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Theoram 9.3
  • 1 answers

Vanshika Tyagi 52 minutes ago

Answer plz
Prove that the perpendicular from the centre of a circle to a chord bisect the chord
  • 1 answers

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 = 90o (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.

Find the ratio ofvtotal surface area of a sphere and a solid hemisphere of same radius
  • 1 answers

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
Rs agarwal ,chapter 11 ,question no 2
  • 3 answers

Muskan Verma 15 hours ago

Sorry. I'm using 1st time n that's why

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

Anjali Mehta 16 hours ago

Don't be so lazy to write the answer
Show that ABCD is a parallelogram
  • 1 answers

Amrita Tiwary 4 hours ago

What is given in the question?😯
1 cubic meter is equal to how many cubic centimeters?
  • 2 answers

Amrita Tiwary 4 hours ago


Aanchal Jaiswal 17 hours ago

1 cubic= 1000000 cubic centimeters..
  • 2 answers

Amish Raj 17 hours ago


Ujjawal Soni 18 hours ago

  • 1 answers

Yogita Ingle 1 day ago

3x +2y - 7 + 6 = 12x
3x - 12x +2y - 1 = 0
-9x + 2y = 1

If a parallelogram and a triangle are on the common base and between the same parallels,then prove that the area of the triangle is equal to half the area of the parallelogram.
  • 1 answers

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)]
ABCD is a rectangle and P,Q,R and S are midpoints of side AB,BC,CD,DA respectively.Show that the quadrilateral PQRS is a rhombus.
  • 3 answers

Mohammad Ali 13 hours ago

How did u insert this picture

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.











PQ = QR = RS = SP


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

             ⇒PQRS is a rhombus.

Bhawna Gupta 1 day, 12 hours ago

Please friends answer me
Ch 10 theroem 10.8 solution
  • 2 answers

Mohammad Ali 13 hours ago

Mohammad Ali 1 day, 13 hours ago

Do u want explanation or example of theroem 10.8
On quadrilateral
  • 0 answers
Semicircular sheet of paper having diameter 35cm is bent to an open conical cup find its radius and height
  • 2 answers

Kumari Divya 1 day, 13 hours ago

R =12.25cm L=17.5cm H=12.49cm

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
R2 + h2 = l2

12.25*12.25 + h2 = 17.5*17.5

150.0625+ h2=306.25



If x^a/b , then find the value of a
  • 1 answers

Vineet Goyal 1 day, 15 hours ago


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