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

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

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...... - (ii) Subtract equation (i) from (ii) 100p/q - p/q = 2343.434343...... 23.434343...... 99p/q = 2320.0 p/q = 2320/99 (iv) 0.6666.... = 6/9 Proceed same as question no. (iii)
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Related questions

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 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

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

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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 :-

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.

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

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

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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

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For three distinct positive numbers p, q and r, if p + q + r = a, then -Maths 9th

Description : For three distinct positive numbers p, q and r, if p + q + r = a, then -Maths 9th

Last Answer : answer:

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

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 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

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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

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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

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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

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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

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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

If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Description : If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

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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

Last Answer : answer:

In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

Description : In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

Last Answer : Let the line 2x + 3y - 30 = 0 divide the join of A(3, 4) and B(7, 8) at point C(p, q) in the ratio k : 1. Then,p = \(rac{7k+3}{k+1}\), q = \(rac{8k+4}{k+1}\)As the point C lies on the line 2x + 3y - 30 ... {3}{2}+1},rac{8 imesrac{3}{2}+4}{rac{3}{2}+1}\bigg)\) = \(\big(rac{27}{5},rac{32}{5}\big)\).

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 .

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 .

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 .

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 .

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.

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.

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

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

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Description : P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Last Answer : Given In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively. To show PRQS is a parallelogram. Proof Since, ABCD is a parallelogram. AB||CD ⇒ AP || QC

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 the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.