Find the distance between points P(8 2) and Q(-1 5) to the nearest tenth.?

Find the distance between points P(8 2) and Q(-1 5) to the nearest tenth.?
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Using the distance formula it is the square root of 61 which is about 7.8 to the nearest tenth
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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.

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.

Points P,Q,R(in this order) divide the line joining the points A(-2,2) and B(2,8) into four equal parts. The coordinates of the point Q are: (a) (-1,7/2) (b) (1,13/2) (c) (0,5) (d) (5,1/2)

Description : Points P,Q,R(in this order) divide the line joining the points A(-2,2) and B(2,8) into four equal parts. The coordinates of the point Q are: (a) (-1,7/2) (b) (1,13/2) (c) (0,5) (d) (5,1/2)

Last Answer : (c) (0,5)

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.

In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Description : In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Last Answer : Solution :-

ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

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Last Answer : Solution :-

The thickness of a laminar boundary layer over a flat plate at two different sections P and Q are 0.8 cm and 2.4 cm respectively. If the section Q is 3.6 m downstream of P, the distance of section P from the leading edge of the plate is (a) 0.32 m (b) 0.22 m (c) 0.40 m (d) 0.53 m

Description : The thickness of a laminar boundary layer over a flat plate at two different sections P and Q are 0.8 cm and 2.4 cm respectively. If the section Q is 3.6 m downstream of P, the distance of section P from the leading edge of the plate is (a) 0.32 m (b) 0.22 m (c) 0.40 m (d) 0.53 m

Last Answer : Ans: (c) Correct Answer is: 0.45 m

A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Description : A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Last Answer : ANSWER: A Explanation: Speed downstream = (28 + 8)km/hr = 36 km/hr Speed upstream = (28 – 8)km/hr = 20 km/hr Let the distance between p and q be ‘x’km Then, x/36 + (x/2)/20 = 38 x/36 +x/40 = 38 19x = 13680 X = 720 km Therefore the distance between p and q is 720km

What is the standard deviation of the following data set rounded to the nearest tenth 52.1 45.5 51 48.8 43.6?

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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.

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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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.

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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 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.

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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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Last Answer : (b) RhombusAB = \(\sqrt{(3-1)^2+(5-1)^2}\) = \(\sqrt{4+16}\) = \(\sqrt{20}\) = \(2\sqrt5\)BC = \(\sqrt{(1-5)^2+(1-3)^2}\) = \(\sqrt{16+4}\) = \(\sqrt{20}\) = \(2\sqrt5\)CD = \ ... = \(6\sqrt2\)Now, AB = BC = CD = AD ⇒ All sides are equal Also, AC ≠ BD ⇒ Diagonals are not equal. ⇒ ABCD is a rhombus.

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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

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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.

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 .

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

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 .

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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

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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).

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

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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

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.

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

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 .

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

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 .

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

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).