GMATprep: geometry

IMO C

h=a*(sqrt3)/2;a=side;h=height
4*sqrt(3)=a*sqrt(3)/2
a=8

perimeter=3*8=24

The answer is C
But the solution is much easier.
When we have a right triangle with one side equal x*3/2 the ather side is equal to x, and the hipotenuse 2x - so PS = 4, PQ = 8 - perimeter is 24

In the figure shown, the length of line segment QS is 4 sqrt(3). What is the perimeter of equilateral triangle PQR?

- 12
- 12 sqrt(3)
- 24
- 24 sqrt(3)
- 48

Since we have an equilateral triangle, the angles are all equal (60°).

Angle QPS = 60
Angle QSP = 90

Therefore, PQS = 180 - 60 - 90 = 30 OR PQS = 1/2 * PQR = 1/2 * 60 = 30

In a 30° - 60° - 90° triangle, the sides have length ratios of 1 : sqrt(3) : 2.

Therefore, the ratios of our triangle are 4 : 4 sqrt(3) : 8.

2*4 + 2*8 = 24

Since this is an equilateral triangle we know that the length of PS = 1/2 the length of PQ. Let's call the length of PQ a.

By Pythagorean Theorem then we know that
PQ^2 = PS^2 + QS^2

Substituting in a for PQ, a/2 for PS and 4 sqrt 3 for QS we get
a^2 = (a/2)^2 + (4 sqrt 3)^2
or
a^2 = (a^2)/4 + 48
solving for a^2
a^2 = 64
or
a = 8

The perimeter of the triangle is thus 3a or 3 times the length of PQ or 24

jdd147 wrote:Since this is an equilateral triangle we know that the length of PS = 1/2 the length of PQ. Let's call the length of PQ a.

By Pythagorean Theorem then we know that
PQ^2 = PS^2 + QS^2

Substituting in a for PQ, a/2 for PS and 4 sqrt 3 for QS we get
a^2 = (a/2)^2 + (4 sqrt 3)^2
or
a^2 = (a^2)/4 + 48
solving for a^2
a^2 = 64
or
a = 8

The perimeter of the triangle is thus 3a or 3 times the length of PQ or 24

What am i doing wrong here ?




PQ^2 = QS^2 + PS^2

PQ^2 = PS^2 + 48 ------ (1)

QR^2 = SR^2 + 48 ------(2)

Adding 1 and 2


PQ^2 + QR^2 = PS^2 + SR ^2 + 96 ----- 3

But, PS^2 + SR ^2 = PR ^2 ------- 4

Substitute 4 in 3....

PQ^2 + QR^2 = PR^2 + 96 -----5

PQ = PR = QR since equilateral triangle

Therefore PQ^2 = 96


PQ = sqrt (96)


But this doesnt yield the correct answer ....

PQS is a 30:60:90 triangle

so PS = 4

PR = 8

Perimeter = 3*8= 24