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Use the Pythagorean theorem to find x.

The triangle is an isosceles triangle. The height is 6 with the altitude bisecting the base. The base is x while the sides of the triangle are unknown. The angles of the triangle on the base are both 60 degrees. I am supposed to use the Pythagorean theorem to find x the base of the triangle.

Hello Guest!

\(Kosinus =\frac{Ankathete}{Hypotenuse}\)

\(cos(60^{\circ})=0.5=\large \frac{\frac{x}{2}}{s}\\ 0.5s= \frac{x}{2}\)   

\(s=x\)

The Pythagorean theorem:

\(\color{blue}s^2=6^2+(\frac{x}{2})^2\\ s=x\\ x^2=36+\frac{x^2}{4}\\ x^2-\frac{x^2}{4}=36\)

\(\frac{3x^2}{4}=36\\ x^2=\frac{4\cdot 36}{3}=48\\ x^2=3\cdot 4^2\)

\(x=4\cdot \sqrt{3}\\ x=6.928\)

  !

In the coordinate plane, \(A=(4,-1)\), \(B=(6,-2)\), and \(C=(1,-2)\).
There exists a point \(Q\) and a constant \(k\) such that for any point \(P\),
\(PA^2 + PB^2 + PC^2 = 3PQ^2 + k\)
Find the constant k.

My attempt:

\(\small{ \begin{array}{|rcll|} \hline \mathbf{PA^2 + PB^2 + PC^2} &=& \mathbf{3PQ^2 + k} \\ \boxed{PA = P-A\\ PB=P-B\\PC=P-C\\ PQ=P-Q } \\ (P-A)^2 + (P-B)^2 + (P-C)^2 &=& 3(P-Q)^2 + k \\ P^2-2PA+A^2 + P^2-2PB+B^2 + P^2-2PC+C^2 &=& 3\left(P^2-2PQ+Q^2 \right)+ k \\ 3P^2+A^2+B^2+C^2-2P(A+B+C) &=& 3P^2-6PQ+3Q^2 + k \\ A^2+B^2+C^2-2P(A+B+C) &=& k+3Q^2-6PQ \\ \boxed{A=\dbinom{4}{-1},\ B=\dbinom{6}{2},\ C=\dbinom{-1}{2} } \\ \dbinom{4}{-1}\dbinom{4}{-1}+\dbinom{6}{2}\dbinom{6}{2}+\dbinom{-1}{2}\dbinom{-1}{2}-2P\Bigg(\dbinom{4}{-1}+\dbinom{6}{2}+\dbinom{-1}{2}\Bigg) &=& k+3Q^2-6PQ \\ (16+1)+(36+4)+(1+4)-2P\Bigg(\dbinom{4+6-1}{-1+2+2}\Bigg) &=& k+3Q^2-6PQ \\ 62-2P \dbinom{9}{3} &=& k+3Q^2-6PQ \\ 62-2P \dbinom{3*3}{3*1} &=& k+3Q^2-6PQ \\ 62-2*3P \dbinom{3}{1} &=& k+3Q^2-6PQ \\ 62-6P \dbinom{3}{1} &=& k+3Q^2-6PQ \\ \text{compare} && \text{compare} \\ {\color{red}62}-6P {\color{blue}\dbinom{3}{1}} &=& {\color{red}k+3Q^2}-6P{\color{blue}Q} \\ \hline \mathbf{{\color{blue}Q}} &=& \mathbf{{\color{blue}\dbinom{3}{1}}} \\\\ {\color{red}k+3Q^2} &=& {\color{red}62} \\ k+3\dbinom{3}{1}\dbinom{3}{1} &=& 62 \\ k+3(3*3+1*1) &=& 62 \\ k+3*10 &=& 62 \\ k+30 &=& 62 \\ k &=& 62-30 \\ \mathbf{k} &=& \mathbf{32} \\ \hline \end{array} }\)

Four years ago Arthur is 11 times as old as his son Arthur Jr,
but four years from now, Arthur will only be 3 times as old as Jr.
What is Arthur’s age when he is twice as old as Arthur Jr?

\(\text{Let Arthur $=a$} \\ \text{Let his son Arthur Jr $=j$} \)

\(\begin{array}{|lrcll|} \hline 1. & \mathbf{a-4} &=& \mathbf{11(j-4)} \\ & a &=& 11(j-4) + 4 \\ & a &=& 11j-44 + 4 \\ & \mathbf{a} &=& \mathbf{11j-40} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 2. & \mathbf{a+4} &=& \mathbf{3(j+4)} \\ & a &=& 3(j+4) - 4 \\ & a &=& 3j+12 - 4 \\ & a &=& 3j+8 \quad | \quad \mathbf{a=11j-40} \\ & 11j-40 &=& 3j+8 \\ & 11j-3j &=& 3j+8+40 \\ & 8j &=& 48 \quad | \quad : 8 \\ & \mathbf{j} &=& \mathbf{6} \\ \hline & \mathbf{a} &=& \mathbf{11j-40} \quad | \quad j=6 \\ & a &=& 11*6 - 40 \\ & a &=& 66-40 \\ & \mathbf{a} &=& \mathbf{26} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3. & \mathbf{a+x} &=& \mathbf{2(j+x)} \\ & a &=& 2(j+x) - x \\ & a &=& 2j + 2x - x \\ & a &=& 2j + x \\ & x &=& a-2j \quad | \quad a=26~j=6 \\ & x &=& 26-2*6 \\ & x &=& 26-12 \\ & x &=& 14 \\ \hline \end{array}\)

14 years from now, Arthur will be \(26+14=\mathbf{40}\) years old.
14 years from now, his son Arthur Jr will be \(6+14=\mathbf{20}\) years old.
Arthur’s age is then twice as old as Arthur Jr

Minor arc TU = 180 - 2*36 = 108 degrees

By the way, that is not the correct answer nor the same question.

My solution: Draw the altitude BP. Then PA = PC = 10. So, we have that 20*BP/2 = 280, so BP = 28. As we have that 10^2+28^2 = 884, the perimeter of ABC is 4√221 + 20.

$$z^3 = 8i$$

$$ \sqrt[3]{z^3} = \sqrt[3]{8i} \Rightarrow \sqrt[3]{8i} = \sqrt[6]{-64}$$

$$= -2i, i - \sqrt{3}, \sqrt{3} + i$$

$\square$

Arthur's age is 50 when he is twice as old as Arthur Jr.

I found another website with the same problemand the answer:

Thank you, a**s!

In the diagram below, ∠PQR = ∠PRQ = ∠STR = ∠TSR, RQ = 8, and SQ = 2. Find PQ.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

QR = QT = 8          XT = QS = 2           QL = 5            QZ = 4

XY = TL = sqrt( QT² - QL² ) = √39

∠PQR = tan^-1(√39 / 3)

PZ = tan∠PQR * QZ = 8.326663998

PQ = sqrt(QZ² + PZ²) = 9.237604326

Quarter = 0.25

Dime = 0.10

Nickel = 0.05

In one set of coins

   3×0.25 + 4×0.10 + 5×0.05

= 0.75 + 0.4 + 0.25

= 1.4

Let x be the number of sets of coins

1.4x = 5.6

x = 4

   4×(3+4+5)

= 48 coins

Area = (x+7)(x+5)-(2x-7)(x+2)

(x+7)(x+5) = x²+12x+35

(2x-7)(x+2) = 2x²-3x-14

(x²+12x+35) - (2x²-3x-14) = -x²+15x+49

Guest power

Once again, you have stolen another person's work...

It is disrespectful and not really good. Just give the link

Thank you so much!!!!!!!!!!!!!!

The probability is 5/18.

Tim can do it twice as fast as hardy- this means when both are working together the total is 3x hardy's speed. 

After 1/3 of packing is done, tim comes to help.

it would take 2 hours normally to pack the remaining 2/3 but at 3x the speed it takes 2/3 HOURS instead of 2 hours, so only 40 minutes?

Alone, it would take 40x3 + 60 = 180, 3 hours for hardy alone. Let me know if anything is errorous.

Hey hihihi!

I think you copied someone else's answer, you and I know that plagiarism is not cool, so next time you find a good answer, just give a link.

Here is the real answer Guest: https://www.wyzant.com/resources/answers/305259/geometry_word_problem