Question

question 5 of 10
the diagram shows squares 1, 2, and 3 constructed on the sides of a right triangle.
which statement about the squares must be true?
a. (area of 1) + (area of 2) = (perimeter of 3)
b. (perimeter of 1) + (perimeter of 2) = (area of 3)
c. (perimeter of 1) + (perimeter of 2) = (perimeter of 3)
d. (area of 1) + (area of 2) = (area of 3)

Explanation:

Step1: Recall the Pythagorean theorem

Let the side lengths of the right - triangle be \(a\), \(b\), and \(c\) (where \(c\) is the hypotenuse). The area of square 1 with side length \(a\) is \(A_1=a^{2}\), the area of square 2 with side length \(b\) is \(A_2 = b^{2}\), and the area of square 3 with side length \(c\) is \(A_3=c^{2}\).

Step2: Apply the Pythagorean theorem

By the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Substituting the area formulas (\(A_1 = a^{2}\), \(A_2=b^{2}\), \(A_3 = c^{2}\)), we get \(A_1+A_2=A_3\).

For perimeter, if the side lengths of the squares are \(a\), \(b\), \(c\), the perimeter of square 1 is \(P_1 = 4a\), the perimeter of square 2 is \(P_2=4b\), and the perimeter of square 3 is \(P_3 = 4c\).
\(P_1+P_2=4(a + b)\) and \(P_3=4c\). Since \(a + b>c\) (triangle inequality, \(a,b,c>0\) and \(a^{2}+b^{2}=c^{2}\)), \(P_1 + P_2
eq P_3\). Also, area and perimeter have different units, so options A and B are incorrect because we can't add an area to a perimeter or a perimeter to an area in a meaningful equality sense (they are different physical quantities).

Answer:

D. (Area of 1) + (Area of 2) = (Area of 3)