Two cards are drawn from a well shuffled pack of 52 cards one after another without replacement. -Maths 9th

Two cards are drawn from a well shuffled pack of 52 cards one after another without replacement. -Maths 9th
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Probability of drawing an ace in the first draw = \(rac{4}{52}.\)Probability of drawing a queen of opposite shade in the second draw = \(rac{2}{51}.\)Probability of drawing a queen in the first draw = \(rac{4}{52}.\)Probability of drawing an ace of opposite shade in the second draw = \(rac{2}{51}.\)∴ Required probability = \(rac{4}{52}\) x \(rac{2}{51}\) + \(rac{4}{52}\) x \(rac{2}{51}\) = \(rac{4}{663}.\)             [‘AND’ and ‘OR’Theorems]
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Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Description : Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Last Answer : (b) \(rac{55}{221}\)S : Drawing 2 cards out of 52 cards ⇒ n(S) = 52C2 = \(rac{|\underline{52}}{|\underline{52}|\underline2}\) = \(rac{52 imes51}{2}\) = 1326A : Event of drawing 2 black cards out of 26 black cards⇒ n ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Last Answer : (d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = ... rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

A card is drawn at random from a well shuffled pack of 52 cards -Maths 9th

Description : A card is drawn at random from a well shuffled pack of 52 cards -Maths 9th

Last Answer : (c) P(X) = P(Y) > P(Z) P(X) = \(rac{26}{52}\) + \(rac{4}{52}\) - \(rac{2}{52}\) = \(rac{28}{52}\) (∵ There are 26 black cards, 4 kings and 2 black kings)P(Y) = \(rac{13}{52}\) + \(rac{ ... }{52}\)(∵ There are 4 aces, 13 diamonds, 4 queens, 1 ace of diamond, 1 queen of diamond) ∴ P(X) = P(Y) > P(Z).

What is the probability a well shuffled pack of 52 cards a card is drawn at random find the probability that it is either a heart or a queen?

Description : What is the probability a well shuffled pack of 52 cards a card is drawn at random find the probability that it is either a heart or a queen?

Last Answer : 41365

Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

Description : Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

Last Answer : (a) \(rac{13}{25}\)Let E : Event of drawing a queen in a single draw the pack of 52 cards. As there are 4 queens in a pack of 52 cards,P(E) = \(rac{4}{52}\) = \(rac{1}{13}\)P(\(\bar{E}\)) = P(not ... {25}\). [Sum of a G.P with infinite terms = \(rac{a}{1-r}\) where a = 1st term, r = common ratio.]

All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is face card. a. 2/23 b. 7/44 c. 3/23 d. 4/25

Description : All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is face card. a. 2/23 b. 7/44 c. 3/23 d. 4/25

Last Answer : c. 3/23

Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings ? -Maths 9th

Description : Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings ? -Maths 9th

Last Answer : Let S : Drawing 2 cards out of 52 card A : Drawing 2 red cards B : Drawing 2 kings A ∪ B : Drawing 2 red cards or 2 kings ∴ n(S) = 52C2 n(A) = 26C2 (∵ There are 26 red cards) n(B) = 4C2 ... \(rac{4 imes3}{52 imes51}\) - \(rac{2}{52 imes51}\) = \(rac{660}{2652}\) = \(rac{55}{221}.\)

Find the probability that the three cards drawn from a pack of 52 cards are all black ? -Maths 9th

Description : Find the probability that the three cards drawn from a pack of 52 cards are all black ? -Maths 9th

Last Answer : Number of ways in which three cards can be drawn from a pack of 52 cards n(S) = 52C3. Let A : Event of drawing all the three cards as black Then, n(A) = 26C3 (∵There are 26 black cards)∴ P(A ... (rac{^{26}C_3}{^{52}C_3}\) = \(rac{26 imes25 imes24}{52 imes51 imes50}\) = \(rac{2}{17}.\)

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. -Maths 9th

Description : One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. -Maths 9th

Last Answer : (c) \(rac{1}{26}\)There is a total of 52 cards n(S) = 52 Let A : Event of drawing a red king Since there are only two red kings in the pack, n(A) = 2 ∴ P(A) = \(rac{2}{52}\) = \(rac{1}{26}\).

Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. -Maths 9th

Description : Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. -Maths 9th

Last Answer : (b) \(rac{23}{26}\)Total number of ways in which 3 letters can be selected from 26 letters = 26C3. If A is not to be included in the choice, there are 25 letters left, so number of ways in which 3 letters can be ... 25}C_3}{^{26}C_3}\) = \(rac{25 imes24 imes23}{26 imes25 imes24}\) = \(rac{23}{26}\).

In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

Description : In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

Last Answer : Distribution of 52 cards can be equally divided among four players. Hence, number of ways is (13!)4! 52! ​ 4!= (13!) 52! ​

In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Description : In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Last Answer : First we divide 52 cards into two sets which contains 1 and 51 cards respectively is 1! 51! 52! Now 51 cards can be divided equally in three sets each contains 17 cards (Here order of sets is not important) in 3!(17!) ... ways Hence, the required number of ways = 1! 51! 52! 3! (17!) 3 51!

