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In my maths assignment, it says:

  1. Construct a triangle with perimeter $115$ mm, altitude $70$ mm and vertical angle $45^\circ$.
  2. Constructing a triangle with altitude $76$ mm and base angles $60$ and $45$ degrees.
  3. Construct an isosceles triangle given $135$ mm as perimeter and altitude as $55$ mm.

I am very confused on what to do from the beginning. How do I construct such Triangles?

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  • $\begingroup$ The second looks like the easiest. Can you see how to get the sides? [Do you know some basic formulae like opposite/hypotenuse = sine?] $\endgroup$
    – almagest
    Jan 25, 2020 at 12:01
  • $\begingroup$ I know how, but the angles and altitude are giving me headaches... $\endgroup$
    – King Royal
    Jan 25, 2020 at 12:08
  • $\begingroup$ If I use 45°for b, 60° will not fit, because of the altitude $\endgroup$
    – King Royal
    Jan 25, 2020 at 12:10
  • $\begingroup$ SOHCAHTOA right? $\endgroup$
    – King Royal
    Jan 25, 2020 at 12:12
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    $\begingroup$ In (1) the perimeter is larger than twice the altitude, when this is inside the triangle. Since 115<2\cdot 70 it must be that the altitude is outside. $\endgroup$ Jan 25, 2020 at 12:25

1 Answer 1

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Call the triangle $ABC$, and the altitude $AD$.

(1) construct a triangle with perimeter length 115 and with the altitude length 70.

That is impossible. $ADB$ is a right-angled triangle, so $AB>AD$. Similarly, $AC>AD$. So the perimeter must be greater than $2AD$.

(2) construct a triangle with $\angle B=45\circ$, $\angle C=60\circ$ and $AD=76$.

We have $AB\sin\angle B=AD$, which gives us $AB$. Similarly $AC\sin\angle C=AD$, which gives us $AC$. Similarly, $AB\cos\angle B+AC\cos\angle C=BC$. So we have all the side lengths and constructing the triangle is trivial.

(3) construct a triangle with perimeter 135, $AD=55$ and $AB=AC$. Suppose $BC=x$. Then $AB=AC=\sqrt{55^2+\left(\frac{x}{2}\right)^2}$. So we have perimeter $135=x+2\sqrt{55^2+\left(\frac{x}{2}\right)^2}$. So we can solve for $x$ (approx 22.7) and hence get the side lengths.

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  • $\begingroup$ Any pictures or videos on how I can construct this? $\endgroup$
    – King Royal
    Jan 25, 2020 at 17:22
  • $\begingroup$ Should I use the perimeter as the base line? $\endgroup$
    – King Royal
    Jan 25, 2020 at 17:23
  • $\begingroup$ @YNK Thanks. Fixed. $\endgroup$
    – almagest
    Jan 25, 2020 at 18:22
  • $\begingroup$ @KingRoyal You keep asking question after question. But, you refuses to answer question posed by people who want to help you. I asked you a question 14 hours ago, which has gone unanswered until now. I am asking the same question again. What are the tools you are supposed to use to do these constructions? Only a compass and a straight-edge? almagest's answer is completely adequate for you to do the constructions. You sound like you are afraid to put your hand to a pen to draw something. Don't be afraid. Start doing your construction. $\endgroup$
    – YNK
    Jan 26, 2020 at 6:50
  • $\begingroup$ I don't see any questions..... I have constructed number 2 and 3, but am very confused in number one. Sorry if I am bothering anyone. It's my first time. $\endgroup$
    – King Royal
    Jan 26, 2020 at 14:04

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