Bhagya

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:
(i) Let the vertices of the triangle be A(2, 3), B(-1, 0) and C(2, – 4)
Here, x₁ = 2, y₁ = 3
x₂ = -1, y₂ = 0
x₃ = 2, y₃ = -4
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [2{0 – (-4)} + (-1){-4 – (3)} + 2{3 – 0}]
= \(\frac{1}{2}\) [2(0 + 4) + (-1)(-4 – 3) + 2(3)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{1}{2}\) [21] = \(\frac{21}{2}\) sq. units

(ii) Let the vertices of the triangle be A(-5, -1), B(3, -5) and C(5, 2)
Here, x₁ = -5, y₁ = -1
x₂ = 3, y₂ = -5
x₃ = 5, y₃ = 2
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [-5{-5 – 2} + 3{2 – (-1)} + 5{-1 – (-5)}]
= \(\frac{1}{2}\) [-5{-7} + 3{2 + 1} + 5{-1 + 5}]
= \(\frac{1}{2}\) [35 + 3(3) + 5(4)]
= \(\frac{1}{2}\) [35 + 9 + 20] = \(\frac{1}{2}\) × 64 = 32 sq. units

Question 2.
In each of the following find the value of’k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:
The given three points will be collinear if the triangle formed by them has zero area.
(i) Let A(7, -2), B(5, 1) and C(3, k) be the vertices of a triangle.
∴ The given points will be collinear, if ar (∆ABC) = 0
or \(\frac{1}{2}\) [7(1 – k) + 5(k + 2) + 3(-2 – 1)] = 0
⇒ 7 – 7k + 5k + 10 + (-6) – 3 = 0
⇒ 17 – 9 + 5k – 7k = 0
⇒ 8 – 2k = 0 ⇒ 2k = 8 ⇒ k = \(\frac{8}{2}\) = 4
The required value of k = 4.
(ii) \(\frac{1}{2}\) [8(- 4 + 5) + k(- 5 -1) + 2(1 + 4)] = 0
⇒ 8 – 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let the vertices of the triangle be A(0, -1), B(2, 1) and C(0, 3).
Let D, E and F be the mid-points of the sides BC, CA and AB respectively.
∴ Coordinates of D are


ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral.
Let us join diagonal BD.

Now, ar(∆ABD)

Question 5.
You have studied in class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(A, -6), B(3, -2) and C(5, 2).
Solution:
Here, the vertices of the triangle are A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
∴ The coordinates of the point D are
\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) or ( 4, 0)
AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD.

ar(∆ADC) = \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= \(\frac{1}{2}\) [-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3
= 3 sq. units (numerically) ………… (2)
From (1) and (2),
ar(∆ABD) = ar(∆ADC)
Thus, a median divides the triangle into two triangles of equal areas.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and 8(3, 7).
Solution:
Let the required ratio be k : 1 and the point C divide them in the above ratio.
∴ Coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)
Since, the point C lies on the given line 2x + y – 4 = 0.
∴ We have \(2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4\) = 0
⇒ 2(3k + 2) + (7k – 2) = 4 × (k + 1)
⇒ 6k + 4 + 7k – 4k – 4 – 2 = 0
⇒ (6 + 7 – 4)k + (-2) = 0 ⇒ 9k – 2 = 0
⇒ k = \(\frac{2}{9}\)
The required ratio = k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.
The points A, B and C will be collinear if area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let the points are A (6, -6), B(3, -7) and C(3, 3)
Let P(x, y) be the centre of the circle. Since, the circle is passing through A, B and C.
∴ AP = BP = CP
Taking AP = BP, we have AP² = BP²
⇒ (x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
⇒ x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² + 14y + 49
⇒ – 12x + 6x + 12y – 14y + 72 – 58 = 0
⇒ – 6x – 2y + 14 = 0
⇒ 3x + y – 7 = 0 ……………….. (1)
Taking BP = CP, we have BP² = CP²
⇒ (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
⇒ x² – 6x + 9 + y² + 14y + 49 = x² – 6x + 9 + y² – 6y + 9
⇒ – 6x + 6x + 14y + 6y + 58 -18 = 0
⇒ 20y + 40 = 0
⇒ y = \(\frac{-40}{20}\) = -2
From (1) and (2), 3x – 2 – 7 = 0
⇒ 3x = 9 ⇒ x = 3
i.e., x = 3 and y = -2
∴ The required centre is (3, -2).

Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let we have a square ABCD such that A(-1, 2) and C(3, 2) are the opposite vertices. Let B(x, y) be an unknown vertex.

