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Syeda answered 1 year(s) ago

AD is a median of triangle ABC and E is the mid-point of AD

Answer.


Given AD is the median of ΔABC and E is the midpoint of AD
Through D, draw DG || BF
In ΔADG, E is the midpoint of AD and EF || DG
By converse of midpoint theorem w

...

Asked by G

Feb 4

  • 3 Answers
  • 6 Likes
  • 68756 Views
Agam answered 1 year(s) , 7 month(s) ago

prove that bisectors of angles of parallelogram form a rectangle


Given: ABCD is a parallelogram. AE bisects ∠BAD. BF bisects ∠ABC. CG bisects ∠BCD and DH bisects ∠ADC
To prove: LKJI is a rectangle
∠BAD + ∠ABC = 180° because adjacent angles of a paralle...

Asked by Aman

Nov 21

  • 3 Answers
  • 7 Likes
  • 61619 Views
Syeda answered 1 year(s) ago

Prove that the sum of the four angles of a quadrilateral is 360 degree?


Consider a quadrilateral PQRS.

Join QS.

To prove: ∠P + ∠Q + ∠R + ∠S = 360º

Proof:

Consider triangle PQS, we have,

⇒ ∠P + ∠PQS

...

Asked by Apurbo

Apr 21

  • 2 Answers
  • 2 Likes
  • 41486 Views
Syeda answered 1 year(s) ago

ABCD is a parallelogram.X and y are mid points of BC and CD


Construction :- Join points B and D.
Since X and Y are the mid points of sides BC and CD respectively,
In ∆BCD, XY|| BD and XY = </...

Asked by Sumit

Feb 18

  • 3 Answers
  • 0 Like
  • 39780 Views
Himanshu answered 2 year(s) , 6 month(s) ago

Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.



Given AC,BD are diagonals of a quadrilateral ABCD are perpendicular.

P,Q,R and S are the mid points of AB,BC, CD and AD respectively.

Proof:
In  ΔABC, P an...

Asked by Priya

Feb 14

  • 2 Answers
  • 3 Likes
  • 38513 Views
Agam answered 1 year(s) , 7 month(s) ago

Show that the line segment joining the midpoint of the opposite side of a quadrilateral bisect each other.


In ΔADC, S and R are the midpoints of AD and DC respectively.

Recall that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and half of i...

Asked by Aryan dey

Mar 3

  • 2 Answers
  • 6 Likes
  • 36408 Views
Syeda answered 1 year(s) ago

E and F are respectively mid points of non parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2(AB + CD).



ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively.

Join CE and produce it to meet BA produced at G.

In ΔEDC and ΔEAG,
<...

Asked by Adithyan

Mar 21

  • 2 Answers
  • 0 Like
  • 35362 Views
Raghunath answered 4 year(s) , 5 month(s) ago

Show that the diagonals of a square are equal and bisect each other at right angles

Sol:

Given that ABCD is a square.

To prove : AC = BD and AC and BD bisect each other at right angles.

Proof: 

(i)  In a Δ ABC and Δ BAD,
...

Asked by Sandhu

Nov 30

  • 1 Answer
  • 5 Likes
  • 34013 Views
Kishore answered 7 year(s) , 6 month(s) ago

Line segment joining the midpoints of the diagonals of a trapezium

Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.

Draw DE and produce it to meet AB at G.

Consider DAEG and DCED

∠AEG = ∠CED (vertically

...

Asked by Rajiv singh

Dec 29

  • 1 Answer
  • 2 Likes
  • 28481 Views
Raghunath answered 4 year(s) , 4 month(s) ago

In a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD

Sol:

Given : ABCD is a quadrilateral. O is a point inside the quadrilateral ABCD.

To prove : OA + OB + OC + OD > AC + BD

...

Asked by Samreen

Jan 18

  • 1 Answer
  • 6 Likes
  • 26154 Views
Akhilesh answered 4 year(s) , 2 month(s) ago

Find the cost of white washing the walls of the room.

total area for white washing =  lateral surface area + l*b
                                                = 2*(l+b)*h + l*b
                                                = 2*(5+4)*3 + 5*4...

