Steps of Construction:1. Draw a line segment AB = 5 cm.2. At A make ∠BAY = 45°.3. Take A as centre and radius = AC = 6 cm (= CA), draw an arc cutting AY at C.4. Join BC to obtain the triangle ABC.5. Draw any ray AX making an acute angle with AB on the side opposite to the vertexC.6. Mark off 6 points (the greater of 6 and 5 in 6/5) A1, A1, A3, A4,A5 and A6 on AX sothatAA1 = A1A2 = A2A3.
2 Answers
Jan 7, 2017
Answer:
Explanation:
As #AD# is drawn perpendicular to #BC# in right angled #DeltaABC#, it is apparent that #DeltaABC# is right angled at #/_A# as shown below (not drawn to scale). As can be seen #/_B# is common in #Delta ABC# as well as #DeltaDBA# (here we have written two triangles this way as #/_A=/_D#, #/_B=/_B# and #/_C=/_BAD#) - as both are right angled (obviously third angles too would be equal) and therefore we have
#Delta ABC~~DeltaDBA# and hence
#(BC)/(AB)=(AB)/(BD)=(AC)/(AD)#..............(1)
therefore, we have #(BC)/(AB)=(AB)/(BD)# or
#AB^2=BCxxBD=9xx4=36#
Hence #AB=6#
Jan 7, 2017
Given that #DeltaABC# is right angled and #AD# is drawn perpendicular to #BC# . So #/_BAC# is right angle.
Given #BC = 9cm and BD =4cm-> CD = BC-BD=5cm#
In #Delta ABC and Delta ABD#
#/_ABD=/_ABC#
#/_ADB=/_BAC= 'right angle'#
#/_BAD=/_ACD ('remaining')#
So #Delta ABC and Delta ABD# are similar
Hence
#(AB)/(BC)=(BD)/(AB)#
#=>AB^2=BDxxBC=4xx9=36#
#=>AB=sqrt36cm=6cm#
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School Subjects/Math and Arithmetic/Algebra/Geometry
Approximately 5.29 centimeters. There may be an easier way tofigure it out, but this is what I came up with. Draw your triangleon a piece of paper. Because you don't know at this point whetherit's an acute triangle (all angles < 90 degrees), a righttriangle (one angle = 90 degrees), or an obtuse triangle (one angle> 90 degrees), you will want to draw all three possibilities.But when you try to draw these triangles, you will be able to tellthat IF any angle is > or = 90 degrees, it will HAVE to be angleC. Starting with the acute triangle: draw a line from point B to apoint, somewhere in the middle of line AC, that forms a 90 degreeangle with line AC. Label this point D. Triangle ABD is now a righttriangle, with a right angle at angle D. Now, there are all kindsof things you can figure out about this triangle, some of which mayhelp you figure out things about the original triangle ABC. One ofthe things we know about any right triangle is that, if, inaddition to the 90-degree right angle, we know the measurement ofat least one other angle, and at least one side, we can figure outall the sides and all the angles. To do so, we need to use thetrigonometric functions - sine, cosine, tangent, etc. Now, we canlook at a table of trigonometric functions and easily determine thesine, cosine, etc. of any angle, provided we know the measurementof that angle. The angle we know is angle A, 60 degrees. But whichfunction do we want to use here? The definition of cosine isadjacent over hypotenuse, or in other words, the length of the sideadjacent to the angle (but not opposite the right angle) divided bythe length of the side that is opposite the right angle. In thiscase, line AB is the hypotenuse, because the right angle is at D.Line AD is the adjacent side. We know the length of the hypotenuse,but not of the adjacent side. But that's okay because we can easilyfigure it out. The formula is cos A = AD/AB. Now substitute what weknow - cos 60 = AD/6. Then look up the cosine of 60 degrees in atrig table, and you get 0.5, so 0.5 = AD/6. Solving for AD yieldsAD = 3. At this point, we need to look back at our 3 differentversions of the triangle ABC. We know now that AD(3 cm) is shorterthan AC (4 cm), and therefore we were right to put point Dsomewhere between A and C. If AD had been longer than AC, then, forvisual purposes, we would want to deal with the obtuse trianglerather than the acute one. What you would do in this case is extendline AC past C until you can draw a line from point B down to theextended line AC to form a right angle, then label this point D.But either way, we can now figure out the length of line CD. If