The sum of the lengths of the legs of a right-angled triangle is $12$, and the hypotenuse is $\sqrt{74}$. Find the total area of the triangle.

Question

I am new to math and I feel stupid; I am trying to fix an issue and I would love it if some one could give me a hand and explain to me step by step. The main thing is I want to understand the solution; I don't just wanna a final number for each case.

Thanks in advance :)

The sum of the lengths of the legs of right-angled triangle is $12$, and the hypotenuse is $\sqrt{74}$. Find the total area of the triangle.

problem 1

Find the minimum value of the function $y=2+x^2(3-x)$ in the intercept $[1;4]$.

Solve the inequality $\log_3(4x^2+1) \ge \log_3(3x^2-4x+1)$.

Answer 1

Hint for Problem 8: we have $$a+b=12$$ squaring this equation we get $$a^2+b^2+2ab=144$$ Now use that $$S=\frac{ab}{2}$$ and $$c^2=a^2+b^2=74$$ Hint: $$y=2+3x^2-x^3$$ then $$y'(x)=6x-3x^2=3x(2-x)$$ and compute the values for $x=0,x=2,x=1$ and $x=4$

Answer 2

Hints for problem $8$. For the rest you will have to show some effort and self work:

If the triangle's legs are denoted $\;x,\,y\;$ , then we have that $$\begin{cases}\;x^2+y^2=74\\{}\\x+y=12\end{cases}\;\;\;\implies x^2+(12-x)^2=74\;$$

Solve the last quadratic in $\;x\;$ and find the numerical value of $\;x\;$ , and then of $\;y\;$ .

Answer 3

For your first problem: you know that the area of a triangle is: $$ \frac{1}{2}\cdot base \cdot height$$ So you need to find this values. By Pythagoras theorem you know that: $a^2 + b^2 = 74$, and it is given that $a+b = 12$. Therefore: $a=12-b$, which can be substituted to obtain: $$(12-b)^2 +b^2= 74$$ $$b^2 - 24b+144 +b^2 = 74$$ $$2b^2 -24b + 70 = 0$$ $$b^2 -12b+35 = 0$$ $$(b-7)(b-5) = 0$$ Hence $b=5$ or $b=7$, and therefore: $a=5$ or $a=7$ So the Area of the triangle when $a=$base and $b=$ height is?

Can you continue from here?

Answer 4

$x+y=12$;

$(x+y)^2= x^2+y^2 +2xy= 144$.

Pythagoras:

$x^2+y^2= (\sqrt{74})^2= 74;$

Hence : $2xy = 70;$

Area= $(1/2)xy= 70/4$. (Why)