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geometry chapter 4 area 1

 
 
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Re: geometry chapter 4 area 1
LensmireJohn - 2024年10月8日 Tuesday 09:34
 

For 4.25, this is the right idea! Since it's a parallelogram we already know that opposite sides of the parallelogram are congruent, so if two sides are lengths  a and b, with a diagonal of length d, then the parallelogram is made up of two congruent triangles with sides a, b, and d. Then Heron's can be used.

To review, Heron's formula says that if a triangle has sides x, y, and z, with semiperimeter s = (a+b+c)/2, it's area is then $\sqrt{s (s-x) (s-y)(s-z)}$.

For 4.23, we want to show that if we have a triangle with sides a, b, and c with a^2+b^2=c^2, then the triangle is right. Consider a triangle with sides a, b, and an angle of 90 degrees (between them). What does the Pythagorean theorem say about the hypotenuse?

In 4.26 we just mean that the length and widths of all the rectangles are integers (whole numbers).

In 4.28, try to think in terms of fractions of the area of ABCD. For example, what fraction of [ABCD] is [AMD] or [CDN]?

Hope these help a bit!