In the rectangular coordinate system, the vertices of a triangle have coordinates (-3, 0), (3, 2), and (0, 11). What is the area of the triangle?
A. 15
B. 18
C. 24
D. 30
E. 36
The OA is D.
Can I solve this PS question using the following formula?
Area of triangle = ½ |y� (x₂ - x₃) + y₂ (x₃ - x�) + y₃ (x� - x₂)| = 1/2 { 27 + 33} = 30. Option D.
Has anyone another strategic approach to solve this PS question? Regards!
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In the rectangular coodinate system, the vertices of a
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Vincen
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Hi AAPL.
Let's find the length of each side.
If A=(-3, 0), B=(3, 2) and C=(0, 11) then: $$AB=\sqrt{\left(3-\left(-3\right)\right)^2+\left(2-0\right)^2}=\sqrt{36+4}=\sqrt{40}.$$ $$AC=\sqrt{\left(0-\left(-3\right)\right)^2+\left(11-0\right)^2}=\sqrt{9+121}=\sqrt{130}.$$ $$BC=\sqrt{\left(0-3\right)^2+\left(11-2\right)^2}=\sqrt{9+81}=\sqrt{90}.$$ Now, we can se that $$AC^2=AB^2+BC^2$$ $$\left(\sqrt{130}\right)^2=\left(\sqrt{40}\right)^2+\left(\sqrt{90}\right)^2.$$ Then we can see that the triangle ABC is a right triangle. Therefore, $$Area=\frac{\sqrt{40}\cdot\sqrt{90}}{2}=\frac{\sqrt{3600}}{2}=\frac{60}{2}=30\ .$$ Hence, the area is 30. This is why the correct answer is the option D.
I hope this answer may help you.
Let's find the length of each side.
If A=(-3, 0), B=(3, 2) and C=(0, 11) then: $$AB=\sqrt{\left(3-\left(-3\right)\right)^2+\left(2-0\right)^2}=\sqrt{36+4}=\sqrt{40}.$$ $$AC=\sqrt{\left(0-\left(-3\right)\right)^2+\left(11-0\right)^2}=\sqrt{9+121}=\sqrt{130}.$$ $$BC=\sqrt{\left(0-3\right)^2+\left(11-2\right)^2}=\sqrt{9+81}=\sqrt{90}.$$ Now, we can se that $$AC^2=AB^2+BC^2$$ $$\left(\sqrt{130}\right)^2=\left(\sqrt{40}\right)^2+\left(\sqrt{90}\right)^2.$$ Then we can see that the triangle ABC is a right triangle. Therefore, $$Area=\frac{\sqrt{40}\cdot\sqrt{90}}{2}=\frac{\sqrt{3600}}{2}=\frac{60}{2}=30\ .$$ Hence, the area is 30. This is why the correct answer is the option D.
I hope this answer may help you.
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Jake@ThePrincetonReview
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A different (and quite easy) strategic approach to solving this problem without getting into square roots, the distance formula, or the Pythagorean theorem is to create a rectangle and subtract the areas of the right triangles not contained in the triangle you're looking for.
Specifically, make points at (-3, 11), (3,11) and (3,0), creating a rectangle with vertices at these three points and (-3,0).
The area of this rectangle is 6 x 11 = 66
Now subtract the area of the right triangle formed by (-3, 0), (-3,11) and (0,11), the area of which is (1/2)(11)(3) = 33/2
Then subtract the area of the right triangle formed by (0,11), (3,11) and (3,2), the area of which is (1/2)(9)(3) = 27/2
Then subtract the area of the right triangle formed by (-3, 0), (3,0), and (3,2), the area of which is (1/2)(6)(2) = 6
Your final answer will be 66 - (33/2 + 27/2 + 6), which is 66 - 36, which equals 30.
Easy, right? Draw it and you'll see.
Specifically, make points at (-3, 11), (3,11) and (3,0), creating a rectangle with vertices at these three points and (-3,0).
The area of this rectangle is 6 x 11 = 66
Now subtract the area of the right triangle formed by (-3, 0), (-3,11) and (0,11), the area of which is (1/2)(11)(3) = 33/2
Then subtract the area of the right triangle formed by (0,11), (3,11) and (3,2), the area of which is (1/2)(9)(3) = 27/2
Then subtract the area of the right triangle formed by (-3, 0), (3,0), and (3,2), the area of which is (1/2)(6)(2) = 6
Your final answer will be 66 - (33/2 + 27/2 + 6), which is 66 - 36, which equals 30.
Easy, right? Draw it and you'll see.
Jake Schiff
GMAT Instructor and Master Trainer
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