If the shaded area is one half the area of the triangle ABC

Question

d1.JPG If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is: A. \frac{1}{2}*w B. \frac{1}{2}*(w+x) C. \sqrt{2x^2+z^2} D. \sqrt{w^2-3y^2} E. \sqrt{y^2+z^2}

Bunuel · Accepted Answer

Property of median:  Each median divides the triangle into two smaller triangles which have the same area.

So, AD is median and thus  y=z , (we could derive this even not knowing the above property. Given:  area_{ABC}=\frac{x(y+z)}{2}=2*area_{ABD}=2*\frac{xy}{2}  -->  y=z ).

Next: AD is hypotenuse in right triangle ABD and thus  AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+(2y)^2-3y^2}=\sqrt{w^2-3y^2}  (as  2y=y+z=BC  and  AC^2=w^2=AB^2+BC^2=x^2+(2y)^2 , so  w^2=x^2+(2y)^2 ).

Answer: D.

Nusa84 · Answer

Quite a tough one, very good explanation, thanks

azule45 · Answer

I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?

azule45 · Answer

i'm sorry, i should have been more specific. i wasn't talking about substituting  4y^2-3y^2  for  y^2 , but rather where did  4y^2-3y^2  come from? i can't seem to figure this out. everything else makes perfect sense, but this. 
much appreciated and sorry if this should be obvious.

Bunuel · Answer

This is just an algebraic operation in order to pair  x^2  and  4y^2 , (added they give  w^2 ).

Question is  AD=?  Well it's simple to get that  AD=\sqrt{x^2+y^2} , but this option is not listed and thus we need to transform it to get the option which is listed (or the other way around we need to transform the options to get this one).

Now, options A and B does not make any sense and we can get rid of them.

Option C:  \sqrt{2x^2+z^2}=\sqrt{2x^2+y^2}  as  y=z  hence it's out too as  \sqrt{2x^2+y^2}
eq{\sqrt{x^2+y^2}} .

Option D:  \sqrt{w^2-3y^2} ,  w  is hypotenuse hence it's equal to  w^2=x^2+(y+z)^2=x^2+(2y)^2=x^2+4y^2 , so  \sqrt{w^2-3y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+y^2}=AD .

Hope it's clear.

azule45 · Answer

ahhh haa, Bunuel. so you worked the problem using the answer choices. got it. smart move. much appreciated for your help.

Bunuel · Answer

Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!

pikwik · Answer

Wouldnt it be easier to assume the triangle to a 3,4,5 triple? AD would thus be sqrt(13) (z=y=2) just plug in the values starting option C. Takes less than a minute. (better than writing the lengthy equations)

Am I missing something here?

Am I missing something here?

jlgdr · Answer

Yes, that is in fact preferred approach See below: OK, so let's say that triangle ABC is 3,4,5. Therefore X=3, W=5 and z+y = 4. Now, smaller triangl to have 1/2 area of larger, then Y=2 and X=2 as well. Now replacing this value in answer choices we can see that only D fits the bill. Therefore OA is D. Hope this helps Peace J

mvictor · Answer

I tried 3-4-5 and 6-8-10 still didn't get to the answer:

y=z=1.5
or y=z=3
y^2/z^2 = 2.25 or y^2/z^2 = 9

x=4 or x=8
x^2 = 16
x^2 = 64

AD^2 = 2.25+16 or 9+64
AD is sqrt(18.25) or sqrt(73)

neither of the answer choices yields any results even close to this.

I doubt such a question would be on the actual gmat...

iqahmed83 · Answer

I am trying to use the formula Area/area=s^2/s^2. Is it possible to use this formula for this problem?

warriorguy · Answer

If I understood your query correctly, then you are trying to use the gist that if two   similar triangles have corresponding sides in proportion  , then their areas will be squared.

Except from the first part of the sentence from which we get z=y, I don't think their is much information left to deduce to use above corollary.

Nunuboy1994 · Answer

If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:

A. 12∗w12∗w
B. 12∗(w+x)12∗(w+x)
C. 2x2+z2−−−−−−−√2x2+z2
D. w2−3y2−−−−−−−√w2−3y2
E. y2+z2−−−−−−√

This question asks us to identify the length of line segment AD; therefore, we must assess the area of triangle ACB, which subsumes triangle ADB. We can plug in hypothetical values in order to derive the answer. For example, assume that we have an isosceles triangle with the ratio 45 : 45 : 90 ( we can make this assumption because we are not limited to a 30 : 60 : 90 triangle although such a triangle could also be used as a demonstration and to derive the answer). The sides of this triangle, considering the ratio of sides in a 45 : 45 : 90 triangle (1 : 1 : √2), are 2 : 2 : √2. We then calculate the area of the triangle (1/2 bh) which equals 2. We can now plug in for the other variables in the triangle: w = 2√2, z= 1 (this is true because if triangle ADB is half the area of triangle ACB then Z and Y must be midpoints of CD and DB) y= 1, x= 2. We can then use the Pythagorean theorem to calculate line segment AD. Whatever this value is, which in this hypothetical scenario happens to be √5, it must equate to the value of the correct equation. Now that we have values for all the variables, we can simply plug in values into each equation and eliminate answer choices until we arrive at a value that equals are hypothetical value of AD.

Fdambro294 · Answer

Rule: If 2 Triangles share the Same Base, then the RATIO of the Areas of the 2 Triangles will be EQUAL = to the Ratio of their Heights

Triangle ADB and Triangle ABC Share Base Leg X.

Area of Triangle ADB : Area of Triangle ABC = 1 : 2

Height of Triangle ADB = y
Height of Triangle ABC = y + z

Applying Rule that because the 2 Triangles share the Same Base, their Area will be in the Ratio of their Heights----

1/2 = y / y + z

y + z = 2y

z = y

We have just proved that y = z. At this point, we can plug in Smart Numbers and Test the Answers:

Let X = 3
Let Y = 2 and Z = 2 (such that Entire Side BDC = 4)
this means the Hypotenuse of W = 5

Let Side AD = D

(D)^2 = (3)'2 + (2)'2

D = sqrt(13)

Testing the A.C.'s with the following smart numbers

X = 3 ------ Y = 2----- Z = 2---- W = 5

when we plug in these Smart Numbers, which answer choice will provide the Length of = sqrt(13)

-D- sqrt( w^2 - 3*y^2)

= sqrt( 5^2 - 3 * (2)^2)

= sqrt ( 25 - 12)

= sqrt(13)

None of the other answer choices match our Target of sqrt(13)

The answer must be -D-

AT1982 · Answer

Please check another way

w^2 = (y + z)^2 + x^2 - 
w^2 = y^2 + z^2 + 2yz + x^2
w^2 = (x^2 + y^2) + 3z^2
Since MID POINT z = y
w^2-3z^2 = AD^2
  = AD

Posted from my mobile device

ZIX · Answer

Here's an easier approach:

Since AD is median, y = z.

Now through pythagoras theorem:

(1)  AD^2 = x^2 + y^2

Also,
 w^2 = x^2 + (y+z)^2 
 w^2 = x^2 + y^2 + z^2 + 2yz

From eq (1) we know x^2 + y^2 = AD^2

w^2 = AD^2 + z^2 + 2yz 
 AD^2 = w^2 - z^2 - 2yz

Since we know y = z,

AD^2 = w^2 - y^2 - 2y^2 
 AD = \sqrt{w^2 - 3y^2}

If the shaded area is one half the area of the triangle ABC

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GMAT Problem Solving (PS) Questions

If the shaded area is one half the area of the triangle ABC

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