ACis longer than AD (as is the case here), then CD = AC - AD = 4 - 3= 1. If AD is longer than AC, then CD = AD - AC. Oh, if AD isexactly equal to AC, then triangle ABC was a right triangle allalong, and we can easily figure out AC by using the sine function.Now, going back to the acute triangle we drew earlier, since we nowknow that ABC is, in fact, an acute triangle. Not only do you havea right triangle in ABD, you have another right triangle defined byBCD. The Pythagorean theorum gives us an easy method fordetermining the length of any side given the other two sides. Butwe only know one side of BCD, CD = 1. Can we figure out BD so thatwe have the necessary two sides' lengths? Yes, we can, if we goback to right triangle ABD, which shares a side (BD) with BCD. Thesine of an angle is defined as the opposite divided by thehypotenuse. In this case, the opposite is line BD, which we do notknow, and the hypotenuse is, again, AB, which we know to be 6. And,from the trig tables, we know that the sine of angle A (60 degrees)is, approximately, 0.866. The formula is sin A = BD/AB, or,substituting what we know, 0.866 = BD/6, then solving for BD yieldsBD = (app) 5.196. So, now we know both BD and CD, the twonon-hypotenuse sides of right triangle BCD. The remaining side, thehypotenuse, is BC, which is the same line we set out to determinethe length of. Now we use the Pythagorean theorum, which says thatthe squares of the lengths of the two non-hypotenuse sides of aright triangle, added together, exactly equal the square of thelength of the hypotenuse. Or a2 + b2 = c2, where c is the length ofthe hypotenuse, and a and b are the length of the other two sides.(Don't get confused with the capital letters A, B, and C, which youused to designate the vertices of the original triangle. The lowercase letters here represent the lengths of the sides of triangleBCD.) So, now, if a = CD = 1, and b = BD = 5.196, then a2 + b2 = 12+ 5.1962 = 1 + 27 = 28. But this is equal to c2. So, to solve forc, just take the square root of 28, which is approximately equal to5.29. Though the question is answered now, I would like toexpand the answer a little bit. Every triangle has three sides andthree angles, for a total of 6 'pieces'. It turns out that, if youknow the measurements of any three of these pieces, and at leastone of them is a side, you can determine the measurements of theother three pieces. We started with three pieces (two sides and anangle), and since at least one of them is a side, we know we canget the rest. We already determined the length of the third side.All that is left to completely define this triangle is themeasurements of angles B and C. These can be obtained as well.We'll start with angle C. Remember that BCD is a right triangle, sowe can use the trig functions again. The cosine of angle C, withrespect to the right triangle BCD, is defined as adjacent dividedby hypotenuse, or CD/BC. In this case, we know both CD and BC, butwe do not know the measurement of angle C. But we can compute thevalue of cos C, which is CD/BC = 1/5.29 =(app) 0.189. Now that weknow this, we can use the trig tables in a backwards manner.Instead of looking up an angle, we'll find the value 0.189 in thebody of the table and trace it to the axes of the table to find themeasurement of the angle. This is called finding the 'inversecosine' of a value, or in other words, the angle measurement whosecosine is equal to the given value. This can also be done on ascientific calculator. The inverse cosine of 0.189 is,approximately, 79 degrees. We now have the measurement of angle C.For angle B, the job is much easier. We know that the sum of allthree angle measurements in any triangle is 180. Thus, computingthe measurement of angle B is simply a matter of subtracting themeasurements of angles A and C from 180. B = 180 - A - C = 180 - 60- 79 = 41. And now we have the measurement of angle B. In summary,the 6 pieces of triangle ABC are (with the ones we computed inbold): * Angle A 60o
* Side AB 6 cm
* Angle B 41o
* Side BC 5.29 cm
* Angle C 79o
* Side AC 4 cm
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