and 
are collinear, prove that 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
be three points such that area of a 
is 4 sq units, find the value of 
.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the area of a ABC having vertices 
and 
is 35 sq units.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 A=2 \cos ^{2} A-1$ Proof: $\text { LHS }=\cos ^{-1}\left(2 x^{2}-1\right) \ldots(1)$ Let $x=\cos A \ldots$...
Answer: AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...
![\left[\begin{array}{ll} 4 & 0 \\ 2 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-00e067d8b3d687152537a9d78c193c7f_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
is a continuous function.Solution: Assume $f(x)=\sec x$ So, $f(x)=\frac{1}{\cos x}$ $f(x)$ is not defined when $\cos x=0$ And $\cos x=0$ when, $x=\frac{\pi}{2}$ and odd multiples of $\frac{\pi}{2}$ like $-\frac{\pi}{2}$...
Solution: It is given that: $f(x)=\left\{\begin{array}{c} \frac{x^{n}-1}{x-1}, \text { when } x \neq 1 \\ n, \text { when } x=1 \end{array}\right.$ L.H.L. and $\mathrm{x}=1$ $\begin{array}{l} \lim...
and 
(ii) 
and 
(i) $\quad \mathrm{A}(9,3)$ and $\mathrm{B}(15,11)$ The given points are $A(9,3)$ and $B(15,11)$. Then $\left(x_{1}=9, y_{1}=3\right)$ and $\left(x_{2}=15, y_{2}=11\right)$ $A...
Solution: The given equations may be written as: $2 x+y-35=0\dots \dots(i)$ $3 \mathrm{x}+4 \mathrm{y}-65=0 \quad \ldots \ldots(ii)$ Here $a_{1}=2, b_{1}=1, c_{1}=-35, a_{2}=3, b_{2}=4$ and...
, 
Solution: The given equations may be written as: $\begin{array}{l} 2 x+5 y-1=0\dots \dots(i) \\ 2 x+3 y-3=0\dots \dots(ii) \end{array}$ Here $a_{1}=2, b_{1}=5, c_{1}=-1, a_{2}=2, b_{2}=3$ and...
, 
Solution: The given equations are: $3 x+2 y+25=0\dots \dots(i)$ $2 \mathrm{x}+\mathrm{y}+10=0 \quad \ldots \ldots(ii)$ Here $\mathrm{a}_{1}=3, \mathrm{~b}_{1}=2, \mathrm{c}_{1}=25, \mathrm{a}_{2}=2,...
Solution: The given equations are: $\begin{array}{l} 6 x-5 y-16=0\dots \dots(i) \\ 7 x-13 y+10=0\dots \dots(ii) \end{array}$ Here $a_{1}=6, b_{1}=-5, c_{1}=-16, a_{2}=7, b_{2}=-13$ and $c_{2}=10$ On...
Solution: \[Arc\text{ }AC\text{ }subtends\angle AOC\]at the centre and \[\angle ADC\]at the remaining part of the circle. Thus, \[\angle AOC\text{ }=\text{ }2\angle ADC\] \[\angle AOC\text{ }=\text{...
Solution: Join \[EB\] Then, in cyclic quad. \[ABEP\] \[\angle APE\text{ }+\angle ABE\text{ }=\text{ }{{180}^{o}}~\ldots ..\text{ }\left( i \right)\][Opposite angles of a cyclic quad. are...
Solution: Given, \[AB\text{ }=\text{ }AD\text{ }=\text{ }DC\text{ }=\text{ }PB\text{ }and~\angle DBC\text{ }=\text{ }{{x}^{o}}\] Join \[AC\text{ }and\text{ }BD\] Proof: \[\angle DAC\text{ }=\angle...
Solution: Join \[DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC\] Given, if \[\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}\] \[I\] is the incentre of the...
Solution: Join \[DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC\] Given, if \[\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}\] \[I\] is the incentre of the...
Solution: Given, In cyclic quad. \[ABCD\] \[AF\text{ }||\text{ }CB\text{ }and\text{ }DA\] is produced to \[E\]such that \[\angle ADC\text{ }=\text{ }{{92}^{o}}~and\angle FAE\text{ }=\text{...
Given, \[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\text{ }and\text{ }D\text{ }and\text{ }E\]are points on \[AB\text{ }and\text{ }AC\]such that \[AD\text{ }=\text{ }AE\] And, \[DE\] is joined....
Solution: In cyclic quadrilateral \[ABEC\] \[\angle BAC\text{ }+\angle BEC\text{ }=\text{ }{{180}^{o}}~\] [Opposite angles of a cyclic quadrilateral are supplementary] \[\angle BAC\text{ }+\text{...
Solution: (i) Given that \[BD\] is a diameter of the circle. And, the angle in a semicircle is a right angle. So, \[\angle BCD\text{ }=\text{ }90{}^\circ \] Also given that, \[\angle DBC\text{...
Let \[ABCD\] be a cyclic quadrilateral and \[PQRS\] be the quadrilateral formed by the angle bisectors of angle \[\angle A,\angle B,\angle C\text{ }and\angle D\] Required to prove: \[PQRS\] is a...
Solution: Join \[OE\] \[Arc\text{ }EC\] subtends \[\angle EOC\]at the centre and \[\angle EBC\]at the remaining part of the circle. \[\angle EOC\text{ }=\text{ }2\angle EBC\text{ }=\text{ }2\text{...
Suppose, \[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\]and circle with \[AB\] as diameter is drawn which intersects the side \[BC\text{ }and\text{ }D\] And, join \[AD\] Proof: It’s seen that,...
Solution: Firstly, join \[OB\text{ }and\text{ }OC\] Proof: \[\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}\] Now, in \[\vartriangle OBC\]...
Solution: Now, \[\angle ABD\text{ }=\angle ACD\text{ }=\text{ }{{30}^{o}}~\][Angles in the same segment] In \[\vartriangle ADB\] by angle sum property we have \[\angle BAD\text{ }+\angle ADB\text{...
Solution: According to the given question, ABC is a triangle with \[AB\text{ }=\text{ }10\text{ }cm,\text{ }BC=\text{ }8\text{ }cm,\text{ }AC\text{ }=\text{ }6\text{ }cm\] Three circles are drawn...
Solution: In the given fig, \[O\]is the centre of the circle and, \[CA\text{ }and\text{ }CB\]are the tangents to the circle from \[C\]Also, \[\angle ACO\text{ }=\text{ }{{30}^{o}}\] \[P\] is any...
Solution: In the given fig, \[O\]is the centre of the circle and, \[CA\text{ }and\text{ }CB\]are the tangents to the circle from \[C\]Also, \[\angle ACO\text{ }=\text{ }{{30}^{o}}\] \[P\] is any...
i) Join \[OC\text{ }and\text{ }OB\] \[AB\text{ }=\text{ }BC\text{ }=\text{ }CD\] And \[\angle ABC\text{ }=\text{ }{{120}^{o}}~\][Given] So, \[\angle BCD\text{ }=\angle ABC\text{ }=\text{...
Solution: In \[\vartriangle ADC\text{ }and\text{ }\vartriangle ABE\] \[\angle ACD\text{ }=\angle AEB~\] [Angles in the same segment] \[AC\text{ }=\text{ }AE\][Given] \[\angle A\text{ }=\angle...
i) In \[\vartriangle ABC\] We know that \[\angle B\text{ }=\text{ }{{90}^{o}}~and\text{ }BC\]is the diameter of the circle. Hence, \[AB\]is the tangent to the circle at \[B\] Now, as \[AB\] is...
Solution: Join \[OC\] \[BCD\]is the tangent and \[OC\]is the radius. As, \[OC\bot BD\] \[\angle OCD\text{ }=\text{ }{{90}^{o}}\] \[\angle OCD\text{ }+\angle ACD\text{ }=\text{ }{{90}^{o}}~\ldots...
Solution: In \[\vartriangle ABD\] \[\angle BAD\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle], \[\angle ABD\text{ }=\text{ }{{60}^{o}}~\] [Given] Or, \[\angle ADB\text{ }=\text{...
Solution: 15. (i) \[PQ\]is a tangent and \[CD\]is a chord. \[\angle DCQ\text{ }=\angle DBC\] [Angles in the alternate segment] \[\angle DBC\text{ }=\text{ }{{40}^{o}}\] \[[As\angle DCQ\text{...
Solution: From \[P,\text{ }AP\] is the tangent And \[PMN\]is the secant for first circle. \[A{{P}^{2}}~=\text{ }PM\text{ }x\text{ }PN\text{ }\ldots .\text{ }\left( 1 \right)\] Again from \[P,\text{...
Join \[AP\text{ }and\text{ }BP\] As \[TPS\] is a tangent And \[PA\]is the chord of the circle. \[\angle BPT\text{ }=\angle PAB~\] [Angles in alternate segments] But, \[\angle PBA\text{ }=\angle...
Join \[OL,\text{ }OM\text{ }and\text{ }ON\] Suppose, \[D\text{ }and\text{ }d\] be the diameter of the circumcircle and incircle. Also, \[R\text{ }and\text{ }r\] be the radius of the circumcircle and...
Solution: Join \[OB\] In \[\vartriangle OBC\] we have \[BC\text{ }=\text{ }OD\text{ }=\text{ }OB\][Radii of the same circle] \[\angle BOC\text{ }=\angle BCO\text{ }=\text{ }{{20}^{o}}\] And ext....
Join \[ED,\text{ }EF\text{ }and\text{ }DF\] And \[BF,\text{ }FA,\text{ }AE\text{ }and\text{ }EC\] \[\angle EBF\text{ }=\angle ECF\text{ }=\angle EDF\text{ }\ldots ..\text{ }\left( i \right)\] [Angle...
Join \[AD\] And \[AB\] is the diameter. We have \[\angle ADB\text{ }=\text{ }90{}^\text{o}~\][Angle in a semi-circle] But, \[\angle ADB\text{ }+~\angle ADC\text{ }=\text{ }180{}^\text{o}~\][Linear...
Given, cyclic quadrilateral \[ABCD\] So, \[\angle A\text{ }+\angle C\text{ }=\text{ }{{180}^{o}}~\][Opposite angles in a cyclic quadrilateral is supplementary] \[3\angle C\text{ }+\angle C\text{...
Solution: \[ABCD\] is a cyclic quadrilateral, \[\angle BCD\text{ }+~\angle BAD\text{ }=\text{ }{{180}^{\circ }}\] [Opposite angles of a cyclic quadrilateral are supplementary] \[\angle BCD\text{...
