Geometry Problem

[frumpychic]

Using the diagram below.
If AB = 5 and BD=3, and the area of ABC is 14, what is the length of AC?[/img]

[neelgandham]

AB =5, BD =3 and my brain says 3,4,5 is a Pytho triplet => AD = 4 as the triangle is acute angled triangle !

Area of ABC = 0.5(BD+DC)*AD = 0.5*(3+DC)*4 = 14 => DC =4

AC is hypotenuse of the right angled triangle ADC => AC = square root(AD^2 + DC^2) = 4* Square Root(2)

[shankar.ashwin]

I am assuming AD is the perpendicular; then AD=4

GIven area = 14 (or) Base*Height = 28

(3+x) (4) = 28
3+x = 7
x = 4 = DC

AC would be 4sqrt(2)

[satishchandra]

Using the diagram below.

Is AD perpendicular to BC?
Data is missing

[neelgandham]

Is AD perpendicular to BC?
Data is missing

The only assumption made is that the triangle is an acute angled triangle(which it is from the diagram). If it is an acute angled triangle, then it is true that AD is perpendicular to BC.You cannot draw a triangle with sides 3,5,X (where X!=4).

[satishchandra]

The only assumption made is that the triangle is an acute angled triangle(which it is from the diagram). If it is an acute angled triangle, then it is true that AD is perpendicular to BC.You cannot draw a triangle with sides 3,5,X (where X!=4).

A triangle is still possible with the sides being 5,3,4.5; 5,3,3.5
I am assuming AD=4.5 or AD=3.5
If i do so, I cant find the answer or I have multiple answers.

[neelgandham]

The only assumption made is that the triangle is an acute angled triangle(which it is from the diagram). If it is an acute angled triangle, then it is true that AD is perpendicular to BC.You cannot draw a triangle with sides 3,5,X (where X!=4).

A triangle is still possible with the sides being 5,3,4.5; 5,3,3.5
I am assuming AD=4.5 or AD=3.5
If i do so, I cant find the answer or I have multiple answers.

Apologies ! I stand corrected !

[GmatMathPro]

The only assumption made is that the triangle is an acute angled triangle(which it is from the diagram). If it is an acute angled triangle, then it is true that AD is perpendicular to BC.You cannot draw a triangle with sides 3,5,X (where X!=4).

Sure you can. If the sides of the triangle are 3,5, and X, X can be any value strictly between 2 and 8,and the triangle will be acute for 4<X<√34

[satishchandra]

)[/b]

Sure you can. If the sides of the triangle are 3,5, and X, X can be any value strictly between 2 and 8,and the triangle will be acute for 4<X<√34

Mathpro I would be interested to know on how you arrived at the range for the triangle to be acute i.e 4<X<√34

[GmatMathPro]

)[/b]

Sure you can. If the sides of the triangle are 3,5, and X, X can be any value strictly between 2 and 8,and the triangle will be acute for 4<X<√34

Mathpro I would be interested to know on how you arrived at the range for the triangle to be acute i.e 4<X<√34

Let the sides of a triangle be a, b, c, with c being the longest side. A consequence of the pythagorean theorem says that the triangle will be acute if and only if c^2<a^2+b^2 and will be obtuse if and only if c^2>a^2+b^2. In this case, the third side can be anything from 2 to 8. If it is between 2 and 5, then 5 will be the biggest side, and for it to be acute we will need 5^2<x^2+3^2 or x^2>16 or x>4. If the third side is bigger than 5, than the third side is the biggest side, in which case we need x^2<5^2+3^2 or x^2<34 or x<√34. Combining the two, we get 4<x<√34.

[satishchandra]

Thanks Mathpro
Good job