Kaplan triangle-third side

[singhsa]

If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?

I 35
II 48
III 56

A. I ONLY
B. 1 AND II ONLY
C. I AND III ONLY
D. II AND III ONLY
E. I,II AND III ONLY.


Here, the third side can obviously be b/w 4 and 20.

oa - b

[GMATGuruNY]

If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?

I 35
II 48
III 56

A. I ONLY
B. 1 AND II ONLY
C. I AND III ONLY
D. II AND III ONLY
E. I,II AND III ONLY.


Here, the third side can obviously be b/w 4 and 20.

oa - b

When given 2 sides of a triangle, we'll get the largest possible area if we make one of the known sides the base and the other known side the height, placing a right angle between them.

Let's say we make 12 the base and 8 the height with a right angle between them. A = 1/2 * 12 * 8 = 48.

If we then change the angle between the two sides so that it's larger or smaller than 90 degrees, the base will remain 12 but the height will be become smaller than 8. Since a smaller height will yield a smaller area, 48 is the largest area possible.

Eliminate C, D and E because they each include 56 as a possible area.
Eliminate A because it doesn't include 48 as a possible area.

The correct answer is B.

[pzazz12]

If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?

I 35
II 48
III 56

A. I ONLY
B. 1 AND II ONLY
C. I AND III ONLY
D. II AND III ONLY
E. I,II AND III ONLY.


answer is d

[singhsa]

When given 2 sides of a triangle, we'll get the largest possible area if we make one of the known sides the base and the other known side the height, placing a right angle between them.

Let's say we make 12 the base and 8 the height with a right angle between them. A = 1/2 * 12 * 8 = 48.

If we then change the angle between the two sides so that it's larger or smaller than 90 degrees, the base will remain 12 but the height will be become smaller than 8. Since a smaller height will yield a smaller area, 48 is the largest area possible.

Eliminate C, D and E because they each include 56 as a possible area.
Eliminate A because it doesn't include 48 as a possible area.

The correct answer is B.

Hey,

I am getting one thing here. We are considering a triangle with a base of 12 and a height of 8. But we're not given the third side. How can we assume the third side to be a hypotenuse.

From the properties of triangles, we know that the third side can only be b/w 4 and 20. So, what if the third side is 10. Won't the area of triangle then become 1/2 * 12*10=60 ?

[GMATGuruNY]

When given 2 sides of a triangle, we'll get the largest possible area if we make one of the known sides the base and the other known side the height, placing a right angle between them.

Let's say we make 12 the base and 8 the height with a right angle between them. A = 1/2 * 12 * 8 = 48.

If we then change the angle between the two sides so that it's larger or smaller than 90 degrees, the base will remain 12 but the height will be become smaller than 8. Since a smaller height will yield a smaller area, 48 is the largest area possible.

Eliminate C, D and E because they each include 56 as a possible area.
Eliminate A because it doesn't include 48 as a possible area.

The correct answer is B.

Hey,

I am getting one thing here. We are considering a triangle with a base of 12 and a height of 8. But we're not given the third side. How can we assume the third side to be a hypotenuse.

From the properties of triangles, we know that the third side can only be b/w 4 and 20. So, what if the third side is 10. Won't the area of triangle then become 1/2 * 12*10=60 ?

If the 3rd side = 10, the area will be smaller than 48 because the base will remain the same while the height becomes smaller.

Please see the attachment, which illustrates why the largest possible area will be achieved if the 2 given sides form the base and the height. Any other configuration will yield a smaller area because the height will become smaller.

[lokesh r]

When given 2 sides of a triangle, we'll get the largest possible area if we make one of the known sides the base and the other known side the height, placing a right angle between them.

Let's say we make 12 the base and 8 the height with a right angle between them. A = 1/2 * 12 * 8 = 48.

If we then change the angle between the two sides so that it's larger or smaller than 90 degrees, the base will remain 12 but the height will be become smaller than 8. Since a smaller height will yield a smaller area, 48 is the largest area possible.

Eliminate C, D and E because they each include 56 as a possible area.
Eliminate A because it doesn't include 48 as a possible area.

The correct answer is B.

Hey,

I am getting one thing here. We are considering a triangle with a base of 12 and a height of 8. But we're not given the third side. How can we assume the third side to be a hypotenuse.

From the properties of triangles, we know that the third side can only be b/w 4 and 20. So, what if the third side is 10. Won't the area of triangle then become 1/2 * 12*10=60 ?

If the 3rd side = 10, the area will be smaller than 48 because the base will remain the same while the height becomes smaller.

Please see the attachment, which illustrates why the largest possible area will be achieved if the 2 given sides form the base and the height. Any other configuration will yield a smaller area because the height will become smaller.

since third side can be anything b/w 4 and 20 exclusive.

how can i say area of triangle will be <48 , if say base is 19?

[GMATGuruNY]

When given 2 sides of a triangle, we'll get the largest possible area if we make one of the known sides the base and the other known side the height, placing a right angle between them.

Let's say we make 12 the base and 8 the height with a right angle between them. A = 1/2 * 12 * 8 = 48.

If we then change the angle between the two sides so that it's larger or smaller than 90 degrees, the base will remain 12 but the height will be become smaller than 8. Since a smaller height will yield a smaller area, 48 is the largest area possible.

Eliminate C, D and E because they each include 56 as a possible area.
Eliminate A because it doesn't include 48 as a possible area.

The correct answer is B.

Hey,

I am getting one thing here. We are considering a triangle with a base of 12 and a height of 8. But we're not given the third side. How can we assume the third side to be a hypotenuse.

From the properties of triangles, we know that the third side can only be b/w 4 and 20. So, what if the third side is 10. Won't the area of triangle then become 1/2 * 12*10=60 ?

If the 3rd side = 10, the area will be smaller than 48 because the base will remain the same while the height becomes smaller.

Please see the attachment, which illustrates why the largest possible area will be achieved if the 2 given sides form the base and the height. Any other configuration will yield a smaller area because the height will become smaller.

since third side can be anything b/w 4 and 20 exclusive.

how can i say area of triangle will be <48 , if say base is 19?

Let's say that the 3 sides of the triangle are 8, 12, and 19.

Any side of a triangle can be deemed the base. We can say that b=8, or b=12, or b=19. Each base will have a corresponding height.
No matter which side is deemed the base, we must always get the same area.
In other words, (1/2)*b*h must always yield the same area -- no matter which side is deemed the base.
So let's say that b=12.
If h=8, then A = (1/2)*8*12 = 48.
But h<8, because if b = 12, the corresponding height must be shorter than each of the other 2 sides (which have lengths of 8 and 19).
Since h<8, A<48.

Even if we say that b=19, we'll have to get the same area as when b=12, so A<48 -- no matter which side we deem the base.

Please see the attachment, which illustrates why 48 is the largest possible area.

Hope this helps!

[frank1]

You showed how the area couldnot be greater than 48 so 56 is out
but
You didn't show why 35 could not be the area.As the question doesnt explicitly says the sides could not be in decimal point
i can see that area can be 35 as well....
any reason for that...

[GMATGuruNY]

You showed how the area couldnot be greater than 48 so 56 is out
but
You didn't show why 35 could not be the area.As the question doesnt explicitly says the sides could not be in decimal point
i can see that area can be 35 as well....
any reason for that...

Please reread my original post. I stated that the correct answer is B, which includes 35 as a possible area. Once I had concluded that 56 could not be the area but 48 could, I was able to eliminate A, C, D and E, so why waste valuable time doing any more work?