The effective resistance between the points A and B in the figure is

The effective resistance between the points A and B in the figure is
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The effective resistance between the points A and B in the figure is A. `5Omega` B. `2Omega` C. `3Omega` D. `4Omega`
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In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

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Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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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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The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

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The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

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In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

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In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

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Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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

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

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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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Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

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Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

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Last Answer : According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

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Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

Description : If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

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In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

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Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

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Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

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Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

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In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

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The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P, Q, R, S and T in such a way that the holes were parallel to each other, perpendicular to the edge and all the holes were still within the lens edge. What is the total number of edges in the lens after the holes were drilled?

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