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In the x y -plane, a triangle has vertices at
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15 Sep 2026, 09:32
15 Sep 2026, 09:32
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50% (02:33) correct
50% (03:01) wrong based on 6 sessions
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In the x y -plane, a triangle has vertices at A(k, k), B(k+3 t, k-t) , and C(k+2 t, k+4 t) , where t>0 . The area of triangle A B C is 70 .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
t |
3 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Re: In the x y -plane, a triangle has vertices at
[#permalink]
01 Oct 2026, 05:03
01 Oct 2026, 05:03
Expert Reply
Area of triangle using vectors from A :
\( B-A=(3 t,-t), \quad C-A=(2 t, 4 t)\)
The area is:
\(\frac{1}{2}|(3 t)(4 t)-(-t)(2 t)|=\frac{1}{2}\left(12 t^2+2 t^2\right)=7 t^2\)
Given area =70:
\(7 t^2=70 \Longrightarrow t^2=10 \Longrightarrow t=\sqrt{10} \approx 3.16\)
So Quantity A is $t \approx 3.16$, which is greater than Quantity B, 3.
\( B-A=(3 t,-t), \quad C-A=(2 t, 4 t)\)
The area is:
\(\frac{1}{2}|(3 t)(4 t)-(-t)(2 t)|=\frac{1}{2}\left(12 t^2+2 t^2\right)=7 t^2\)
Given area =70:
\(7 t^2=70 \Longrightarrow t^2=10 \Longrightarrow t=\sqrt{10} \approx 3.16\)
So Quantity A is $t \approx 3.16$, which is greater than Quantity B, 3.
gmatclubot
Re: In the x y -plane, a triangle has vertices at [#permalink]
01 Oct 2026, 05:03
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