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# The figure above shows a row of triangular tiles each with two sides

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Math Expert
Joined: 02 Sep 2009
Posts: 55150
The figure above shows a row of triangular tiles each with two sides  [#permalink]

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09 Aug 2017, 02:13
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Difficulty:

15% (low)

Question Stats:

96% (01:15) correct 4% (02:59) wrong based on 39 sessions

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The figure above shows a row of triangular tiles each with two sides 15 centimeters long and one side 18 centimeters long. What is w, the width of the row, in centimeters?

(A) 10
(B) 12
(C) 15
(D) 9√2
(E) 9√3

Attachment:

2017-08-09_1258.png [ 11.61 KiB | Viewed 732 times ]

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Re: The figure above shows a row of triangular tiles each with two sides  [#permalink]

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09 Aug 2017, 02:18
Bunuel wrote:

The figure above shows a row of triangular tiles each with two sides 15 centimeters long and one side 18 centimeters long. What is w, the width of the row, in centimeters?

(A) 10
(B) 12
(C) 15
(D) 9√2
(E) 9√3

Attachment:
2017-08-09_1258.png

$$w = \sqrt{(15^2) - (18/2)^2} = \sqrt{225-81} = \sqrt{144} = 12$$
B
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Re: The figure above shows a row of triangular tiles each with two sides  [#permalink]

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09 Aug 2017, 03:06

Formula used :
Area = 1/2 * Base * Height
Area = $$\sqrt{s(s-a)(s-b)(s-c)}$$ where s is the semiperimeter($$\frac{a+b+c}{2}$$) and a,b, and c are the three sides of the triangle

Given data : a=b=15, c=18
We can compute the semiperimeter s = $$\frac{a+b+c}{2} = \frac{48}{2} = 24$$
Therefore, the area of the triangle = $$\sqrt{24(24-15)(24-15)(24-18)}$$ = $$\sqrt{24*(9)(9)(6)}$$ = $$\sqrt{2^3*3*(3^2)(3^2)(2*3)}$$ = $$\sqrt{2^4*3^6} = 2^2 * 3^3$$

The area of the triangle is also equal to $$\frac{1}{2} * 18 * w = 9w$$

We can equate the area of the triangle : $$3^2(w) = 2^2 * 3^3 => w = 2^2 * 3 = 12$$(Option B)
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Re: The figure above shows a row of triangular tiles each with two sides  [#permalink]

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09 Aug 2017, 03:43
1
In an isosceles triangle, altitude is a perpendicular bisector of the base.
Thus, $$w^2 + 9^2 = 15^2$$
$$w^2 = 225 - 81$$
$$w = \sqrt{144} = 12$$ (Choice B)
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The figure above shows a row of triangular tiles each with two sides  [#permalink]

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09 Aug 2017, 09:34
Bunuel wrote:

The figure above shows a row of triangular tiles each with two sides 15 centimeters long and one side 18 centimeters long. What is w, the width of the row, in centimeters?

(A) 10
(B) 12
(C) 15
(D) 9√2
(E) 9√3

Attachment:
2017-08-09_1258.png

Each triangle contains a 3-4-5 right triangle.

Width of row = altitude of one triangle.

Altitude of the isosceles triangle is a perpendicular bisector of the base; intercept point creates two right angles and divides base in half.

Base = 9. Hypotenuse = 15. Multiple of 3, and . . . Multiple of 5 . . .

If 3-4-5 right triangle is familiar, no need to calculate; 9-12-15 is common.

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The figure above shows a row of triangular tiles each with two sides   [#permalink] 09 Aug 2017, 09:34
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