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No you cannot because LN and MN are not perpendicular to each. The formula you used is valid only if one of the side is considered as base and other one is considered as perpendicular on the base. Hope this makes it clear otherwise let me know.

Start by calculating the distance between the 3 points 1) (5,1) & (1,3) = 2 sqrt(5) 2) (5,1) & (3,5) = 2sqrt(5) 3) (1,3) & (3,5) = 2sqrt(2)

As you can see this is an isosceles triangle. If you don't know the direct formula for the area of the isosceles triangle, you can calculate the height by applying Pythagoras theorem sqrt(2 sqrt(5)^2-1/2(2 sqrt(2)^2)) = 3 sqrt(2). area = 1/2 * base*height = 1/2 *2sqrt(2) * 3 sqrt(2) = 6. So the answer is 6. _________________

1. Do such question appear on the gmat? 2. a,b,c=2 root5.

Applying the formula for P= 2rt5+2rt5+2rt5/2 =3rt5

Therefore area=sq.rt. of 3rt5(rt5)(rt5)(rt5) 5 root3 which is approx 8.66 :(

I think this is a very gmat like question. A difficult one.

Here go two different approaches:

1) My approach

. Make the diagram. . Find the sides. -> LM= sqrt [ (5-1)^2 + (3-1)^2 ] = sqrt(20). -> MN = sqrt(20). -> LN = sqrt(8) -> this is our base . Find the height -> to find the height divide the triangle in two and apply Pythagoras -> sqrt(20)^2 = sqrt(4)^2 + h^2 -> h=sqrt(18) => Area = sqrt(8) * sqrt(18) / 2 = 6

2) The gmatclub approach (which is a great shortcut)

Draw a triangle, then limit it by a rectangle, find the area of the rectangle and subtract the areas of the right triangles surrounding the triangle in question. Plot the vertices. Create a rectangle with points in A(1,1) , B(5, 1) , C(5, 5) , and D(1, 5) . The area is 16.

Now the other 3 triangles have areas of: LAB = \frac{1}{2}*2*4 = 4 . BNC = \frac{1}{2}*2*4 = 4 , and LDN = \frac{1}{2}*2*2 = 2 . Total is 10. 16 - 10 = 6 .

When you have all the 3 coordinate of triangle...you can solve it by matrx way... assuming coordinates as (x1, y1) , (x2, y2), (x3, y3) matrix will be x1 x2 x3 y1 y2 y3 1 1 1

now area will be half the matrix value i.e 1/2(x1(y2-y3)-x2(y1-y3)+x3(y1-y2))

Hi all, I tried solving it using the formulas of matrices using the formula \frac{1}{2}[x1 (y2-y3) + x2( y3-y1) + x3(y1-y2)] and got the answer as 4. Took the values for x1, x2, x3 and y1, y2, y3 as the values of the three co-ordinates of the vertices.

Can you please tell me where am I making a mistake?

D Area of square around triangle (4 x 4 = 16) minus the 3 triangles from the corners of square to get desired triangle: .5x2x2 = 2 .5x2x4 = 4 .5x2x4 = 4 Overall: 16 - 2 - 4 - 4 = 6 _________________

Another way to solve this problem is to draw the diagram and find the area of triangle. Area LMN = Area ALMB + Area BMNC - Area ALNC = 1/2(3+5)(2) + 1/2(5+1)(2) - 1/2(3+1)(4) = 8 + 6 - 8 = 6

Attachments

A.png [ 7.55 KiB | Viewed 6883 times ]

_________________

"You can do it if you believe you can!" - Napoleon Hill "Insanity: doing the same thing over and over again and expecting different results." - Albert Einstein

my pick is D. Plotted a diagram with 3 given vertices, to represent the triangle in bold lines. In addition, plotted 3 more triangles in dotted lines, by connecting each of the vertices with the adjacent vertices as a right triangle. This makes 4 triangles. (Calculated the area of rectangle, comprising 4 triangles) - (Area of the 3 right triangles) = area of the required/ center triangle

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

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:

Attachment:

Triangle.PNG [ 14.56 KiB | Viewed 5737 times ]

Notice that the area of the blue square is 4^2=16 and the area of the red triangle is 16 minus the areas of 3 little triangles which are in the corners (2*2/2, 4*2/2 and 4*2/2). Therefore the area of a triangle LMN=16-(2+4+4)=6.

It is easy to calculate if you count the lenthg of sides of the triangle it is clear that LM=MN so basic of height point M is situated in the midle of the line LN, this point has coordinate (2,4).

The length of the height is (3^2 +3^2)^1/2 = 18^1/2 The length of the LN is 8^1/2

The square solution is a pretty easy one, however even if it does not strike to you, one can always see that because of the similarity in the digits used as co-ordinates, its pretty easy to see that the triangle is isosceles, and hence, taking the points L(1,3), M(5,1), N(3,5) in to consideration.

ML=2 \sqrt{5} MN=2 \sqrt{5} and NL = 2\sqrt{2}

Since, ML=MN, the perpendicular dropped from M to the base NL would divide the base in half, and hence the co-ordinates of the mid point will be say D(2,4) Now you can easily calculate the length of MD which will turn out to be 3\sqrt{2}

And hence, area = 1/2 * base * height = 1/2 * 2 \sqrt{2} * 3 \sqrt{2} = 6 which is the answer, Hence D.

PS: Like my approach? Help me with some Kudos. _________________

PS: Like my approach? Please Help me with some Kudos.

Hi all, I tried solving it using the formulas of matrices using the formula \frac{1}{2}[x1 (y2-y3) + x2( y3-y1) + x3(y1-y2)] and got the answer as 4. Took the values for x1, x2, x3 and y1, y2, y3 as the values of the three co-ordinates of the vertices.

Can you please tell me where am I making a mistake?

May be you did some calculation mistake perhaps? I applied the formula 1/2 * (5 (5-3) + 3 * (3-1) + 1 * (1-5))

=1/2 * (10+6-4) =6

Hence, D

PS: Like my approach? Help me with some Kudos. _________________

PS: Like my approach? Please Help me with some Kudos.