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Posted by Rahul R 15 minutes ago

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ANSWER

Posted by Jaishree Sanodiya an hour ago

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Posted by Avi Tuteja an hour ago

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Posted by Om Agrawal an hour ago

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Posted by Tisha Sharma 2 hours ago

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Gaurav Seth 2 hours ago

**Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.**

Given: A circle with centre O. AC is a chord and OB ⊥ AC.

To prove: AB = BC.

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

∠OBA = ∠OBC = 90^{o} (Since OB ⊥ AC)

OA = OC (Radii of the same circle)

OB = OB (Common side)

ΔOBA ≅ ΔOBC (By RHS congruence rule)

⇒ AB = BC (Corresponding sides of congruent triangles)

Thus, OB bisects the chord AC.

Hence, the theorem is proved.

Posted by Lakshy Choudary 2 hours ago

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Posted by Baona Paan 13 hours ago

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Posted by Shashank Jaswai 15 hours ago

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Posted by Rajeev Yadav 15 hours ago

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Saloni Kumar 15 hours ago

T. S. A Of sphere With radius r is 4*22/7r*r
T. S. A Of hemisphere with radius r is 3*22/7r*r
Ratio should be 4:3

Posted by Muskan Verma 16 hours ago

- 3 answers

Anjali Mehta 15 hours ago

Opp. Side of a 11gm are equal .
Ab=cd given
In ∆abd and bdc
Bd=db,ang.abd=ang.bdc.
∆abd=~bdc
So,ad=bc=7cm
We can say that abcd is a llgm

Posted by Muskan Verma 16 hours ago

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Posted by Swarnadeep Saha 17 hours ago

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Posted by Debapratim Kundu 17 hours ago

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Posted by Sandeep Pal 17 hours ago

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Posted by Sanika Deshmukh 18 hours ago

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Posted by Jangya Seni Nayak 20 hours ago

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Posted by Ankit Kumar 1 day ago

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Posted by Tanmay Ku Behera 1 day, 1 hour ago

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Posted by Rahul Kumar 1 day, 11 hours ago

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Jayavel Arumugam 23 hours ago

Let ΔABP and a parallelogram ABCD be on the same base AB and between the same parallels AB and PC.
To Prove : ar( ΔPAB ) = (1/2)ar( ABCD)
Draw BQ ||AP to obtain another parallelogram.ABQP and ABCD are on the same base AB and between the same parallels AB and PC.
There fore, ar(ABQP) = ar(ABCD)
But ΔPAB ≅ ΔBQP( Diagonals PB divides parallelogram ABQP into two congruent triangles. So ar (PAB) = ar(BQP) -----------(2) ∴ ar (PAB) = (1/2)ar(ABQP) -----------------(3) [ from (2)]
This gives ar (PAB) = (1/2)ar(ABCD) [ from (1) and (3)]

Posted by Bhawna Gupta 1 day, 12 hours ago

- 3 answers

Gaurav Seth 1 day, 12 hours ago

ABCD is a rectangle 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 rhombus.

Sol: In a rectangle ABCD, P is the mid-point of AB, Q is the mid-point of BC, R is the mid-point of CD, S is the mid-point of DA AC is the diagonal.

From (1) and (2), we get

PQ = SR and PQ || SR

Similarly, by joining BD, we have

PS = QR and PS || QR

i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.

∴ PQRS is a parallelogram.

Now, in ΔPAS and ΔPBQ,

∴Their corresponding parts are equal.

⇒

PS = PQ

Also

PS = QR

[Proved]

and

PQ = SR

[Proved]

PQ = QR = RS = SP

i.e. PQRS is a parallelogram having all of its sides equal.

⇒PQRS is a rhombus.

Posted by Vishakha Sharma 1 day, 14 hours ago

- 2 answers

Posted by Manav Ramchandani 1 day, 15 hours ago

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Posted by Lavania Lavania 1 day, 15 hours ago

- 2 answers

Ram Kushwah 1 day, 14 hours ago

diameter =d = 35 cm

Radius= r = 17.5 cm

Circumference of semi circle = πr

= (22/7) x 17.5

= 77 cm

so the circumference of base of cone

2πR = 77

R = 77 x (7/44)

R = 12.25 cm

The radius of semi circular sheet = slant height of conical cup

which is

l = 17.5 cm

As,

R^{2} + h^{2 }= l^{2}

12.25*12.25 + h^{2} = 17.5*17.5

150.0625+ h^{2}=306.25

h^{2}=156..1875

h=12.49

Posted by Vineet Goyal 1 day, 15 hours ago

- 0 answers

Posted by Hurriya Usmani 1 day, 15 hours ago

- 1 answers

Posted by Abhimanyu S 1 day, 15 hours ago

- 1 answers

Soumya Mishra 1 day, 15 hours ago

Just divide both the measurement...you'll get the answer

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