How to crack this SAT Geometry problem? Thanks :)

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2 Answers
Feb 1, 2018

D. 65

Explanation:

We shall start with the center triangle, we are given an angle, and the length of the side opposite angle.

First, we want the hypotenuse of the triangle. We know the opposite side.
sin(30)=O /H =5/H
H=5/sin(30)=10

Given an angle 30^circ, a hypotenuse 10, and a side x, we need to find x.

cos(30)=A/H=x/10
x=10cos(30)=5sqrt(3)

Now we need y:
Going back to our center triangle. We need the other side now, we already have 5 and 10, so we just use Pythagorus to find the remaining side.

a=sqrt(10^2-5^2)=5sqrt(3)

"Hypotenuse of left triangle"=5sqrt(3)

y="adjacent"

cos(30)=A/H=y/(5sqrt(3))

y=5sqrt(3)cos(30)=7.5

"Area of rectangle"=xy=7.5*5sqrt(3)~~64.95~~65cm^2-=D

Feb 1, 2018

Answer is (D).

Explanation:

The diagram has been redrawn naming the points.

enter image source here

The diagram is divided in four triangles DeltaAPD, DeltaPQD, DeltaQCD andDeltaBQP. They are all similar and their angles are 30^@, 60^@ and 90^@.

Before we proceed further, it may be noted that if smallest side in such triangles is a, the hypotenuse is 2a and third side is sqrt3a. Further, let the unit be centimeter and for ease we do not mention centimeter, each time we find length of any line.

Therefore as in DeltaPQD we have PD=5, DP=5sqrt3 and DQ=10. Now, as DQis also hypotenuse of DeltaQCD, QC=5 and CD=5sqrt3 i.e. x=5sqrt3.

In DeltaAPD, we have seen hypotenuse DP=5sqrt3, therefore AP being smallest side is (5sqrt3)/2 and AD=(5sqrt3)/2xxsqrt3
= (5xx3)/2=15/2 i.e. y=15/2.

Hence area of tile is 5sqrt3xx15/2=75sqrt3/2=37.5xx1.732=64.95~=65

Hence, answer is (D).