Question 674652
What is the number of square units in the area of a triangle whose sides are 5, 6 and &#8730;<span style="text-decoration: overline">61</span>.  Explain answers in simplest form.  (I think it's 21, but i think it's wrong.)


You never bothered to share with us how you arrived at 21 sq units. 


However, you need to use the law of cosines formula, {{{a^2 = b^2 + c^2 - 2bc Cos A}}}

After substituting the sides (based on their angles), into the above-formula, you will derive the degree measure of angle A. 


Using the degree measure of angle A, you can then determine the area of the triangle by applying the following formula for a non-right triangle: {{{A = (1/2)bc Sin A}}}


This should give you an area of {{{15.001}}} &#8776; {{{highlight_green(15)}}} sq units


OR


It can be seen that the triangle is actually a right-angled one, so you can use the pythagorean formula: {{{a^2 + b^2 = c^2}}}, with c being the longest side, or the hypotenuse of the triangle, represented by {{{sqrt(61)}}}. 


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