The lengths of a triangle are 5cm, 7cm and x cm. What are the possible values of x?
I don't know where to start with this question as i don't thin enough information has been given. With an answer can an explanation please be given. Thanks!
$\endgroup$5 Answers
$\begingroup$If $a,b,c$ be sides of a triangle then
$a+b>c$
and $a-b<c$
Given sides are 5 and 7 hence
$(7-5)<x<(7+5)$ or,
$2<x<12$
$\endgroup$ $\begingroup$Ah, it's enough information. Suppose you are given 2 sticks of length 5cm and 7cm. Then someone gives you a stick of length 100000 meters. can you make a triangle by putting the ends of the sticks together? Well...no.
You have to figure out what the longest stick you can use to make a triangle with the first 2 (5cm and 7cm) would be. Assume that a flat triangle (zero area) still counts as a triangle.
$\endgroup$ $\begingroup$You can use the triangle inequality, i.e. for three points $A,B$ and $C$ and the distances $AB, AC$ and $BC$ we have $AB + BC \geq AC$ thus here we can substitute in your values for the sides and get $5+7\geq x$ so $x\leq12$.
But the triangle inequality must hold for other arangements of sides as well, i.e. $AC + CB \geq AB$ so using the values above this gives $x + 7 \geq 5$ which implies $x>-2$ (now since $x>0$ this doesn't really help)
Consider the final arrangement $BA + AC \geq BC$, i.e. $5 + x \geq 7$ gives $x\geq2$ so combining all of these gives $$ 2 \leq x \leq 12$$
$\endgroup$ $\begingroup$I have a clock. The hour hand is $5$ cm long. The minute hand is $7$ cm long. $x$ cm is the distance between the end of the hour hand and the end of the minute hand. (Imagine a rubber band connecting the two ends, making a triangle.) As the hands go around, how close together can the ends of the hands get? How far apart can they get?
$\endgroup$ $\begingroup$Given any triangle, if a, b, and c are the lengths of the sides, the following is always true:
$a + b > c$
$a + c > b$
$b + c > a$
Let $a=5,b=7$
Then
$5+7>c$
$5+c>7$
$7+c>5$
$c<12$
$c>2$
$c>-2$
Hence
c belongs to [3,11].