Since, all sides of a square are equal.
∴ AB = BC ⇒ AB² = BC2²
⇒ (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
⇒ x² + 2x + 1 + y² – 4y + 4
⇒ x² – 6x + 9 + y² – 4y + 4
⇒ 2x + 1 = -6x + 9
⇒ 8x = 8 ⇒ x = 1  …………………. (1)
Since, each angle of a square = 90°.
∴ ∆ABC is a right angled triangle.
∴ Using Pythagoras theorem, we have AB² + BC² = AC²
⇒ (x + 1)² + (y – 2)²] + [(x – 3)² + (y – 2)²]
= [(3 + 1)² + (2 – 2)²]
⇒ [x² + 2x + 1 + y² – 4y + 4] + [x² – 6x + 9 + y² – 4y + 4]
= [4² + 0²]
⇒ 2x² + 2y² + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x² + 2y² – 4x – 8y + 2 = 0
⇒ x² + y² – 2x – 4y + 1 = 0 …………….. (2)
Substituting the value of x from (1) into (2), we have
1 + y² – 2 – 4y + 1 = 0
⇒ y² – 4y + 2 – 2 = 0
⇒ y² – y = 0
⇒ y(y – 4) = 0
⇒ y = 0 or y = 4
Hence, the required other two vertices are (1, 0) and (1, 4).

Question 5.
The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Solution:
(i) By taking A as the origin and AD and AB as the coordinate axes. We have P(4, 6), Q( 3, 2) and R( 6, 5) as the vertices of ∆PQR.
(ii) By taking C as the origin and CB and CD as the coordinate axes, then the vertices of ∆PQR are P(-12, – 2), Q(-13, – 6) and R(- 10, – 3)
Case I: When P(4, 6), Q(3, 2) and R(6, 5) are the vertices.
∴ ar(∆PQR) = \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= \(\frac{1}{2}\) [-12 – 3 + 24] = \(\frac{9}{2}\) sq. units
Case II: When P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices.
∴ ar(∆PQR)
= \(\frac{1}{2}\) [-12(- 6 + 3) + (-13)(- 3 + 2) + (-10)(-2 + 6)]
= \(\frac{1}{2}\) [-12(-3) + (-13)(-1) + (-10) × (4)]
= \(\frac{1}{2}\) [36 + 13 – 40] = \(\frac{9}{2}\) sq. units
Thus, in both cases, the area of ∆PQR is the same.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]
Solution:

Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meets BCat D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
We have the vertices of ∆ABC as A (4, 2), B(6, 5) and C(1, 4).
(i) Since AD is a median

∴ Coordinates of D are


Also, CR : RF = 2 : 1 i.e., the point R divides CF in the ratio 2 : 1
∴ Coordinates of R are

(iv) We observe that P, Q and R represent the same point.
(v) Here, we have A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC. Also AD, BE and CF are its medians.
∴ D, E and F are the mid points of BC, CA and AB respectively.
We know, the centroid is a point on a median, dividing it in the ratio 2 : 1.
Considering the median AD, coordinates of

Question 8.
ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
We have a rectangle whose vertices are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1).
∵ P is mid-point of AB
∴ Coordinates of P are


We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.
∴ It can be a square or a rhombus.
But PR ≠ QS i.e., its diagonals are not equal.
∴ PQRS is a rhombus.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:
We can calculate the mean as follows :

∴ Mean = \(\overline{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{162}{20}=8.1\)
Thus, mean number of plants per house is 8.1 Since, values of x_i and f_i are small, so we have used the direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let the assumed mean, a = 150
∵ Class size, h = 20
∴ \(u_{i}=\frac{x_{i}-a}{h}=\frac{x_{i}-150}{20}\)
∴ We have the following table:

Now, \(\overline{x}\) = a + h × {\(\frac{1}{N}\) Σf_iu_i}
= 150 + 20 × \(\left(\frac{-12}{50}\right)\) = 150 – \(\frac{24}{5}\)
= 150 – 4.8 = 145.20

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Solution:
Let the assumed mean, a = 18
∵ Class size, h = 2
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-18}{2}\)
Now, we have the following table:

⇒ [f + 44] (0) = 2[f – 20]
⇒ 2[f – 20] = 0 ⇒ f = 20
Thus, missing frequency is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:
Let the assumed mean, a = 75.5
∵ Class size, h = 3
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-75.5}{3}\)
Now, we have the following table:

Thus, the mean heart beats per minute is 75.9.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Let the assumed mean, a = 57
∴ d_i = x_i – 57
Now, we have the following table:

Thus, the average number of mangoes per box = 57.19. We choose assumed mean method.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:
Let the assumed mean, a = 225
∵ Class size, (h) = 50
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-225}{50}\)
Now, we have the following table:

Thus, the mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of S0₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of S0₂ in the air.
Solution:
Let the assumed mean, a = 0.14
∵ Class size, (h) = 0.04
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-0.14}{0.04}\)
Now, we have the following table:

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:
Using the direct method, we have the following table:

Thus, mean number of days a student remained absent = 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:
Let the assumed mean, a = 70
∵ Class size, (h) = 10
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-70}{10}\)
Now, we have the following table:

Thus, the mean literacy rate is 69.43%

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of the construction:

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction :
I. Draw a circle of radius 6 cm. Let its centre be O.
II. Take a point P such that OP = 10 cm. Join OP.

