Question

How to crack this SAT Geometry problem? Thanks :)

Answer 1

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`

Answer 2

Answer is (D).

Explanation:

The diagram has been redrawn naming the points.

The diagram is divided in four triangles `DeltaAPD`, `DeltaPQD`, `DeltaQCD` and`DeltaBQP`. 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 `DQ`is 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).