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Post a LessonAnswered on 16/04/2024 Learn CBSE - Class 10/Mathematics/UNIT IV: Geometry/Circles
Sadika
To prove that the parallelogram circumscribing a circle is a rhombus, we first need to establish some properties:
Now, let's consider a parallelogram ABCD circumscribing a circle with center O. Let P, Q, R, and S be the points of tangency between the circle and the sides of the parallelogram, as shown below:
P-----Q
/ \
/ \
A-----------B
| O |
| |
D-----------C
\ /
\ /
R-----S
Answered on 16/04/2024 Learn CBSE - Class 10/Mathematics/UNIT IV: Geometry/Circles
Sadika
Let's denote the midpoint of BC as M. To prove that the tangent to the circle at P bisects BC we need to show that BM=MC.
Answered on 16/04/2024 Learn CBSE - Class 10/Mathematics/UNIT IV: Geometry/Circles
Sadika
Let's denote the center of the circle as O, and the points of contact of the two parallel tangents as P and Q. The two tangents are parallel, so OP and OQ are parallel as well.
Answered on 16/04/2024 Learn CBSE - Class 10/Mathematics/UNIT IV: Geometry/Circles
Sadika
In the given scenario, let's denote:
Answered on 16/04/2024 Learn CBSE - Class 10/Mathematics/UNIT IV: Geometry/Circles
Sadika
Let's denote:::: the above question
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