Area of Triangle ABC - June 24-28, 1996
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In triangle ABC, AC=18 and D is the point on AC for which AD=5. Perpendiculars drawn from D to AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?
Different approaches to solving this problem came in, and we've chosen to feature a variety of them. Take a look at solutions by Thomas Kuo, Jason Yeung, Brian Gordon, Ken Duisenberg, and Dave Peterson.Thomas S. Kuo School: Murray Junior High School, Ridgecrest, California Grade: 7th A * H ** * D * * * G * * * * * * * * * * * * B * * * * * * * * * C F E AC = 18, AD = 5, DC = 13 DH and CG are perpendicular to AB AF and DE are perpendicular to BC DH = 4, DE = 5 x = AF, y = CG, z = BC The area of the triangle is 51.43 unit^2. 1. Find the length of AF or x Compare similar triangles AFC and DEC: x/5 = 18/13, x = AF = 90/13 ~ 6.92 2. Find the angles angle A = angle BAC = arc tan (4/3) ~ 53.13 degree angle C = angle ACB = arc tan (5/12) ~ 22.62 degree angle B = angle ABC = 180 - angle A - angle C ~ 104.25 degree 3. Find the length of CG or y Compare similar triangles AGC and AHD: y/4 = 18/5, y = CG = 72/5 ~ 14. 4. Find the length of BC or z In right triangle GBC: sin B = CG/BC = y/z, z = BC = CG/sin B = 14.4 / sin(104.25) ~ 14.86 5. Find the area of triangle ABC Area = AF * BC /2 = x * z / 2 = (14.86)*(6.92)/2 ~ 51.43
Jason Yeung Grade: 10 School: Iolani School I cannot think of a very geometric approach, so I will use trig. My method is the law of sines, and my solution is approx. square root of 2644.8984, or 51.42857. The triangle is actually divided into 3 different figures: a 3-4-5 right triangle, a 5-12-13 right triangle, and a 4-sided figure. I plan to use Heron's Formula to find the area, so I must find all 3 sides of the triangle. With the 3-4-5 right triangle, using the law of sines, I was able to find the measurement of angle A, which is approx. 53.130102. I did the same with triangle 5-12-13 and got an angle C of 22.619865. With those angles, I was able to figure out angle B, which is 104.25002. From there, I was able to use Law of Sines to find sides AB and BC. Finally,, I used Heron's formula to find the area of the triangle.
Brian Gordon School: Dartmouth '92 The answer is 51 3/7 square units. I did it analytically. Place A at (0,0) and C at (18,0), so that D is at (5,0). Then the base of ABC is 18, and if I can find the y-coordinate of B, that will be the height, and so I can use A=.5*18*h Let E be the point on AB where the perpendicular from D is drawn. Let F be the point on CB where the perpendicular from D is drawn. From the given, DE=4 and DF=5. There are little right triangles formed by these altitudes, which specifically give us the lengths AE=3 and CF=12. (These are from 3-4-5 and 5-12-13 triangles) I next went to find the slopes of AB and CB. I got these from the right triangles AED and CFD. The easiest way to see the slopes for the lines is to draw altitudes-on-hypotenuse for both. Drawing these creates even smaller right triangles which are SIMILAR to the original ones. The slopes for the lines are then the same ratio as the legs of the original right triangles: AB has slope 4/3 and passes through (0,0) CB has slope -5/12 and passes through (18,0) The two equations of the lines are y=(4/3)x and y=(-5/12)x + 15/2 Solving them simultaneously for B gives x=30/7, y=40/7 With a y-coordinate of 40/7, that is the height of the triangle ABC, so the area is 1/2 * 18 * 40/7 = 360/7 = 51 3/7 --bri
Ken Duisenberg Grade: Hewlett Packard Engineer School: Stanford University '91 Answer: Answer: 360/7 ~= 51.429 Solution: Given: Triangle ABC, AC=18. D on AC: AD=5. Distance from D to AB=4, Distance from D to BC = 5. Find: Area of ABC. Let E be the point on AB perpendicular to D (DE = 4). Let F be the point on BC perpendicular to D (DF = 5). Then, using the Pythagorean Theorem: AE^2 + 4^2 = 5^2 --> AE = 3. CF^2 + 5^2 = 13^2 --> CF = 12. So, sin(A)=4/5, cos(A)=3/5, sin(C)=5/13, cos(C)=12/13. Then, using the Sine Law for triangles: sin(B) sin(A) ------ = ------ 18 BC So, (18 * 4) (1) BC = -------- 5*sin(B) Angle B = 180 - A - C, so using the following formulae, find sin(B): sin(X +/- Y) = sin(X)*cos(Y) +/- cos(X)*sin(Y) cos(X +/- Y) = cos(X)*cos(Y) -/+ sin(X)*sin(Y) (2) sin(B) = sin( (180 - A) - C) = sin(180-A)*cos(C) - cos(180-A)*sin(C) = [sin(180)*cos(A)-cos(180)*sin(A)]*cos(C) - [cos(180)*cos(A)+sin(180)*sin(A)]*sin(C) = [ 0 - (-1)(4/5)]*(12/13) - [(-1)(3/5) + 0](5/13) = (4/5)(12/13) + (3/5)(5/13) sin(B) = 63/65 Substituting (2) into (1), BC = 104/7 Using BC as the base of the triangle, the height is h = 18 * sin(C) = 18*5/13 = 90/13 And the area of ABC is base*height/2: Area = (104/7) * (90/13) / 2 = (13*2*4/7) * (90/13) / 2 = 4*90/7 Area = 360/7
Date: Fri, 28 Jun 1996 16:35:43 -0400 Dave Peterson, Rochester NY To find the area of the triangle, we need the altitude h from B to AC. Draw this altitude and call the segment BX. From a sketch of the triangle, X is on the segment AC. (This is important later.) B /\ /| \ / | \ E/ | \ F /\ | /\ /\/ \| /\/ \ / |\ / \ / | \ / \ / _| \ / \ /______|_|_____\/____________________\ A X D C We are given that AC = 18 and AD = 5, so CD = 13; DE = 4 and DF = 5, so by the Pythagorean theorem AE = 3 and CF = 12. Triangles AED and AXB are similar, so BX/DE = AX/AE, i.e. h/4 = AX/3. Triangles CFD and CXB are similar, so BX/DF = CX/CF, i.e. h/5 = CX/12. The sides AX and CX are unknown, but we know their sum; so solve for them and add: AX = h*3/4 CX = h*12/5 AC = AX + CX = h*(3/4 + 12/5) = h*63/20 Now solve h * 63/20 = 18 to get h, and use it to get the area: h = 18*20/63 = 360/63 = 40/7 Area = 18*h/2 = 360/7 My first solution also gave me the lengths of AB and BC (using five simultaneous equations!), but I don't need them to get the area. If you want them, the similar triangles give them easily. Just for fun, I used the same method with variables, to get a general formula for this sort of problem. If AD = a, CD = b, DE = c, and DF = d, the area is c*d*(a + b)^2 ------------------------------------- 2*(sqrt(a^2 - c^2) + sqrt(b^2 - d^2)) But now, suppose that when you say "perpendiculars drawn from D to AB and CB," you mean to the lines, not necessarily to the segments! Then I could draw the triangle ADE with E below AC, and B would be far to the left, making an obtuse triangle. X would be on line AC but off to the left, and instead of AX+CX=AC, I would have to use CX-AX=AC. Then h*(12/5 - 3/4) = h*33/20 = 18, and h = 360/33 = 120/11. The area would be 1080/11. I don't think that's what you meant.
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