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If AB=20 and BC=25, what is the length of AD in the figure above?
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01 Aug 2017, 13:21
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If AB=20 and BC=25, what is the length of AD in the figure above?
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01 Aug 2017, 20:21
Given data: AB = 20, BC = 25 Using Pythagoras theorem in right triangle ABC, \(BC^2 = AB^2 + AC^2\) > \(AC^2 = BC^2  AB^2\) Substituting values, we will get \(AC^2 = 625  400\) > \(AC = \sqrt{225} = 15\) The area of the triangle is \(\frac{1}{2}\) * AB * AC. It is also equal to \(\frac{1}{2}\) * BC * AD If AD = x, \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\) Therefore, the side AD's length is 12(Option C)
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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01 Aug 2017, 14:00
ADxBC = AC x AB ADx25=15x20 AD=13 Sent from my MotoG3TE using GMAT Club Forum mobile app



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If AB=20 and BC=25, what is the length of AD in the figure above?
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02 Aug 2017, 08:10
Bunuel wrote: 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 These triangles (ABC and DAB) are similar 345 right triangles. Both have one 90° angle, and angle B is common to both. If two angles are equal, the third is equal. (AA = AAA = similar triangles.) Whenever a right triangle has longer leg length that is a multiple of 4 (20), and a hypotenuse that is a multiple of 5 (25), the figure is a 345 right triangle. Or find AC, and the 3x4x5x ratio will be clear. Let AC = a, AB = b, and BC = c. By Pythagorean theorem, \(a^2 + b^2 = c^2\) \(a^2 + 20^2 = 25^2\) \(a^2 + 400 = 625\) \(a^2 = 225\) \(a = 15\) = AC 152025 divided by 5 throughout = 345 side ratio Smaller triangle DAB must also be a 345 right triangle. Its hypotenuse AB = 20. To find length of shorter leg AD, find multiplier, using hypotenuse AB. 3x: 4x: 5x > 5x is hypotenuse, 3x is shorter leg. 5x = 20 x = 4 = multiplier AD corresponds with AC. Both are the shorter leg, i.e., 3x. 3x = (3)(4) = 12 Answer C
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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04 Jun 2018, 03:36
This question can be answered in less than 30 secs by a simple method The sides are given are  25, and 20. If you remember the Pythagorean triplet  345, this is a piece of cake for you. The ratio of AB and BC suggests the length of AC as  5*5 = 25 (AB) 5*4 = 20 (BC) AC should be 5*3 = 15. Further, The second Right triangle also holds the multiples of the Pythagorean triplet 543 as its sides. Hence, AC  5*3 = 15 AD  4*3 = 12 CD  3*3 = 9. EasyPeasy!



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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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23 Jun 2018, 14:03
AC = 15. (from Pythagorean triplet) 1/2*(AB)*(AC)=1/2*(BC)*(AD)  from area formula. base height change 20*15=25*AD AD=20*15/25=12
Answer C



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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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11 Jul 2018, 03:06
pushpitkc wrote: Given data:AB = 20, BC = 25 Since the triangle is a right angled triangle, Using Pythagoras theorem \(BC^2 = AB^2 + AC^2\) => \(AC^2 = BC^2  AB^2\) \(AC^2 = 625  400\) \(AC = \sqrt{225} = 15\) The area of the triangle is 1/2 * AC * BDThe area if also equal to 1/2 * BC * AD Therefore, AC*BD = BC*AD Let AD be x \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\)= 12(Option C) Not able to understand the highlighted part. How is the area 1/2 * AC * BD? It seems there is no link between AC and BD. Please explain.
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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11 Jul 2018, 03:46
AkshdeepS wrote: pushpitkc wrote: Given data:AB = 20, BC = 25 Since the triangle is a right angled triangle, Using Pythagoras theorem \(BC^2 = AB^2 + AC^2\) => \(AC^2 = BC^2  AB^2\) \(AC^2 = 625  400\) \(AC = \sqrt{225} = 15\) The area of the triangle is 1/2 * AC * BDThe area if also equal to 1/2 * BC * AD Therefore, AC*BD = BC*AD Let AD be x \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\)= 12(Option C) Not able to understand the highlighted part. How is the area 1/2 * AC * BD? It seems there is no link between AC and BD. Please explain. Hi AkshdeepSThanks for informing. It was a typo and have corrected it now
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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21 Jul 2018, 03:57
pushpitkc wrote: Given data: AB = 20, BC = 25 Using Pythagoras theorem in right triangle ABC, \(BC^2 = AB^2 + AC^2\) > \(AC^2 = BC^2  AB^2\) Substituting values, we will get \(AC^2 = 625  400\) > \(AC = \sqrt{225} = 15\) The area of the triangle is \(\frac{1}{2}\) * AB * AC. It is also equal to \(\frac{1}{2}\) * BC * AD If AD = x, \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\) Therefore, the side AD's length is 12(Option C) Hi pushpitkc, how is your gmat prep ? hope bumpy roads and pit stops are already behind you, and there is just highway in front of you you write the area of the triangle \(\frac{1}{2}\) * AB * AC. is equal to the area of this traingle \(\frac{1}{2}\) * BC * AD the above is followed by equation If AD = x, \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\) I cant recognize formula of area of triangle the \(\frac{1}{2}\) is absent here By the way why do you assume that both triangles are of equal area ? is there some rule ? many thanks
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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22 Jul 2018, 03:48
dave13 wrote: pushpitkc wrote: Given data: AB = 20, BC = 25 Using Pythagoras theorem in right triangle ABC, \(BC^2 = AB^2 + AC^2\) > \(AC^2 = BC^2  AB^2\) Substituting values, we will get \(AC^2 = 625  400\) > \(AC = \sqrt{225} = 15\) The area of the triangle is \(\frac{1}{2}\) * AB * AC. It is also equal to \(\frac{1}{2}\) * BC * AD If AD = x, \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\) Therefore, the side AD's length is 12(Option C) Hi pushpitkc, how is your gmat prep ? hope bumpy roads and pit stops are already behind you, and there is just highway in front of you you write the area of the triangle \(\frac{1}{2}\) * AB * AC. is equal to the area of this traingle \(\frac{1}{2}\) * BC * AD the above is followed by equation If AD = x, \(20*15 = 25*x\) Solving for x, \(x = \frac{20*15}{25}\) I cant recognize formula of area of triangle the \(\frac{1}{2}\) is absent here By the way why do you assume that both triangles are of equal area ? is there some rule ? many thanks Hey dave13Sorry for the delay in replying to your doubt. And yes, the bumpy roads are finally beginning to disappear. The area of the triangle is \(\frac{1}{2}\) the product of the base and the height. For the triangle ABC, if BC is the height of the triangle, then AD is the length of the base. Similarly, if AB is the height of the triangle, then AC is the length of the base. Now, we have two areas of the same triangle and this can be equated. Now,\(\frac{1}{2}*AD*BC = \frac{1}{2}*AB*AC\) > \(\frac{1}{2}*2*AD*BC = AB*AC\) > \(AD*BC = AB*AC\) Hope this clears your confusion!
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?
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22 Jul 2018, 06:29
Area of the triangle remains constant (Area = 1/2*base*height) Therefore 1/2*AC*AB=1/2*CB*AD
Also, because of pythagoras theorem, AC^2 = CB^2  AB^2
Substituting, [Root(625  400)]*20=25*AD AD=15*20/25 AD=12




Re: If AB=20 and BC=25, what is the length of AD in the figure above? &nbs
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