The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th
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Value of 1.999 is given below .
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The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Description : The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Last Answer : Value of 1.999 is given below .

Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Last Answer : Let x = 0.7Bar ⇒ x = 0.777.......... .........(1) Multiplying (1) by 10 ⇒ 10x = 7.7...... = 7.777 ...........(2) Subtracting (1) from (2) ⇒ 10x - x = 9x ⇒ 7.777 - 0.777 = 7 ... and then solving it, we get. ⇒ (594 + 770 + 470)/990 ⇒ 1834/990 ⇒ 917/495 Hence, 0.6 + 0.7Bar + 0.47Bar = 917/495

Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Last Answer : Let x=0.666....... (1) Multiply equation (1 by 10 10x = 6.666....... (2) Subtract equation (1) from (2) x=6/9 Similarly 0.7bar =7/9 and 0.47bar = 47/99. 6/9+7/9+47/99=190/99

Express 0.6 in the form p/q, where p and q are integers and q is not equals to 0. -Maths 9th

Description : Express 0.6 in the form p/q, where p and q are integers and q is not equals to 0. -Maths 9th

Last Answer : Let x = 0.6 recurring Then, x = 0.666..... ....(i) implies 10x = 6.666........ .....(ii) Substracting (i) from (ii),we get 9x = 6 implies x = 6/9 implies x = 2/3

Express 0.00323232... in the form p/q, where p and q are integers and q is not equals 0. -Maths 9th

Description : Express 0.00323232... in the form p/q, where p and q are integers and q is not equals 0. -Maths 9th

Last Answer : Solution :-

Express 0.35777... in the form p/q,where p and q are integers and q is not equals to 0. -Maths 9th

Description : Express 0.35777... in the form p/q,where p and q are integers and q is not equals to 0. -Maths 9th

Last Answer : Solution :-

Express:2.0151515... in the p/q form, where p and q are integers and q is not equals to 0. -Maths 9th

Description : Express:2.0151515... in the p/q form, where p and q are integers and q is not equals to 0. -Maths 9th

Last Answer : Solution :-

When the integers from 1 to 999 are written on a paper, the number of zeros used in writing it is: a) 192 b) 189 c) 200 d) 203 e) 216

Description : When the integers from 1 to 999 are written on a paper, the number of zeros used in writing it is: a) 192 b) 189 c) 200 d) 203 e) 216

Last Answer : From 1 to 10 : 1 zero From 11-100 : 10 zeros (100 has 2 zeros) From 101 – 900 : 8×20 = 160 zeros From 901-999 : we have 18 zeros Total = 1+10+160+18 = 189 zeros Answer: b)

If the roots of the equation x^3 – ax^2 + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b ? -Maths 9th

Description : If the roots of the equation x^3 – ax^2 + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b ? -Maths 9th

Last Answer : Let the roots of the equation x3 – ax2 + bx – c = 0 be (α – 1), α, (α + 1) ∴ S2 = (α – 1)α + α(α + 1) + (α + 1) ( ... ; 1 = b ⇒ 3α2 – 1 = b ∴ Minimum value of b = – 1, when α = 0.

Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

Let p and q be the roots of the quadratic equation x^2 – (a – 2)x – a – 1 = 0. What is the minimum possible value of p^2 + q^2 ? -Maths 9th

Description : Let p and q be the roots of the quadratic equation x^2 – (a – 2)x – a – 1 = 0. What is the minimum possible value of p^2 + q^2 ? -Maths 9th

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If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

If pqr = 1, the value of (1)/([1+p+q^-1]) + (1)/([1+q+r^-1])+ (1)/([1+r+p^-1]) will be equal to : -Maths 9th

Description : If pqr = 1, the value of (1)/([1+p+q^-1]) + (1)/([1+q+r^-1])+ (1)/([1+r+p^-1]) will be equal to : -Maths 9th

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How many integers are there between two successive interesting? -Maths 9th

Description : How many integers are there between two successive interesting? -Maths 9th

Last Answer : No integers are there between two successive integers

How many integers are there between two successive interesting? -Maths 9th

Description : How many integers are there between two successive interesting? -Maths 9th

Last Answer : This answer was deleted by our moderators...

