Grade 10 Maths (due tomrrow)

[Deathsnova]

A rectangular room measuring 4m by 8m is to be used for a home theatre. The walls of the room are 3m high, The projector is to be mounted on the ceiling halfway between the longer side walls, and facing one of the 4m wide walls where the screen will be mounted. The screen measures 1.6m by 0.9 m and is set up with its centre in the centre of the 4m wide wall. if the optimum distance for the projector is 7m from the centre of the screen, how far along hte celing from the wall on which the screen is mounted should the projector be installed?


Yeah its a bitch of a question.

 

[malakadang]

This question itself is a tad ambiguous.

It's simple Pythagoras.

You construct a Right Angle triangle, the right angle being from the ceiling. Down from where the screen lays on the wall is 2m. This is the center point, and assuming the projector shines on this center point at a 7m optimal length, we know the hypotenuse must be 7m.

With this, we have two lengths, and simple Pythagoras takes over. 7² - 2² = x². x being the length the projector lies from the screen. Therefore, x² = 45 and the square root of that is 6.71 2dp.

The screen projector is therefore 6.71m away from the screen, or, 1.29m away from the opposite wall.

 

[Regiregi90]

^ what he said i hated pythagoras the crappiest module within maths

useful if you know how to do it
if you dont well then your F*cked

 

[malyce]

Someone beat me to it

Malakadong is right but it seems the question is a trick question due to you also have screen dimensions, which don't really mean anything.

 

[lol jud]

Draw a diagram, always helps