Given, \[ABCD\] is a cyclic quadrilateral in which \[AD\text{ }||\text{ }BC\] And, \[\angle ADC\text{ }=\text{ }{{110}^{o}},\angle BAC\text{ }=\text{ }{{50}^{o}}\] We know that, \[\angle B\text{...
We have, \[AB\]as the diameter and \[AC\]as the chord. Now, draw \[OL\bot AC\] Since\[OL\bot AC\] and hence it bisects \[AC\] \[O\] is the centre of the circle. Therefore, \[OA\text{ }=\text{...
Given: \[AB\text{ }and\text{ }AC\] are two equal chords of \[C\text{ }\left( O,\text{ }r \right)\] Required to prove: Centre, \[O\] lies on the bisector of \[\angle BAC\] Construction: Join \[BC\]...
We know that, If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, \[AB\text{ }=\text{ }\left( 5\text{ }-\text{ }3 \right)\text{...
(i) Given, \[radius\text{ }=\text{ }10\text{ }cm\] In rhombus \[OABC\] \[OC\text{ }=\text{ }10\text{ }cm\] So,...
Given: \[A\] circle with center \[O\] and radius \[r\] \[AB\text{ }and\text{ }CD\]are two chords such that \[AB\text{ }>\text{ }CD\] Also, \[OM\bot AB\text{ }and\text{ }ON\bot CD\] Required to...
Solution: Let’s draw a tangent \[TS\text{ }at\text{ }P\]to the circles given. As \[TPS\]is the tangent and \[PD\]is the chord, we have \[\angle PAB\text{ }=\angle BPS\text{ }\ldots .\text{ }\left( i...
Solution: Join \[OA,\text{ }OB,\text{ }OA,\text{ }OB\text{ }and\text{ }OO\] \[CD\]is the tangent and \[AO\]is the chord. \[\angle OAC\text{ }=\angle OBA\text{ }\ldots \text{ }\left( i \right)\] ...
Let \[DE\]be the tangent to the circle at \[P\] And, \[DE\text{ }||\text{ }QR\][Given] \[\angle EPR\text{ }=\angle PRQ\][Alternate angles are equal] \[\angle DPQ\text{ }=\angle PQR\][Alternate...
(v) Number of favorable outcomes for a diamond or a spade = 13 + 13 = 26 So, number of favorable outcomes = 26 Hence, P(getting a diamond or a spade) = 26/52 = 1/2
(iii) Favorable outcomes for jack or queen of hearts = 1 jack + 1 queen So, the number of favorable outcomes = 2 Hence, P(jack or queen of hearts) = 2/52 = 1/26 (iv) Number of favorable outcomes for...
Solution: We have, Total possible outcomes = 52 (i) Number queens of red color = 2 Number of favorable outcomes = 2 Hence, P(queen of red color) = 2/52 (ii) Number of black cards = 26 Number of...
(v) Favorable outcomes for a number less than or equal to 9 are 1, 2, 3, 4, 5, 6, 7, 8, 9 So, number of favorable outcomes = 9 Hence, P(the pointer will be at a number less than or equal to 9) =...
(iii) Favorable outcomes for a prime number are 2, 3, 5, 7, 11 So, number of favorable outcomes = 5 Hence, P(the pointer will be at a prime number) = 5/12 (iv) Favorable outcomes for a number...
Solution: We have, Total number of possible outcomes = 12 (i) Number of favorable outcomes for 6 = 1 Hence, P(the pointer will point at 6) = 1/12 (ii) Favorable outcomes for an even number are 2, 4,...
(iii) Number of favourable outcomes for neither Re 1 nor Rs 5 coins = Number of favourable outcomes for Rs 2 coins = 50 = n(E) Hence, probability (neither Re 1 nor Rs 5 coin) = n(E)/ n(S) = 50/100 =...
Solution: We have, Total number of coins = 20 + 50 + 30 = 100 So, the total possible outcomes = 100 = n(S) (i) Number of favourable outcomes for Re 1 coins = 30 = n(E) Probability (Re 1 coin) =...
(iii) Number of favorable outcomes for white or green ball = 16 + 8 = 24 = n(E) Hence, probability for drawing a white or green ball = n(E)/ n(S) = 24/34 = 12/17
Solution: Total number of possible outcomes = 10 + 16 + 8 = 34 balls So, n(S) = 34 (i) Favorable outcomes for not a red ball = favorable outcomes for white or green ball So, number of favorable...
Solution: We know that, P(do not have the same birthday) + P(have same birthday) = 1 0.897 + P(have same birthday) = 1 Thus, P(have same birthday) = 1 – 0.897 P(have same birthday) =...
Solution: We have, Total possible outcomes = number of red balls. (i) Number of favourable outcomes for black balls = 0 Hence, P(black ball) = 0 (ii) Number of favourable outcomes for red balls =...
Solution: We know that, P(E) + P(not E) = 1 So, 0.59 + P(not E) = 1 Hence, P(not E) = 1 – 0.59 = 0.41
(iii) As 0 ≤ 37 % = (37/100) ≤ 1 Thus, 37 % can be a probability of an event. (iv) As -2.4 < 0 Thus, -2.4 cannot be a probability of an event.
Solution: We know that probability of an event E is 0 ≤ P(E) ≤ 1 (i) As 0 ≤ 3/7 ≤ 1 Thus, 3/7 can be a probability of an event. (ii) As 0 ≤ 0.82 ≤ 1 Thus, 0.82 can be a probability of an...
