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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...

Oct 30

• 4 Likes
• 68626 Views

Oct 30

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

...

Mar 3

• 3 Likes
• 37862 Views

Mar 3

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

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.

...

Jan 4

• 3 Likes
• 36603 Views

Jan 4

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...

Oct 24

• 4 Likes
• 33307 Views

Oct 24

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

In ΔAED and ΔBFC,
DE = CF    (Distance between parallel sides is same)
∠AED = ∠BFC = 90°
ΔAED ≅ ΔBFC  (RHS Congruence criterion)
Hence ∠DAE = ∠...

Mar 8

• 2 Likes
• 33098 Views

Mar 8

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...

Mar 5

• 3 Likes
• 25358 Views

Mar 5

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

...

Feb 18

• 0 Like
• 25126 Views

Feb 18

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

...

Mar 2

• 2 Likes
• 22865 Views

Mar 2

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

...

Feb 24

• 2 Likes
• 21591 Views

Feb 24

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...

Mar 21

• 0 Like
• 21042 Views

Mar 21

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:

To Prove: AB parallel to CD
Construction: Join BD
...

Feb 22

• 2 Likes
• 19162 Views

Feb 22

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

Dec 17

• 6 Likes
• 19092 Views

Dec 17

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

ABC is an isosceles triangle inscribed in a circle

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...

Mar 9

• 0 Like
• 15706 Views

Mar 9

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

Mar 5

• 0 Like
• 15388 Views

Mar 5

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...

Mar 3

• 3 Likes
• 13457 Views

Mar 3

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
...

Nov 10

• 0 Like
• 12212 Views

Nov 10

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...

Jan 13

• 4 Likes
• 9795 Views

Jan 13

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...

Mar 12

• 1 Like
• 9590 Views

Mar 12

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°...

Dec 25

• 1 Like
• 9388 Views

Dec 25

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...

Jan 5

• 0 Like
• 9184 Views

Jan 5

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