+0

0
468
4

1. The diagonals of rectangle PQRS intersect at point X. If PS = 10 and RS=24, then what is $$\cos\angle PXS$$

2. In a triangle ABC, take point D on $$\overline{BC}$$ such that $$DB = 14, DA = 13, DC = 4,$$and the circumcircles of triangles ADB and ADC have the same radius. Find the area of triangle ABC

3. Let $\mathcal{R}$ denote the circular region bounded by $x^2 + y^2 = 36$. The lines $x = 4$ and $y = 3$ partition $\mathcal{R}$ into four regions $\mathcal{R}_1$, $\mathcal{R}_2$, $\mathcal{R}_3$, and $\mathcal{R}_4$. Let $[\mathcal{R}_i]$ denote the area of region $\mathcal{R}_i$. If $[\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4]$, then compute $[\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]$

Apr 11, 2019
edited by yasbib555  Apr 11, 2019

#1
+1

1.  See the image X  = (12, 5)

Length of SX  = sqrt (12^2 + 5^2 ) = sqrt (169) = 13

Length of PX  = sqrt (12^2 + (10 - 5)^2 )  = sqrt (12^2 + 5^2) = sqrt (169) = 13

Using the Law of Cosines

RS^2 = PX^2 + SX^2 - 2(PX)(SX)cos(PXS)

10^2 = 13^2 + 13^2 - 2(13)(13) cos (PXS)

100 = 2(169) - 2(169)cos(PXS)

100 - 338

_________  = cos(PXS)

-338

-238

____  = cos (PXS)

-338

238                            119

___  = cos(PXS)  =   ____

338                            169   Apr 12, 2019
#3
0

thank you so much!!

yasbib555  Apr 12, 2019
#2
+2

Since No.1 has been solved by CPhill, I will solve No.2.

Solution No.2: $$\angle ABC$$ = $$\angle ACB$$ which implies that $$ABC$$ is isosceles. We drop the perpendicular $$M$$ from $$A$$ to $$BC$$; we get that $$MC = 9$$ and so $$MD = 5$$ . Now by Pythagoras on $$AMD$$..., $$AM = 12$$. Thus $$ABC$$'s areo equals to $$\frac{1}{2}(12*18 = 216) = \boxed{108}$$ .

RB - ∃

Apr 12, 2019
edited by RobertBoyle  Apr 12, 2019
#4
0

thank you so much too!!!

yasbib555  Apr 12, 2019