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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?

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.

 Jan 8, 2019
edited by dietwater  Jan 10, 2019
 #1
avatar+3994 
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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.

smileysmiley

 Jan 8, 2019
 #3
avatar+7 
0

This was wrong for me.

dietwater  Jan 10, 2019
 #2
avatar+98172 
+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

 

 

cool cool cool

 Jan 8, 2019

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