Four cards are drawn from a full pack of cards. Find the probability that : -Maths 9th

Description : Four cards are drawn from a full pack of cards. Find the probability that : -Maths 9th

Last Answer : 4 cards can be drawn from a pack of cards in 52C4 ways ∴ Exhaustive number of cases = n(S) = 52C4 (a) There are 4 suits, each containing 13 cards. Let A : Event of drawing one card from each suit ⇒ Favourable number of ... = \(rac{15229}{54145}\) (∵ P(Event) + P(complement of event) = 1)

Two cards are drawn at random from a pack of 52 cards. What is the probability that both of them are either black or queen cards? a) 55/442 b) 54/221 c) 55/221 d) 51/221

Description : Two cards are drawn at random from a pack of 52 cards. What is the probability that both of them are either black or queen cards? a) 55/442 b) 54/221 c) 55/221 d) 51/221

Last Answer : c) 55/221

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are Red or both are king? A) 52/221 B) 55/190 C) 55/221 D) 19/221

Description :  Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are Red or both are king? A) 52/221 B) 55/190 C) 55/221 D) 19/221

Last Answer : Answer: C) We have n(s) = 52C2 = 1326. Let A = event of getting both red cards B = event of getting both king A∩B = event of getting king of red cards n(A) = 26C2 = 325, n(B)= 4C2= 6 and n(A∩B) = 2C2 = 1 P(A ... S) = 1/1326 P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls, one by one without replacement. -Maths 9th

Description : Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls, one by one without replacement. -Maths 9th

Last Answer : Given, 3 white (3 W), 3 red (3 R), 4 green (4 G), 4 black (4 B) balls Total no. of balls = 3 + 3 + 4 + 4 = 14 Two balls are to be drawn, one by one without replacement. There are 4 possibilities.First BallSecond ... }{13}\) = \(rac{33+33+40+40}{14 imes13}\) = \(rac{146}{182}\) = \(rac{73}{91}.\)

A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. -Maths 9th

Description : A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. -Maths 9th

Last Answer : Let A : Event of getting at least 3 black balls Then n(A) = 5C3 x 11C1 + 5C4 (∵ Besides 5 black balls, there are 11 other balls)(3 black + others) (4 black)= \(rac{5 imes4}{2}\) x 11 + 5 = 115Total numbers of ways ... = 1820∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{115}{1820}\) = \(rac{23}{364}.\)

A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. -Maths 9th

Description : A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. -Maths 9th

Last Answer : (d) \(rac{9}{20}\)Let S be the sample space for drawing 2 cards out of 4 aces, 4 kings, 4 queens and 4 jacks i.e, 16 cards. Then n(S) = 16C2 P(Drawing at least one ace) = 1 - P(Drawing no ace) Let E : Event of ... \(rac{11}{20}\)∴ P(drawing at least one ace) = 1 - \(rac{11}{20}\) = \(rac{9}{20}\) .

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Description : A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Last Answer : Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) ... x \(rac{11}{16}\) x \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)

What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement? a) 4/664 b) 8/52 c) 4/663 d) 4/52

Description : What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement? a) 4/664 b) 8/52 c) 4/663 d) 4/52

Last Answer : Answer: C)  Probability of drawing a jack = 4/52 = 1/13 After drawing one card, the number of cards are 51. Probability of drawing a queen = 4/51. Now, the probability of drawing a jack and queen consecutively is 1/13 * 4/51 = 4/663

Three cards are drawn at random from an ordinary pack of cards. Find out the probability that they will consist of a king, aqueen and an ace?

Description : Three cards are drawn at random from an ordinary pack of cards. Find out the probability that they will consist of a king, aqueen and an ace?

Last Answer : Answer: 64/2210.

A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, -Maths 9th

Description : A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, -Maths 9th

Last Answer : Let W1 and W2 denote the events of drawing a white ball from the first and one from the second bag respectively. Let B1 and B2 denote the events of drawing black balls from the two bags in the same order. Then P ... }\) = \(rac{14}{27}.\) (By addition theorem for mutually exclusive events.

Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Description : Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Last Answer : (c) \(rac{3n^2-3n+2}{(3n-1)(3n-2)}\)In 3n consecutive natural numbers, (i) n numbers are of the form 3p (ii) n numbers are of the form 3p + 1 (iii) n numbers are of the form 3p + 2 For the ... ) We can select one number from each set.∴ Favourable number of cases = nC3 + nC3 + nC3 + (nC1 nC1 nC1)

A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Description : A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Last Answer : Answer: B) Probability of drawing 1 blue marker pen =8/16 Probability of drawing another blue marker pen = 7/15 Probability of drawing 1 red marker pen = 3/14 Probability of drawing 2 blue marker pens and 1 red marker pen = 8/16*7/15*3/14=1/20

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Last Answer : The area of the triangle is

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Last Answer : The area of the triangle is

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

Description : Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

Last Answer : Solution :- Let, O and O' be the centres of two congruent circles. As, AB is the common chord of these circles. ∴ ACB = ADB As congruent arcs subtent equal angles at the centre. ∠AOB = ∠AO'B ⇒ 1/2∠AOB = 1/2∠AO'B ⇒ ∠BPA = ∠BQA ⇒ BP = BQ (Sides opposite to equal angles)

Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ Hence proved.

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. -Maths 9th

Description : A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. -Maths 9th

Last Answer : (c) \(rac{111}{240}\)P(Drawing of two balls of different colours from one of the bags)= P(choosing the 1st bag) P(Drawing 1 green out 5 green and 1 out of 11 blue balls) + P(choosing the 2nd bag) P(Drawing 1 green out ... (rac{11}{48}\) + \(rac{7}{30}\) = \(rac{55+56}{240}\) = \(rac{111}{240}\).

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Let each side of ㎝ equilateral triangle ABC be ′a′㎝ Now, ar△OAB=21 AB OP=21 a 14=7a㎠→1 ar△OBC= BC OQ =21 a 10=5a㎠→2 ar△OAC=21 AC OR=21 a 6=3a㎠→3 ∴ar△ABC=1+2+3=7a+5a+3a=15a㎠ Also area of equilateral ... ABC=43 a2 Now, 43 a2=15a⇒a=3 15 4 3 3 =3603 =203 ㎝ Now, ar△ABC=43 (203 )2=3003 ㎠

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : NEED ANSWER

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Area of triangle =

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

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