III. Bisect OP and let M be its midpoint.
IV. Taking M as centre and MP or MO as radius draw a circle.
Let the new circle intersects the given circle at A and B. Join PA and PB.
Thus, PA and PB are the required tangents. By measurement, we have : PA = PB = 8 cm.
Justification:
Join OA and OB
Since PO is a diameter.
∴ ∠OAP = 90° = ∠OBP [Angles in a semicircle]
Also, OA and OB are radii of the same circle.
⇒ PA and PB are tangents to the circle.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:
Steps of Construction :
I. Draw two concentric circles with centre O and radii 4 cm and 6 cm.
II. Take any point P on outer circle.
III. Join PO and bisect it and let the midpoint of PO is represented by M.
IV. Taking M as centre and OM or MP as radius, draw a circle such that this circle intersects the circle (of radius 4 cm) at A and B.
V. Join AP.
Thus, PA is the required tangent.
By measurement, we have PA = 4.5 cm

Justification:
Join OA. As PO is diameter
∴ ∠PAO = 90° [Angle in a semi-circle]
⇒ PA⊥OA
∵ OA is a radius of the inner circle.
∴ PA has to be a tangent to the inner circle.
Verification:
In right ∆PAO, PO = 6 cm, OA = 4 cm.

Hence both lengths are approximately equal.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction :
I. Draw a circle of radius 3 cm with centre O and draw a diameter.
II. Extend its diameter on both sides and cut OP = OQ = 7 cm
III. Bisect PO such that M be its mid-point.
IV. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at A and B.
V. Join PA and PB.
Thus, PA and PB are the two required tangents from P.
VI. Now bisect OQ such that N is its mid point.
VII. Taking N as centre and NO as radius, draw a circle. Let it intersect the given circle at C and D.
VIII. Join QC and QD.
Thus, QC and QD are the required tangents from Question

Justification:
Join OA to get ∠OAP = 90°
[Angle in a semi-circle]
⇒ PA⊥OA
⇒ PA is a tangent.
Similarly, PB⊥OB
⇒ PB is a tangent.
Now, join OC to get ∠QCO = 90°
[Angle in a semi-circle]
⇒ QC⊥OC ⇒ QC is a tangent.
Similarly, QD⊥OD
⇒ QD is a tangent.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction :
I. With centre O and radius = 5 cm, draw a circle.
II. Taking a point A on the circle draw ∠AOB = 120°.

III. Draw a perpendicular on OA at A.
IV. Draw another perpendicular on OB at B.
V. Let the two perpendiculars meet at C.
Thus CA and CB are the two required tangents to the given circle which are inclined to each other at 60°.
Justification:
In a quadrilateral OACB, using angle sum
property, we have
120° + 90° + 90° + ∠ACB = 360°
⇒ 300° + ∠ACB = 360°
⇒ ∠ACB = 360° – 300° = 60°.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction:
I. Draw a line segment AB = 8 cm
II. Draw a circle with centre A and radius 4 cm, draw another circle with centre B and radius 3 cm.
III. Bisect the line segment AB. Let its mid point be M.
IV. With centre as M and MA (or MB) as radius, draw a circle such that it intersects the two circles at points P, Q, R and S.
V. Join BP and BQ.
Thus, BP and BQ are the required two tangents from B to the circle with centre A.
VI. Join RA and SA.
Thus, RA and SA are the required two tangents from A to the circle with centre B.

Justification:
Let us join A and P.
∵ ∠APB = 90° [Angle in a semi-circle]
∴ BP⊥AP
But AP is radius of the circle with centre A.
⇒ BP has to be a tangent to the circle with centre A.
Similarly, BQ has to be tangent to the circle with centre A.
Also AR and AS are tangents to the circle with centre B.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of construction :
I. Draw ∆ABC such that AB = 6 cm, BC = 8 cm and ∠B = 90°.
II. Draw BD⊥AC. Now bisect BC and let its midpoint be O.
So O is centre of the circle passing through B, C and D.

III. Join AO
IV. Bisect AO. Let M be the mid-point of AO.
V. Taking M as centre and MA as radius, draw a circle intersecting the given circle at B and E.
VI. Join AB and AE. Thus, AB and AE are the required two tangents to the given circle from A.
Justification:
Join OE, then ∠AEO = 90° [Angle in a semi circle]
∴ AE⊥OE.
But OE is a radius of the given circle.
⇒ AE has to be a tangent to the circle.
Similarly, AB is also a tangent to the given circle.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction:
I. Draw the given circle using a bangle.
II. Take two non parallel chords PQ and RS of this circle.

