BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Geometry:

Post your questions about customer service issues, GMAT exam policies, and GMAT exam structure. Get an official answer from GMAC, the organization behind the GMAT test!
This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Wed Jan 21, 2015 8:08 am

Geometry:

by Joshua » Wed Jan 21, 2015 8:15 am
Uhm i have a very difficult question here can you help me to answer it? Here it is
A side of a triangle measures 3cm. A line segment is drawn parallel to the side forming a trapezoid whose areas is 2/3 of the area of tthe triangle. How long is the line segment ?
Top
Source: — Ask the Test Maker |
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Thu Jan 22, 2015 8:58 am
Joshua wrote:Uhm i have a very difficult question here can you help me to answer it? Here it is
A side of a triangle measures 3cm. A line segment is drawn parallel to the side forming a trapezoid whose area is 2/3 of the area of the triangle. How long is the line segment ?
Use the simplest possible example, a right triangle with base 3 and vertical side 6. So the height is also 6.

Area is 9.

So 2/3 of the area is 6 and 1/3 is 3.

Draw a line parallel to the base that cuts across to create a triangle above and a trapezoid below. The small triangle needs to have area 3 to make the trapezoid area 6.

Ratio of small triangle's base and height will be same as that of the original triangle's base and height, which is 1:2.

Base of small triangle is x.

Area of small triangle will be (b*h)/2 = (1x*2x)/2 = 3

So 1x*2x = 6 x^2 = 3 x = sqrt3

So the new line segment will have length sqrt3.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Wed Jan 21, 2015 8:08 am

by Joshua » Fri Jan 23, 2015 4:25 am
Uhm Sir Murray Thank you for the answer but I'm still wondering how you get the area of the trapezoid and the triangle. Also, this is not sarcastic but wHat is the relation of the Example Right Triangle that has an area of 9 to the Problem if the Given Base is Located to the Base of the trapezoid? I mean at the base of the original triangle which is 3 cm and when a Line Segment is drawn, The Trapezoid has now a base of 3 cm?
Sorry for too much questions, I just want to clarify it ... Thank you In advance :D
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Fri Jan 23, 2015 6:02 am
The underlying idea of my approach is that what works for one triangle, line segment and trapezoid will work for any triangle, line segment and trapezoid. The only thing you defined was that one side of the triangle has to be 3cm long. So I chose to make the triangle a right triangle with base 3 and height 6, so that I would have easy numbers and ratios with which to work.

So I have a triangle with one side 3, as you said, and that side is the base. Then, to form the trapezoid, I drew a line parallel to the base. Now the base of the triangle is also the base of the trapezoid and the line segment is the top of the trapezoid.

I didn't use a formula to get the area of the trapezoid. I used a formula to get the area of the little triangle which has as its base the line segment I just drew.

Since the original triangle is a right triangle, the height is the same as the length of the vertical side, which is 6. So the area of the triangle is (b*h)/2 = (3*6)/2.

Therefore, the big, original triangle has area 9. So the little triangle needs to have area 3. This way the remaining area, which is the area of the trapezoid below the line, is 6, which is 2/3 of 9.


Image
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Wed Jan 21, 2015 8:08 am

by Joshua » Fri Jan 23, 2015 6:16 am
Thank you very much Sir Marty for all of your time given to my problem :D :D :D

I just one last question, where did you get the 6cm or the height of the original triangle also the vertical side? And pardon for repeating again but how did you get the square root of 3?
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Fri Jan 23, 2015 6:32 am
Joshua wrote:Thank you very much Sir Marty for all of your time given to my problem :D :D :D

I just one last question, where did you get the 6cm or the height of the original triangle also the vertical side?
I made it up.

While I am not all that familiar with the relationships between triangles and trapezoids, given the way the problem is written, I went with the assumption that any triangle with one side length 3 would work. So I was free to chose the shape of the triangle and the lengths of the other sides. So I chose right triangle with height 6, because that would be an easy triangle with which to work.

Then, if you can believe it, I tried this with other triangles, including an equilateral triangle and an isosceles right triangle, and the same method worked in each case. I guess the reason it works is that when you draw a line segment parallel to the base of a triangle, you always create a similar triangle with the ratio of b:h for the original triangle being the same as the ratio of b:h of the new smaller triangle. So essentially to get a triangle with area 1/3 the area of the original triangle, you are always drawing a line segment to create a smaller triangle whose height = (height of the original triangle)/sqrt3.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Fri Jan 23, 2015 6:48 am
To get sqrt3, I did the following.

I got the area of the original triangle, with one side 3, as defined, and height and side 6, as I chose.

That area is bh/2 = 18/2 = 9

The problem said that the trapezoid has to have 2/3 the area of the original triangle.

So the area of the trapezoid is 2/3 * 9 = 6

To get from 9 to 6, we need to cut off the top of the original triangle by making a smaller triangle with area 3.

The original triangle has a b:h ratio of 3:6 or 1:2. So any triangle created by drawing a line segment parallel to the base of the original triangle will also have a b:h ratio of 1:2.

To answer the question, we are looking for the length of the base of that little triangle. Call that x.

The height of that little triangle will be 2x, because the b:h ratio is 1:2.

We already know that the area of the little triangle is 3.

a = bh/2. So 3 = (x * 2x)/2

Solve for x. 3 = 2x^2/2 3 = x^2 sqrt3 = x

So x is the length of the base of the little triangle and of the top of the trapezoid and it's sqrt3.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Wed Jan 21, 2015 8:08 am

by Joshua » Fri Jan 23, 2015 6:49 am
Uhm Sir Marty Murray , This time I just want to say THANK YOU SO MUCH FOR THIS OPPORTUNITY , for the Time you gave for me , for helping me to answer this difficult question. Actually I try my best to answer this several times and I think I Consumed 24 hours to answer this Thank you so much :D :D :D :D :D :D :D GODbless :D :D
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Fri Jan 23, 2015 7:06 am
The funny part is I just got it.

To get one third the area of the original triangle you need to divide by 3.

So when you draw the line segment you are creating a triangle similar to the original triangle and the new triangle has area A/3.

A = bh/2.

In creating similar triangles, you always divide everything by the same amount.

A 6:8:10 triangle is similar to a 3:4:5 triangle, and the heights can also be divided by that same amount, which in this case is 2.

So to get similar triangle with 1/3 the area you divide both b and h by sqrt3.

A = bh/2 A/3 = (b/sqrt3 * h/sqrt3)/2 = (bh/3)/2

How cool.

Been fun working with you.

Rock on.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Wed Jan 21, 2015 8:08 am

by Joshua » Fri Jan 23, 2015 7:17 am
Uhm Sir Murray thank you for the additional info. you have given a while ago , i wanna Ask you something if it is okay for you, uhm what kind of topic or title of this problem in geometry? i mean it is similarities if it is , what kind or specific ?

Thank you in advance :D :D :D
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Fri Jan 23, 2015 7:55 am
I guess the topic is Similar Triangles, or Similar Triangles - Ratio of Areas.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
Post new topic Post Reply
• Page 1 of 1