(iii) All the outcomes are favourable to the event E = ‘sum of two numbers ≤ 12’. Thus, P(E) = n(E)/ n(S) = 36/36 = 1
Solution: We have, the number of possible outcomes = 6 × 6 = 36. (i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ = E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} The number...
Solution: We have, Total number of shirts = 50 Total number of elementary events = 50 = n(S) (i) As, trader accepts only good shirts and number of good shirts = 44 Event of accepting good shirts =...
Solution: Total result = 0 sec to 40 sec Total possible outcomes = 40 So, n(S) = 40 Favourable results = 0 sec to 15 sec Favourable outcomes = 15 So, n(E) = 15 Hence, the probability that the music...
(iii) Number of black cards left = 23 cards (13 club + 10 spade) Event of drawing a black card = E = 23 So, n(E) = 23 Hence, probability of drawing a black card = n(E)/ n(S) = 23/49
Solution: We have, Total number of cards = 52 If 3 face cards of spades are removed Then, the remaining cards = 52 – 3 = 49 = number of possible outcomes So, n(S) = 49 (i) Number of black face cards...
Solution: We have, Total number of balls in the box = 7 + 8 + 5 = 20 balls Total possible outcomes = 20 = n(S) (i) Event of drawing a white ball = E = number of white balls = 5 So, n(E) = 5 Hence,...
Solution: Out of the two friends, A’s birthday can be any day of the year. Now, B’s birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely. So,...
Solution: (i) We know that, The probability of winning of A + Probability of losing of A = 1 And, Probability of losing of A = Probability of winning of B Therefore, Probability of winning of A +...
(v) Numbers less than 8 = { 2, 3, 4, 5, 6, 7} Event of drawing a card with number less than 8 = E = {6H cards, 6D cards, 6S cards, 6C cards} So, n(E) = 24 Thus, probability of drawing a card with...
(iii) Event of drawing a queen of black colour = {Q(spade), Q(club)} = E So, n(E) = 2 Thus, probability of drawing a queen of black colour = n(E)/ n(S) = 2/52 = 1/26 (iv) Event of drawing a card...
Solution: We have, the total number of possible outcomes = 52 So, n(S) = 52 (i) No. of face cards in a deck of 52 cards = 12 (4 kings, 4 queens and 4 jacks) Event of drawing a face cards = E = (4...
Solution: The number of possible outcomes when dice is thrown = {1, 2, 3, 4, 5, 6} So, n(S) = 6 (i) Event of getting a number greater than 2 = E = {3, 4, 5, 6} So, n(E) = 4 Thus, probability of...
(iii) Probability of not drawing a yellow ball = 1 – Probability of drawing a yellow ball Thus, probability of not drawing a yellow ball = 1 – 1/8 = (8 – 1)/ 8 = 7/8 (iv) Neither yellow ball nor red...
Solution: The total number of balls in the bag = 3 + 4 + 1 = 8 balls So, the number of possible outcomes = 8 = n(S) (i) Event of drawing a yellow ball = {Y} So, n(E) = 1 Thus, probability of drawing...
(iii) E = event of getting both heads or both tails = {HH, TT} n(E) = 2 Hence, probability of getting both heads or both tails = n(E)/ n(S) = 2/4 = ½
Solution: We know that, when two coins are tossed together possible number of outcomes = {HH, TH, HT, TT} So, n(S) = 4 (i) E = event of getting both heads = {HH} n(E) = 1 Hence, probability of...
Solution: In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 For two dice, n(S) = 6 x 6 = 36 Favorable cases where the sum is 10 or more with 5 on 1st die = {(5, 5), (5,...
Solution: We know that, Number of pages in the book = 85 Number of possible outcomes = n(S) = 85 Out of 85 pages, pages that sum up to 8 = {8, 17, 26, 35, 44, 53, 62, 71, 80} So, pages that sum up...
(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6} So, n(E) = 6 Hence, probability of getting a number less than 8 = n(E)/ n(S) = 6/6 = 1 (iv) On a dice, numbers greater than 6 = 0 So, n(E) =...
Solution: We know that, In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 (i) On a dice, numbers less than 3 = {1, 2} So, n(E) = 2 Hence, probability of getting a number...
(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15} So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a multiple of 3 and...
Solution: We know that, there are 25 cards from which one card is drawn. So, the total number of elementary events = n(S) = 25 (i) From numbers 1 to 25, there are 8 numbers which are multiple of 3...
(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47} So, favorable number of events = n(E) = 47 Hence, probability of selecting a card with a number less...
(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e. {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59} So, favorable number of events = n(E)...
Solution: We kwon that, there are 100 cards from which one card is drawn. Total number of elementary events = n(S) = 100 (i) From numbers 1 to 100, there are 20 numbers which are multiple of 5...
(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6 So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a number...
Solution: We know that, there are totally 9 cards from which one card is drawn. Total number of elementary events = n(S) = 9 (i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10...
Solution: (i) Winning of Geeta is a complementary event to winning of Ritu Thus, P(winning of Ritu) + P(winning of Geeta) = 1 P(winning of Geeta) = 1 – P(winning of Ritu) P(winning of Geeta) =...
Solution: (i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1. P(A) + P(B) = 1 (ii) P(A) = 0.46 Let P(B) be...
