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# geometry helppppp

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Intern
Joined: 06 Oct 2011
Posts: 10

Kudos [?]: 6 [0], given: 2

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17 Aug 2012, 03:56
1)In the figure above, AC=6 and BC=3. Point P lies on AB between A and B such that CP is perpendicular to AB. Which of the following could be the length of CP?

A) 2
B) 4
C) 5
D) 7
E) 8

Picture Reference:http://24.media.tumblr.com/tumblr_m8wbaeMGRS1rd5jmyo1_1280.jpg

2) In the figure below, E is the midpoint of AC. AC is perpendicular to AB, and AD=DB. If BC=4 cm, what is the value of BEsquare+CDsquare?

A) 25

B) 24

C) 20

D) 16

E) none

Picture Reference: http://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg

Kudos [?]: 6 [0], given: 2

Director
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1057 [0], given: 43

WE: Science (Education)

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17 Aug 2012, 04:50
arifaisal wrote:
1)In the figure above, AC=6 and BC=3. Point P lies on AB between A and B such that CP is perpendicular to AB. Which of the following could be the length of CP?

A) 2
B) 4
C) 5
D) 7
E) 8

Picture Reference:http://24.media.tumblr.com/tumblr_m8wbaeMGRS1rd5jmyo1_1280.jpg

Triangle PBC is a right triangle (PC is perpendicular to AB), therefore CP being a leg is shorter than the hypotenuse, which is BC.
Only 2 is acceptable from the given list of answers.

2) In the figure below, E is the midpoint of AC. AC is perpendicular to AB, and AD=DB. If BC=4 cm, what is the value of BEsquare+CDsquare?

A) 25

B) 24

C) 20

D) 16

E) none

Picture Reference: http://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg

Using Pythagoras's theorem, $$BE^2+CD^2=AB^2+(\frac{AC}{2})^2+AC^2+(\frac{AB}{2})^2=\frac{5}{4}(AB^2+AC^2)=\frac{5}{4}BC^2=\frac{5}{4}*4^2=20.$$

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Kudos [?]: 1057 [0], given: 43

Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 622

Kudos [?]: 763 [0], given: 59

Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)

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17 Aug 2012, 05:55
Q1

IN Triangle CPB
CP is the base and CB is the hypotenuse. And Base will always be less than Hypotenuse.
So CP < 3

Q2

In Right Triangle BAE
BE^2 = AB^2 + AE^2
= AB^2 + (AC/2)^2 [because AE = AC/2]
BE^2 = AB^2 + (AC^2)/4 ... (1)

= AC^2 + (AB/2)^2 [as AD= DB = AB/2]
CD^2 = AC^2 + (AB^2)/4 ...(2)

(1) + (2) gives us
BE^2 + CD^2 = AB^2 + (AC^2)/4 + AC^2 + (AB^2)/4 = (5/4) (AB^2 + AC^2)
= (5/4)* (BC^2) [ As in RIght Triangle BAC AB^2 + AC^2 = BC^2]

= (5/4) * 4^2 = 20

Hope it helps!
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Kudos [?]: 763 [0], given: 59

Re: geometry helppppp   [#permalink] 17 Aug 2012, 05:55
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