Book C: Pythagoras worksheet 3 - Solutions and Commentary
Introduction
Pythagoras' theorem can be used to solve engineering problems (unbelievable!).
In the 2 problems below, draw a sketch to illustrate the scenario, adding the values to the sketch.
Lastly calculate the solution showing line by line working to 2 decimal places.
Question 1
You have a 3.5m ladder. It is to be placed against a wall with a safe distance out from that wall at the ladder foot of 0.8m. How far up the wall will it reach?
Solution
In normal use, a ladder is lent against a wall, at an angle to the vertical. Assuming the ground and wall are at right angles to each other, and that the wall is vertical (this should be a reasonable assumption in most cases), the ladder will form the hypotenuse.
The ladder itself is 3.5 m long.
The ladder protrudes 0.8 m horizontally from the wall at its base.
A sketch of the arrangement is shown below.
The problem is solved using the formula \( a = \sqrt{c^2 - b^2} \), where \( c \) is the length of the ladder, and \( b \) is the distance that the ladder protrudes from the wall at its base. Plugging this in yields the answer.
\( a = \sqrt{c^2 - b^2} = \sqrt{3.5^2 - 0.8^2} = \sqrt{11.61} \) = 3.41 m (2 dp) #.
In other words, the ladder will go 3.41 m up the wall.
Question 2
A rectangular wall needs a diagonal brace to be fitted from the right bottom corner to the left top corner. The wall is 3.2m high and 6.8m long. Will a 7.3m brace be long enough?
Solution
The naming of the diagonal brace instantly identifies it as the hypotenuse, assuming the frame horizontals and uprights are at right angles to each other (this should be a reasonable assumption in most cases).
The length of the brace is 7.3 m long. For the brace to be suitable, the diagonal distance must be 7.3 m or less.
The wall is 3.2 m high and 6.8 m long.
The required brace length is calculated, and compared to the length of the brace that is on hand. If the brace length is at least the diagonal distance between the corners of the wall, the brace is suitable. If it is less (i.e. the brace is too short), it is not suitable.
A sketch of the arrangement is shown below.
The required length is found using the formula \( c = \sqrt{a^2 + b^2} \), where \( c \) is the required length of the brace, and \( a \) and \( b \) are the horizontal and vertical dimensions of the wall. Plugging this in yields the answer.
\( c = \sqrt{a^2 + b^2} = \sqrt{6.8^2 + 3.2^2} = \sqrt{56.48} \) = 7.51 m (2 dp).
The required length is 7.51 m. The brace is 7.3 m long.
The brace is 0.21 m too short, so the brace is not long enough #.