Question:
In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is
- 2.5
- 3.5
- 0.5
- 1.5
Correct Answer: Option: 3
Refer to the diagram below:
Applying chord chord power theorem
$AE\times BE=CE\times DE$
$\Rightarrow AE\times BE=7\times 15=105$ …(1)
Also, it is given that AE+BE=20.5 ...(2)
${{\left( AE-BE \right)}^{2}}={{\left( AE+BE \right)}^{2}}-4\times AE\times BE$
$\begin{align} & \Rightarrow {{\left( AE-BE \right)}^{2}}={{\left( 20.5 \right)}^{2}}-4\times 105 \\ & \Rightarrow {{\left( AE-BE \right)}^{2}}=420.25-420=0.25 \\ & \Rightarrow \left( AE-BE \right)=0.5 \\ \end{align}$CAT Online Course
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