The lengths of the sides of an obtuse-angled triangle are x, y, and z,

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The lengths of the sides of an obtuse-angled triangle are x, y, and z, where x, y and z are integers. If xy = 4, what is the value of z?

A. 1
B. 2
C. 3
D. 4
E. 5

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ArunSharma12 · Answer

The lengths of the sides of an obtuse-angled triangle are x, y, and z, where x, y and z are integers. If xy = 4, what is the value of z?

A. 1
B. 2
C. 3
D. 4
E. 5

possible values of (x,y)
 (1,4) - 3<z<5; Z can be 4 but triangle wont be obtuse
(2,2) - 0<z<4; z can be 1,2 or 3. z =3 makes the triangle obtuse 
Ans: C

eakabuah · Answer

We are given that x,y, and z are the sides of an obtuse-angled triangle and that x,y, and z are integers. Since xy=4, then possible values of x and y are either 2 and 2 or 1 and 4. An obtuse-angled triangle only means that one of the interior angles of the triangle is more than 90.
But from the property of triangles x+y>z, x+z>y, and z+y>x, x=y=2 is the only viable possibility.
The only possible value of z given that x=y=2 is for z=3. This way 2+2>3 and 2+3>2. x,y, and z now form an isosceles triangle with the sine of half the obtuse angle, a/2, sine(a/2)=1.5/2=0.75. But we know that sine(45) =0.707 hence we can clearly infer that a/2>45. Therefore a>90.

The answer is C.

freedom128 · Answer

Sides x, y and z are all integers. Triangle has an obtuse angle and x.y = 4
--> Potential (x,y) set = (1,4), (2,2)

If (x,y)=(1,4), then 3<z<5.
Possible integer z = 4.
However, x^2 + y^2 > z^2 --> acute-angled triangle (NO!).

If (x,y)=(2,2), then 0<z<4. Possible integer z = 1,2,3.
- For z=1, x^2 + y^2 > z^2 --> acute-angled triangle (NO!)
- For z=2, x^2 + y^2 > z^2 --> acute-angled or equilateral triangle (NO!)
- For z=3, x^2 + y^2 < z^2 --> obtuse-angled triangle (YES!)

If the triangle has an obtuse angle, then x=2, y=2, and z=3

FINAL ANSWER IS (C)

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firas92 · Answer

(x,y)=(1,4) or (2,2) or (4,1) For an obtuse triangle, If z is the longest side, x^2+y^217 so z must be at least 5 But if z is 5 , it is greater than the sum of lengths the other two sides which is not possible for a triangle If (x,y)=(2,2) we have z^2>8 so z must be 3 . Note that a (2,2,3) triangle is possible but a (2,2,z>3) triangle is not possible for integer values of z If z is not the longest side, 4 must be the longest side and so z^2+1<16 or z^2<15 or z is at most 3 but again, a (1,3,4) or (1,2,4) or (1,1,4) triangle is not possible So our obtuse triange must be (x,y,z)=(2,2,3) and so z=3 Answer is (C) Posted from my mobile device

exc4libur · Answer

note: sides must be a-bx^2+y^2, z^2>2^2+2^2, z^2>8, z>2.7 = {3} Ans (C)

unraveled · Answer

The lengths of the sides of an obtuse-angled triangle are x, y, and z, where x, y and z are integers. If xy = 4, what is the value of z?

A. 1
B. 2
C. 3
D. 4
E. 5

xy = 4
1*4 = 4 OR 4*1 = 4 OR 2*2 = 4
So z can't be either 1 or 4 since the triangle would not remain an obtuse angled triangle as:
  =   is an isosceles triangle not obtuse angled triangle 
  =   is an isosceles triangle not obtuse angled triangle 
 Strike out option A, and D.

z also can't be 2 since the triangle becomes an equilateral triangle as
  =   is an equilateral triangle not obtuse angled triangle 
 Strike out option B.

Finally, z can't be 5 since it would invalidate properties of triangle and triangle formation is not possible as
  =   - the largest side has to be < 4
 Strike out option E.

So, possible sides of triangle are 2,2,3. Hence C.

Answer C.

ankitgoswami · Answer

Since it is given that x,y &z are integers. Therefore, from xy =4, we can say that the possible values are 2 x 2 or 4 x 1.

Case 1 : 4 x 1

In this case the third side can only be 4

But this doesn't satisfy the condition of
 a^2 + b^2 < c^2

Case 2 : 2 x 2 
Third side in this case can take three values = 1,2,3

Can't be 2 - will become equilateral
Can't be 1
Its 3 as it satisfies the above equation as well

So it's C

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Fdambro294 · Answer

Concept 1: given 2 known sides (A and B) and 1 unknown side (X), the triangle inequality theorem holds that

(A - B) < X < (A + B)

Concept 2: given any obtuse triangle in which one angle is greater than 90 degrees and the other two angles are less than 90 degrees - given that L = length of longest side across from the obtuse angle and P and Q are the lengths of the 2 shorter sides, the following inequality must hold true for an obtuse triangle:

(L)^2 > (P)^2 + (Q)^2

Given that X, Y, and Z are integer side lengths and X * Y = 4, we have two possible scenarios:

Scenario 1: X = 1 and Y = 4 and Z is the unknown side

Triangle inequality theorem ——>
(4 - 1) < Z < (4 + 1)

3 < Z < 5

The only integer value for Z must be 4

Z = 4; X = 1; Y = 4

Inequality for Obtuse Triangle:

(4)^2 > (1)^2 + (4)^2 ——— does NOT hold true. Thus, the sides of X and Y are not 1 and 4

Scenario 2. X = 2 and Y = 2 and Z is unknown

(2 - 2) < Z < (2 + 2)

0 < Z < 4

Z can equal 1, 2, or 3

In order to be an obtuse triangle, the only possible value for Z is 3 because then Z would be the longest side across from the obtuse angle and:

(3)^2 > (2)^2 + (2)^2

9 > 8 ———- satisfied

If Z = 1 or 2, then we can not satisfy this inequality for an obtuse triangle

Answer: Z = 3

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The lengths of the sides of an obtuse-angled triangle are x, y, and z,

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