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)
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
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
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