MHB Area of Triangle with Given Data

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The area of triangle AEF can be calculated using the formula A = 1/2 * base * height, where the base AF is 7 and the height AB is 5, resulting in an area of 35/2. It is clarified that a triangle does not need to be a right triangle to apply this formula. The discussion also highlights that the lines are not drawn to scale, but this does not affect the area calculation since the height is known. There is a mention of using Heron's formula, but the simpler method is preferred here. The key takeaway is that the area can be determined without knowing all side lengths, using just the base and height.
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For b) area of AEF so one side is 7 - don't know how to get other 2 sides

not sure if right triangle; don't think so

how to use the data given since two of sides are slantedView attachment 6374
 

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A triangle doesn't have to be a right triangle to use the area formula:

$$A=\frac{1}{2}bh$$

For $\triangle AEF$, we have $b=\overline{AF}=7$ and $h=\overline{AB}=5$

Thus, we get:

$$A=\frac{1}{2}\cdot7\cdot5=\frac{35}{2}$$

Suppose you aren't convinced we can use this formula except for right triangles. We could then drop a vertical line from point $E$ to $\overline{AD}$ and label the intersection $G$. $G$ will be to the right of $F$ and we'll say $\overline{FG}=x$

Now, we have the right triangle $AEG$, whose area is:

$$A_1=\frac{1}{2}(5+x)7$$

We also have the right triangle $FEG$ whose area is:

$$A_2=\frac{1}{2}(x)7$$

We can now find the area of $AEF$ by taking $A_1$, the area of the larger right triangle, and subtracting $A_2$, the area of the smaller right triangle:

$$A=A_1-A_2=\frac{1}{2}(5+x)7-\frac{1}{2}(x)7=\frac{1}{2}\cdot7\left((5+x)-x\right)=\frac{1}{2}\cdot7\cdot5=\frac{35}{2}$$
 
Thanks the line AB equals EF - what threw me off was EF looks at an angle - but I guess I assume since not drawn to scale - the 2 lines are the same.
 
alextrainer said:
Thanks the line AB equals EF - what threw me off was EF looks at an angle - but I guess I assume since not drawn to scale - the 2 lines are the same.

I would say, going by the diagram, that:

$$\overline{EF}>\overline{AB}$$

But, as I showed, we don't need to know $\overline{EF}$ since $\overline{AB}$ is the altitude of the triangle.
 
alextrainer said:
Thanks the line AB equals EF - what threw me off was EF looks at an angle - but I guess I assume since not drawn to scale - the 2 lines are the same.

They are not drawn to scale, but it doesn't mean that both are equal. Are you trying to find the area through herons formula? Or it is just that you were unaware about the simpler method?
 
You don't need to know the lengths of the sides of a triangle to find its area. The area of a triangle is (1/2)*base*height. Here the base is AF which we are told has length 7 and the height is 5.
 
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