III. Draw the perpendicular bisectors of PQ and RS such that they intersect at O. Therefore, O is the centre of the given circle.
IV. Take a point N outside this circle.
V. Join ON and bisect it. Let M be the mid-point of ON.
VI. Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at A and B.
VII. Join NA and NB. Thus, NA and NB are the required two tangents.
Justification:
Join OA and OB.
Since ∠OAN = 90° [Angle in a semi circle]
∴ NA⊥OA.
Also NA is a radius.
∴ NA has to be a tangent to the given circle.
Similarly, NB is also a tangent to the given circle.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x + 1 )² = 2(x – 3)
(ii) x² – 2x = (-2)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1 )(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1 )(x – 3) = (x + 5)(x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³
Solution:
(i) We have, (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x² + 2x + 1 – 2x + 6 = 0
⇒ x² + 7 = 0
Since, x² + 7 is a quadratic polynomial.
∴ (x + 1)² = 2(x – 3) is a quadratic equation.

(ii) We have, x² – 2x = (-2) (3 – x)
⇒ x² – 2x = – 6 + 2x
⇒ x² – 2x – 2x + 6 = 0
⇒ x² – 4x + 6 = 0
Since, x² – 4x + 6 is a quadratic polynomial
∴ x² – 2x = (-2)(3 – x) is a quadratic equation.

(iii) We have, (x – 2)(x + 1) = (x – 1) (x + 3)
⇒ x² – x² = x² + 2x³
⇒ x² – x – 2 – x² – 2x + 3 = 0
⇒ – 3x +1 = 0
Since, – 3x + 1 is a linear polynomial.
∴ (x – 2)(x + 1) = (x – 1)(x + 3) is not a quadratic equation.

(iv) We have, (x – 3) (2x + 1) = x(x + 5)
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0
Since, x² – 10x – 3 is a quadratic polynomial.
∴ (x – 3)(2x + 1) = x(x + 5) is a quadratic equation.

(v) We have, (2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² – 7x + 3 = x² + 4x – 5
⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Since, x² – 11x + 8 is a quadratic polynomial
∴ (2x – 1)(x – 3) = (x + 5)(x -1) is a quadratic equation.

(vi) We have, x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ x² + 3x + 1 – x² + 4x – 4 = 0
⇒ 7x – 3 = 0
Since, 7x – 3 is a linear polynomial.
x² + 3x + 1 = (x – 2)² is not a quadratic equation.

(vii) We have, (x + 2)³ = 2x(x² – 1)
⇒ x³ + 3x²(2) + 3x(2)² + (2)³ = 2x³ – 2x
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ x³ + 6x² + 12x + 8 – 2x³ + 2x = 0
⇒ -x³ + 6x² + 14x + 8 = 0
Since, -x³ + 6x² + 14x + 8 is a polynomial of degree 3.
∴ (x + 2)³= 2x(x² – 1) is not a quadratic equation.

(viii) We have, x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ + 3x²(-2) + 3x(-2)² + (-2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0 ⇒ 2x² – 13x + 9 = 0
Since, 2x² – 13x + 9 is a quadratic polynomial.
∴ x³ – 4x² – x + 1 = (x – 2)³ is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let the breadth = x metres
∵ Length = 2(Breadth) + 1
∴ Length = (2x + 1)metres
Since, Length × Breadth = Area
∴ (2x + 1) × x = 528 ⇒ 2x² + x = 528
⇒ 2x² + x – 528 = 0
Thus, the required quadratic equation is 2x² + x – 528 = 0

(ii) Let the two consecutive positive integers be x and (x + 1).
∵ Product of two consecutive positive integers = 306
∴ x(x + 1) = 306 ⇒ x² + x = 306
⇒ x² + x – 306 = 0
Thus, the required quadratic equation is x² + x – 306 = 0

(iii) Let the present age of Rohan be x years
∴ Mother’s age = (x + 26) years
After 3 years,
Rohan’s age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
= (x + 29) years
According to the condition,
(x + 3) × (x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 29x + 3x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Thus, the required quadratic equation is x² + 32x – 273 = 0

(iv) Let the speed of the train = u km/hr Distance covered = 480 km Distance

In second case,
Speed = (u – 8) km/hour

⇒ 480u – 480(u – 8) = 3u(u – 8)
⇒ 480u – 480u + 3840 = 3u² – 24u
⇒ 3840 – 3u² + 24u = 0
⇒ u² – 8u – 1280 = 0
Thus, the required quadratic equation is u² – 8u – 1280 = 0

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 1.
Find the sum of the following APs:
(i) 2,7,12, ….. to 10 terms
(ii) -37, -33, -29, …. to 12 terms
(iii) 0.6, 1.7, 2.8, …. to 100 terms
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}\), ……
Solution:
(i) Here, a = 2, d = 7 – 2 = 5, n = 10

Thus, the sum of first 10 terms is 245.

