Area of Gold scoring region = π(10.5)² cm²
= \(\frac{22}{7} \times\left(\frac{105}{10}\right)^{2} \mathrm{cm}^{2}=\frac{22}{7} \times \frac{105}{10} \times \frac{105}{10} \mathrm{cm}^{2}\)
= \(\frac{22 \times 15 \times 105}{100}\) cm² = 346.50 cm²
Since, each band is 10.5 cm wide.
∴ Radius of Red scoring region
= 10.5 cm + 10.5 cm
= 21 cm
Area of Red scoring region
= π(10.5 + 10.5)² cm² – π(10.5)² cm²
= [π(21)² – π(10.5)²] cm²
= π[(21)² – (10.5)²] cm²

Area of White scoring region
= π[(42 + 10.5)² – (42)²] cm²
= π[(52.5)² – (42)²] cm²
= π[(52.5 + 42)(52.5 – 42)] cm²
= \(\frac{22}{7}\) × 94.5 × 10.5 = 22 × \(\frac{945}{10} \times \frac{15}{10}\)
= 3118.5 cm²

Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution:
Diameter of a wheel = 80 cm
∴ Radius of the wheel = \(\frac{80}{2}\) cm = 40 cm
So, circumference of the wheel
= 2πr = 2 × \(\frac{22}{7}\) × 40 cm
⇒ Distance covered by a wheel in one revolution = \(\frac{2 \times 22 \times 40}{7}\) cm
Distance travelled by the car in 1 hour (i.e., in 60 mins)
= 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes

Thus, the required number of revolutions = 4375

Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units
Solution:
(A): We have
Numerical area of the circle
= Numerical circumference of the circle
⇒ πr² = 2πr
⇒ πr² – 2πr = 0
⇒ r² – 2r = 0
⇒ r(r – 2) = 0
⇒ r = 0 or   r = 2
But r cannot be zero
∴ r = 2
Thus, the radius of circle is 2 units.