Question

Which sets of three numbers could represent the lengths of the sides of a right triangle?

• A. $7,9,11$
• B. $8,15,17$
• C. $10,15,20$
• D. $12,18,22$

Answer 1

The correct option is B

$8,15,17$

Explanation for the correct answer:

According to the converse of the Pythagorean theorem, if the square of the longest side is equal to the sum of square of other two sides, then it is a right triangle.

Option (B): $8,15,17$

Square of the longest side i.e ${\left(17\right)}^2=289$

Sum of squares of represented as $x^2+y^2$.

Substitute $x$ as $8$ and $y$ as $15$ in the expression $x^2+y^2$.

$$\begin{gathered} x^2+y^2={\left(8\right)}^2+{\left(15\right)}^2 \\ \Rightarrow x^2+y^2=64+225 \\ \therefore x^2+y^2=289 \end{gathered}$$

Square of the longest side $=$ Sum of squares of other two sides.

So, option B is correct.

Explanation for the incorrect answer:

Option (A): $7,9,11$

Square of the longest side ${\left(11\right)}^2=121$.

Sum of squares of represented as $x^2+y^2$.

Substitute $x$ as $7$ and $y$ as $9$ in the expression $x^2+y^2$.

$$\begin{gathered} x^2+y^2={\left(7\right)}^2+{\left(9\right)}^2 \\ \Rightarrow x^2+y^2=49+81 \\ \therefore x^2+y^2=130 \end{gathered}$$

Square of the longest side $\ne$ Sum of squares of other two sides

So, option A is incorrect.

Option (C): $10,15,20$

Square of the longest side i.e. ${\left(20\right)}^2=400$

Sum of squares of represented as $x^2+y^2$.

Substitute $x$ as $10$ and $y$ as $15$in the expression $x^2+y^2$.

$$\begin{gathered} x^2+y^2={\left(10\right)}^2+{\left(15\right)}^2 \\ \Rightarrow x^2+y^2=100+225 \\ \therefore x^2+y^2=325 \end{gathered}$$

Square of the longest side $\ne$ Sum of squares of other two sides

So, option C is incorrect.

Option (D): $12,18,22$

Square of the longest side i.e. ${\left(22\right)}^2=484$

Sum of squares of represented as $x^2+y^2$

Substitute $x$ as $12$ and $y$ as $18$ in the expression $x^2+y^2$.

$$\begin{gathered} x^2+y^2={\left(12\right)}^2+{\left(18\right)}^2 \\ \Rightarrow x^2+y^2=144+324 \\ \therefore x^2+y^2=468 \end{gathered}$$

Square of the longest side $\ne$ Sum of squares of other two sides

So, option D is incorrect.

Hence, option B is correct.