(ii) Here, a = -37, d = -33 – (-37) = 4, n = 12

Thus, the sum of first 12 terms is -180.

(iii) Here, a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100

Thus, the required sum of first 100 terms is 5505.

(iv)

Thus, the required sum of first 11 terms is \(\frac{33}{20}\)

Question 2.
Find the sums given below:
(i) 7+10\(\frac{1}{2}\) + 14 + …… + 84
(ii) 34 + 32 + 30+ …….. + 10
(iii) -5 + (-8) + (-11) + ……. + (-230)
Solution:
(i) Here,

Let n be the number of terms.

(ii) Here, a = 34, d = 32 – 34 = -2, l = 10
Let n be the number of terms

(iii) Here, a = -5,d = -8- (-5) = -3,1 = -230
Let n be the number of terms.

Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50,find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) given a_n = 4, d = 2,S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) Here, a = 5, d = 3 and a_n = 50 = l

Question 4.
How many terms of the AP : 9,17, 25, ….. must be taken to give a sum of 636?
Solution:
Here, a = 9, d = 17 – 9 = 8, S_n = 636

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Question 6.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
We have, first term (a) = 17, last term (l) = 350 = T_n and common difference (d) = 9
Let the number of terms be n.

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.
Solution:
Here, n = 22, T₂₂ = 149 = l, d = 7
Let the first term of the AP be a.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, we have S₇ = 49 and S₁₇ = 289
Let the first term of the AP be a and d be the common difference, then

Question 10.
Show that a₁, a₂,…… a_n,………, form an AP where a_n is defined as below:
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3^rd, the 10^th and the n^th terms.
Solution:
We have, S_n = 4n – n²
S₁ = 4(1) – (1)² = 4 – 1 = 3 ⇒ First term (a) = 3

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 are 6,12,18, ……. , (6 × 40)
And, these numbers are in AP, such that a = 6,
d = 12 – 6 = 6 and _n = 6 × 40 = 240 = l

= 20[12 + 39 × 6] = 20[12 + 234]
= 20 × 246 = 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are 8, 16, 24, 32, ….., 120.
These numbers are in AP, where, a = 8 and l = 120

Thus, the sum of first 15 multiples of 8 is 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, …… , 49
These numbers are in AP such that a = 1 and l = 49

Thus, the sum of odd numbers between 0 and 50 is 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Here, penalty for delay on 1^st day = ₹ 200
2^nd day = ₹ 250
3^rd day = ₹ 300
……………………………………….
……………………………………….

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Sum of all the prizes = ₹ 700
Let the first prize be a
2^nd prize = (a – 20)
3^rd prize = (a- 40)
4^th prize = (a – 60)
The above prizes form an AP
Now, we have, first term = a
Common difference = d = (a – 20) – a = -20

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students ?
Solution:
Number of classes = 12
∵ Each class has 3 sections.
∴ Number of trees planted by class I = 1 × 3 = 3
Number of trees planted by class II =2 × 3 = 6
Number of trees planted by class III = 3 × 3 = 9
Number of trees planted by class IV = 4 × 3 = 12
………………………………………………………………….
Number of trees planted by class XII = 12 × 3 = 36
The numbers 3, 6, 9,12, ……. , 36 form an AP
Here, a = 3, d = 6 – 3 = 3 and n = 12

Question 18.
A spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles? (Take \(\pi=\frac{22}{7}\))

Solution:

Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint :To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Solution:
Here, number of potatoes = 10
The to and fro distance of the bucket:

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4\(\sqrt{3}\)x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) We have, 2x² – 7x + 3 = 0
Dividing both sides by 2, we get

(ii) We have 2x² + x – 4 = 0
Divide both sides by 2, we get

(iv) We have, 2x² + x + 4 = 0
Dividing both sides by 2, we get

Since, the square of a number cannot be negative.
∴ There is no real value of x satisfying the given equation.

Question 2.
Find the roots of the following quadratic equations, using the quadratic formula:
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4\(\sqrt{3}\)x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) Comparing the given equation with ax² + bx + c = 0, we get a = 2, b = -7, c = 3
∴ b² – 4ac = (-7)² – 4(2)(3) = 49 – 24 = 25 > 0
Since b² – 4ac > 0, therefore the given equation has real roots, which are given by

Taking negative sign, x = \(\frac{7-5}{4}=\frac{2}{4}=\frac{1}{2}\)
Thus, the roots of the given equation are 3 and \(\frac{1}{2}\).

