Answer:
Yes, the triangle is a right triangle.
Step-by-step explanation:
A triangle with a base of 8 meters and longest side is 17 meters.
Area of triangle is [tex]60\text{ m}^2.[/tex]
Let third side be x m
Semi-perimeter of triangle, [tex]s=\dfrac{8+17+x}{2}=\dfrac{25+x}{2}[/tex]
Using Heron's formula to find area of triangle.
[tex]\text{Area of triangle }=\sqrt{s(s-a)(s-b)(s-c)}[/tex]
[tex]60=\sqrt{\dfrac{25+x}{2}\left(\dfrac{25+x}{2}-8\right)\left(\dfrac{25+x}{2}-17\right)\left(\dfrac{25+x}{2}-x\right)}[/tex]
squaring both sides
[tex]3600=\dfrac{25+x}{2}\times \dfrac{x+9}{2}\times \dfrac{x-9}{2}\times \dfrac{25-x}{2}[/tex]
[tex]3600=\dfrac{-50625+706x^{2}-x^{4}}{16}[/tex]
[tex]x^4-706x^2+108225=0[/tex]
[tex](x^2-225)(x^2-481)=0[/tex]
[tex]x^2=225\text{ or }x^2=481[/tex]
[tex]x=\pm 15,\pm\sqrt{481}[/tex]
x should be positive integer.
So, [tex]x=15\text{ m}[/tex]
Using pthagoreous theorem:
[tex]17^2=15^2+8^2[/tex]
[tex]289=225+64[/tex]
[tex]289=289[/tex]
Therefore, the given triangle is a right triangle.