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Topic: **Area of triangles problem solving****Question:**
I'm making a "thinking" math problem for my students ("thinking" because it's kind of a tricky question), and I got one from my friend but I don't know how to solve it! XD
So if you could please solve it for me and tell me how you got the answer, that would be greatly appreciated.
Problem: If a triangle has a perimeter of 3, what is the largest possible area?

July 20, 2019 / By Dacey

First of all, we know that the answer must always be less that 1 square unit. We know this because the area of a triangle is base times height. Given a base of 1 unit, the remaining perimeter measure can only total 2 units. Dividing that in half to gain the greatest possible height gives us sides of 1 unit each, which when brought together at a common vertex give us a height of less that 1 unit. In order to achieve a height of 1 unit and still maintain a perimeter of 3 units, we would have to shorten the base to less than 1 unit. Therefore, our answer will always be less that 1 square unit. The maximum area given a perimeter of 3, or for any given perimeter, is achieved by forming an equilateral triangle. In this case an area of .866 square units is formed.

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First of all, we know that the answer must always be less that 1 square unit. We know this because the area of a triangle is base times height. Given a base of 1 unit, the remaining perimeter measure can only total 2 units. Dividing that in half to gain the greatest possible height gives us sides of 1 unit each, which when brought together at a common vertex give us a height of less that 1 unit. In order to achieve a height of 1 unit and still maintain a perimeter of 3 units, we would have to shorten the base to less than 1 unit. Therefore, our answer will always be less that 1 square unit. The maximum area given a perimeter of 3, or for any given perimeter, is achieved by forming an equilateral triangle. In this case an area of .866 square units is formed.

if this means 3 unit, the largest could only be 1/2 units squared, right? for the total of the sides should always be equal to three, and however you put it, the area would always be the same.

if this means 3 unit, the largest could only be 1/2 units squared, right? for the total of the sides should always be equal to three, and however you put it, the area would always be the same.

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You want to set up an equation for each of the ways. The distance does not change for both ways so we can set the same unknown variable for both ways (lets go with x). The time travelled for one particular way is also unknown so lets set that as t and since the time travelled for the other way is t less than the total time we can set it as (1hr - t) Now set up your equations: (speed)=(distance)/(time) to school: 3=x/t from school: 5=x/(1-t) Now isolate the x so we can set the equations equal to each other to school: x=3t from school: x=5(1-t) therefore we get the equation 3t=5(1-t). Now solve for t and you get t=5/8 Go back to what you were trying to find. The distance(x) right? So plug our newly found info into one of the equations so we can find x e.g. x=3*5/8=15/8 Remember x represents only one way so multiply that by 2 and you get 30/8=3 6/8= 3 3/4

Start with distance d = rate x time. So 3m/h x t1 = d and 5mph x t2 = d. Then your second equation is t1 + t2 = 1. The first part means 3t1 = 5t2 or t1 = 5/3t2. From the second part t2 = 1-t1, so sub that back into the first part. Then solve for t1, which = 5/8. Multiply by 3 (m/h), then 2 for the total distance back and forth and you get 3 3/4.

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