Asked by Adithyan

Mar 22

  • 2 Answers
  • 0 Like
  • 20589 Views
Raghunath answered 4 year(s) , 6 month(s) ago

BM and CN are perpendiculars to a line passing throughthe vertex A of a triangle ABC.If L is the mid-point of BC, prove that LM=LN.

Sol:

Given: In a ΔABCl is a straight line passing through the vertex A . BM ⊥ l and CN ⊥ l. L is the mid point of BC.

To prove: LM = LN

C

...

Asked by Shatoddruh

Dec 6

  • 1 Answer
  • 2 Likes
  • 17656 Views
Agam answered 1 year(s) , 10 month(s) ago

2 parallel lines l and m intersect by transversal p.


Given l||m and p is the transversal
To prove: PQRS is a rectangle
Proof:
RS, PS, PQ and RQ are bisectors of interior angles formed by the transversal with the parallel lines....

Asked by Hariharan

Oct 18

  • 2 Answers
  • 3 Likes
  • 16645 Views
Kishore answered 5 year(s) , 7 month(s) ago

In a parallelogram ABCD, E and F are the midpoints


Given ABCD is a parallelogram
Hence AB || CD
⇒ AE || FC
Also AB = CD (Opposite sides of parallelogram ABCD)
⇒ AE = FC (Since E and F are midpoints of AB and CD)
In...

Asked by Merlin

Nov 14

  • 1 Answer
  • 0 Like
  • 14327 Views
Raghunath answered 4 year(s) , 3 month(s) ago

ABCD is a cyclic quadrilateral whose diagonal intersects at E. if ∠BAC = 30° and ∠DBC = 70° . find∠BCD. if AB=BC find ∠ECD.

ABCD is a cyclic quadrilateral whose diagonal intersect at E.


∠CBD = ∠CAD (Angles in the same segment are equal)

∠CAD = 70°

∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°

∠BCD +

...

Asked by Garvita

Mar 10

  • 1 Answer
  • 0 Like
  • 13831 Views
Ramesh answered 4 year(s) , 8 month(s) ago

Prove that (i) AE = BE (II) ∠DAE = 15°.



Given : ABCD is a square and EDC is an equilateral triangle. AD = BC, DE = CE
To Prove : i) AE = BE
               ii) ∠DAE = 15°
Construction : Join A to E and B to E. <...

Asked by Padmini

Oct 4

  • 2 Answers
  • 2 Likes
  • 13518 Views
Agam answered 1 year(s) , 10 month(s) ago

ABCD is a trapezium in which AB is parallel to CD.


Given in trapezium ABCD, AB||CD and AD = BC
Draw perpendiculars DP and CQ on AB
Consider triangles APD and BQC
∠P = ∠Q = 90°
DP = CQ [Distance between parallel sides is sa...

Asked by Fareeda

Jan 25

  • 2 Answers
  • 1 Like
  • 13263 Views
Subhashri commented 3 year(s) , 3 month(s) ago

In the below figure, ABCD is a parallelogram and DAB = 60 . If the bisector AP and BP of angles A and B respectively meet P on CD. Prove that P is the midpoint of CD.

Since AB∥DC and Ap is a transversal,

∠PAB=∠DPA      (Alternate angles)
but,∠PAB=∠DAP      (Given)
∠DPA=∠DAP=30º
so, DP=DA   (Isosceles triangle Property)
similarl...

Asked by Soumya R

Feb 18

  • 2 Answers
  • 1 Like
  • 12502 Views
Ramesh answered 4 year(s) , 5 month(s) ago

Prove that AF=2AB.

Sol:


ABCD is a parallelogram. E is the midpoint of BC. So, BE = CE.
DE produced meets the AB produced at F.
Consider the triangles CDE and BFE.
BE = CE [Given]...

Asked by Prince

Dec 13

  • 1 Answer
  • 1 Like
  • 9641 Views
Kishore answered 5 year(s) , 3 month(s) ago

Find the angles of the parallelogram.


Given ABCD is a parallelogram in which DP⊥AB and AQ ⊥BC.
Given ∠PDQ = 60°
In quad. DPBQ, by angle sum property we have
∠PDQ + ∠DPB + ∠B + ∠BQD = 360°
60° + 90° + ∠B + 90°...

  • 1 Answer
  • 3 Likes
  • 8536 Views
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