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In triangle ABC above, does (AB)^2 + (BC)^2 = (AC)^2 ? 1) x + z = 100

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In triangle ABC above, does (AB)^2 + (BC)^2 = (AC)^2 ? 1) x + z = 100 [#permalink] New post   27 Oct 2021, 03:30
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70% (00:46) correct 30% (01:06) wrong based on 96 sessions
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In triangle ABC above, does \((AB)^2 + (BC)^2 = (AC)^2\) ?

(1) \(x + z = 100\)

(2) \(x = z\)



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In triangle ABC above, does (AB)^2 + (BC)^2 = (AC)^2 ? 1) x + z = 100 [#permalink] New post   27 Oct 2021, 04:05
Option (A) is the right answer.

Question is basically asking us, is \(y = 90^{\circ}\)?

Statement 1: \(x+z=100\)
This gives y = 80\(^{\circ}\).
Thus \(y\) \(\neq{}\) \(90\)\(^{\circ}\).
Sufficient.

Statement 2: \(x=z\)
This gives, \(2x + y = 180\)
\(y = 180 - 2x\)
\(Y\) can multiple values for different values of '\(x\)'.
Insufficient.
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Re: In triangle ABC above, does (AB)^2 + (BC)^2 = (AC)^2 ? 1) x + z = 100 [#permalink] New post   27 Oct 2021, 05:04
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Bunuel wrote:

In triangle ABC above, does \((AB)^2 + (BC)^2 = (AC)^2\) ?

(1) \(x + z = 100\)

(2) \(x = z\)


Solution:

For the above triangle to follow \((AB)^2 + (BC)^2 = (AC)^2\), it has to be a right angle triangle, right-angled at \(B\).

Or \(\angle B=y\) has to be \(90^o\)


Statement 1: \(x + z = 100\)

If \(x + z = 100\), then \(y=180-(x+z)=180-100=80\).

We get that \(y=80\) and thus we can be sure that \((AB)^2 + (BC)^2 = (AC)^2\) will not stand.

Statement 1 alone is sufficient to answer the question. We can eliminate options B, C and E.


Statement 2: \(x = z\)

If \(x = z\),

then\(y=180-(x+z)=180-2x\) or \(180-2z\).

Now we cannot be sure if \(180-2z\) or \(180-2x\) is equal to 90 or not. It is possible when \(x=z=45\) but that is unknown.

Thus, statement 2 alone is not sufficient.


Hence the right answer is Option A.
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Re: In triangle ABC above, does (AB)^2 + (BC)^2 = (AC)^2 ? 1) x + z = 100 [#permalink] New post   27 Oct 2021, 07:51
In triangle ABC above, does (AB)^2+(BC)^2=(AC)^2 ?

(1) x+z=100
Clearly angle y = 80
Clearly does not satisfy the equation.
Sufficient

(2) x=z
if x=45 and z=45
y=90 then
(AB)^2+(BC)^2=(AC)^2

if x=30 and z =30
then y=120
then it s=does not satisfy the equation.

IMO A
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Re: In triangle ABC above, does (AB)^2 + (BC)^2 = (AC)^2 ? 1) x + z = 100 [#permalink]
27 Oct 2021, 07:51
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Bunuel
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