Category: Class 10

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.2

In questions 1 to 3, choose the correct option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
(A): ∵ QT is a tangent to the circle at T and OT is radius

∴ OT⊥QT
Also, OQ = 25 cm and QT = 24 cm
∴ Using Pythagoras theorem, we get
OQ² = QT² + OT²
⇒ OT² = OQ² – QT² = 25² – 24² = 49
⇒ OT = 7
Thus, the required radius is 7 cm.

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°
∴ OP⊥PT and OQ⊥QT
⇒ ∠OPT = 90° and ∠OQT = 90°
Now, in the quadrilateral TPOQ, we get
∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360° – 290° = 70°

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) : Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.
∴ OA⊥AP and OB⊥BP
⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have
∠BPA + ∠PAO + ∠AOB + ∠OBP = 360°
⇒ 80° + 90° + ∠AOB + 90° = 360°
⇒ 260° + ∠AOB = 360°
⇒ ∠AOB = 360° – 260° ⇒ ∠AOB = 100°
In right ∆OAP and right ∆OBP, we have
OP = OP [Common]
∠OAP = ∠OBP [Each 90°]
OA = OB [Radii of the same circle]
∴ ∆OAP ≅ ∆OBP [By RHS congruency]
⇒ ∠POA = ∠POB [By CPCT]
∴ ∠POA = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 100° = 50°

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
In the figure, PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since, the tangents at a point to a circle is perpendicular to the radius through the point.

∴ PQ⊥AB
⇒ ∠APQ = 90°
And PQ⊥CD
⇒ ∠PQD = 90° ⇒ ∠APQ = ∠PQD
But they form a pair of alternate angles.
∴ AB || CD
Hence, the two tangents are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
In the figure, the centre of the circle is O and tangent AB touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.
Join OP.
Since, tangent at a point to a circle is perpendicular to the radius through that point.

∴ OP⊥AB
⇒ ∠OPB = 90° ……….. (1)
But by construction, PQ⊥AB
⇒ ∠QPB = 90° ………….. (2)
From (1) and (2),
∠QPB = ∠OPB
which is possible only when O and Q coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
∵ The tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OTA = 90°
Now, in the right ∆OTA, we have
OA² = OT² + AT² [Pythagoras theorem]

⇒ OT² = 5² – 4²
⇒ OT² = (5 – 4)(5 + 4)
⇒ OT² = 1 × 9 = 9 = 3²
⇒ OT = 3
Thus, the radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
Since, OP is the radius of the smaller circle.
∴ OP⊥AB ⇒ ∠APO = 90°
Also, radius perpendicular to a chord bisects the chord.

∴ OP bisects AB
⇒ AP = \(\frac{1}{2}\) AB
Now, in right ∆APO,
OA² = AP² + OP²
⇒ 5² = AP² + 3² ⇒ AP² = 5² – 3²
⇒ AP² = 4² ⇒ AP = 4 cm
⇒ \(\frac{1}{2}\) AB = 4 ⇒ AB = 2 × 4 = 8 cm
Hence, the required length of the chord AB is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC
Solution:
Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
∴ AP = AS, BP = BQ,
DR = DS and CR = CQ
Adding them, we get
(AP + BP) + (CR + RD) = (BQ + QQ) + (DS + SA)
⇒ AB + CD = BC + DA

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

Prove that ∠AOB = 90°.
Solution:
∵ The tangents drawn to a circle from an external point are equal.
∴ AP = AC ……… (1)
Join OC.
In ∆PAO and ∆CAO, we have
AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC [From (1)]
⇒ ∆PAO ≅ ∆CAO [SSS congruency]
∴ ∠PAO = ∠CAO
⇒ ∠PAC = 2∠CAO …………. (2)
Similarly, ∠CBQ = 2∠CBO ……………… (3)

Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
2∠CAO + 2∠CBO = 180° [From (2) and (3)]
⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90° …………… (4)
Also, in ∆AOB,
∠BAO + ∠OBA + ∠AOB = 180° [Sum of angles of a triangle]
⇒ ∠CAO + ∠CBO + ∠AOB = 180°
⇒ 90° + ∠AOB = 180° [From (4)]
⇒ ∠AOB = 180° – 90°
⇒ ∠AOB = 90°

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:
Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ∆OAP and right ∆OBP, we have
PA = PB [Tangents to circle from an external point]
OA = OB [Radii of the same circle]
OP = OP [Common]
⇒ ∆OAP ≅ ∆OBP [By SSS congruency]
∴ ∠OPA = ∠OPB [By CPCT]
and ∠AOP = ∠BOP
⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP
In right ∆OAP,
∠AOP + ∠OPA + ∠PAO = 180°
⇒ ∠AOP = 180° – 90° – ∠OPA
⇒ ∠AOP = 90° – ∠OPA
⇒ 2∠AOP = 180° – 2∠OPA
⇒ ∠AOB = 180° – ∠APB
⇒ ∠AOB + ∠APB = 180°

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.
Since, tangents to a circle from an external point are equal in length

∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding, we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
But AB = CD [Opposite sides of parallelogram]
and BC = AD
∴ AB + CD = AD + BC ⇒ 2AB = 2BC
⇒ AB = BC
Similarly, AB = DA and DA = CD
Thus, AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
Here ∆ABC circumscribes the circle with centre O. Also, radius = 4 cm
Let AC and AB touches the circle at E and F, respectively and join OE and OF.
∵ The sides BC, CA and AB touches the circle at D, E and F respectively.
∴ BF = BD = 8 cm
[ ∵ Tangents to a circle from an external point are equal]
CD = CE = 6 cm
AF = AE = x cm (say)

∴ The sides of the ∆ABC are 14 cm, (x + 6) cm and (x + 8) cm
Perimeter of ∆ABC
= [14 + (x + 6) + (x + 8)] cm
= [14 + 6 + 8 + 2x] cm
= (28 + 2x) cm
⇒ Semi perimeter of ∆ABC,
s = \(\frac{1}{2}\) [28 + 2x] cm = (14 + x) cm
∴ s – a = (14 + x) – (8 + x) = 6
s – b = (14 + x) – (14) = x
s – c = (14 + x) – (6 + x) = 8
where, a = AB, b = BC, c = AC

Squaring both sides, we get
(14 + x)² = (14 + x)3x
⇒ 196 + x² + 28x = 42x + 3x²
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ (x – 7)(x + 14) = 0
⇒ x – 7 = 0 or x + 14 = 0
⇒ x = 7 or x = -14
But x = -14 is rejected.
∴ x = 7
Thus, AB = 8 + 7 = 15 cm, BC = 8 + 6 = 14 cm and CA = 6 + 7 = 13 cm

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.
Join OP, OQ, OR and OS.
We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2
∠3 = ∠4
∠5 = ∠6 and ∠7 = ∠8
Also, the sum of all the angles around a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∴ 2(∠1 + ∠8 + ∠5 + ∠4) = 360°
⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° …………. (1)
and 2(∠2 + ∠3 + ∠6 + ∠7) = 360°
⇒ (∠2 + ∠3 + ∠6 + ∠7) = 180° ……………. (2)
Since, ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD, ∠1 + ∠8 = ∠AOD and ∠4 + ∠5 = ∠BOC
∴ From (1) and (2), we have
∠AOD + ∠BOC = 180°
and ∠AOB + ∠COD = 180°

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
Here, the number of all the possible outcomes = 5 × 5 = 25
(i) For both customers visiting on same day:
Favourbale outcomes are (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)
∴ Number of favourable outcomes = 5
∴ Required probability = \(\frac{5}{25}=\frac{1}{5}\)

(ii) For both the customers visiting on consecutive days:
Favourable outcomes are (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Wed., Tue.), (Thu., Wed.), (Fri., Thu.)
∴ Number of favourable outcomes = 8
∴ Required probability = \(\frac{8}{25}\)

(iii) For both the customers visiting on different days:
We have probability for both visiting on same day = \(\frac{1}{5}\)
∴ Probability for both visiting on different days = 1 – [Probability for both visiting on the same day]
= \(1-\left[\frac{1}{5}\right]=\frac{5-1}{5}=\frac{4}{5}\)
∴ The required probability = \(\frac{4}{5}\).

