Thread: [SOLVED] SAT geometry

In the figure below, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the darkened region is

a) $\displaystyle 8+3pi$

b)$\displaystyle 10+3pi$

c)$\displaystyle 14+3pi$

d)$\displaystyle 1+6pi$

e)$\displaystyle 12+6pi$

Can you please show the steps? Thanks in advance

hey i forgot a major info. The graph stated that SR is 6, sorry i didn't include it on the graph!!

Originally Posted by fabxx

In the figure below, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the darkened region is

a) $\displaystyle 8+3pi$

b)$\displaystyle 10+3pi$

c)$\displaystyle 14+3pi$

d)$\displaystyle 1+6pi$

e)$\displaystyle 12+6pi$

Can you please show the steps? Thanks in advance

i will start you off.

Let $\displaystyle RC$ be $\displaystyle w$, that is, the width of the rectangle.
Let $\displaystyle AR$ be $\displaystyle l$, that is, the length of the rectangle.
Let $\displaystyle P$ be the perimeter of the darkened region.

Note that the diagonal $\displaystyle AC = 6$

we are given $\displaystyle l + w = 8$ ............(1)

by Pythagoras' theorem, $\displaystyle l^2 + w^2 = 6^2$ .............(2)

you have two simultaneous equations with two unknowns, you can solve for $\displaystyle l$ and $\displaystyle w$. the rest should be straight forward

$\displaystyle P = AS + AC + CT + \text{ arc}SBT$

now, $\displaystyle AS + l = 6 \implies \boxed{AS = 6 - l}$

$\displaystyle \boxed{AC = 6}$

$\displaystyle CT + w = 6 \implies \boxed{CT = 6 - w}$

$\displaystyle \boxed{\text{arc}SBT = \frac {90}{360} \cdot 2 \pi r}$, and you know $\displaystyle r$.

so you can find all the components and hence the perimeter

Hey i forgot to include a major info in the graph. The radius is 6 is meant that SR is 6. I already corrected it in my previous thread. (:

Originally Posted by fabxx

Hey i forgot to include a major info in the graph. The radius is 6 is meant that SR is 6. I already corrected it in my previous thread. (:

as you can see, we already figured that out. you told us that R was the center and the circle had a radius of 6 (RT = 6 also ). nothing changes in what i did

Fabxx,

This is SATs. Pythagoras is too long for SAT. You are supposed to get this answer in not more than 30 seconds, without a calculator. The worst thing a candidate make in a SAT exam is take it with a Texas Instruments. Really, this is not a joke.

To solve this question, follow me carefully:
It is given that AR + RC =8 ...(1)
SR=RB=CA=6 ...(2)

Perimeter = SA + arc ST + TC + CA (From above, we already know CA=6)

SA = 6-AR
TC= 6 - RC

Add the two equations above:
SA + TC = 12 - (AR+RC) = 12 - 8 =4

Therefore,

Perimeter = arc ST + 4 + 6 = $\displaystyle \frac{1}{4}*2*\pi*6$ + 4 +6 = 10+3$\displaystyle \pi$


As you can see, you save time by not solving simultaneous equations and not evaluating pythagoras expression.

Hope that helps.

Originally Posted by shailen.sobhee

Fabxx,


Perimeter = arc ST + 4 + 6 = $\displaystyle \frac{1}{4}*2*\pi*6$ + 4 +6 = 10+3$\displaystyle \pi$

how did you get that arc ST is $\displaystyle \frac{1}{4}*2*\pi*6$ ?

Thanks in advance (:

Originally Posted by fabxx

how did you get that arc ST is $\displaystyle \frac{1}{4}*2*\pi*6$ ?

Thanks in advance (:

$\displaystyle 2 \pi * 6$ gives the circumference. you have 1/4 of it (it's a quarter circle)

Thanks to shailen.sobhee and Jhevon!

Originally Posted by shailen.sobhee

Fabxx,

SR=RB=CA=6 ...(2)

How did you get that CA=6? CA isn't the radius, and from the graph, you can easily see that CA≠SR

Thanks in advance

Originally Posted by fabxx

How did you get that CA=6? CA isn't the radius, and from the graph, you can easily see that CA≠SR

Thanks in advance

you are correct. but RB is the radius, and CA = RB since they are both diagonals of the rectangle (sobhee mentioned this)