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?

Write your answer as a common fraction (or an integer)

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.

dietwater Jan 8, 2019

#1**+2 **

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.

tertre Jan 8, 2019

#2**+2 **

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

CPhill Jan 8, 2019