(v) There are 26 red cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens and 2 jacks). The number of favourable outcomes for the event of drawing a face card of red color = 6...
(iii) Number of red cards in a deck = 26 The number of favourable outcomes for the event of drawing a red card = 26 Then, probability of drawing a red card = 26/52 = ½ (iv) There are 52 cards...
Solution: We know that, Total number of cards = 52 So, the total number of outcomes = 52 There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are...
(iii) If E = event of getting an odd number = {1, 3, 5} So, n(E) = 3 Then, probability of a getting an odd number = n(E)/ n(s) = 3/6 = ½
Solution: Here, the sample space = {1, 2, 3, 4, 5, 6} n(s) = 6 (i) If E = event of getting an even number = {2, 4, 6} n(E) = 3 Then, probability of a getting an even number = n(E)/ n(s) = 3/6 = ½...
(iii) E = event of getting a number not greater than 4 = {1, 2, 3, 4} So, n (E) = 4 Then, probability of getting a number not greater than 4 = n(E)/ n(s) = 4/6 = 2/3
Solution: Here, the sample space = {1, 2, 3, 4, 5, 6} So, n (s) = 6 (i) If E = event of getting a number greater than 4 = {5, 6} So, n (E) = 2 Then, probability of getting a number greater than 4 =...
(v) There are 3 + 2 = 5 balls which are not black So, the number of favourable outcomes = 5 Thus, P(getting a white ball) = 5/10 = ½
(iii) There are 3 white balls So, the number of favourable outcomes = 3 Thus, P(getting a white ball) = 3/10 = 3/10 (iv) There are 3 + 5 = 8 balls which are not red So, the number of favourable...
Solution: Total number of balls = 3 + 5 + 2 = 10 So, the total number of possible outcomes = 10 (i) There are 5 black balls So, the number of favourable outcomes = 5 Thus, P(getting a black ball) = ...
Solution: Total number of balls = 3 + 5 + 2 = 10 So, the total number of possible outcomes = 10 (i) There are 5 black balls So, the number of favourable outcomes = 5 Thus, P(getting a black ball) = ...
Solution: Here, the sample space = {H, T} i.e. n(S) = 2 (i) If A = Event of getting a tail = {T} Then, n(A) = 1 Hence, the probability of getting a tail = n(A)/ n(S) = 1/2 (ii) Not getting a tail As...
Diameter of cylindrical tank = 2.8 m radius = 1.4 m Height = 4.2 m \[\begin{array}{*{35}{l}} Volume\text{ }of\text{ }water\text{ }filled\text{ }in\text{ }it\text{ }=\text{ }\pi {{r}^{2}}h \\...
FIGURE: Side of square (a) = 7m the radius of semi-circle = 7/2 m Length of the tunnel = 80 m \[\begin{array}{*{35}{l}} Area\text{ }of\text{ }cross\text{ }section\text{ }of\text{ }the\text{...
FIGURE: Side of square (a) = 7m the radius of semi-circle = 7/2 m Length of the tunnel = 80 m \[\begin{array}{*{35}{l}} Area\text{ }of\text{ }cross\text{ }section\text{ }of\text{ }the\text{...
Length of the platform = 22 m Circumference of semi-circle (c) = 11 m \[\begin{array}{*{35}{l}} radius\text{ }=\text{ }\left( c\text{ x }2 \right)/\text{ }\left( 2\text{ x }\pi \right)\text{...
Radius of the base of poles (r) = 6 cm Height of the cylindrical part (h1) = 110 cm Height of the conical part (h2) = 9 cm \[\begin{array}{*{35}{l}} Total\text{ }volume\text{ }of\text{ }the\text{...
Let edge of the cube = a Then, volume of the cube = a x a x a = a3 The sphere that exactly fits in the cube will have radius ‘a/2’ \[\begin{array}{*{35}{l}} Volume\text{ }of\text{ }sphere\text{...
Diameter of the sphere = 6 cm Radius = 3 cm \[~volume\text{ }=\text{ }4/3\text{ }\pi {{r}^{3~}}=\text{ }4/3\text{ x }22/7\text{ x }3\text{ x }3\text{ x }3\text{ }=\text{ }792/7\text{...
Radius of both cone and hemisphere = 8 cm And, height = 8 cm \[\begin{array}{*{35}{l}} Volume\text{ }of\text{ }the\text{ }solid\text{ }=\text{ }Volume\text{ }of\text{ }cone\text{ }+\text{...
Diameter of the cylinder = 12 cm => radius = 6 cm Height of the cylinder = 15 cm Diameter of the cone = 6 cm =>radius = 3 cm Height of the cone = 12 cm Radius of the hemisphere = 3 cm Let the...
Radius of the largest sphere that can be formed inside the cylinder will be equal to the radius of the cylinder. radius of the largest sphere = 7 cm \[\begin{array}{*{35}{l}} Volume\text{ }of\text{...
Diameter of solid metallic sphere = 6 cm So, its radius = 3 cm Height of solid metal cone = 45 Diameter of metal cone = 12 cm Its radius = 6 cm Let the number of solid metallic spheres be ‘n’...
Height of the cone (h) = 24 cm Height of the cylinder (H) = 36 cm Radius of the cone (r) = twice the radius of the cylinder = 10 cm Radius of the cylinder (R) = 5 cm \[\begin{array}{*{35}{l}}...
