Question

How to solve for the variables of the figures below?

Answer 1

For Figure A `x=11/2` and `y=11sqrt(3)/2`.

For Figure B `x=16`, and `y=8sqrt(2)`.

Explanation:

Figure A shows a 30-60-90 triangle with a hypotenuse of length 11. The short leg is 1/2 the length of the hypotenuse, so `x=11/2`. The long leg is the square root of 3 times the length of the short leg so `y=11sqrt(3)/2`.

Figure B shows a 45-45-90 triangle with a leg length of `8sqrt(2)`. The legs of the triangle have equal length, so `y=8sqrt(2)`. We can get `y` from the Pythagorean Theorem.

`(8sqrt(2))^2+(8sqrt(2))^2=x^2`

`256=x^2`

`x=16`