(ii) Comparing the given equation with ax² + bx + c = 0, we get a = 2, b = 1, c = – 4
b² – 4ac = (1)² – 4(2)(-4) = 1 + 32 = 33 > 0
Since b² – 4ac > 0, therefore the given equation has real roots, which are given by

(iii) Comparing the given equation with
ax² + bx + c = 0, we get
a = 4, b= 473, c = 3
b² – 4ac = (473)² – 4(4)(3)
= (16 × 3) – 48 = 48 – 48 = 0
Since b² – 4ac = 0, therefore the given equation has real and equal roots, which are given by

(iv) Comparing the given equation with ax² + bx + c = 0, we get a = 2, b = 1, c = 4
b² – 4ac = (1 )² – 4(2)(4) = 1 – 32 = -31 < 0 Since
b² – 4ac < 0, therefore the given equation does not have real roots.

Question 3.
Find the roots of the following equations:

Solution:


Taking negative sign, x = \(\frac{3-1}{2}\)
Thus, the required roots of the given equation are 2 and 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be x years.
3 years ago, Rehman’s age = (x – 3) years
5 years later, Rehman’s age = (x + 5) years
According to the condition,

⇒ 3(2x + 2) = x² + 2x – 15
⇒ 6x + 6 = x² + 2x – 15
⇒ x² + 2x – 6x – 15 – 6 = 0
⇒ x² – 4x – 21 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0,
we get a = 1, b = -4, c = -21
b² – 4ac = (-4)² – 4(1)(-21) = 16 + 84 = 100

Since, age cannot be negative.
∴ x ≠ -3 ⇒ x = 7
So, the present age of Rehman = 7 years

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics = x
∴ Marks in English = 30 – x
According to the condition,
(x + 2) × [(30 – x) – 3] = 210
⇒ (x + 2) × (30 – x – 3) = 210
⇒ (x + 2)(- x + 27) = 210
⇒ -x² + 25x + 54 = 210
⇒ -x² + 25x + 54 – 210 = 0
⇒ -x² + 25x – 156 = 0
⇒ x² – 25x + 156 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get a = 1,b = -25, c = 156

When x = 12, then 30 – x = 30 – 12 = 18 Thus, marks in Mathematics = 13, marks in English = 17 or
Marks in Mathematics = 12, Marks in English = 18.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side i.e., breadth = x metres

∴ The longer side i.e., length = (x + 30) metres and diagonal = (x + 60) metres
In a rectangle,
(diagonal)² = (breadth)² + (length)²
⇒ (x + 60)² = x² + (x + 30)²
⇒ x² + 120 x + 3600 = x² + x² + 60x + 900
⇒ x² + 120x + 3600 = 2x² + 60x + 900
⇒ 2x² – x² + 60x – 120x + 900 – 3600 = 0
⇒ x² – 60x – 2700 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get a = 1, b = -60, c = -2700
∴ b² – 4ac = (-60)² – 4(1)(-2700)
⇒ b² – 4ac = 3600 + 10800
⇒ b² – 4ac = 14400

Since breadth cannot be negative,
x ≠ -30 ⇒ x = 90
∴ x + 30 = 90 + 30 = 120
Thus, the shorter side is 90 m and the longer side is 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger number be x.
Since, (smaller number)² = 8(larger number)
⇒ (smaller number)² = 8
⇒ smaller number = \(\sqrt{8 x}\)
According to the condition,

Thus, the smaller number = 12 or -12
Thus, the two numbers are 18 and 12 or 18 and -12.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:


Thus, speed of the train is 40 km/hr.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the smaller tap fills the tank in x hours
∴ The larger tap fills the tank in (x – 10) hours.
Amount of water flowing through both the taps in one hour


[ ∵ Time cannot be negative]
x = 25 ⇒ x – 10 = 25 – 10 = 15
Thus, time taken to fill the tank by the smaller tap alone is 25 hours and by the larger tap alone is 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of the passenger train be x km/h.
∴ Average speed of the express train = (x + 11) km/h
Total distance covered = 132 km

Comparing equation (1) with ax² + bx + c = 0, we get a = 1, b = 11, c = -1452

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of the smaller square be x m.
⇒ Perimeter of the smaller square = 4x m
∴ Perimeter of the larger square = (4x + 24) m
⇒ Side of the larger square

Area of the smaller square = x² m²
Area of the larger square = (x + 6)² m²
According to the condition,
x² + (x + 6)² = 468
⇒ x² + x² + 12x + 36 = 468
⇒ 2x² + 12x – 432 = 0
⇒ x² + 6x – 216 = 0 …(1)
[Dividing both sides by 2] Comparing equation (1) with ax² + bx + c = 0, we get
a = 1, b = 6, c = -216
b² – 4ac = (6)² – 4(1)(-216) = 36 + 864 = 900

