In the largest triangle, the missing side has length
√((11 + 5)² - <em>x</em> ²) = √(256 - <em>x</em> ²)
But it's also the hypotenuse of the triangle with side lengths 11 and <em>y</em>, so that its length can also be written as
√(11² + <em>y</em> ²) = √(121 + <em>y</em> ²)
so that
√(256 - <em>x</em> ²) = √(121 + <em>y</em> ²)
or, by taking the squares of both sides,
256 - <em>x</em> ² = 121 + <em>y</em> ²
<em>y</em> ² = 135 - <em>x</em> ²
In the smallest triangle, you have
5² + <em>y</em> ² = <em>x</em> ² ==> <em>x</em> ² = 25 + <em>y</em> ²
Substitute this into the previous equation and solve for <em>y</em> :
<em>y</em> ² = 135 - (25 + <em>y</em> ²)
<em>y</em> ² = 110 - <em>y</em> ²
2<em>y</em> ² = 110
<em>y</em> ² = 55
<em>y</em> = √55
Answer:
Step-by-step explanation:
so we see the two line of the same length of 14, which tells us that the lengths of 5x and 7x-8 have to be the same length. so we know we can set them equal to each other. Often times that's the key to math problems , finding where you can set things equal.
5x = 7x-8 ( subtract 5x from both sides and then add 8 to both sides )
8 = 2x ( divide both sides by 2 )
4 = x
nice this looks right too , both sides add to 20 :)
Answer:
49π
Step-by-step explanation:
Circle area: πr^2
Since A to C is 14, and they're both in the center, A to B and B to C will both be 7 as a radius.
Since we know the radius, we can put it in the formula, πr^2.
We only need to solve for one circle, since both circles are the same. By solving the circle, we will put in 7 for the radius, giving us π7^2, which is 49π.
