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In rectangle ABCD below, and Find the following. AB=5,AF=7,
[#permalink]
26 May 2019, 01:51
26 May 2019, 01:51
1
Expert Reply
In rectangle ABCD below, AB=5,AF=7, FD=3. Find the following.
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
#GREexcercise In rectangle ABCD below.jpg [ 16.07 KiB | Viewed 14383 times ]
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
Attachment:
#GREexcercise In rectangle ABCD below.jpg [ 16.07 KiB | Viewed 14383 times ]
Show: :: OA
(a) 50 (b) 17.5 (c) 5 \(\sqrt{5}\) (d) 30
Math Review
Question: 10
Page: 260
Difficulty: medium
Question: 10
Page: 260
Difficulty: medium
Re: In rectangle ABCD below, and Find the following. AB=5,AF=7,
[#permalink]
27 May 2019, 12:49
27 May 2019, 12:49
5
Expert Reply
Carcass wrote:
In rectangle ABCD below, AB=5,AF=7, FD=3. Find the following.
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
Attachment:
#GREexcercise In rectangle ABCD below.jpg
Show: :: OA
(a) 50 (b) 17.5 (c) 5 \(\sqrt{5}\) (d) 30
Math Review
Question: 10
Page: 260
Difficulty: medium
Question: 10
Page: 260
Difficulty: medium
GIVEN: AB = 5, AF = 7, and FD = 3
Add the information to the diagram
(a) Area of ABCD
Area of rectangle = (base)(height) = (10)(5) = 50
(b) Area of triangle AEF
Area of triangle = (base)(height)/2
= (7)(5)/2 = 17.5
(c) Length of BD
Since ∆ABD is a RIGHT triangle, we can apply the Pythagorean Theorem.
We get: 5² + 10² = (side BD)²
Simplify: 25 + 100 = (side BD)²
Simplify: 125 = (side BD)²
Solve: side BD = √125 = √25 x √5 = 5√5
(d) Perimeter of ABCD
Add lengths of 4 sides
Perimeter = 5 + 10 + 5 + 10 = 30
Cheers,
Brent
_________________
General Discussion
Re: In rectangle ABCD below, and Find the following. AB=5,AF=7,
[#permalink]
06 Aug 2021, 23:31
06 Aug 2021, 23:31
GreenlightTestPrep wrote:
Carcass wrote:
In rectangle ABCD below, AB=5,AF=7, FD=3. Find the following.
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
Attachment:
#GREexcercise In rectangle ABCD below.jpg
Show: :: OA
(a) 50 (b) 17.5 (c) 5 \(\sqrt{5}\) (d) 30
Math Review
Question: 10
Page: 260
Difficulty: medium
Question: 10
Page: 260
Difficulty: medium
GIVEN: AB = 5, AF = 7, and FD = 3
Add the information to the diagram
(a) Area of ABCD
Area of rectangle = (base)(height) = (10)(5) = 50
(b) Area of triangle AEF
Area of triangle = (base)(height)/2
= (7)(5)/2 = 17.5
(c) Length of BD
Since ∆ABD is a RIGHT triangle, we can apply the Pythagorean Theorem.
We get: 5² + 10² = (side BD)²
Simplify: 25 + 100 = (side BD)²
Simplify: 125 = (side BD)²
Solve: side BD = √125 = √25 x √5 = 5√5
(d) Perimeter of ABCD
Add lengths of 4 sides
Perimeter = 5 + 10 + 5 + 10 = 30
Cheers,
Brent
How do you assume that the length of EF is the same as AB that is 5. I think that it will only be the case when EF is perpendicular to AD.
Re: In rectangle ABCD below, and Find the following. AB=5,AF=7,
[#permalink]
07 Aug 2021, 05:09
07 Aug 2021, 05:09
2
Expert Reply
taskforce wrote:
How do you assume that the length of EF is the same as AB that is 5. I think that it will only be the case when EF is perpendicular to AD.
Be careful. I'm not saying that EF has length 5; I'm saying that triangle AEF has height 5 (since AB = 5).
Here's what I mean:
For more on finding the area of a triangle, you can watch this video (starting at 5:07):
_________________
Re: In rectangle ABCD below, and Find the following. AB=5,AF=7,
[#permalink]
07 Aug 2021, 08:22
07 Aug 2021, 08:22
GreenlightTestPrep wrote:
taskforce wrote:
How do you assume that the length of EF is the same as AB that is 5. I think that it will only be the case when EF is perpendicular to AD.
Be careful. I'm not saying that EF has length 5; I'm saying that triangle AEF has height 5 (since AB = 5).
Here's what I mean:
For more on finding the area of a triangle, you can watch this video (starting at 5:07):
Got it! I think I have found a hole in my understanding
gmatclubot
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