Question about forming an equation to calculate area of rectangle

frank.

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Triangle ABC is isosceles with AB=AC=20 cm and BC=24 cm. A rectangles PQRS is drawn inside the triangle with PQ on BC, and S and R on AB and AC respectively (a rectangle inside of an isosceles triangle).

If PQ = 2x cm, show that the area A cm² of the rectangle is given by A = 8x(12-x)/3.


So I think the key to solving this is finding an expression for PS. I've worked out the height of the isosceles triangle using Pythagoras, but that didn't get me anywhere. Then I realised (or I think) that right angled triangle BPS is a similar triangle to one half of the isosceles. So I worked out the scale factor, which is 12/(12-x). Hence, the length of BS should be 20/scale factor, which is (60-5x)/3. I then tried to worked out PS using Pythagoras (because I know what BS and BP are in terms of x). Then the equation got really messy and I'm stuck. Not really sure if what I did is right because it's a lot of work given the amount of marks available for this part of the question.

Any help would be appreciated, thanks.
 
Triangle ABC is isosceles with AB=AC=20 cm and BC=24 cm. A rectangles PQRS is drawn inside the triangle with PQ on BC, and S and R on AB and AC respectively (a rectangle inside of an isosceles triangle).

If PQ = 2x cm, show that the area A cm² of the rectangle is given by A = 8x(12-x)/3.


So I think the key to solving this is finding an expression for PS. I've worked out the height of the isosceles triangle using Pythagoras, but that didn't get me anywhere. Then I realised (or I think) that right angled triangle BPS is a similar triangle to one half of the isosceles. So I worked out the scale factor, which is 12/(12-x). Hence, the length of BS should be 20/scale factor, which is (60-5x)/3. I then tried to worked out PS using Pythagoras (because I know what BS and BP are in terms of x). Then the equation got really messy and I'm stuck. Not really sure if what I did is right because it's a lot of work given the amount of marks available for this part of the question.

Any help would be appreciated, thanks.
Assuming the base of the triangle BC = 24, the height of the triangle is 16.

Using similar triangles, PS = 16/12 * (12-x)

So area of the rectangle = PQ * PS ..... and continue....
 
Assuming the base of the triangle BC = 24, the height of the triangle is 16.

Using similar triangles, PS = 16/12 * (12-x)

So area of the rectangle = PQ * PS ..... and continue....

Oh dear God, I eventually realised that I had equated 20² to 200 in my calculations, which was why the height I worked out turned out to be a surd. The question actually told me that A = 8x(12-x)/3, so I tried to find a way to avoid the surd. Of course, I should have recognised the Pythagorean triplet in the first place. I then ended up making y/12-x = 16-y/x where y=PS. Then substituted y into A=2xy. Your answer looks much simpler though; don't know how I missed that. Thanks for the help.
 
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