Answer:
We conclude that [tex]\sqrt{5}, \sqrt{5},\sqrt{10}[/tex] is an isosceles right triangle.
Step-by-step explanation:
Given the sides of a triangle
[tex]\sqrt{5}, \sqrt{5},\sqrt{10}[/tex]
Given that the two sides are equal. Thus, it must be an isosceles triangle.
Let us check whether it is an isosceles right triangle or not.
We know that for a right-angled triangle with sides a, b and the hypotenuse c is defined as:
[tex]c=\sqrt{a^2+b^2}[/tex]
Given
[tex]a=\sqrt{5}[/tex]
[tex]b=\sqrt{5}[/tex]
now substituting [tex]a=\sqrt{5}[/tex] and [tex]b=\sqrt{5}[/tex] in the equation
[tex]c=\sqrt{a^2+b^2}[/tex]
[tex]c=\sqrt{\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^2}[/tex]
[tex]c=\sqrt{10}[/tex]
Thus,
[tex]a=\sqrt{5}[/tex]
[tex]b=\sqrt{5}[/tex]
[tex]c=\sqrt{10}[/tex]
Which satisfies the given side lengths [tex]\sqrt{5}, \sqrt{5},\sqrt{10}[/tex]
Therefore, we conclude that [tex]\sqrt{5}, \sqrt{5},\sqrt{10}[/tex] is an isosceles right triangle.
