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1. Two squares with side lengths 5 and 4 are placed next to each other as shown in the figure below. What is the area of the shaded region?
https://docs.google.com/drawings/d/1-Gf1dNe3ADs-ixHxNNKpr2_ANW5ncMSh4IMOWHcQoNo/edit?usp=sharing (image link)

2 .Equilateral triangle \(\triangle DEF\) overlays square \(ABCD\) so that \(B\) is the midpoint of \(EF.\) What is the ratio of the area of \(ABCD\) to that of \(\triangle DEF?\)

https://docs.google.com/drawings/d/1zgvBS_XDR79c7MgZ5hlkmOqbe8fXJ5uoO6d8uzW0NjI/edit?usp=sharing (image link)
 

3. In \(\triangle ABC\) shown in the figure, the length of \(AB, BC\) and \(CA\) are respectively \(6, 8\) and \(10,\) the angle bisector of \(\triangle BAC\) meets \(BC\) at \(D.\) Find the area of \(\triangle ADC\)

https://docs.google.com/drawings/d/1w_VckTc1ljm6frQCKD2ZvKr8vYS4KtsPCLPIVyHXVlg/edit?usp=sharing (image link)

4. Let \(\triangle ABC\) be an equilateral triangle. There is a point \(O\) inside the triangle such that \(\angle AOB=115^{\circ}\) and \(\angle BOC=125^{\circ}.\)
If a triangle is constructed with sides with length \(AO, BO,\) and \(CO,\) find the degree measure of the largest interior angle of this triangle.
https://docs.google.com/drawings/d/1F-W0NVc6idJkiP4vuJLyIEK-Njo8PyQpXEanatZAzvQ/edit?usp=sharing (image link)

 Apr 20, 2021
edited by mworkhard222  Apr 20, 2021
 #1
avatar+118611 
+2

1)  The  toal area  of  boh squares = 4^2  + 5^2  =  16 +  25  = 41

 

We  have three squares outside  the  shaded  region

 

One is a  right triangle with  legs  of  1  and  5.....its area  =  (1/2) 5 * 1 =  5/2  =  2.5

 

Another  is a triangle  ith a height of  5 and a  base of  9.....its  area  = (1/2)(9) (5)  = 45/2  = 22.5

 

The last  triangle is also a right triangle with legs of  4...its area = (1/2) (4) (4)  = 16/ 2  =  8

 

Shaded  region  =   41   -  [ 2.5  + 22.5  + 8]  =    8  (units)^2

 

 

cool cool cool

 Apr 20, 2021
 #2
avatar+118611 
+2

2)  Let  the side  of  the  square  =  S.....so [ ABCD ]  =  S^2

 

The  heght of  the  equilateral  triangle =  Ssqrt (2)

 

The  side  of  the equilateral  triangle  =  Ssqrt (2) * 2 / sqrt (3) =  (2/3)sqrt  (6)  S

 

Area  of  the  equilateral  trinagle = (1/2)  ( (2/3) sqrt (6 S))^2 sqrt (3)  / 2    =  (2sqrt(3) / 3 *  S^2

 

Ratio  of      area  of    square  to  area  of  equilateral triangle =   S^2  /   [ 2sqrt (3)  / 3 * S^2  ]   =

 

3 / [2 sqrt (3) ]   =   sqrt (3)  / 2

 

 

cool cool cool

 Apr 20, 2021
 #3
avatar+118611 
+2

3)  Let DC =  x     and  BD  =  8  -   x

 

Since  AD  is  an angle  bisector, we  have  the following relationship

 

AB/ BD =  AC / DC

 

6/ (8 - x)  =  10 / x               cross-multiply

 

6x  =  10  (8- x)

 

6x  =  80   - 10x

 

6x + 10x  =  80

 

16x  =  80

 

x = 80 / 16   =   5  =   DC

 

The  area  of  triangle   ADC  =  (1/2) (AB) (DC)  = (1/2)(6)(5)  = 15

 

 

cool cool cool

 Apr 20, 2021

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