But the length of a square cannot be negative,
∴ x ≠ -18 ⇒ x = 12
Length of the smaller square = 12 m
and the length of the larger square = x + 6
= 12+ 6 = 18 m

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

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

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Here, r = 1 cm and h = 1 cm.
Volume of the conical part = \(\frac{1}{3}\) πr²h and volume of the hemispherical part = \(\frac{2}{3}\) πr³h

∴ Volume of the solid shape
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³h = \(\frac{1}{3}\) πr²h[h + 2r]
= \(\frac{1}{3}\) π(1)² [1 + 2(1)] cm³
= (\(\frac{1}{3}\) π × 1 × 3) cm³ = π cm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Here, diameter = 3 cm
⇒ Radius (r) = \(\frac{3}{2}\) cm
Total height = 12 cm
Height of a cone (h₁) = 2 cm

∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h₂) = (12 – 4) cm = 8 cm.
Now, volume of the cylindrical part = πr²h₂
Volume of both conical parts = 2[\(\frac{1}{3}\) πr²h₁]
∴ Volume of the whole model

Question 3.
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:
Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.

Diameter = 2.8 cm
⇒ Radius (r) = 1.4 cm
∴ Length of the cylindrical part (h)
= 5 cm – (1.4 + 1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Now, volume of the cylindrical part = πr²h and volume of both the hemispherical ends
= 2(\(\frac{2}{3}\) πr³) = \(\frac{4}{3}\) πr³
∴ Volume of a gulab jamun

Volume of 45 gulab jamuns

Question 4.
A pen stand made up of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Solution:
Dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × \(\frac{35}{10}\) cm³
= 525 cm³
Since, each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of each depression
= \(\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times\left(\frac{5}{10}\right)^{2} \times \frac{14}{10} \mathrm{cm}^{3}=\frac{11}{30} \mathrm{cm}^{3}\)
Since, there are 4 depressions
∴ Total volume of 4 depressions = \(\frac{44}{30}\) cm³
Now, volume of the wood in entire stand = [Volume of the wooden cuboid] – [Volume of 4 depressions]
= 525cm³ – \(\frac{44}{30}\) cm³
= \(\frac{15750-44}{30} \mathrm{cm}^{3}=\frac{15706}{30} \mathrm{cm}^{3}\)
= 523.53 cm³

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Height of the conical vessel (h) = 8 cm
Base radius (R) = 5 cm
Volume of water in conical vessel = \(\frac{1}{3}\) πR²h

Thus, the required number of lead shots = 100

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Solution:
Height of the big cylinder (h) = 220 cm
Base radius (r) = \(\frac{24}{2}\) cm = 12cm
∴ Volume of the big cylinder
= πr²h = (π(12)² × 220) cm³
Also, height of smaller cylinder (h₁) = 60 cm

Base radius (r₁) = 8 cm
∴ Volume of the smaller cylinder πr₁²h₁
= (π(8)² × 60) cm³
∴ Volume of iron pole = [Volume of big cylinder] + [Volume of the smaller cylinder]
= (π × 220 × 12² × π × 60 × 8²) cm³
= 3.14[220 × 12 × 12 + 60 × 8 × 8] cm³

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Height of the conical part = 120 cm
Base radius of the conical part = 60 cm
Volume of the conical part
= \(\frac{1}{3}\) πr²h = [\(\frac{1}{3} \times \frac{22}{7}\) × (60)² × 120] cm³
Radius of the hemispherical part = 60 cm.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of the cylindrical part = πr²h
= (3.14 × 1² × 8) cm³ = \(\frac{314}{100}\) × 8 cm³

∴ Total volume of the glass vessel = [Volume of cylindrical part] + [Volume of spherical part]

⇒ Volume of water in the vessel = 346.51 cm³
Since the child finds the volume as 345 cm³
∴ The child’s answer is not correct.
The correct answer is 346.51 cm³.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos48° – sin42°
(iv) cosec31°- sec59°
Solution:

(iii) cos 48° – sin 42°
cos 48° = cos (90° – 42°) = sin 42°
[∵ cos (90° – A) = sin A]
∴ cos 48° – sin 42° = sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
cosec 31° = cosec (90° – 59°) = sec 59° [ ∵ cosec (90° – A) = sec A]
∴ cosec 31° – sec 59° = sec 59° – sec 59° = 0

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution;
(i) L.H.S. = tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan 23° tan 42° tan (90° – 23°)
= cot 42° tan 23° tan 42° cot 23° [ ∵ tan (90° – A) = cot A]

(ii) L.H.S. = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin38° sin(90° – 38°)
= cos 38° sin 38° – sin 38° cos 38° [ ∵ sin(90° – A) = cosA and cos(90° – A)= sinA]
= 0 = R.H.S.
⇒ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot(A – 18°)
cot(90 – 2A) = cot (A – 18)
90 – 2A = A – 18
90 + 18 = A + 2A
3A= 108
A = \(\frac {108}{3}\)
A = 36°.

