11.7

Hey guys. This was an easy section. Yeah.

One was of determining the ratio of the areas of two figures is to calculate the quotient of the two areas.

As in, lets say a triangle has a height of 8 and a base of 12. And yes, I'm way too lazy to actually find a picture or draw one that corresponds to this. And theres another figure, a parallelogram with a height of 9 and a base of 12.

Everyone here should know the formulas for these areas, but in case you needed to be reminded:
A of Triangle= 1/2BH, where B=Base and H=Height
A of Parallelogram= BH, where B=Base and H=Height

So, if you wanted to compare area of the Parallelogram to the area of the triangle, you would first find the area of the parallelogram and then divide it by the area of the triangle.

A of Triangle= 1/2(8X12)=48
A of Parallelogram= 10X9= 90
which would be 90 over 48, and that simplifies to 15 over 8, or 15:8

One theorem we discussed in class is theorem 109, which states:
If two figures are similar, then the ratio of their areas equals the square of the ratio of corresponding segments. (Similar-Figures Theorem)

In other words, its basically this:

A1/A2=(S1/S2) squared.
where A1 and A2 are areas and S1 and S2 are measures of corresponding segments.

Another theorem we learned in the bum rush that is the end of class is Theorem 110, which states:
A median of a triangle divides the triangle into two triangles with equal areas.

In other words, Imagine a triangle PQS, nothing special about it.
Then, imagine a line from vertice P to seg. QS. The line bisects the segment at point R.
This theorem states that the Area of triangle PQR is = to the Area of triangle PRS

Imagination is better than knowledge. Or pictures. Or something.

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Last two lines are the best lyrics ever, by the way.