Need urgent help for this question for maths (sc) (1 Viewer)

[goodie10]

http://www4.boardofstudies.nsw.edu.au/testResults.cfm?testID=14121949&testUUID=7CFF3F57-A22B-09D0-3DC54925186FBF19&question_id=2912&question_num=4&correctAnswer=B&user_answer=C&courseID=5010&testQuestionID=282836429

PLEASE HELP spent like half an hour trying to solve this

 

[Coookies]

Could you use pythagoras on one of the small triangles to find AD or BC and then find the area of the rectangle?

 

[goodie10]

Could you use pythagoras on one of the small triangles to find AD or BC and then find the area of the rectangle?

thats what i thought, but there wasnt anything that could give me a measurement for the base

 

[goodie10]

thats what i thought, but there wasnt anything that could give me a measurement for the base

and there's no measurement for the side

 

[dionb2014]

think about proving that triangles ade and ecb are similar. the areas will be the s^2 for the triangles.

 

[goodie10]

think about proving that triangles ade and ecb are similar. the areas will be the s^2 for the triangles.

Well, how would you solve this question?

 

[lath]

let x = AD
let y = AE
therefore EB = (10-y)

x^2 + y^2 = 6^2
x^2 + (10-y)^2 = 8^2

solve simulteously and you get y = 3.6
then sub back into equation, x = 4.8

therefore area = 4.8 x 10.

 

[fiesycal]

Well, how would you solve this question?

The rectangle is twice the area of the triangle DEC. If you visualise it trianlge EBC and triangle AED fit perfectly above triangle DEC. The area of DEC is 6x8/2 = 24. Therefore the area of the whole rectangle is twice this ie. 48. So the answer is B.

 

[lath]

The rectangle is twice the area of the triangle DEC. If you visualise it trianlge EBC and triangle AED fit perfectly above triangle DEC. The area of DEC is 6x8/2 = 24. Therefore the area of the whole rectangle is twice this ie. 48. So the answer is B.

you can't make inferences like that.

 

[fiesycal]

you can't make inferences like that.

It's not an inference, just a fact of rectangles. Any triangle with two of the vertices touching two corners of a rectangle is half the rectangles area.

 

[dionb2014]

Another way to solve it would be to use the similar polygon ratio. If you prove the simmilarity among the three triangles you can then say that 6:10 = 6^2:10^2 then as you have the area of the middle triangle you can sub it into the ratio. You can then do the same for the other triangle. This allows you to find the are of each of the three and add them together for the entire rectangle.

 

[dionb2014]

I also thought about solving them simultaneously lath.

 

[lath]

It's not an inference, just a fact of rectangles. Any triangle with two of the vertices touching two corners of a rectangle is half the rectangles area.

ahh ok...didn't know about that

 

[michaeljennings]

you can't make inferences like that.

You can lol

 

[dionb2014]

@michaeljennings
am I right?

 

[goodie10]

Thanks guys, I understand it now. Cause by finding the area of the centre triangle you are able to find the area of the whole of the triangle by using simple algebra and hence forth. Thanks for the help !

 

[michaeljennings]

@michaeljennings
am I right?

Yeah if you can prove they are similar, but it is a multiple choice question meaning that you shouldnt be spending too long on this question. Fiesycal's method is much more efficient

 

[RealiseNothing]

Every question should have an elegant way to doing it, in this case the triangle is half the area of the rectangle.

 

[michaeljennings]

Every question should have an elegant way to doing it, in this case the triangle is half the area of the rectangle.

+1

 

[nataaleee]

another question, for the ways to work out either of these:

http://www4.boardofstudies.nsw.edu....swer=D&courseID=5010&testQuestionID=282615477

http://www4.boardofstudies.nsw.edu....swer=C&courseID=5010&testQuestionID=282615479

thank you!