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Difficulty:
15%
(low)
Question Stats:
85% (01:50) correct
15%
(02:33)
wrong
based on 73
sessions
History
Date
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If AB=20 and BC=25, what is the length of AD in the figure above?
A. 6
B. 9.6
C. 12
D. 20
E. 24
Attachment:
1.png [ 3.41 KiB | Viewed 4503 times ]
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Posts: 1,052
Explanation:
AC^2 = BC^2 - AB^2
AC^2 = 625 - 400
AC^2 = 225
AC = 15
1/2 × Base AC × Height AB = 1/2 × Base BC × Height AD
15 × 20 = 25 × Height AD
Height AD = 3×4 = 12
IMO-C
Posted from my mobile device
AC^2 = BC^2 - AB^2
AC^2 = 625 - 400
AC^2 = 225
AC = 15
1/2 × Base AC × Height AB = 1/2 × Base BC × Height AD
15 × 20 = 25 × Height AD
Height AD = 3×4 = 12
IMO-C
Posted from my mobile device
Triangle ABC and DBA are similar
\(\frac{AD}{AC}\) = \(\frac{AB}{BC}\)
AC = \(\sqrt{(25^2-20^2)}\) = 15
\(AD = 15*\frac{20}{25} =12 \\
\)
\(\frac{AD}{AC}\) = \(\frac{AB}{BC}\)
AC = \(\sqrt{(25^2-20^2)}\) = 15
\(AD = 15*\frac{20}{25} =12 \\
\)
Bunuel
asishron29181
Joined: 04 May 2020
Last visit: 04 Nov 2022
Posts: 185
Location: India
Schools: Great Lakes (I)
GMAT 1: 610 Q47 V27
WE:Programming (Computer Software)
Given, AB=20 and BC=25
BC^2 = AB^2 + AC^2
625= 400 + AC^2
AC = 15
Now in the above figure,
Area of triangle ABC = 0.5 * AB * AC = 0.5 * BC * AC
20*15 = 25x
x= 12
AD=12
IMO C
BC^2 = AB^2 + AC^2
625= 400 + AC^2
AC = 15
Now in the above figure,
Area of triangle ABC = 0.5 * AB * AC = 0.5 * BC * AC
20*15 = 25x
x= 12
AD=12
IMO C
AB = 20, BC = 25 => AC = 15
Area of triangle ABC = (1/2) x 20 x 15 = (1/2) x 25 x AD
Hence, AD = 3 x 4 = 12 (C)
Area of triangle ABC = (1/2) x 20 x 15 = (1/2) x 25 x AD
Hence, AD = 3 x 4 = 12 (C)
Posts: 731
Pythagores triplet.
3,4,5
5,12,13
15,20,25
etc...
So other side is 15;
AB x AC = BC x AD, Since right angle triagle.
So AD=12
IMO C
Posted from my mobile device
3,4,5
5,12,13
15,20,25
etc...
So other side is 15;
AB x AC = BC x AD, Since right angle triagle.
So AD=12
IMO C
Posted from my mobile device
Posts: 1,101
AC^2 = 25^2 - 20^2
or, AC^2 = 225 or, AC = 15.
AD/AC = AB/BC or, AD/15 = 20/25 or, AD = 12.
So, I think C.
or, AC^2 = 225 or, AC = 15.
AD/AC = AB/BC or, AD/15 = 20/25 or, AD = 12.
So, I think C.
Posts: 57
Hello!
We are given two sides of a right triangle.
If you notice the two sides, if divided by 5, fit the 3, 4, 5 right triangle formula, you could determine that AC is 15
But if not, or if you want to be sure, let's use the Pythagorean Theorem
\(a^2 + b^2 = c^2\)
\(20^2 + b^2 = 25^2\)
400 + \(b^2\) = 625
\(b^2\) = 225
b = 15
You don't have to learn all the squares from 1-100 but knowing popular ones like \(25^2\) and \(15^2\) is very useful
So now that we know that AC is indeed 15, how do we use this information?
The area of a triangle is \(\frac{1}{2}\)(base)(height)
In this case it is \(\frac{1}{2}\)(15)(20) = 15(10) = 150
Ok, so what?
Well, now you can use BC as the base and AD as the height and plug it into the equation to determine the length of AD
1/2(25)(AD) = 150
25(AD) = 300
AD = 12
The answer is (C)
We are given two sides of a right triangle.
If you notice the two sides, if divided by 5, fit the 3, 4, 5 right triangle formula, you could determine that AC is 15
But if not, or if you want to be sure, let's use the Pythagorean Theorem
\(a^2 + b^2 = c^2\)
\(20^2 + b^2 = 25^2\)
400 + \(b^2\) = 625
\(b^2\) = 225
b = 15
You don't have to learn all the squares from 1-100 but knowing popular ones like \(25^2\) and \(15^2\) is very useful
So now that we know that AC is indeed 15, how do we use this information?
The area of a triangle is \(\frac{1}{2}\)(base)(height)
In this case it is \(\frac{1}{2}\)(15)(20) = 15(10) = 150
Ok, so what?
Well, now you can use BC as the base and AD as the height and plug it into the equation to determine the length of AD
1/2(25)(AD) = 150
25(AD) = 300
AD = 12
The answer is (C)
Bunuel
AC = \(\sqrt{25^2 - 20^2} = 15\)
Now, ABC and ADB are similar triangles:
So,\(\frac{AB}{BC} = \frac{AD}{AC}\)(Why? They would be in similar ratio)
AD = 15 * 20/25 = 12, (C) IMO!
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