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.
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.