Question

practice & problem solving
leveled practice in 7 and 8, find the missing side length of each triangle.
7
9
12
$\square^2 + 12^\square = c^2$
$\square + \square = c^2$
$\square = c^2$
$\sqrt{\square} = \sqrt{\square}$
$c = \square$
the length of the hypotenuse is $\square$ units.
8.
6 in.
8 in.
$\square^2 + b^2 = \square^2$
$\square + b^2 = \square$
$b^2 = \square$
$\sqrt{\square} = \sqrt{\square}$
$b \approx \square$
the length of leg $b$ is about $\square$ inches.

Explanation:

Step1: Apply Pythagorean theorem (Q7)

$9^2 + 12^2 = c^2$

Step2: Calculate squared terms (Q7)

$81 + 144 = c^2$

Step3: Sum the values (Q7)

$225 = c^2$

Step4: Take square roots (Q7)

$\sqrt{225} = \sqrt{c^2}$

Step5: Solve for c (Q7)

$c = 15$
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Step1: Apply Pythagorean theorem (Q8)

$6^2 + b^2 = 8^2$

Step2: Calculate squared terms (Q8)

$36 + b^2 = 64$

Step3: Isolate $b^2$ (Q8)

$b^2 = 64 - 36 = 28$

Step4: Take square roots (Q8)

$\sqrt{b^2} = \sqrt{28}$

Step5: Solve for b (Q8)

$b \approx 5.3$

Answer:

For Triangle 7:

$9^2 + 12^2 = c^2$
$81 + 144 = c^2$
$225 = c^2$
$\sqrt{225} = \sqrt{c^2}$
$c = 15$
The length of the hypotenuse is 15 units.

For Triangle 8:

$6^2 + b^2 = 8^2$
$36 + b^2 = 64$
$b^2 = 28$
$\sqrt{b^2} = \sqrt{28}$
$b \approx 5.3$
The length of leg $b$ is about 5.3 inches.