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10th Maths 7.3

NCERT Class 10th solution of Exercise 7.1 NCERT Class 10th solution of Exercise 7.2 NCERT Class 10th solution of Exercise 8.1 NCERT Class 10th Maths Projects Exercise 7.3 Q1. Find the area of the triangle whose vertices are: i)

$(2, 3), (-1, 0), (2, -4)$ ii)

$(-5, -1), (3, -5), (5, 2)$ Sol. : i) Given: Let

$A(2, 3), B(-1, 0),$ and

$C(2, -4)$ Solve: Area of

$triangle=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$ar(ABC)=1/2[2(0+4)+(-1)(-4-3)+2(3-0)]$

$ar(ABC)=1/2[2(4)+(-1)(-7)+2(3)]$

$ar(ABC)=1/2[8+7+6]$

$ar(ABC)=21/2$ square units Answer: The area of

$triangleABC$ is

$21/2$ square units. ii) Given: Let

$A(-5, -1), B(3, -5),$ and

$C(5, 2)$ Solve: Area of

$triangle=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$ar(ABC)=1/2[-5(-5-2)+3(2+1)+5(-1+5)]$

$ar(ABC)=1/2[-5(-7)+3(3)+5(4)]$

$ar(ABC)=1/2[35+9+20]$

$ar(ABC)=1/2[64]$

$ar(ABC)=32$ square units Answer: The area of

$triangleABC$ is

$32$ square units. Q2. In each of the following find the value of

$k$ , for which the points are collinear. i...

NCERT Class 10th solution of Exercise 7.1 NCERT Class 10th solution of Exercise 7.3 NCERT Class 10th Maths Projects Exercise 7.2 Q1. Find the coordinates of the point which divides the join of

$(-1, 7)$ and

$(4, -3)$ in the ratio

$2 : 3$ . Sol. : Given: Let

$x_1=-1, y_1=7, x_2=4, y_2=3$ and

$m_1=2, m_2=3$ To find: The coordinates of the point. Solve: Section formula

$x=(m_1x_2+m_2x_1)/(m_1+m_2)$

$x=(2times4+3times(-1))/(2+3)$

$x=(8-3)/5$

$x=5/5$

$x=1$

$y=(m_1y_2+m_2y_1)/(m_1+m_2)$

$y=(2times(-3)+3times7)/(2+3)$

$y=(-6+21)/5$

$y=15/5$

$y=3$ Answer: The coordinates of the point is

$(1, 3)$ . Q2. Find the coordinates of the points of trisection of the line segment joining

$(4, -1)$ and

$(-2, -3)$ . Sol. : Given: Let

$x_1=4, y_1=-1, x_2=-2, y_2=-3$ . To Find: The coordinates of the points. Solve: For

$P$

$m_1=1, m_2=2$

$x'=(m_1x_2+m_2x_1)/(m_1+m_2)$

$x'=(1times(-2)+2times4)/(1+2)$

$x'=(-2+8)/3$

$x'=6/3$

$x'=2$

$y'=(m_1y_2+m_2y_1)/(m_1+m_2)$ `y'=(1times(-3)+2times(-...

NCERT Class 10th solution of Exercise 7.2 NCERT Class 10th solution of Exercise 7.3 NCERT Class 10th Maths Projects Exercise 7.1 Q1. Find the distance between the following points: i)

$(2, 3), (4,1)$ ii)

$(-5,7), (-1, 3)$ iii)

$(a, b), (-a, -b)$ Sol. : i) Given: Let

$P(2, 3), Q(4, 1)$ Solve: Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((4-2)^2+(1-3)^2)$

$PQ=sqrt((2)^2+(-2)^2)$

$PQ=sqrt(4+4)$

$PQ=sqrt8$

$PQ=2sqrt2$ units. Answer: Distance

$2sqrt2$ units. ii) Given: Let

$P(-5, 7), Q(-1, 3)$ Solve: Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((-1+5)^2+(3-7)^2)$

$PQ=sqrt((4)^2+(-4)^2)$

$PQ=sqrt(16+16)$

$PQ=sqrt(32)$

$PQ=16sqrt2$ units Answer: Distance

$16sqrt2$ units. iii) Given: Let

$P(a, b), Q(-a, -b)$ Solve: Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((-a-a)^2+(-b-b)^2)$

$PQ=sqrt((-2a)^2+(-2b)^2)$

$PQ=sqrt(4a^2+4b^2)$

$PQ=sqrt(4(a^2+b^2))$

$PQ=2sqrt(a^2+b^2)$ units Answer: Distance

$2sqrt(a^2+b^2)$ units. Q2. Find the distance between the points ...

NCERT Class 10th solution of Exercise 6.1 NCERT Class 10th solution of Exercise 6.2 NCERT Class 10th solution of Exercise 6.3 NCERT Class 10th solution of Exercise 6.4 NCERT Class 10th solution of Exercise 7.1 NCERT Class 10th Maths Projects Exercise 6.5 Q1. Sides of triangles are given below. Determine which of them are right triangles In case of a triangle, write the length of its hypotenuse. i)

$7cm, 24cm, 25cm$ ii)

$3cm, 8cm, 6cm$ iii)

$50cm, 80cm, 100cm$ iv)

$13cm, 12cm, 5cm$ . Sol. : i) We use Pythagoras Theorem

$(25)^2=(24)^2+(7)^2$

$625=576+49$

$625=625$ So, it is a right triangle with hypotenuse is

$25$ . Answer: Yes,

$25$ . ii) We use Pythagoras Theorem

$(8)^2=(6)^2+(3)^2$

$64=36+9$

$64≠45$ No, it is not a right triangle. Answer: No. iii) We use Pythagoras Theorem

$(100)^2=(80)^2+(50)^2$

$10000=6400+2500$

$10000≠8900$ No, it is not a right triangle. Answer: No. iv) We use Pythagoras Theorem

$(13)^2=(12)^2+(5)^2$

$169=144+25$

$169=169$ Yes, it is a right triangle with hypot...

Chapter 15 Probability NCERT Class 9th solution of Exercise 14.1 NCERT Class 9th Maths Projects Exercise 15.1 Q1. In a cricket match, a batswoman hits a boundary

$6$ times out of

$30$ balls she plays, Find the probability that she did not hit a boundary. Sol. : Given: Total number of balls

$n(S)=30$ Number of hits

$6$ To find: Probability of not hits. Solve: Number of not hits

$n(E)=30-6=24$ Probability of not hits

$P(E)=(n(E))/(n(S)) =24/38=4/5$ Answer: Probability of not hits

$4/5$ . Q2.

$1500$ families with

$2$ children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having Number of girls in a family

$2$

$1$

$0$ Number of families

$475$

$814$

$211$ i)

$2$ girls ii)

$1$ girl iii) No girl. Also, check whether the sum of these probabilities is

$1$ . Sol. : Given: `n(E_1)=475, n(E_2)=814, ...

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10th Maths 7.3

10th Maths 7.2

10th Maths 7.1

10th Maths 6.5

9th Maths 15.1