Answer:
Option A
Step-by-step explanation:
Here we need to tell which could be the side lengths of a right angled triangle . As we know that in a right angled triangle , the sum of square two sides along the right angle is equal to the square of side opposite to right angle .
That is ,
[tex]\longrightarrow a^2 = b^2+ c^2 [/tex]
[tex]\begin{picture}(8,8)\setlength{\unitlength}{1cm} \thicklines\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){3}}\put(4,3){\line(-4,-3){4}}\put(2,-0.5){$\sf \sqrt{48}in$}\put(4.4,1.5){$\sf \sqrt{12}in. $} \put(1.2,2){$\sf \sqrt{60}in. $}\end{picture}[/tex]
We can start by looking at the options , first option is ,
[tex]\longrightarrow \sqrt{48}\ in , \sqrt{12}\ in , \sqrt{60} in [/tex]
So by squaring first two sides and adding them , we have ,
[tex]\longrightarrow (\sqrt{48})^2+(\sqrt{12})^2\\[/tex]
[tex]\longrightarrow 48+12 = 60[/tex]
And by squaring the third side we have ,
[tex]\longrightarrow \sqrt{60}^2 = 60 [/tex]
Hence here the sum of square of two sides is equal to the square of third side .
Therfore the triangle with sides √48 in , √12in and √60 forms a right angle ∆.
