What Are the Possible Values of c for a Triangle with Area 5?

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In summary, the length of side AB of the triangle is 5 units, the slope of side BC is (c-4)/-1, the area of the triangle is 3c/2 square units, to determine if the triangle is a right triangle we can use the Pythagorean Theorem, and the value of c that would make this triangle an equilateral triangle can be found by setting the equations for two equal sides equal to each other and solving for c.
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anemone
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Here is this week's POTW:

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A triangle with vertices $A(0,\,0),\,B(3,\,4)$ and $C(2,\,c)$ has area 5 units². Find all possible values of $c$.

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Congratulations to skeeter for his correct solution, which you can find below:

for $0 \le c \le 4$, maximum area = 4

for $c > 4$ ...

$\vec{AB} \times \vec{AC} = 10$

\[ \begin{vmatrix} 1 & 1 & 1\\ 3 & 4 & 0\\ 2 & c & 0 \end{vmatrix}=10 \implies 3c-8 = 10 \implies c=6 \]

for $c < 0$ ...

$\vec{AC} \times \vec{AB} = 10$

\[ \begin{vmatrix} 1 & 1 & 1\\ 2 & c & 0\\ 3 & 4 & 0 \end{vmatrix}=10 \implies 8-3c = 10 \implies c=-\dfrac{2}{3} \]
 

FAQ: What Are the Possible Values of c for a Triangle with Area 5?

What is the length of side AB?

The length of side AB can be calculated using the distance formula, which is √[(x2-x1)^2 + (y2-y1)^2]. In this case, the coordinates for A are (0,0) and the coordinates for B are (3,4). Plugging these values into the formula, we get √[(3-0)^2 + (4-0)^2] = √(9+16) = √25 = 5. Therefore, the length of side AB is 5 units.

What is the slope of side BC?

The slope of a line can be calculated using the formula (y2-y1)/(x2-x1). In this case, the coordinates for B are (3,4) and the coordinates for C are (2,c). Plugging these values into the formula, we get (c-4)/(2-3) = (c-4)/(-1) = 4-c. Therefore, the slope of side BC is 4-c.

What is the area of the triangle?

The area of a triangle can be calculated using the formula A = (1/2)bh, where b is the base of the triangle and h is the height. In this case, the base of the triangle is the length of side AB, which we calculated to be 5 units. To find the height, we can use the slope of side BC. Since the slope is 4-c, the height is the distance from point C to the line AB, which is the y-coordinate of point C. Therefore, the height is c units. Plugging these values into the formula, we get A = (1/2)(5)(c) = (5c)/2. Therefore, the area of the triangle is (5c)/2 square units.

What is the perimeter of the triangle?

The perimeter of a triangle is the sum of the lengths of all three sides. In this case, we already know the length of side AB, which is 5 units. To find the length of side AC, we can use the distance formula again. The coordinates for A are (0,0) and the coordinates for C are (2,c). Plugging these values into the formula, we get √[(2-0)^2 + (c-0)^2] = √(4+c^2). Therefore, the length of side AC is √(4+c^2) units. The length of side BC is the distance between points B and C, which is the y-coordinate of point C. Therefore, the length of side BC is c units. Adding these three lengths together, we get the perimeter of the triangle to be 5 + √(4+c^2) + c units.

Is this triangle a right triangle?

A right triangle is a triangle with one angle measuring 90 degrees. To determine if this triangle is a right triangle, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the length of side AB is 5 units and the length of side BC is c units. Therefore, using the Pythagorean theorem, we get 5^2 + c^2 = (2^2 + c^2) + (3^2 + 4^2). Simplifying, we get 25 + c^2 = 4 + c^2 + 9 + 16. This simplifies to 25 = 29, which is not true. Therefore, this triangle is not a right triangle.

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