In the given figure, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF then the radius fo the circle is
(a) 3 cm (b) 4 cm
(c) 5 cm (d) 6 cm
In the given figure, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF then the radius fo the circle is
(a) 3 cm (b) 4 cm
(c) 5 cm (d) 6 cm
Given : DE and DF are perpendicular to each other. DE= 5 cm. DE and DF are tangents to the circle with centre A.
Now since DE and DF are tangents to the circles
DE⊥ AE and DF ⊥ AF. ( The tangent at any point of a circle is perpendicular to the radius through the point of contact )
All the four angles of the quadrilateral AEFD are of 90°.Therefore AEFD is a rectangle.
And since opposite sides of arectangle are equal.
∴ AF = ED = 5cm = radius
Hence radius of the circle is 5 cm.
One thing more can be proved from this...........
Also since AF= AE= 5cm. and opposite side of AE is FD
Therefore AE=FD= 5 cm.
Hence all the sides of the rectangle AEFD are equal. Therefore it is proved that AEDF is a square
Given : DE and DF are perpendicular to each other. DE= 5 cm. DE and DF are tangents to the circle with centre A.
Now since DE and DF are tangents to the circles
DE⊥ AE and DF ⊥ AF. ( The tangent at any point of a circle is perpendicular to the radius through the point of contact )
All the four angles of the quadrilateral AEFD are of 90°.Therefore AEFD is a rectangle.
And since opposite sides of arectangle are equal.
∴ AF = ED = 5cm = radius
Hence radius of the circle is 5 cm.
One thing more can be proved from this...........
Also since AF= AE= 5cm. and opposite side of AE is FD
Therefore AE=FD= 5 cm.
Hence all the sides of the rectangle AEFD are equal. Therefore it is proved that AEDF is a square