Diameter of the base = 3.5 m So, its radius = 3.5/2 m = 1.75 m = 7/4 m Height of cylindrical part = 4 + 2/3 = 14/3 (i) \[\begin{array}{*{35}{l}} Capacity\text{ }\left( volume \right)\text{ }of\text{...
Diameter of cylindrical boiler = 3.5 m So, the radius (r) = 3.5/2 = 35/20 = 7/4 m Height (h) = 2m Radius of hemisphere (R) = 7/4 m \[\begin{array}{*{35}{l}} Total\text{ }volume\text{ }of\text{...
Height of the cylindrical part = H = 8 m Height of the conical part = h = (13 – 8) m = 5 m Diameter = 24 m Its radius \[\begin{array}{*{35}{l}} =\text{ }24/2\text{ }=\text{ }12\text{ }m \\...
Radius of the cylindrical part of the tent (r) =105/2 m Slant height (l) = 80 m the total curved surface area of the tent \[\begin{array}{*{35}{l}} =\text{ }2\pi r\text{ }h\text{ }+\text{ }\pi rl ...
Radius of solid cylinder (R) = 12 cm And, Height (H) = 16 cm \[\begin{array}{*{35}{l}} Volume\text{ }=\text{ }\pi {{R}^{2~}}h\text{ }=\text{ }22/7\text{ }x\text{ }12\text{ }x\text{ }12\text{...
Height of the cylinder (h) = 10 cm And radius of the base (r) = 6 cm Volume of the cylinder = πr2 h Height of the cone = 10 cm Radius of the base of cone = 6 cm \[\begin{array}{*{35}{l}}...
Height of cone = 8 cm Radius = 5 cm \[\begin{array}{*{35}{l}} Volume\text{ }=~1/3\text{ }\pi {{r}^{2~}}h \\ =\text{ }1/3\text{ x }22/7\text{ x }5\text{ x }5\text{ x }8\text{ }c{{m}^{3}} \\ =\text{...
Since, the diameter of the largest hemisphere that can be placed on a face of a cube of side 7 cm will be 7 cm. radius = r = 7/2 cm It’s curved surface area = 2 πr2 = 2 x 22/7 x 7/2 x 7/2 = 77...
\[\left( iii \right)\text{ }Weight\text{ }of\text{ }material\text{ }drilled\text{ }out\text{ }=\text{ }1232\text{ x }7g\text{ }=\text{ }8624g\text{ }=\text{ }8.624\text{ }kg\]
Dimensions of rectangular solid l = 42 cm, b = 30 cm and h = 20 cm Conical cavity’s diameter = 14 cm its radius = 7 cm Depth (height) = 24 cm (i) Total surface area of cuboid...
Radius of the hemisphere part (r) = 3.5 m = 7/2 m Volume of hemisphere \[\begin{array}{*{35}{l}} =\text{ }2/3\text{ }\pi {{r}^{3}} \\ =\text{ }2/3\text{ x }22/7\text{ x }7/2x\text{ }7/2\text{ x...
Height of the cone = 15 cm Diameter of the cone = 7 cm its radius = 3.5 cm Radius of the hemisphere = 3.5 cm Volume of the solid = Volume of the cone + Volume of the hemisphere...
Radius of hemispherical bowl = 9 cm \[Volume\text{ }=\text{ }2/3\text{ }\pi {{r}^{3}}~=\text{ }2/3\text{ }\pi {{9}^{3}}~=\text{ }2/3\text{ }\pi \text{ }x\text{ }729\text{ }=\text{ }486\text{ }\pi...
Volume of rectangular block \[=\text{ }49\text{ x }44\text{ x }18\text{ }c{{m}^{3}}~=\text{ }38808\text{ }c{{m}^{3}}~\ldots \ldots \text{ }\left( i \right)\] Let r be the radius of sphere....
Let the radius of the smaller cone be r cm Diameter of bigger cone = 40 cm the radius = 20 cm and height = 9 cm \[\begin{array}{*{35}{l}} Volume\text{ }of\text{ }larger\text{ }cone\text{ }=\text{...
Height of the solid right circular cone = 32 cm Internal radius metallic spherical shell = 3 cm External radius spherical shell = 5 cm the volume of the spherical shell \[\begin{array}{*{35}{l}}...
External diameter of the hollow sphere = 8 cm radius (R) = 4 cm Internal diameter of the hollow sphere = 4 cm radius (r) = 2 cm Then, the volume of metal used in hollow sphere = Also given, Diameter...
Radius of the sphere = R = 15 cm So, the volume of sphere melted \[=\text{ }4/3\text{ }\pi {{R}^{2}}~=\text{ }4/3\text{ x }\pi \text{ x }15\text{ x }15\text{ x }15\] Radius of each cone recasted = r...
The volume of first sphere = 27 x volume of second sphere Let the radius of the first sphere = r1 and, radius of second sphere = r2 (i) Then, according to the question we have...
Radius of metallic sphere = 2mm = 1/5 cm \[\begin{array}{*{35}{l}} Volume\text{ }=\text{ }4/3\text{ }\pi {{r}^{3}}~=\text{ }4/3\text{ x }22/7\text{ x }1/5\text{ x }1/5\text{ x }1/5\text{ }=\text{...
