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by arifaisal » Fri Aug 17, 2012 2:52 am
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:https://24.media.tumblr.com/tumblr_m8wba ... 1_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: https://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg
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by rijul007 » Fri Aug 17, 2012 3:00 am
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

Image
CP is perpendicular on AB.
This means CPB is a Right Triangle.
Here BC is a hypotenuse.

As you know, hypotenuse is the largest side in a right triangle, length of CP is less than length of BC.
or
CP < 3

Option A is the only choice that satisfies it and hence is the correct answer choice.
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by rijul007 » Fri Aug 17, 2012 3:11 am
arifaisal wrote: 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

Image
AE = EC = y
AD = DB = x
BC = 4

(BC)² = (AC)² + (AB)² = (2x)² + (2y)² = 4x² + 4y² = 16
4x² + 4y² = 16
x² + y² = 4

(BE)² = (AE)² + (AB)² = y² + (2x)² = 4x² + y²
(CD)² = (AD)² + (AC)² = x² + (2y)² = x² + 4y²

(BE)² + (CD)² = 4x² + y² + x² + 4y² = 5x² + 5y² = 5(x² + y²) = 5*4 = 20

Option C
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by GMATGuruNY » Fri Aug 17, 2012 6:02 am
arifaisal wrote: 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
Image

The correct answer must work for ANY triangle that satisfies the given conditions.

Let ∆ABC be a 45-45-90 triangle.
In a 45-45-90 triangle, the sides are proportioned x : x : x√2.
Since BC=4, AB=AC=4/√2.
Since D is the midpoint of AB, AD = 2/√2.

Since ∆ACD is a right triangle, AD² + AC² = CD².
Thus:
(2/√2)² + (4/√2)² = CD²
2+8 = CD².
CD² = 10.

Since AB=AC and D and E are both midpoints, BE=CD and BE² = CD².
Thus, BE² + CD² = 10+10 = 20.

The correct answer is C.
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