Question

1. a 15-ft long ladder leans up against the side of a house. it reaches 12 ft up the side of the house. how far is the base of the ladder from the base of the house to the nearest tenth of a ft? draw a labelled diagram and show your calculation.
2. a rectangular sports field measures 45m long and 35m wide. what is the length of the diagonal distance from one corner to the farthest opposite corner to the nearest metre?

draw a labelled diagram and show your calculation.

Explanation:

Problem 4:

Step1: Identify the triangle type

This is a right - triangle problem, where the ladder is the hypotenuse (\(c = 15\) ft), the height on the house is one leg (\(a=12\) ft), and the distance from the base of the ladder to the house is the other leg (\(b\)). We use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\).

Step2: Rearrange the formula to solve for \(b\)

We get \(b^{2}=c^{2}-a^{2}\). Substitute \(c = 15\) and \(a = 12\) into the formula: \(b^{2}=15^{2}-12^{2}=225 - 144=81\).

Step3: Solve for \(b\)

Take the square root of both sides: \(b=\sqrt{81} = 9\) ft.

Step1: Identify the triangle type

For a rectangle, the diagonal forms a right - triangle with the length and width of the rectangle. The length (\(l = 45\) m) and width (\(w = 35\) m) are the legs of the right - triangle, and the diagonal (\(d\)) is the hypotenuse. We use the Pythagorean theorem \(d^{2}=l^{2}+w^{2}\).

Step2: Substitute the values

Substitute \(l = 45\) and \(w = 35\) into the formula: \(d^{2}=45^{2}+35^{2}=2025 + 1225=3250\).

Step3: Solve for \(d\)

Take the square root of 3250: \(d=\sqrt{3250}\approx57\) m (rounded to the nearest metre).

Answer:

The base of the ladder is 9.0 ft from the base of the house.

Problem 5: