Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts Dedicated counsellor for each student Detailed Performance Evaluation view all coursesPage 2Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts Dedicated counsellor for each student Detailed Performance Evaluation Text Solution 10.5 cm9.5 cm8.5 cm7.5 cm Answer : C Solution : Join AC . Then,<br> `AE: CE = DE : BE` <br> `:. AE xx BE = CE xxDE` <br> Let CD `= x cm `. Then, <br> `AE = (AB+ BE) = ( 11+3) cm = 14 cm, BE = 3 cm, CE = ( x + 3.5 ) cm ` and `DE = 3.5 cm` <br> `:. 14 xx 3 = ( x+ .3.5) implies x + 3.5 = (14 xx 3)/( 3.5) = 12` <br> `implies x = (12 - 3.5 ) = 8.5 cm` <br> `:. CD = 8.5 cm` Two chords AB and CD of a circle intersect each other at a point E outside the circle. Question: Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD = ?
Solution: (c) 8.5 cm Then AE : CE = DE : BE (Intersecting secant theorem) $\Rightarrow x+3.5=\frac{14 \times 3}{3.5}=\frac{42}{3.5}=12$ ⇒ x = (12 – 3.5) cm = 8.5 cm Open in App Suggest Corrections |