Question

find the length of the missing side of the right triangle. round to three decimal places, if necessary. the legs of the right triangle are represented by a and b, and the hypotenuse is represented by c.

1. a = 6, b = 8

1. a = 5, c = 13

solve the problem. if necessary, round to the nearest tenth.

1. on a sunny day, a flag pole and its shadow form the sides of a right triangle. if the hypotenuse is 52 m long and the shadow is 48 m, how tall is the flag pole?

1. a car dealer advertised a big sale by stretching a string of banners from the top of the building to the edge of the driveway. if the building is 21 m high and the driveway is 43 m from the building, how long is the string of banners?

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle, \(c^{2}=a^{2}+b^{2}\) (when \(c\) is the hypotenuse) or \(b^{2}=c^{2}-a^{2}\) (when \(b\) is a leg).

Problem 1:

Given \(a = 6\), \(b = 8\), and \(c\) is the hypotenuse.

$$c=\sqrt{a^{2}+b^{2}}=\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100}$$

Problem 2:

Given \(a = 5\), \(c = 13\), and \(b\) is a leg.

$$b=\sqrt{c^{2}-a^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}$$

Problem 3:

Let the height of the flag - pole be \(h\), hypotenuse \(c = 52\) m, shadow \(a=48\) m.

$$h=\sqrt{c^{2}-a^{2}}=\sqrt{52^{2}-48^{2}}=\sqrt{(52 + 48)(52 - 48)}=\sqrt{100\times4}=\sqrt{400}$$

Problem 4:

Let the length of the string of banners be \(l\), height of the building \(a = 21\) m, distance of the driveway from the building \(b = 43\) m.

$$l=\sqrt{a^{2}+b^{2}}=\sqrt{21^{2}+43^{2}}=\sqrt{441+1849}=\sqrt{2290}$$

Step2: Calculate the values

Problem 1:

\(c=\sqrt{100}=10\)

Problem 2:

\(b=\sqrt{144}=12\)

Problem 3:

\(h=\sqrt{400}=20\)

Problem 4:

\(l=\sqrt{2290}\approx47.8\)

Answer:

1. \(10\)
2. \(12\)
3. \(20\) m
4. Approximately \(47.8\) m