In the fig above angle BAC and angle ADC is a right angle, If CD = 3 , AD = ??
please refer attachment
D
geomatry
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I'm trying to put together the actual solution to find AD just for reference, but regardless the below question can be solved without doing that (and you should aim to solve similar questions without finding the actual answer, but rather just find out if it can be solved)
We have a triangle with essentially 6 segments of various lengths: AB,AC,BC (main triangle), AD (the line through it which is also the height),BD,CD (that make up BC).
Now realistically, we only really have four variables: AB,AC,AD,BD
This is because CD = 3 and BC = BD + CD = BD + 3, so we can substitute for those two.
In order to solve any equation with four variables we need four equations (and in general - to solve an equation with X variables, we'd need X dissimilar equations). We currently have three, because we have three right triangles:
1. AC^2 + AB^2 = BC^2 --> you can replace that with AC^2 + AB^2 = (BD+3)^2
2. AD^2 + 3^2 = AC^2
3. AD^2 + BD^2 = AB^2 (we know angle ADB is a right angle because angle ADC is a right angle)
Therefore, all we need is one more equation, or one value for either of the variables in our equations. Since either choice gives us a value for one of these variables, the answer is indeed D
We have a triangle with essentially 6 segments of various lengths: AB,AC,BC (main triangle), AD (the line through it which is also the height),BD,CD (that make up BC).
Now realistically, we only really have four variables: AB,AC,AD,BD
This is because CD = 3 and BC = BD + CD = BD + 3, so we can substitute for those two.
In order to solve any equation with four variables we need four equations (and in general - to solve an equation with X variables, we'd need X dissimilar equations). We currently have three, because we have three right triangles:
1. AC^2 + AB^2 = BC^2 --> you can replace that with AC^2 + AB^2 = (BD+3)^2
2. AD^2 + 3^2 = AC^2
3. AD^2 + BD^2 = AB^2 (we know angle ADB is a right angle because angle ADC is a right angle)
Therefore, all we need is one more equation, or one value for either of the variables in our equations. Since either choice gives us a value for one of these variables, the answer is indeed D
Please feel free to private message me with any questions; I don't check the forums regularly, whereas PMs go straight to my email.
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A height drawn through the right angle of a triangle forms 3 SIMILAR TRIANGLES.
Proof:
Let ∠CAB = x and ∠BAD = y.
Since ∠BAC is a right angle, x+y=90.
The result is that all 3 triangles -- ∆ACD, ∆BAD, and ∆ABC -- have the same combinations of angles:
x-y-90.
Triangles that have the same combination of angles are SIMILAR.
The legs of similar triangles are in the SAME RATIO.
Thus, in all 3 triangles:
(leg opposite x) : (leg opposite y) = (leg opposite x) : (leg opposite y).
In ∆ACD, (leg opposite x) : (leg opposite y) = CD : AD.
In ∆BAD, (leg opposite x) : (leg opposite y) = AD : BD.
Since the two ratios are equal, we get:
CD / AD = AD / BD
(AD)² = (BD)(CD).
Statement 2: BD = 16/3
Since BD = 16/3, CD=3, and (AD)² = (BD)(CD), we get:
(AD)² = (16/3)(3)
(AD)² = 16
AD=4.
SUFFICIENT.
Statement 1: AB = 20/3
Since the 2 statements cannot contradict each other, it must be possible that BD=16/3 and AD=4 in statement 1.
To confirm that these values imply that AD=20/3, apply the Pythagorean theorem to ∆BAD:
BD² + AD² = AB²
(16/3)² + 4² = (20/3)²
256/9 + 144/9 = 400/9
400/9 = 400/9.
This works.
Moreover, NO VALUES OTHER than BD=16/3 and AD=4 will yield that AB=20/3.
As shown above, (AD)² = (BD)(CD).
Thus:
If the value of BD decreases, so will the value of AD, with the result that AB<20/3.
If the value of BD increases, so will the value of AD, with the result that AB>20/3.
Thus, in order that AB=20/3, it must be true that BD=16/3 and AD=4, as indicated in Statement 1.
SUFFICIENT.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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