Question

the right triangle below has legs of length a = 11 and b = 6. the hypotenuse has length c. answer the questions below to find how a, b, and c are related. part 1: compute the total combined area of the four triangles: part 2: compute the area of the large (outer) square: part 3: using your answers in parts 1 and 2, find the area of the small (inner) square. c² = part 4: we are given the side lengths a = 11 and b = 6. compute a² + b². a² + b² = part 5: use <, >, or = to complete the statement below. a² + b² c²

Explanation:

Step1: Calculate area of one triangle

The area of a right - triangle is $A=\frac{1}{2}ab$. Here $a = 11$ and $b = 6$, so the area of one triangle is $\frac{1}{2}\times11\times6=33$. The total combined area of the four triangles is $4\times33 = 132$.

Step2: Calculate area of large square

The side - length of the large (outer) square is $a + b=11 + 6=17$. So the area of the large square is $(a + b)^2=(11 + 6)^2=289$.

Step3: Calculate area of small square

The area of the small (inner) square is the area of the large square minus the total area of the four triangles. So $c^{2}=(a + b)^2-4\times\frac{1}{2}ab=289-132 = 169$.

Step4: Compute $a^{2}+b^{2}$

$a^{2}=11^{2}=121$ and $b^{2}=6^{2}=36$. Then $a^{2}+b^{2}=121 + 36=157$.

Step5: Compare $a^{2}+b^{2}$ and $c^{2}$

Since $a^{2}+b^{2}=157$ and $c^{2}=169$, we have $a^{2}+b^{2}

Answer:

Part 1: 132
Part 2: 289
Part 3: 169
Part 4: 157
Part 5: $<$