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  • 1 answers

Yogita Ingle 4 years, 10 months ago

Statement : There is one and only one circle passing through three given noncollinear points.

Given : AB and CD are two equal chords of the circle.

OM and ON are perpendiculars from the centre at the chords AB and CD.

To prove : OM = ON.
Construction : Join OA and OC.

Proof :

In ΔAOM and ΔCON,

OA = OC . (radii of the same circle)
MA = CN . (since OM and ON are perpendicular to the chords and it bisects the chord and AM = MB, CN = ND)

∠OMA = ∠ONC = 90°
ΔAOM ≅ ΔCON (R. H. S)
OM = ON (c. p. c. t.)

Equal chords of a circle are equidistant from the centre.

  • 2 answers

Abu Hamza 3 years, 8 months ago

7246.8=7000.00+200.00+40.00+ 6.00+.80. OR 7246.8=7000+200+40+6+8/10.

Yogita Ingle 4 years, 10 months ago

7246.8 = 7000 + 200 + 40 + 6 + 8/10

= 7 × 103 + 2 ×  102 + 4 ×  101 + 6 ×  100 + 8 ×  10-1

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Mr Indian Kashyap 4 years, 10 months ago

X cccccccccxxxx
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Swapna Tripathydash 4 years, 8 months ago

When u construct a quadrilateral atleast 5 measurement u will need. You given the 4 measurements.
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  • 5 answers

Swapna Tripathydash 4 years, 8 months ago

2(l+b)

Parveen Ali 4 years, 11 months ago

Question 1

Anand Soni 4 years, 11 months ago

Mo

Shivani Sharma 5 years ago

2( lenght + breadth )

Yogita Ingle 5 years ago

The perimeter of a rectangle is defined as the sum of all the sides of a rectangle.

From the definition of the perimeter we know, the perimeter of a rectangle, p = 2 ( l+b) units

where

“l” is the length of the rectangle

“b” is the breadth of the rectangle

  • 3 answers

Sourav Kumar 4 years, 10 months ago

A+b=11 (A+b)^2=11^2 A^2+b^2+2ab=121 2ab=121-61 Ab=60÷2 Ab=30

Sourav Kumar 4 years, 10 months ago

30

《Shivam 》 5 years ago

39
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Shubham Mishra 4 years, 3 months ago

|5|+|6|=11
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Swapna Tripathydash 4 years, 8 months ago

= 3{9x^2-25y^2}. = 3{(3x)^2-(5y)^2} = 3{(3x+5y)(3x-5y)}. Using the identity of a^2-b^2=(a+b)(a-b) = 3(3x+5y)(3x-5y) (ANS)
Hbb
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NAks
X+2
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X+2 is the answer because there is no same variables are there.
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