This is a sketch of the given info. \(\overline{AD}\) is an altitude and \(\overline{AC}\cong\overline{AB}\), and the triangle is right.
Because \(\overline{AC}\cong\overline{AB}\) by the given info, the isosceles triangle theorem tells us that the angles opposite of their sides are congruent, so \(\angle B\cong\angle C\). We can now figure out \(m\angle B\quad\text{and}\quad m\angle C\).
\(m\angle B+m\angle C+m\angle BAC=180^{\circ}\) | Now, substitute in the known values. |
\(m\angle B+m\angle C+90=180\) | \(m\angle B=m\angle C\) |
\(m\angle B+m\angle B+90=180\) | Now, solve for both angles. |
\(2m\angle B+90=180\) | |
\(m\angle B+45=90\) | |
\(m\angle B=45^{\circ}\) | |
\(m\angle B=m\angle C=45^{\circ}\) | |
Since an altitude always bisects the angle of an isosceles triangle, \(m\angle CAD=\frac{1}{2}*90=45^{\circ}\).
Since \(m\angle CAD=m\angle C=45^{\circ}\), by the inverse of the isosceles triangle theorem, \(AD=DC=4\sqrt{2}\)
The entire base, however, is 2 times this length, so \(BC=2*4\sqrt{2}=8\sqrt{2}\). The altitude is \(4\sqrt{2}\). Now, let's use the area formula for a triangle.
\(\frac{1}{2}bh=\frac{1}{2}(8\sqrt{2})(4\sqrt{2})\) | Now, simplify here. |
\(\frac{1}{2}(32*\left(\sqrt{2}\right)^2)\) | |
\(\frac{1}{2}(32*2)\) | The 1/2 and 2 cancel out here. |
\(32\) | This is the area of the triangle. |
The length of \(\overline{AC}\) can be found using the law of cosines. The law of cosines relates the remaining side by knowing two sides and its opposite included angle. I think this picture sums it up nicely.
Knowing this information, we can now find the missing length.
\(c^2=a^2+b^2-2ab\cos C\) | Now, plug in the values we already know in the diagram and solve for the missing side. |
\(AC^2=1^2+3^2-2(1)(3)\cos40^{\circ}\) | Now, take the principal square root of both sides since the negative answer is nonsensical in the context of geometry. |
\(AC=\sqrt{1^2+3^2-2(1)(3)\cos40^{\circ}}\approx2.32\) | No units are given in the problem. |
Just like in the previous problem you posted, factor out the GCF of each term in the parentheses.
\((8x^4+8x^3)+(27x+27)\) | Now, factor out the GCF. |
\(8x^3(x+1)+27(x+1)\) | (x+1) is common to both factors, so we can rewrite this. |
\((8x^3+27)(x+1)\) | (8x^3+27) happens to be a sum of cubes and must be dealth with as such. |
\(a=\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\\ b=\sqrt[3]{27}=3\) | Now that a and b have been identified expand the sum of cubes. |
\((2x+3)((2x)^2-(3)(2x)+3^2)(x+1)\) | Now, simplify. |
\((2x+3)(4x^2-6x+9)(x+1)\) | 4x^2-6x+9 is an irreducible trinomial and nothing else can be reduced further. |