MATH SOLVE

3 months ago

Q:
# Am I correct ??? I don’t really know

Accepted Solution

A:

What you have here is a situation with two similar triangles. The triangle in the lower left is similar to the triangle in the upper right - I've included an image with "cutouts" of those triangles so you can see the similarities. Similar triangles have a very important property: the ratios of their corresponding sides are equivalent. Here, we can set up a ratio between the sides of length 64 and x on the larger triangle, and the corresponding sides of length x and 36 on the smaller triangle. Setting the two equal to each other, we have

[tex] \frac{64}{x}= \frac{x}{36} [/tex]

Multiplying both sides of the equation by 36 and x, we get

[tex]64\cdot36=x^2[/tex]

finally, we take the square root of both sides of the equation to find x:

[tex]\sqrt{64\cdot36}=x\\ \sqrt{64}\sqrt{36}=x\\ 8\cdot6=x\\ 48=x[/tex]

[tex] \frac{64}{x}= \frac{x}{36} [/tex]

Multiplying both sides of the equation by 36 and x, we get

[tex]64\cdot36=x^2[/tex]

finally, we take the square root of both sides of the equation to find x:

[tex]\sqrt{64\cdot36}=x\\ \sqrt{64}\sqrt{36}=x\\ 8\cdot6=x\\ 48=x[/tex]