Question

Part A: What is the approximate length of rod PR? Round your answer to the nearest hundredth. Explain how you found your answer, stating the theorem you used. Show all your work.
Part B: The length of rod PR is adjusted to 16 feet. If width PQ remains the same, what is the approximate new height QR of the scaffold? Round your answer to the nearest hundredth. Show all your work.

Answer 1

Let AC be the ladder and A be the window.

Given: AC=15m, AB=12m, CB=am

In right angled triangle ACB,

(Hypotenuse)2=(Base)2+[Perpendicular)2 [By Pythagoras theorem]

⇒(AC)2=(CB)2+(AB)2

⇒(15)2=(a)2+(12)2

⇒225=a2+144

⇒a2=225−144=81

⇒a=81=9cm

Thus, the distance of the foot of the ladder from the wall is 9m.

Answer 2

Part A:

Let the length of rod PR be x ft

6^(2) + 14^(2) = x^(2) (Pythagorean theorem)

x^(2) = 232

x = sqrt 232

x = 15.2315462117...

x = 15 (Rounded to the nearest hundredth)

Therfore, the length of rod PR is 15 ft.

Part B:

Let the length of height QR be y ft.

y^(2) + 14^(2) = 16^(2) (Pythagorean theorem)

y^(2) = 16^(2) - 14^(2)

y^(2) = 60

y = sqrt 60

y = 7.7459666924...

y = 7 (Rounded to the nearest hundredth)

Therfore, the new height QR is 7 ft.