Question

Which of the following lengths make a right triangle?

• A. $\sqrt{5},3,\sqrt{14}$
• B. $6,2\sqrt{3},8$
• C. $3\sqrt{2},4\sqrt{2},5\sqrt{2}$
• D. $2\sqrt{3},2\sqrt{2},2\sqrt{5}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\sqrt{5},3,\sqrt{14}$
C $3\sqrt{2},4\sqrt{2},5\sqrt{2}$
D $2\sqrt{3},2\sqrt{2},2\sqrt{5}$

If in a triangle, the square of the longest side equals the sum of the squares of the other two sides, then the triangle is right-angled.

$(\sqrt{5}{)}^2+3^2=5+9=14=(\sqrt{14}{)}^2$

Thus, the lengths $\sqrt{5},3$ and $\sqrt{14}$ form a right-triangle.

Similarly, $(3\sqrt{2}{)}^2+(4\sqrt{2}{)}^2=18+32=50=(5\sqrt{2}{)}^2$
shows that the lengths $3\sqrt{2},4\sqrt{2}$ and $5\sqrt{2}$ form the sides of a right-angled triangle.

Similarly, $(2\sqrt{3}{)}^2+(2\sqrt{2}{)}^2=12+8=20=(2\sqrt{5}{)}^2$ shows that the lengths $2\sqrt{3},2\sqrt{2}$ and $2\sqrt{5}$ form the sides of a right-angled triangle.

But, $6^2+(2\sqrt{3}{)}^2=36+12=48\ne 64=8^2)$

Thus, $6,2\sqrt{3}$ and $8$ would not form a right triangle.