Question

1. refer to the map of the woodlands camp at the right.

round your answers to the nearest tenth.
a. how far is it from sycamore cabin to oak cabin?
b. a camper in hickory cabin wants to visit a friend in
elm cabin. how much farther is it if she walks to the
mess hall first?

1. justify conclusions rodrigo is buying a ( 5 \frac { 1 } { 2 } ) -foot-long fishing rod for

his father for his birthday. he wants to put it in a box so that his dad will not be
able to guess what is in the box. the box he wants to use is 4 feet long and
4 feet wide. will the pole fit in the box? justify your reasoning.

Explanation:

5. a.

Step1: Use the Pythagorean theorem

The distance from Sycamore cabin to Oak cabin forms a right - triangle with legs \(30\) yd and \(50\) yd. The Pythagorean theorem is \(d=\sqrt{a^{2}+b^{2}}\), where \(a = 30\) and \(b = 50\).

$$d=\sqrt{30^{2}+50^{2}}=\sqrt{900 + 2500}=\sqrt{3400}$$

Step2: Calculate the square root

\(\sqrt{3400}\approx58.3\) yd

5. b.

Step1: Calculate the distance via Mess Hall

The distance from Hickory to Mess Hall is \(60\) yd, and from Mess Hall to Elm is \(40\) yd. The total distance is \(60 + 40=100\) yd.

Step2: Calculate the direct distance

Using the Pythagorean theorem for the right - triangle with legs \(60\) yd and \(40\) yd. \(d=\sqrt{60^{2}+40^{2}}=\sqrt{3600+1600}=\sqrt{5200}\approx72.1\) yd

Step3: Find the difference

The difference is \(100-72.1 = 27.9\) yd

6.

Step1: Calculate the length of the diagonal of the box

Using the Pythagorean theorem for a rectangle (the box) with length \(l = 4\) ft and width \(w = 4\) ft. The diagonal \(D\) of the box is \(D=\sqrt{l^{2}+w^{2}}\), where \(l = 4\) and \(w = 4\)

$$D=\sqrt{4^{2}+4^{2}}=\sqrt{16 + 16}=\sqrt{32}\approx5.7$$

ft

Step2: Compare the length of the rod and the diagonal

The length of the rod is \(5\frac{1}{2}=\frac{11}{2}=5.5\) ft. Since \(5.5\lt5.7\)

Answer:

a. The distance from Sycamore cabin to Oak cabin is approximately \(58.3\) yd.
b. It is approximately \(27.9\) yd farther to walk via the Mess Hall.

1. Yes, the pole will fit in the box. Because the length of the diagonal of the box (\(\approx5.7\) ft) is greater than the length of the fishing rod (\(5.5\) ft).