In the xy-plane, the Vertices of a Triangle have Coordinates (0,0),(3,3), and (7,0) GMAT Problem Solving

Question: In the xy-plane, the vertices of a triangle have coordinates (0,0),(3,3), and (7,0). What is the perimeter of the triangle?

1. 13
2. (14)
3. (16)
4. (17)
5. 12+\(3\sqrt2\)

How to draw a graph on this?

‘In the xy-plane, the vertices of a triangle have coordinates (0,0),(3,3), and (7,0).’ - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide 2022”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

Recalling frequent triangle sides is an effective technique to solve the problem.
- We know that the length of one of the sides is 7 because it's the distance between coordinates (0,0) and (7,0).
Now We can split the triangle into 2 smaller triangles
Triangle 1 coordinates (0,0), (3,3) and (3,0)
Triangle 2 coordinates (3,0), (3,3) and (7,0)

Triangle 1 appears to have sides that are extremely similar to the typical triangle, which has sides (1):(1):(\(\sqrt2\)). therefore its sides can be written as (3):(3):(3√2) hence one side of the original triangle is (3√2).
Triangle 2 appears to have sides that are exactly the same as the common triangle with sides (3):(4):(5) so, one side of the original triangle is (5).
Triangle perimeter is (7) + (3√2) + (5) = 12 +3√2
The answer is E which is 12 +3√2

Correct Answer: E

Approach Solution 2:

 there is another approach to solve this question which is pretty easy

And uses algebra method
The method for calculating the separation between two points
x1,y1 and x2, y2 is
d= \(\sqrt{(x1-x2)^2+(y1-y2)^2}\)

Distance between (0, 0) and (7, 0) is d=7
Distance between (0, 0) and (3, 3) is

d=\( \sqrt{(0-3)^2+(0-3)^2}\)

= \(\sqrt{(x1-x2)^2+(y1-y2)^2}\)

= \(\sqrt{18}\)

= \(3\sqrt2\)

Distance between (3, 3) and (7, 0) is

d=\(\sqrt{(3-7)^2+(3-0)^2}\)

= 5

Triangle perimeter is (7) + (3√2) + (5) = 12 +3√2
The answer is E which is 12 +3√2

Correct Answer: E

Approach Solution 3:

Get distance between (0,0) and (3,3):√(3−0)^2+(3−0)^2=√(18)=3√(2)
Get distance between (3,3) and (7,0): √(7−3)^2+(3−0)^2=√(16+9)=√25=5
Get distance between (0,0) and (7,0): 7

P=7+5+3√2=12+3√2
Correct Answer: E

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