Question 2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score in the two throws:

What ist the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table as follows

∴ Number of all possible outcomes = 36
(i) For total score being even:
Favourable outcomes = 18
[∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8,12]
∴ The required probability = \(\frac{18}{36}=\frac{1}{2}\)

(ii) For total score being 6 :
In list of scores, we have four 6’s.
∴ Favourable outcomes = 4
∴ Required probability = \(\frac{4}{36}=\frac{1}{9}\)

(iii) For toal score being at least 6:
The favourable scores are : 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12
∴ Number of favourable outcomes = 15
∴ Required probability = \(\frac{15}{36}=\frac{5}{12}\)

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Let the number of blue balls in the bag be x.
Total number of balls = x + 5 Number of possible outcomes = (x + 5).
For a blue ball, favourable outcomes = x
Probability of drawing a blue ball = \(\frac{x}{x+5}\)
Similarly, probability of drawing a red ball = \(\frac{5}{x+5}\)
Now, we have \(\frac{x}{x+5}=2\left[\frac{5}{x+5}\right]\)
⇒ \(\frac{x}{x+5}=\frac{10}{x+5}\) ⇒ x = 10
Thus the required number of blue balls = 10.

Question 4.
A box contains 12 balls out of whichxare black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
∵ The total number of balls in the box = 12
∴ Number of possible outcomes = 12
Case – I: For drawing a black ball
Number of favourable outcomes = x
∴ Probability of getting a black ball = \(\frac{x}{12}\)

Case – II: When 6 more black balls are added
Then, the total number of balls = 12 + 6 = 18
⇒ Number of possible outcomes = 18
Now, the number of black balls = (x + 6)
∴ Number of favourable outcomes = (x + 6)
∴ Required probability = \(\frac{x+6}{18}\)
According to the given condition,
\(\frac{x+6}{18}=2\left(\frac{x}{12}\right)\)
⇒ 12 (x + 6) = 36x ⇒ 12x + 72 = 36x
⇒ 36x – 12x = 72 ⇒ 24x = 72
⇒ x = \(\frac{72}{24}\) = 3
Thus, the required value of x is 3.

Question 5.
Ajar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\) Find the number of blue marbles in the jar.
Solution:
There are 24 marbles in the jar.
∴ Number of possible outcomes = 24.
Let there are x blue marbles in the jar.
∴ Number of green marbles = 24 – x
⇒ Favourable outcomes = (24 – x)
∴ Required probability for drawing a green marbles \(\frac{24-x}{24}\)
Now, according to the given condition,
\(\frac{24-x}{24}=\frac{2}{3}\)
⇒ 3(24 – x) = 2 × 24 ⇒ 72 – 3x = 48
⇒ 3x = 72 – 48 ⇒ 3x = 24 ⇒ x = \(\frac{24}{3}\) = 8
Thus, the required number of blue marbles is 8.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
A circle can have an infinite number of tangents.

Question 2.
Fill in the blanks :
(i) A tangent to a circle intersects it in _________ point(s).
(ii) A line intersecting a circle in two points is called a ______.
(iii) A circle can have ______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.
Solution:
(i) Exactly one
(ii) Secant
(iii) Two
(iv) Point of contact

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm
Solution:
In right ∆QPO,
OQ² = OP² + PQ²

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
We have the required figure, as shown

Here, l is the given line and a circle with centre O is drawn.
Line n is drawn which is parallel to l and tangent to the circle. Also, m is drawn parallel to line 1 and is a secant to the circle.

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