Triangle RST lies in the xy-plane. Point R has coordinates (a,b), poin

Can someone try this? And may be come up with a better solution or easier explanation?

Thank you!

Given:
1. Triangle RST lies in the xy-plane.
2. Point R has coordinates (a,b), point S is at the origin, and point T has coordinates (c,d), such that bd + ac = 0.

Asked: What is the area of triangle RST?

bd + ac = 0
bd = - ac
\(\frac{b-0}{a-0} * \frac{d-0}{c-0} = \frac{b}{a} *\frac{d}{c} = \frac{bd}{ac} = -1\)

Line joining R(a, b) & S(0,0) is perpendicular to line joining T(c, d) & S (0,0)
Angle RST is 90

Area of right angled triangle RST \(= \frac{1}{2} * Base * Height = \frac{1}{2} * \sqrt{(a-0)^2 +(b-0)^2} * \sqrt{(c-0)^2 +(d-0)^2} = \frac{1}{2} * \sqrt{a^2 +b^2} * \sqrt{c^2 +d^2}\)


(1) \((\sqrt{a^2+b^2}) (\sqrt{c^2+d^2})=48\)
Area of triangle RST \(= \frac{1}{2} * (\sqrt{a^2+b^2}) (\sqrt{c^2+d^2}) = \frac{1}{2} * 48 = 24\)
SUFFICIENT

(2) \(\sqrt{a^2+b^2} + \sqrt{c^2+d^2}=14\)
We need product of \(\sqrt{a^2+b^2}\ and\ \sqrt{c^2+d^2}\) to calculate the area and NOT sum
NOT SUFFICIENT

IMO A

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.