What is the area of a triangle with the following vertices

Question

What is the area of a triangle with the following vertices L(1, 3), M(5, 1), and N(3, 5) ?

A. 3
B. 4
C. 5
D. 6
E. 7

M06-09

Bunuel · Accepted Answer

There is a direct formula to calculate the are of a triangle based on coordinates of its vertices and one could use it to solve this problem. Though if you make a diagram minimum simple calculations will be needed: graph1.PNG Take note that the area of the blue square is 4^2 = 16 . The area of the red triangle is the area of this blue square minus the combined areas of three smaller triangles situated at the corners. These little triangles have areas of \frac{1}{2}*2*2 , \frac{1}{2}*2*4 , and \frac{1}{2}*4*2 . Hence, the area of triangle LMN is calculated as 16 - (2 + 4 + 4) = 6 . Answer: D. Hope it's clear.

Bunuel · Answer

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

If the vetices 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}| .

So for:  L(1, 3) ,  M(5, 1) , and  N(3, 5)  the area will be:  area=|\frac{1(1-5)+5(5-3)+3(3-1)}{2}|=6 .

lansac · Answer

Plot the 3 points.
The base is distance between A(1,3) and B(3,5) = sqrt(8)
Height is distance between C(5,1) and (2,4). 2,4 is the midpoint. So height is sqrt(18)
Area =1/2*sqrt(8)* sqrt(18)=6

LalaB · Answer

NM=sqroot((5-3)^2+(1-5)^2)=2sqroot5
LN=sqroot((1-3)^2+(3-5)^2)=2sqroot2
LM=sqroot((5-1)^2+(1-3)^2)=2sqroot5

height of the triangle=Sqroot(LM^2-(LN/2)^2))=3sqroot2

so,area=1/2*LN*height=1/2*2sqroot2*3sqroot2=2*3=6

gmatpunjabi · Answer

Hey Buunel what is the formula for the area of the Triangle Based on the Vertices?

gmatpunjabi · Answer

THank you for the equation!!

shadabkhaniet · Answer

Formula is good.

+1

sharmila79 · Answer

In the above question, two sides are same that is 2root5 and the third side is 2root 2. Can't this be an isosceles triangle? I know that I am missing something, please help

Kris01 · Answer

Hello Sharmila,

Yes the above triangle can is an isolsceles triangle and the area can be calculated using the formula (1/2)*b*h where h=(hypotenuse^2(1/4)(base)^2).

h=(2*sqrt(5))^2-((1/4)(2*sqrt(2))^2

It would be great if you could highlight your doubt here.

sharmila79 · Answer

Hi Kris,
Thanks for the detailed explanation. As soon as I got the lengths I decided that it is a 45:45:90 right triangle and was stuck with that. I was aware that the short leg and long leg (base and height) are longer than the hypotenuse, but still couldn't bring the diagram (your attachment) into picture. My doubt is cleared. I have one more question, if two sides of a triangle are of same length and that length is shorter than the third side, then it has to be a 45:45:90 isosceles triangle. Right?

Kris01 · Answer

Hello Sharmila,

To find whether the angles in a triangle are 45-45-90 you would also need to confirm whether the ratio of the sides is 1:1:sqrt(2).

For example, imagine an isosceles triangle with equal sides of 5 cm in length and forming a 120 degree angle between them. The equal angles would be 30 deg each. The third side will still be longer than the equal sides.

Hope this helps! Let me know if I can help you any further.

sharmila79 · Answer

Thanks Kris! I was actually using that 1:1:root2, but did not make any sense with the values I got.

Sweety04 · Answer

Thanks for the formula. I guess its the best way to solve this type of problem. Any suggestions?

Posted from my mobile device

Naveen Reddy · Answer

basically vertices are L(1, 3), M(5, 1), and N(3, 5), since only area is asked here we can subtract 1 from each of the x and y co-ordinates which makes the new points as L(0, 2), M(4, 0), and N(2, 4) which makes it a right angled traingle framed on the x and y axes with corner at N(2, 4). i t wouldn't be hard to calculate the area of new traingle as 4sq.units.

MarkusKarl · Answer

Nice solution. Thanks.

I started my process by calculating the base as LM and were going to calculate the perpendicular line to LM that went through point N to get the height. But at that point I realized it would be time consuming and I had to be doing something wrong.

kanthaliya · Answer

formula for the triangle,whose vertices are given:half of determinant of the three vertices with 1 added to each set as a third parameter--->1/2*det((1,3,1), (5,1,1), (3,5,1))

himanshu1105 · Answer

A = 1/2 | x1 y1 | 
 |x2 y2 | 
 |x3 y3|
 |x1 y1|
 => Cross multiply A = 1/2  
 insert values => 1/2   = 6 Answer !

SiddharthR · Answer

This question can easily be solved using the Heron's formula --> sqrt(s(s-a)(s-b)(s-c))

s = (a + b + c)/2 ; a, b, c are the distances between any two points say LM, MN and LN. This distance can be calculated using the formula -->

sqrt(x2 - x1)^2 + (y2 - y1)^2)

Once you have each of the values a, b and c, plug it into the Heron's formula to get the area of the triangle

adhya9 · Answer

IMO D

Area of a triangle when coordinates are given = | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | / 2

Area = | -4 +10 + 6 | / 2 = 6

What is the area of a triangle with the following vertices

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What is the area of a triangle with the following vertices

Prep Toolkit

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