The angle bisector, median, and altitude are all the same things in an isosceles triangle.
That means that the legs of the smaller triangles created by the altitude are 4 root 2.
By special properties of an right triangle, the hypotenuse is 8.
Since the legs of the original right triangle are congruent, the area of the original triangle is 8*8/2.
Now we just simplify:
8*8/2=64/2=32
Answer:
It is irrational because it can not be represented as a fraction of two integers
Step-by-step explanation:
Given
![H = \sqrt[3]5](https://tex.z-dn.net/?f=H%20%3D%20%5Csqrt%5B3%5D5)
Required
Why is the area irrational?
First, we need to calculate the area
![Area = \frac{1}{2}(3.6 + 12\frac{1}{3}) * \sqrt[3]5](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%283.6%20%2B%2012%5Cfrac%7B1%7D%7B3%7D%29%20%2A%20%5Csqrt%5B3%5D5)
<em>It is irrational because it can not be represented as a fraction of two integers</em>
Answer:
4x and -2x
Step-by-step explanation:
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For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14
For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so y + x = 6
Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:
(3y - x = 14) + (y + x = 6) => 4y = 20, y = +5; Substitute this into y + x = 6: 5 + x = 6, x = +1
<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>
Answer:
The equivalent ratios are B and C.
