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15 Nov 2016, 01:01
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Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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15 Nov 2016, 01:03
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You sure this has not been discussed? Seems like it would have been. Please use search feature and continue the discussion there.
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15 Nov 2016, 06:22
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Najaf1991
Hi,
The solution given is wrong and maybe if other answers are given in similar manner, you may think to dump this book..
As ∆ABP cannot be right angle as ∆ABD is right angled at A...
Having said this, the answer is ofcourse correct...
I'll give you two methods..
1) simple and estimate..
The top line is 3 and lower AP is 4, so BP should be> than CP...
BP is \(\sqrt{81}\), so D and E are out..
A \(\sqrt{49}\) is too less as compared to 9, so can be ruled out..
You are left with B and C... Now it is slightly tough but C stands a better chance as you should realise that CP should almost equal to BP..
2) Correct method
Draw a perpendicular from P on AD at E and on BC at F...
Now take ∆s APE and DPE...
Check connected image...
Hope it helps
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