I is the set of integers. Describe the following relations in words, giving its domain and range. -Maths 9th

Description : I is the set of integers. Describe the following relations in words, giving its domain and range. -Maths 9th

Last Answer : R = {(0, 0), (1, – 1), (2, – 2), (3, – 3) ...} = {(x, y) : y = – x, x ∈ W} Domain = {0, 1, 2, 3, ....} = W, Range = {...,– 3, – 2, – 1, 0}

Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Description : Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Last Answer : (c) equivalence relationGiven, a R b = a = 2k .b for some integer. Reflexive: a R a ⇒ a = 20.a for k = 0 (an integer). True Symmetric: a R b ⇒ a = 2k b ⇒ b = 2-k . a ⇒ b R a as k, -k are both ... = 2k1 + k2 c, k1 + k2 is an integer. ∴ a R b, b R c ⇒ a R c True ∴ R is an equivalence relation.

An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

If a, b, c are distinct positive integers, then ax^(b–c) + bx^(c–a) + cx^(a–b) is -Maths 9th

Description : If a, b, c are distinct positive integers, then ax^(b–c) + bx^(c–a) + cx^(a–b) is -Maths 9th

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Let ABCD be a parallellogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn -Maths 9th

Description : Let ABCD be a parallellogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn -Maths 9th

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If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Description : If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Last Answer : (d) 226 × 52C26 | 104C26Since there are 52 distinct cards in a deck and each distinct card is 2 in number.∴2 decks will also contain only 52 distinct cards, two each.∴ Probability that the player gets all distinct cards = \(rac{^{52}C_{26} imes2^{26}}{^{104}C_{26}}\).

Express 1.27 bar in p/q form. -Maths 9th

Description : Express 1.27 bar in p/q form. -Maths 9th

Last Answer : (1) x = 1.2727272727......... (2) 100x = 127.27272727......... From subtracting (1) and (2 ) We get : 99x = 126.0000000 x = 126/99 = 14/11.

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

Express 1.27 bar in p/q form. -Maths 9th

Description : Express 1.27 bar in p/q form. -Maths 9th

Last Answer : (1) x = 1.2727272727......... (2) 100x = 127.27272727......... From subtracting (1) and (2 ) We get : 99x = 126.0000000 x = 126/99 = 14/11.

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Description : Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Last Answer : (i) 0.15 = 15/100 = 3/20 (ii) 0.00026 = 26/100000 = 13/50000 (iii) 23.43434343.... Let p/q = 23.434343...... - (i) Multiply both side by 100:- 100 * p/q = 100 * 23.434343...... 100p/q = 2343. ... .434343...... 99p/q = 2320.0 p/q = 2320/99 (iv) 0.6666.... = 6/9 Proceed same as question no. (iii)

Express 1.23(bar 23) in the form of p/q -Maths 9th

Description : Express 1.23(bar 23) in the form of p/q -Maths 9th

Last Answer : Let X = 1.23(bar on 23) 100X = 123.23(bar on 23) 100X-X = 123.23+1.23(bar on 23) 100X-99x = 122(bar on 23) 99X = 122 x=122/99

Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Description : Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Last Answer : (i) 0.15 = 15/100 = 3/20 (ii) 0.00026 = 26/100000 = 13/50000 (iii) 23.43434343.... Let p/q = 23.434343...... - (i) Multiply both side by 100:- 100 * p/q = 100 * 23.434343...... 100p/q = 2343. ... .434343...... 99p/q = 2320.0 p/q = 2320/99 (iv) 0.6666.... = 6/9 Proceed same as question no. (iii)

Express 1.23(bar 23) in the form of p/q -Maths 9th

Description : Express 1.23(bar 23) in the form of p/q -Maths 9th

Last Answer : Let X = 1.23(bar on 23) 100X = 123.23(bar on 23) 100X-X = 123.23+1.23(bar on 23) 99X = 122(bar on 23) X = 122/99

0.0037 bar on 37 in the form of p/q -Maths 9th

Description : 0.0037 bar on 37 in the form of p/q -Maths 9th

Last Answer : Let 0.0037 be a 0.0037x10000=a x 10000 37.37=10000a 37.37x100=10000a x100 3737.37=1000000a 3737.37-37.37=1000000a-10000a 3700=990000a 3700/990000=a 37/9900=a

Express the following number in the form of p\q. (a) -25. 6875 -Maths 9th

Description : Express the following number in the form of p\q. (a) -25. 6875 -Maths 9th

Last Answer : This answer was deleted by our moderators...

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

SIMplify (2x+p-q)raise to power 2 -(2x-p+q)raise to power2 -Maths 9th

Description : SIMplify (2x+p-q)raise to power 2 -(2x-p+q)raise to power2 -Maths 9th

Last Answer : 2p-2q+2 is answer of the question

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.