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

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If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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01 Aug 2017, 12:21
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25% (medium)

Question Stats:

78% (02:21) correct 22% (02:24) wrong based on 230 sessions

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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:

T6019.png [ 4.76 KiB | Viewed 2773 times ]

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If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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01 Aug 2017, 19:21
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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?  [#permalink]

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01 Aug 2017, 13:00
1

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If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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02 Aug 2017, 07:10
2
1
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
Attachment:
T6019.png

These triangles (ABC and DAB) are similar 3-4-5 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 3-4-5 right triangle.

Or find AC, and the 3x-4x-5x 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

15-20-25 divided by 5 throughout = 3-4-5 side ratio

Smaller triangle DAB must also be a 3-4-5 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

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Re: If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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04 Jun 2018, 02:36
1
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 - 3-4-5, 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 -5-4-3 as its sides.
Hence,
AC - 5*3 = 15
CD - 3*3 = 9.

Easy-Peasy!
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Re: If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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23 Jun 2018, 13:03
AC = 15. (from Pythagorean triplet)
1/2*(AB)*(AC)=1/2*(BC)*(AD) - from area formula. base height change

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Re: If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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11 Jul 2018, 02:06
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 * BD
The area if also equal to 1/2 * BC * AD

$$20*15 = 25*x$$

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.

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Re: If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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11 Jul 2018, 02:46
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 * BD
The area if also equal to 1/2 * BC * AD

$$20*15 = 25*x$$

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.

Hi AkshdeepS

Thanks 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?  [#permalink]

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21 Jul 2018, 02:57
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?  [#permalink]

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22 Jul 2018, 02:48
1
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 dave13

Sorry 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,$$\frac{1}{2}*AD*BC = \frac{1}{2}*AB*AC$$ -> $$\frac{1}{2}*2*AD*BC = AB*AC$$ -> $$AD*BC = AB*AC$$

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Re: If AB=20 and BC=25, what is the length of AD in the figure above?  [#permalink]

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22 Jul 2018, 05:29
Area of the triangle remains constant (Area = 1/2*base*height)

Also, because of pythagoras theorem, AC^2 = CB^2 - AB^2

Re: If AB=20 and BC=25, what is the length of AD in the figure above? &nbs [#permalink] 22 Jul 2018, 05:29
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