Q5 of 59 Page 82

In figure 3.81, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE × GE = 4r2


In ∆DEF,


∠DFE = 90° {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}


Given: EF = diameter of the circle.


DE2 = DF2 + EF2 {Using Pythagoras theorem}


⇒ DE2 = DF2 + (2r)2


⇒ DE2 = DF2 + 4r2


⇒ DF2 = DE2- 4r2


Also, DE × DG = DF2


This property is known as tangentsecant segments theorem.


⇒DE × DG = DE2 - 4r2


⇒ DE2- DE × DG = 4r2


⇒DE(DE – DG) = 4r2


⇒ DE × EG = 4r2


Hence, proved.


More from this chapter

All 59 →
3

In figure 3.79, O is the centre of the circle and B is a point of contact. seg OE seg AD, AB = 12, AC = 8, find


(1) AD


(2) DC


(3) DE.


 4

In figure 3.80, if PQ = 6, QR = 10, PS = 8 find TS.


 1

Four alternative answers for each of the following questions are given. Choose the correct alternative.

Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centers?


 1

Four alternative answers for each of the following questions are given. Choose the correct alternative.

Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle?