Q. 3 A5.0( 3 Votes )

<span lang="EN-US

Answer :

Given: A circle with center P. CB tangent and line AC intersect a circle in point D

Construction: Join BD.



To Prove: ADB =90° [Angle inscribed in semicircle]


PBC = 90° [Tangent perpendicular to the radius]


i.e. ABC =90°


In Δ ACB and Δ ABD


ABC = ADB [Each is of 90°]


CAB = DAB [Common angle]


ΔACB ΔABD [AA property]



AC × AD = (AB)2(1)


AP = PB …(radii of the same circle)


AB = AP +PB


AB = 2AP


Substituting value of AB in equation (1)


AC × AD = (2AP)2


AC × AD = 4(AP)2


AC × AD =4 (radius)2


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Solve any five suMaharashtra Board - Geometry Papers

Solve any five suMaharashtra Board - Geometry Papers

<span lang="EN-USMaharashtra Board - Geometry Papers

<span lang="EN-USMaharashtra Board - Geometry Papers

Solve any four suMaharashtra Board - Geometry Papers

<span lang="EN-USMaharashtra Board - Geometry Papers

<span lang="EN-USMaharashtra Board - Geometry Papers

<span lang="EN-USMaharashtra Board - Geometry Papers

Four alternative MHB - Mathematics Part-2

Four alternative MHB - Mathematics Part-2