I am new to math and I feel like stupid am trying to fix an issue I would love if some one give me a hand and explain to me by steps, the main thing is I wanna understand the solution I don't just wanna a finale number for each case,

thank 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

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problem 2

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

problem 3

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


closed as off-topic by Namaste, Did, Martin Sleziak, user284331, Saad Apr 8 '18 at 0:33

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    $\begingroup$ Welcome to MSE. Since you are new here, let me tell you some of the rules and customs of the site. Please ask one question at a time. Please type your questions using MathJax instead of posting images. Images are not searchable and not accessible to people using screen readers for example. You will got a lot more help and a fewer votes to close if you show that you have made some effort. What have you done on the problems? Where are you stuck? $\endgroup$ – saulspatz Apr 7 '18 at 20:03
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  • $\begingroup$ $$ \frac{1}{2}ab = \frac{1}{4}\left[(a+b)^2-(a^2+b^2)\right].$$ $\endgroup$ – Jack D'Aurizio Apr 7 '18 at 20:59
  • $\begingroup$ Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. $\endgroup$ – Martin Sleziak Apr 7 '18 at 23:56
  • $\begingroup$ For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have typed formulas from the pictures in your post. $\endgroup$ – Martin Sleziak Apr 7 '18 at 23:57

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 $x=4$

  • $\begingroup$ Once you get $a^2+b^2+2ab=144$ and $a^2+b^2=74,$ it's not so many more steps to find $\frac12ab.$ $\endgroup$ – David K Apr 7 '18 at 20:17
  • $\begingroup$ thank you ! this step a2+b2+2ab=144 was the breaker $\endgroup$ – sasha Apr 7 '18 at 20:27
  • $\begingroup$ That is nice, have a nice evening! $\endgroup$ – Dr. Sonnhard Graubner Apr 7 '18 at 20:30

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?

  • $\begingroup$ yes i can :) thank you $\endgroup$ – sasha Apr 7 '18 at 20:28

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 $\;y\;$ .

  • $\begingroup$ awesome ! i got the idea now , thanks buddy :) $\endgroup$ – sasha Apr 7 '18 at 20:28


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


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

Hence : $2xy = 70;$

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


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