Question

1. the diagram shows a right - triangle and three squares. the area of the largest square is 36 units². which combination could be the areas of the smaller squares? there could be more than one answer. (a) 6 & 30 (b) 8 & 28 (c) 4 & 16 b) for each answer you selected above, how long would the side lengths of the triangle be?

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with squares on its sides, if the areas of the squares on the two legs are $a$ and $b$, and the area of the square on the hypotenuse is $c$, then $a + b=c$. The area of the largest square is 36, so $c = 36$.

Step2: Check option (a)

If $a = 6$ and $b = 30$, then $a + b=6 + 30=36$. So this is a valid combination.

Step3: Check option (b)

If $a = 8$ and $b = 28$, then $a + b=8 + 28=36$. So this is a valid combination.

Step4: Check option (c)

If $a = 4$ and $b = 16$, then $a + b=4+16 = 20
eq36$. So this is not a valid combination.

Step5: Find side - lengths for valid options

For option (a):
If the area of a square $A_1 = 6$, then the side - length $s_1=\sqrt{6}$; if $A_2 = 30$, then the side - length $s_2=\sqrt{30}$; if $A_3 = 36$, then the side - length $s_3 = 6$.
For option (b):
If the area of a square $A_1 = 8$, then the side - length $s_1=\sqrt{8}=2\sqrt{2}$; if $A_2 = 28$, then the side - length $s_2=\sqrt{28}=2\sqrt{7}$; if $A_3 = 36$, then the side - length $s_3 = 6$.

Answer:

(a) The side - lengths are $\sqrt{6}$, $\sqrt{30}$, and 6.
(b) The side - lengths are $2\sqrt{2}$, $2\sqrt{7}$, and 6.