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Kishore
answered
5 year(s)
, 7 month(s) ago

The height of a cone is 30 cm. A small cone is cut off

Given height of the cone = 30 cm

Let the small cone is cut off at a height ‘h’ from the top

Let the radius of big cone be r
_{1} and small cone be r
_{2}.

Volume...

- 1 Answer
- 4 Likes
- 68626 Views

Raghunath
answered
4 year(s)
, 3 month(s) ago

In a circle of radius 5 cm , AB and AC are two chords such that AB = BC = 6 cm. Find the length of chord BC.

** Given** AB and AC are two equal words of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

⇒ OA is the bisector of ∠BAC.

Again, the internal bi

...- 2 Answers
- 3 Likes
- 37862 Views

Syeda
answered
1 year(s)
ago

Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic?

Answer.

Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH

and DF of angles A, B, C and D respectively and the points E, F, G

and H form a quadrilateral EFGH.

- 2 Answers
- 3 Likes
- 36603 Views

Agam
answered
1 year(s)
, 1 month(s) ago

AB and CD are two chords of a circle, such that AB=6 cm, CD =12 cm

Given chords AB=6 cm, CD =12 cm and AB||CD

Draw OP⊥ AB. Let it intersect CD at Q and AB at P

∴ AP = PB = 3 cm and CQ = DQ = 6 cm [Since perpendicular draw from the centre of the chord bis...

- 2 Answers
- 4 Likes
- 33307 Views

Syeda
answered
1 year(s)
ago

prove that if non parallel sides of a trapezium are equal than prove that it is cyclic

In ΔAED and ΔBFC,

AD = BC (Given)

DE = CF (Distance between parallel sides is same)

∠AED = ∠BFC = 90°

ΔAED ≅ ΔBFC (RHS Congruence criterion)

Hence ∠DAE = ∠...

- 2 Answers
- 2 Likes
- 33098 Views

Kishore
answered
5 year(s)
, 3 month(s) ago

A chord of a circle is equal to the radius of the circle

In ΔOAB,

AB = OA = OB (radii)

Hence ΔOAB is an equilateral triangle

That is each angle of ΔOAB is 60°

∴ ∠AOB = 60°

∠AOB = 2 ∠ACB [Angle subtended by an arc of a ci...

- 2 Answers
- 3 Likes
- 25358 Views

Raghunath
answered
4 year(s)
, 3 month(s) ago

ABC is a right angled triangle, right angled at A. a circle is inscribed in it. the length of two sodes containing angle A is 12 cm and 5 cm. find the radius.

Consider ABC be the right angled triangle such that ∠A = 90° and AB = 5cm, AC = 12 cm.

And O be the centre and r be the radius of the incircle.

AB, BC and CA are tangents to the cir

...- 2 Answers
- 0 Like
- 25126 Views

Raghunath
answered
4 year(s)
, 3 month(s) ago

Prove that there is one and only one circle passing trough three given non-collinear points.

**Theorem**: There is one and only one circle passing through three given non-collinear points.

**Given**: Three non collinear points P, Q and R

**To **

- 1 Answer
- 2 Likes
- 22865 Views

Syeda
answered
1 year(s)
ago

circle drawn on any one of the equal sides of an isosceles triangle

**Given**: ΔABC is an isosceles triangle with AB = AC. A circle is drawn taking AB as the diameter which intersects the side BC at D.

**To prove:** BD = DC

**Construction**: Join AD<

Asked by Pavan kumar shukla

Feb 24

- 4 Answers
- 2 Likes
- 21591 Views

Raghunath
answered
4 year(s)
, 2 month(s) ago

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres

Let OO' intersect AB at M

Now Draw l...

- 1 Answer
- 0 Like
- 21042 Views

Ramesh
answered
5 year(s)
, 1 month(s) ago

if a pair of opposite sides of a cyclic quadrilateral are equal, then prove that the other two sides are parallel

Sol:

**Given:** Cyclic quadrilateral ABCD such that AD = BC

**To Prove:** AB parallel to CD

**Construction:** Join BD

...

- 1 Answer
- 2 Likes
- 19162 Views

Ramesh
answered
5 year(s)
, 2 month(s) ago

PROVE THAT THE SUM OF THE OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL ARE SUPPLEMENTARY??????

Sol :

Given : Let ABCD is cyclic quadrilateral.

To prove : ∠A + ∠C = 180° and ∠B + ∠D = 180°.

Construction : join OB and OD.

Proof : ∠BOD = 2 ∠BAD

∠BAD = 1...

- 1 Answer
- 6 Likes
- 19092 Views

Bhawna
answered
4 year(s)
, 2 month(s) ago

ABC is an isosceles triangle inscribed in a circle

Draw AD⊥BC

Hence D is the midpoint of BC

That is BD = DC = (1/2) BC

Therefore, BD = DC = 7 cm

In right triangle ADB, by Pythagoras theorem we have

AB
^{2}...

- 2 Answers
- 0 Like
- 15706 Views

Agam
answered
1 year(s)
, 10 month(s) ago

Prove that ,in a cyclic trapezium,the non-parallel sides are equal

We know that the angle at the circumference is 1/2 of the centre angle .similarly ab is parallel to dc

- 5 Answers
- 0 Like
- 15388 Views

Raghunath
answered
4 year(s)
, 3 month(s) ago

Prove that the bisectors of angles formed by the producing opposite sides of a cyclic quadrilateral (provided that they are not ll) intersect arch other at right angle.

A cyclic quadrilateral ABCD whose opposite sides when produced intersect at the [points P and Q]. The bisectors PF and QF of ∠P and ∠Q respectively at F

produce QF to meet AB at E

In ΔQDG...

- 2 Answers
- 3 Likes
- 13457 Views

Kishore
answered
5 year(s)
, 7 month(s) ago

Equal chords of a circle are equidistant from the centre or centres

Given a circle with centre O and chords AB = CD

Draw OP⊥ AB and OQ ⊥ CD

Hence AP = BP = (1/2)AB and CQ = QD = (1/2)CD

Also ∠OPA = 90° and ∠OQC = 90°

Since AB = CD

...

- 1 Answer
- 0 Like
- 12212 Views

Ramesh
answered
4 year(s)
, 4 month(s) ago

Find angle BOC and BDC.

Sol:

∠ABC = 50° ⇒ ∠ACB = 50° [Isosceles triangle property]

∠BAC = 180° - (50° + 50°) = 80°

∠BOC = 2 x 80° = 160° [Angle made by an arc at the centre of a circle is twice...

- 1 Answer
- 4 Likes
- 9795 Views

Shivam
answered
4 year(s)
, 2 month(s) ago

O is the centre of circle, angle bco=30º

We know that angle subtended by an arc of a circle at the centre is double the angle subtended by it on remaining part of the circle.

Arc CD subtends ∠COD at centre and subtends ∠BCD at B on the circle...

- 3 Answers
- 1 Like
- 9590 Views

G
commented
5 year(s)
, 5 month(s) ago

Prove that the circle drawn with any side of a rhombus as diameter

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O.

Recall that, diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°...

- 1 Answer
- 1 Like
- 9388 Views

Jay
commented
5 year(s)
, 5 month(s) ago

Two tangents segment PA and pB are drawn to a circle

here, in figure,

angle APB= 120°

so,angle APO= 60 °

Here, triangle APO is right angle triangle.

By the formulas of trigonometric ratios:-

AP/ OP = cos 60 °=1/2...

- 3 Answers
- 0 Like
- 9184 Views

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