What am I doing wrong here?

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Ilikemeat321: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Thu Jul 24, 2008 7:33 am

What am I doing wrong here?

by Ilikemeat321 » Fri Jul 25, 2008 8:15 am

I can't link the Polygons but you can probably figure it out what it looks like? If not, can someone show me how to link screenshots?

here's the question:

ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

A. 2√ 2
B. 2√ 3
C. 4
D.2√ 5
E.6

mysolution was:

to find a right triangle first

Find the altitude which is BA or AB (however you want to call it)
Find the Base would be AD- BC
Then find CD

to find the Altitude

Area=28 for a rectangle, side L= 8(given AD) Altitude = 3.5

Since BC is parallel to AD my base for the triangle would be 8-6=2

now I have base + altitude I get

3.5^2+2^2=CD

16.25=cd? I do not see this answer can someone tell me which step is wrong?

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Source: Beat The GMAT — Problem Solving | Fri Jul 25, 2008 8:15 am

GMAT/MBA Expert

Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

Re: What am I doing wrong here?

by Ian Stewart » Fri Jul 25, 2008 9:00 am

Ilikemeat321 wrote: ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

A. 2? 2
B. 2? 3
C. 4
D.2? 5
E.6

mysolution was:

to find a right triangle first

Find the altitude which is BA or AB (however you want to call it)
Find the Base would be AD- BC
Then find CD

to find the Altitude

Area=28 for a rectangle, side L= 8(given AD) Altitude = 3.5

Since BC is parallel to AD my base for the triangle would be 8-6=2

now I have base + altitude I get

3.5^2+2^2=CD

16.25=cd? I do not see this answer can someone tell me which step is wrong?

I hope I'm understanding the picture correctly. I think there are two errors in the above- the area, 28, is the area not just of the rectangular portion of the quadrilateral; it also includes the triangular portion. Let's call the height (the length of AB) 'h'. We can divide the picture into a rectangle, with a base of 6, and a right angled triangle, with a base of 2. We know that the area of the rectangle plus the area of the triangle is 28:

6*h + 2*h/2 = 28
7*h = 28
h = 4

You've also left off a square (^2) on the right side of the Pythagorean formula at the end of your solution. If we use h = 4, and b = 2 for the right angled triangle, we should have:

CD^2 = 4^2 + 2^2
CD^2 = 20
CD = root(20) = 2*root(5)

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

Quote

Ilikemeat321: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Thu Jul 24, 2008 7:33 am

Re: What am I doing wrong here?

by Ilikemeat321 » Fri Jul 25, 2008 9:55 am

Ian Stewart wrote:
Ilikemeat321 wrote: ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

A. 2? 2
B. 2? 3
C. 4
D.2? 5
E.6

mysolution was:

to find a right triangle first

Find the altitude which is BA or AB (however you want to call it)
Find the Base would be AD- BC
Then find CD

to find the Altitude

Area=28 for a rectangle, side L= 8(given AD) Altitude = 3.5

Since BC is parallel to AD my base for the triangle would be 8-6=2

now I have base + altitude I get

3.5^2+2^2=CD

16.25=cd? I do not see this answer can someone tell me which step is wrong?
I hope I'm understanding the picture correctly. I think there are two errors in the above- the area, 28, is the area not just of the rectangular portion of the quadrilateral; it also includes the triangular portion. Let's call the height (the length of AB) 'h'. We can divide the picture into a rectangle, with a base of 6, and a right angled triangle, with a base of 2. We know that the area of the rectangle plus the area of the triangle is 28:

6*h + 2*h/2 = 28
7*h = 28
h = 4

You've also left off a square (^2) on the right side of the Pythagorean formula at the end of your solution. If we use h = 4, and b = 2 for the right angled triangle, we should have:

CD^2 = 4^2 + 2^2
CD^2 = 20
CD = root(20) = 2*root(5)

Thanks I got it! I when 28= 8 * L I forgot that it will include an extra right triangle in there.

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zhrmghg: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Jul 28, 2008 10:26 pm

by zhrmghg » Mon Jul 28, 2008 11:07 pm

i suggest you to go on

Quote

artistocrat: Master | Next Rank: 500 Posts; Posts: 152; Joined: Wed Mar 12, 2008 4:36 pm; Thanked: 8 times; Followed by:2 members

by artistocrat » Mon Sep 01, 2008 6:45 am

You guys are making this far too difficult. The question is what is the height when we know the area of a trapezoid. The area of a trapezoid is (b1+b2)H/2, where Height (H) is the unknown. (8+6)H/2=28. H=4