Circle NCERT Exercise 10.2 solution part five Class Ten Mathematics

Circle

Exercise 10.2 Part 5

Question: 10 - Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Answer: Construction: Draw a circle with centre O. Tangent PA and PB are drawn to circle.


10 circle exercise solution

To Prove: `∠APB + ∠AOB = 180°`

`∠OAP = ∠OBP = 90°`

In quadrilateral OAPB

`∠AOB + ∠APB + ∠OAP + ∠OBP = 360°` (sum of all angles of a quadrilateral)

Or, `∠AOB + ∠APB + 90° + 90° = 360°`

Or, `∠AOB + ∠APB = 360° - 180° = 180°` proved


Question: 11 - Prove that the parallelogram circumscribing a circle is a rhombus.

Answer: Construction: Draw a circle with centre O. Draw a parallelogram ABCD which touches the circle at P, Q, R and S.

10 circle exercise solution

Given; AB || DC

AD || BC

To Prove: ABCD is a rhombus


In ΔAOB and ΔDOC;

`AB = DC` (given that AD || BC)

`∠AOB = ∠DOC` (Vertically opposite angles)

`∠BAO = ∠DCO` (Alternate angles)

Hence; `ΔAOB ≅ ΔDOC`

Hence; `AO = CO` and `BO = DO`

Since diagonals are bisecting each other, so given parallelogram is a rhombus.



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