In the xy plane, the vertices of a triangle have coordinates

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To answer this question ... all you need is distance between vertices :

(0,0) & (7,0) : 7
(0,0) & (3,3) : \(3\sqrt{2}\)
(7,0) & (3,3) : \(\sqrt{16+9}=5\)

Perimeter = sum of these three sides = \(12 + 3\sqrt{2}\)

Answer : e

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?

(A) 13
(B) 34
(C) root 43
(D) 7+root3
(E) 12 +3root2

Get distance between (0,0) and (3,3): \(\sqrt{(3 - 0)^2 + (3-0)^2} = \sqrt{(18)} = 3 \sqrt{(2)}\)
Get distance between (3,3) and (7,0): \(\sqrt{(7-3)^2 + (3-0)^2} = \sqrt{16 + 9}= \sqrt{25}=5\)
Get distance between (0,0) and (7,0): 7

\(P = 7 + 5 + 3\sqrt{2} = 12 + \sqrt{2}\)

Answer: E

Hi Bunnuel,

Is there any formula to calculate area of triangle when all coordinates are known?

Please help. Many thanks in advance for your kind response

Regards
Megha

I really doubt that you'll need it for the GMAT but here you go:

If the vertices of a triangle are: \(A(a_x, a_y)\), \(B(b_x, b_y)\) and \(C(c_x,c_y)\) then the area of ABC is:

\(area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}|\).

Ohh,

A day before I saw question like that on gmat club. Anyways many thanks for your response and telling me its highly unlikely to come.

Regards
Megha

Relatively straight-forward problem, though I'd suggest plotting it so you can visualize the sides and what two side lengths you will need to calculate.

Base length is 7

(1) Distance between (0,0) & (3,3) --> sqrt[(3^2)+(3^2)] = 3sqrt(2)

(2) Distance between (3,3) & (7,0) --> sqrt[(3-0)^2 + (3-7)^2] = sqrt(9+16) = sqrt(25) =5

Add them all up --> 12+ 3sqrt(2)

E

An efficient way of solving this problem is by recalling common sides of a triangle.

- We know that the length of one of the sides is 7 simply because it's the distance between coordinates (0,0) and (7,0).

- 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 has sides that appear to be very similar to the common triangle with sides (1):(1):(√2). In fact, its sides can be written as (3):(3):(3√2) and now we know that one side of the original triangle is (3√2).

- Triangle 2 has sides that are exactly the same as the common triangle with sides (3):(4):(5) and now we know that one side of the original triangle is (5).


Our perimeter is (7) + (3√2) + (5) = 12 +3√2

Looking at the answer choices, we can also tell that the only answer choice that contains (3√2) is answer choice E

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