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What is the area of the triangle ABC?
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Bunuel
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What is the area of the triangle ABC? [#permalink] New post  20 Jun 2019, 23:14
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What is the area of the triangle ABC?


(A) 15

(B) \(15\sqrt{3}\)

(C) \(\frac{5\sqrt{119}}{2}\)

(D) 32.5

(E) 36


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Lampard42
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Re: What is the area of the triangle ABC? [#permalink] New post  20 Jun 2019, 23:54
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Bunuel wrote:
Image
What is the area of the triangle ABC?


(A) 15

(B) \(15\sqrt{3}\)

(C) \(\frac{5\sqrt{119}}{2}\)

(D) 32.5

(E) 36


Show: ::
Attachment:
Capture1 (1).JPG


The altitude to the base from A will form a 30-60-90 triangle with Hypo as 12. So Other altitude using 1-root3-2 will be 6root3 (as opposite 60degrees) so area is 15root3 IMO B
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Dillesh4096
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Re: What is the area of the triangle ABC? [#permalink] New post  21 Jun 2019, 01:18
1
Bunuel wrote:
Image
What is the area of the triangle ABC?


(A) 15

(B) \(15\sqrt{3}\)

(C) \(\frac{5\sqrt{119}}{2}\)

(D) 32.5

(E) 36

Show: ::
Attachment:
The attachment Capture1 (1).JPG is no longer available



Image

Draw a perpendicular from point C to meet side AB at point D

CDB is a 30-60-90 triangle with
∠C = 30 deg, ∠D = 90 deg & ∠B = 60 deg

--> BD: CD: BC = 1: \(\sqrt{3}\): 2
--> BD = k, CD = \(\sqrt{3}\)k, BC = 2k
Given BC = 2k = 5
--> k = 5/2

Altitude = \(\sqrt{3}\)k = 5/2*\(\sqrt{3}\)

Area of triangle ABC = 1/2*base*height = 1/2*AB*CD
= 1/2*12*5/2*\(\sqrt{3}\)
= 15\(\sqrt{3}\)

IMO Option B

Pls Hit Kudos if you like the solution
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LeoN88
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Re: What is the area of the triangle ABC? [#permalink] New post  21 Jun 2019, 01:44
1
Bunuel wrote:
Image
What is the area of the triangle ABC?


(A) 15

(B) \(15\sqrt{3}\)

(C) \(\frac{5\sqrt{119}}{2}\)

(D) 32.5

(E) 36


Show: ::
Attachment:
The attachment Capture1 (1).JPG is no longer available


AD= 6 Root(3) by 60-90-30 Triangle Properties

Now, Area= (1/2) * 5 * 6(Root 3)
Attachment:
Capture.PNG
Capture.PNG [ 44.3 KiB | Viewed 596 times ]
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