Diameter of bigger ball = 8 cm radius of bigger ball = 4 cm \[Volume\text{ }=\text{ }4/3\text{ }\pi {{r}^{3}}~=\text{ }4/3\text{ }\pi \text{ }{{4}^{3}}~=\text{ }265\pi /3\text{ }c{{m}^{3}}\] Radius...
Let the radius of the spherical ball = r the volume = 4/3 πr3 And, the radius of smaller ball = r/2 Volume of smaller ball \[=\text{ }4/3\text{ }\pi {{\left( r/2 \right)}^{3}}~=\text{ }4/3\text{...
Volume of the sphere = 38808 cm3 Let the radius of the sphere \[\begin{array}{*{35}{l}} =\text{ }r \\ 4/3\text{ }\pi {{r}^{3}}~=\text{ }38808 \\ 4/3\text{ x }\left( 22/7 \right)\text{ x...
Surface area of the sphere = 2464 cm2 Let the radius of the sphere be r. surface area of the sphere \[\begin{array}{*{35}{l}} =\text{ }4\pi {{r}^{2}} \\ 4\pi {{r}^{2}}~=\text{ }2464 \\ 4\text{ x...
Let slant height of the first cone = l So, the slant height of the second cone = 2l Radius of the first cone = r1 And, the radius of the second cone = r2 Curved surface area of first cone = πr1l...
Let radius of each cone = r Given that, ratio between their slant heights = 5: 4 Let slant height of the first cone = 5x And slant height of second cone = 4x So, curved surface area of the first...
Let radius of cone y = r radius of cone x = 3r volume of cone y = V Then, volume of cone x = 2V Let h1 be the height of x and h2 be the height of y. \[\begin{array}{*{35}{l}} Volume\text{ }of\text{...
The ratio between radius and height = 5: 12 Volume of the right circular cone = 2512 cm3 Let its radius (r) = 5x, its height (h) = 12x and slant height = l \[\begin{array}{*{35}{l}}...
Circumference of the base (c) = 66 m Height of the conical tent (h) = 12 m Radius \[=\text{ }c/2\pi \text{ }=\text{ }66/\text{ }2\pi \text{ }=\text{ }\left( 33\text{ x }7 \right)/22\text{ }=\text{...
Curved surface area of the cone = 12320 cm2 Radius of the base = 56 cm Let the slant height be ‘l’ Curved surface area \[\begin{array}{*{35}{l}} =\text{ }\pi rl\text{ }=\text{ }12320\text{...
Slant height of the cone (l) = 17 cm Base radius (r) \[\begin{array}{*{35}{l}} =\text{ }8\text{ }cm \\ {{l}^{2}}~=\text{ }{{r}^{2}}~+\text{ }{{h}^{2}} \\ {{h}^{2}}~=\text{ }{{l}^{2}}~-\text{...
Height of the cylinder box = h = 35 cm Base radius of the cylinder box = r = 10 cm Width of metal sheet = 1m = 100 cm Area of metal sheet required = total surface area of the box Length x width =...
Let the original dimensions of the solid right cylinder be radius (r) and height (h) in cm. Then its volume = πr2h cm3and curved surface area = 2πrh Now, after the changes the new dimensions are:...
The radius of a solid right cylinder (r) = 100 cm let the height of a solid right circular cylinder (h) = 100 cm The volume (original) of a solid right circular cylinder \[\begin{array}{*{35}{l}}...
The ratio between height and radius of a cylinder = 3:1 Volume = 1029π cm3 …….(i) the radius of the base = r Then, it’s height will be = 3r Volume \[\begin{array}{*{35}{l}} =\text{ }\pi...
Length of the open pipe = 50 cm Its external diameter = 20 cm It’s external radius (R) = 10 cm Its internal diameter = 6 cm => It’s internal radius (r) = 3 cm Surface area of pipe open from both...
Internal radius of the cylindrical container = 10 cm = r Height of water = 7 cm = h So, the surface area of the wetted surface \[\begin{array}{*{35}{l}} =\text{ }2\pi rh\text{ }+\text{ }\pi...
Diameter of cylindrical container = 42 cm it’s radius (r) = 21 cm Dimensions of rectangular solid = 22cm x 14cm x 10.5cm The volume of solid = 22 x 14 x 10.5 cm3 ….. (i) Let the height of water = h...
Inner radius of the pipe = r = 5/2 = 2.5 cm External radius of the pipe = R = Inner radius of the pipe + Thickness of the pipe \[=\text{ }2.5\text{ }cm\text{ }+\text{ }0.5\text{ }cm\text{ }=\text{...
The diameter of the cylinder = 20 cm the radius (r) = 10 cm and the curved surface area = 100 cm2 Height = h cm (i) Curved surface area \[\begin{array}{*{35}{l}} =\text{ }2\pi rh \\ 2\pi rh\text{...
External diameter of hollow cylinder = 20 cm So, it’s radius = 10 cm = R Thickness = 0.25 cm the internal radius \[=\text{ }\left( 10\text{ }-\text{ }0.25 \right)\text{ }=\text{ }9.75\text{...
Radius of the well = 2.8/2 = 1.4 m Depth of the well = 28 m Hence, the volume of earth dug out \[\begin{array}{*{35}{l}} =~\pi {{r}^{2}}h \\ =\text{ }\left( 22/7 \right)\text{ x }1.4\text{ x...