Question 4.
If tan A = cotB, prove that A + B = 90°.
Solution:
tan A = cot B and cot B = tan (90° – B) [∵ tan (90° – θ) = cot θ]
∴ A = 90° – B ⇒ A + B = 90°

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A= cosec (A – 20)
cosec (90 – 4A) = cosec (A – 20)
90 – 4A = A – 20
90 + 20 =A + 4A
110 = 5A
A = \(\frac{110}{5}\)
A = 22°.

Question 6.
If A, 6 and C are interior angles of a triangle ABC, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)
Solution:
Since, sum of the angles of ∆ABC is 180° i.e.,
A + B + C = 180°
∴ B + C = 180° – A
Dividing both sides by 2, we get

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90 – 23) + cos (90 – 15)
[sin(90 – θ) = cosθ
cos(90 – θ) = sinθ]
= cos 23 + sin 15

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 1.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution:
Let the required point be P(x, y).
Here the end points are (-1, 7) and (4, – 3)
Ratio = 2 : 3 = m₁ : m₂
∴ x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
= \(\frac{(2 \times 4)+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)
And y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)
= \(\frac{2 \times(-3)+(3 \times 7)}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Thus, the required point is (1, 3)

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution:
Let the given points be A(4, -1) and B(-2, -3)

Let the points P and Q trisects AB.
i.e., AP = PQ = QB
i.e., P divides AB in the ratio of 1 : 2 and Q divides AB in the ratio of 2 : 1
Let the coordinates of P be (x, y)

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the figure. Niharika runs 1/4^th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5^th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag ?.

Solution:
Let us consider ‘A’ as origin, then
AB is the x-axis.
AD is the y-axis.
Now, the position of green flag-post is (2, \(\frac{100}{4}\)) or (2, 25)
And the position of the red flag-post is (8, \(\frac{100}{5}\)) or (8, 20)
Distance between both the flags

Let the mid-point of the line segment joining the two flags be M(x, y).

or x = 5 and y = 22.5
Thus, the blue flag is on the 5th line at a distance 22-5 m above AB.

Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the given points are A(-3, 10) and B(6, -8).
Let the point P(-1, 6) divides AB in the ratio m₁ : m₂.
∴ Using the section formula, we have

⇒ -1(m₁ + m₂) = 6m₁ – 3m₂
and 6(m₁ + m₂) = – 8m₁ + 10m₂
⇒ -m₁ – m₂ – 6m₁ + 3m₂ = 0
and 6m₁ + 6m₂ + 8m₁ – 10m₂ = 0
⇒ -7m₁ + 2m₂ = 0 and 14m₁ – 4m₂ = 0 or 7m₁ – 1m₂ = 0
⇒ \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\) and \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\)
⇒ 2m₂ = 7m₁ and 7m₁ = 2m₂
⇒ m₁ : m₂ = 2 : 7 and m₁ : m₂ = 2 : 7
Thus, the required ratio is 2 : 7.

Question 5.
Find the ratio in which the line segment joining 4(1, -5) and B(-4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:
The given points are A( 1, -5) and B(-4, 5). Let the required ratio = k:l and the required point be P(x, y).
Since the point P lies on x-axis,
∴ Its y-coordinate is 0.

Question 6.
If (1, 2), (4, y), (x, 6) and (3,5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let the given points are A( 1, 2), B(4, y), C(x, 6) and D(3, 5)

Since, the diagonals of a parallelogram bisect each other.
∴ The coordinates of P are :

∴ The required values of x and y are 6 and 3 respectively.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and Bis (1,4).
Solution:
Let centre of the circle is 0(2, -3) and the end points of the diameter be A(x, y) and B(1, 4)

Since, the centre of a circle bisects the diameter.
∴ 2 = \(\frac{x+1}{2}\) ⇒ x + 1 = 4 or x = 3
And -3 = \(\frac{y+4}{2}\) ⇒ y + 4 = -6 or y = -10
Thus, the coordinates of A are (3, -10)

Question 8.
If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and Plies on the line segment AB.
Solution:
Here, the given points are A(-2, -2) and B(2, -4)
Let the coordinates of P are (x, y)
Since, the point P divides AB such that
AP = \(\frac{3}{7}\)AB or \(\frac{A P}{A B}\) = \(\frac{3}{7}\)
⇒ AB = AP + BP
⇒ \(\frac{A P+B P}{A P}\) = \(\frac{7}{3}\)

Question 9.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Here, the given points are A(-2, 2) and B(2, 8)
Let P₁, P₂ and P₃ divide AB in four equal parts.

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
[Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals)]
Solution:
Let the vertices of the given rhombus are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1)