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# Help on HW

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1) Square ABCD and rectangle ACEG have the same area. Find the ratio AC: CE.

2) Right triangle ABC has legs AC = 21 and BC = 28 and hypotenuse AB = 35. A square is inscribed in triangle ABC as shown below. Find the side length of the square.

3) A large square is divided into 4 small congruent rectangles and a small square as shown. The areas of the large and small squares are 25 and 7, respectively. What is the length of a diagonal of a small rectangle?

4) A rectangle and a square have the same perimeter. One side-length of the rectangle is 25% longer than the other. What is the ratio between the areas of the rectangle and the square?

5) Shown below is rectangle EFGH. Its diagonals meet at Y. Let X be the foot if an altitude is dropped from E to line FH. If EX = 24 and GY = 25, find the perimeter of rectangle EFGH.

Jan 8, 2019
edited by dietwater  Jan 10, 2019

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1. Imagine the side lengths of the square are (s, s, s, s) . The area is the same since s*s=s^2 Now, label the figures correctly and you would see that AC is the diagonal of the square, which is $$s\sqrt{2}.$$ Now, we have for rectangle AC*CE=$$s^2$$, but we already know AC, so we have $$CE=\frac{s^2}{s\sqrt{2}}$$ . Rationalizing the denominator, we get $$CE=s\sqrt{\frac{5}{2}}$$ . Now, dividing AC by CE, we get an answer of $$\frac{2}{\sqrt{5}}$$ , which can be changed into   $$\boxed{\frac{2\sqrt{5}}{5}}.$$

Do you have a diagram for problem 2?

Hope this helped,

tertre.

Jan 8, 2019
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This was wrong for me.

dietwater  Jan 10, 2019
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2) Right triangle ABC has legs AC = 21 and BC = 28 and hypotenuse AB = 35. A square is inscribed in triangle ABC as shown below. Find the side length of the square.

Let the side of the square = x

We will have two smaller right triangles formed within the larger triangle as well as the area of the square

One triangle will have an area of  (1/2) (21 - x) * x   and the other will have an area of (1/2)(28 - x) * x

So....the sum of these three areas will equal the area of the large right triangle

So we have that

(1/2) 21 * 28  =  (1/2) x [ (21 - x) + (28 - x) ] + x^2

21 * 28 =  x [ 49 - 2x ] + 2x^2

588 = 49 x - 2x^2 + 2x^2

588 = 49x

x = 588/49   =   12  units

Jan 8, 2019