Base circumference of cylinder (c) = 8 cm So, the radius \[=\text{ }c/2\pi \text{ }=\text{ }\left( 8\text{ x }7 \right)/\text{ }\left( 2\text{ x }22 \right)\text{ }=\text{ }14/11\text{ }cm\] Length...
The inner radius of the pipe = 2.1 cm Length of the pipe = 12 m = 1200 cm Volume of the pipe \[\begin{array}{*{35}{l}} =\text{ }\pi {{r}^{2}}h\text{ }=\text{ }22/7\text{ x }{{2.1}^{2}}~x\text{...
Since, a circular cylinder whose Height, h = 20 cm and base radius, r = 7 cm (i) Volume of cylinder \[\begin{array}{*{35}{l}} =\text{ }\pi {{r}^{2}}h\text{ }=\text{ }22/7\text{ }x\text{...
Steps for Construction: i) Draw a line segment BC = 5.6 cm ii) With centres B and C, draw two arcs of 5.6 cm radius each which intersect each other at A. iii) And, join AB and AC. iv) Draw angle...
Steps for construction: i) Draw a line segment BC = 6 cm ii) Draw two arcs of radius 6 cm with centres B and C, which intersect each other at A. iii) Then, join AC and AB. iv) Draw perpendicular...
Steps for construction: i) Draw a line segment BC = 6 cm. ii) Draw an arc with centre B and radius 8 cm. iii) Draw another arc with centre C and radius 5 cm which intersects the first arc at A. iv)...
iii) OC is the bisector of angle C Thus, ∠ACO = ∠BCO
i) O is called the incentre of the incircle of ΔABC. ii) OR and OQ are the radii of the incircle and OR = OQ.
iii) Yes, the perpendicular bisector of BC will pass through O.
i) O is called the circumcentre of circumcircle of ΔABC. ii) OA, OB and OC are the radii of the circumcircle.
Steps for Construction: i) Draw a line segment BC = 4 cm. ii) At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm. iii) Draw another perpendicular line EY from E. iv) From C, draw a...
Steps for construction: i) Draw a line segment AB = 6 cm ii) Draw a ray at A, making an angle of 60o with BC. iii) With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C. iv)...
Steps for Construction: i) Draw a line segment BC = 5 cm ii) With centres as B and C, draw two arcs of radius 5 cm each which intersect each other at A. iii) Then, join AB and AC. iv) Draw angle...
Steps for construction: A. Construction of triangle: a) Draw a line segment BC = 7 cm b) At B, draw a ray BX making an angle of 45o and cut off BE = AB – AC = 1 cm c) Join EC and draw the...
Steps for construction: i) Draw a line segment BC = 4.5 cm ii) With centres B and C, draw two arcs of radius 4.5 cm which intersect each other at A. iii) And join AC and AB. iv) Draw the...
Steps for Construction: i) draw a circle with centre O, with radius BC = 4.5 cm ii) Draw arcs making an angle of 180º – 60º = 120º at O such that ∠AOB = 120º iii) At A and B, draw two rays making...
Steps for construction: i) draw a circle with centre O,with radius BC = 5 cm ii) Draw arcs making an angle of 180º – 45º = 135º at O such that ∠AOB = 135º iii) At A and B, draw two rays making an...
Steps for construction: i) Taking O as the centre, draw a circle of diameter 9 cm (radius = 4.5 cm). ii) Mark a point P outside the circle, such that PO = 7.5 cm. iii) Taking OP as the diameter,...
Steps for construction: i) Draw a circle with centre O and radius 3 cm. ii) Take a point P from O, such that OP = 5 cm. iii) Now, draw a bisector of OP which intersects OP at M. iv) With centre M,...
Join \[OC\] \[\angle BCD\text{ }=\angle BAC\text{ }=\] \[{{30}^{o}}~\] [Angles in the alternate segment] It’s seen that, arc \[BC\]subtends \[\angle DOC\]at the center of the circle And \[\angle...
Solution: (i) As \[PQ\] is the tangent and \[OR\]is the radius. So, \[OR\bot PQ\] \[\angle ORT\text{ }=\text{ }{{90}^{o}}\] \[\angle TRQ\text{ }=\text{ }{{90}^{o}}-\text{ }{{30}^{o}}~=\text{...
Solution: As \[BD\] is the diameter, we have \[\angle BCD\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle] Now in \[\vartriangle BCD\] \[\angle CDB\text{ }+\angle CBD\text{ }+\angle BCD\text{...
Solution: (i) Given, \[PAQ\]is a tangent and \[AB\]is the chord \[\angle QAB~=\angle ADB\text{ }=\text{ }{{30}^{o}}~\] [Angles in the alternate segment] (ii) \[OA\text{ }=\text{ }OD\][radii of the...
Solution: (i) \[PA\text{ }=\text{ }AB\text{ }+\text{ }BP\text{ }=\text{ }\left( AB\text{ }+\text{ }4 \right)\text{ }cm\] \[PC\text{ }=\text{ }PD\text{ }+\text{ }CD\] \[=\text{ }5\text{ }+\text{...
Solution: As \[PAB\]is the secant and \[PT\]is the tangent, we have \[P{{T}^{2}}~=\text{ }PA\text{ }x\text{ }PB\] \[{{12.5}^{2}}~=\text{ }10\text{ }x\text{ }PB\] \[PB\text{ }=\text{ }\left(...