How do you work out this maths question? Please help :)?

The area of a right angled triangle is 20cm(squared)


The other angle is 45 degrees.


How do you work out how long the longest side of the triangle is?How do you work out this maths question? Please help :)?
If it's a right triangle, and one angle is 45 degrees, that means the third angle is 45 degrees as well since we know the sum of the angles must be 180 degrees. This means it is an isosceles right triangle.





In this case, we know the two shorter sides are of equal length. Let's imagine the triangle sitting on one of these shorter sides as its base, and we'll say the base is x cm. This means the vertical edge is the other identical side, so the height is x cm as well. If the area is 1/2 * base * height, then we have area = (x^2)/2.





Now, since we know the area is 20 sq. cm., we set these two equal and solve for x.





(x^2)/2 = 20





Multiply both sides by 2:


x^2 = 40





Take the square root of each side:


x = sqrt(40)





Note that normally we'd need the positive and negative square root of 40, but since we're dealing with the length of a triangle's side we can ignore the negative solution.





Now that we know what x is, we can find the third side (the longest side, the hypotenuse) using the Pythagorean Theorem:


c^2 = a^2 + b^2





In this case, a and b are both x. Thus:


c^2 = (sqrt(40))^2 + (sqrt(40))^2





So:


c^2 = 40 + 40 = 80





And again taking the square root:


c = sqrt(80) = 4*sqrt(5) after simplifying.





The shortcut in this problem is that if we know that we have an isosceles right triangle, we can multiply the length of the sides by sqrt(2) to get the length of the hypotenuse.


--%26gt; sqrt(2)*sqrt(40) = sqrt(80)