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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).Is AD perpendicular to BC?
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A triangle is still possible with the sides being 5,3,4.5; 5,3,3.5neelgandham wrote: 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).
Apologies ! I stand corrected !satishchandra wrote:A triangle is still possible with the sides being 5,3,4.5; 5,3,3.5neelgandham wrote: 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).
I am assuming AD=4.5 or AD=3.5
If i do so, I cant find the answer or I have multiple answers.
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<√34neelgandham wrote: 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).
Mathpro I would be interested to know on how you arrived at the range for the triangle to be acute i.e 4<X<√34GmatMathPro wrote:)[/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
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 wrote:Mathpro I would be interested to know on how you arrived at the range for the triangle to be acute i.e 4<X<√34GmatMathPro wrote:)[/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
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