Beware of the Sneaky Options! | Geometry

Test takers often ask for tips and short cuts to cut the work. So the question writer might want to award the test taker who pays attention to details and puts in the required effort. Today, we will look at an example of this – if it seems to be too easy, it is a trap!

Question: In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) ${8\times\sqrt2}$

(B) ${24\times\sqrt3}$

(D) ${144\times\sqrt2}$

(E) 384

Solution:

My first thought on seeing this question: equilateral triangle so I am thinking area $= \frac{s^2 \times \sqrt3}{4}$ and/or altitude = $\frac{s\times\sqrt3}{2}$ .

The irrational number in play is $\sqrt3$ . There is only one option with $\sqrt3$ in it. So will it be the answer?

Now here is what makes me uncomfortable – there is only one option with $\sqrt3$ . So it seems that the answer has been served on a plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the answer would be that obvious.

Next step would be to think a bit harder:

The area of the triangle has $\sqrt3$ in it so the side would be a further square root of $\sqrt3$ . So the actual irrational number would be the fourth root of 3 but we don’t have any option that has the fourth root of 3.

Then let’s go deeper and actually solve the question.

Area of the equilateral triangle $= Side^2 \times \frac{\sqrt3}{4} = 48$

$Side^2 = \frac{48\times4}{\sqrt3}$

$Side^2 = \frac{4\times4\times4\times3}{\sqrt3}$

Side $= 8 \times \sqrt[4]3$

Now note that the side of the equilateral triangle is the side of the square too. Hence all sides of the three squares will be of length $8 \times \sqrt[4]3$ .

All nine sides of the figure are the sides of squares. Hence,

Perimeter of the nine sided figure $= 9 \times 8 \times \sqrt[4]3$

Perimeter of the nine sided figure $= 72 \times \sqrt[4]3$

Now look at the options: We have an option that has $72 \times \sqrt2$ . The other options are either much smaller or much greater.

Think about it: $\sqrt[4]3 = \sqrt{\sqrt3} = \sqrt {1.732}$

$\sqrt2$ will be very similar to $\sqrt {1.732}$ .

Number properties will help you figure this out – Squares of smaller numbers (but greater than 1) are only a bit larger than the numbers: For example,

$(1.1)^2 = 1.21$ ,