Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Multilateration wikipedia , lookup
Dessin d'enfant wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
Euclidean geometry wikipedia , lookup
Integer triangle wikipedia , lookup
This topic was given by Mrs Kalyani, we were meant to understand the basic facts and thus summarise it in our own words. We also had to demonstrate some sums. I have extracted some information from the internet, understood it and then written it in my own words. However, other shapes and figures were copied from many different sites, and also some questions were extracted from other sites. I have also used some questions from the book in order to demonstrate my ability. Thank you: Deip. D. Majithia Grade 11. In my project, I am assigned to write about Similarities. I will write everything that I have understood in my own words, and will demonstrate certain sums. Most of the information I have got is from the internet, I have also acknowledged some of the information from my mathematics teacher and lastly I have used sources like my text book for extra sums and more important information. I will demonstrate the sums and prove them. Similarities: Similarities could mean two things which are the same. They could be similar but not identical (NOT COMPLETELY SAME). Obviously this topic is to do with maths (subtopic: Geometry). Two Figures are similar if they have the same shape. Shapes that are similar have all corresponding (matching) sides equal in the same proportion. This means they have the same shape but different sizes. If we say that a triangle A is similar to a triangle B then we write : A ~ thing. - B (~) this is the notation used to show that something is similar to another In the above bolded statement, if shapes are similar their corresponding sides are equal and in the same proportion. 6cm e.g.1 3cm Rectangle B 2cm Rectangle A 4cm When I say corresponding sides, I mean sides 2cm and 4cm and sides 3cm and 6cm. Their Ratio is the same. 4/2 = 6/3 or vice versa is equal to 2. SIMILAR POLYGONS: Polygons are said to be similar only if: (1) Their corresponding angles are the same (2) Their corresponding sides are in a ratio (1) This above shape was extracted from a site, so that I could use it as an example for my project. In the above shape. Quadrilateral ABCD ~ ( is similar to) EFGH. This means that sides AB and EF, AD and EH, BC and FG and DC and HG are proportional. In order to find out what is the ratio of their proportion. We can divide the second (new) figure’s side by the first (old). If we take Quad ABCD as our first figure then: Ratio= EH/AD is the ratio of proportion. 2/4. The ratio is ½. Therefore we can say that: AB/EF = AD/EH= BC/FG = DC/HG (2) And in the second theorem, I stated that their corresponding angles are the same. This means angle A = angle E , angle B = angle F, angle D= angle H and lastly angle C = angle G. NOTE: In any figure, if corresponding angles are equal but corresponding sides are not proportional then we cannot call that figure similar. A B In the above shape, we cannot say Parallelogram A is similar to Parallelogram B, this is because the corresponding angles are the same ( all are right angles) but sides are not proportional. And therefore this parallelogram is not similar. ---------------------------------------------------------------------------------------------------------------------------------- SIMILAR TRIANGLES Just like polygons, in Triangles if two angles of the triangle are the same, as the two angles in another triangle then, the two triangles are therefore similar. This above image was extracted from www.cliffsnotes.com . In this image, we can see that angle B is equal to angle E. In figure ABC, if B is 100 degrees, C is 20 degrees then A obviously has to be 60 degrees, this is because: Angle A + Angle B + Angle C = 180 degrees Also in figure, DEF, we can easily claim that F is 20 degrees, the same theorem used above, can be used in this shape, and we can prove that F is 20 degrees as: Angle D + Angle E + Angle F = 180 degrees All triangles add up to 180 degrees. Let us use the same example as the above. In TRIANGLE ABC AB= 3cm BC= 5 cm AC= 4 cm In TRIANGLE DEF DE= 1.5cm EF= 2.5cm DF= 2 cm Taking in the Ratio, AB/DE = BC/EF = AC/DF Let us assume that we haven’t got DF, therefore DF= x , if we take a common ratio, then in order to find x, we can take AB/DE = AC/x SOLN: 3cm/1.5cm = 4cm/ x – If we cross multiply, then 3x is equal to (1.5 * 4): Therefore 3x= 6. X= 6/3 and x which is DF= 2cm, this proves that the ratio analysis I have used is correct. P 5 3 Q 2 T 3 R S Z cm. In the above question, extracted from Maths Text book. We have to find side Z, in this figure. We know that PQT ~ PRS. For that reason, the ratios will be equal. PQ/PR = QT/RS = PT/PS. So in order to get RS, we can use either of the ratios. For that reason PT/PS = QT/RS SOLN: PT/PS = QT/RS 5/8 = 3/Z Cross Multiplication: 5 x Z = 8 x 3 5Z = 24 Z= 24/5 = 4.8cm Question 3: C In question 3, we have to find X cm and W cm. We know that Triangle ABE ~ Triangle ACD. And therefore: AB/AC= AE/AD= BE/CD. In order to get w that is BC, we can find AC as that is in the ratio and subtract 3cm from AC. Soln: AB/AC = BE/CD 3/w = 4/6 w= (6 x 3) divided by 4 w 6cm B 3cm 4 A X cm E 2 D In the figure on the left, such figures are often called Side Splitters, we can say that AB/DB = AC/DE = BC/BE however, AB/DB becomes AB-DB/ DB so therefore actually DB/AD =EB/EC = AC/DE. Question 1. In the figure on left, we are assigned to find X, in this figure, i can say the ABC ~ DBE. Therefore AB-DB /DB = BC-BE/BE = AC/DE. x/12 = 15/10 10x=180 Therefore, x= 18 PERIMETER AND AREA OF SIMILAR TRIANGLES In similar triangles, that is corresponding angles are the same and corresponding sides are proportional then, ratio of corresponding sides = ratio of their perimeter. So lets take an example that a triangle, ABC with sides 3cm , 6cm and 5 cm respectively and another triangle DEF, with sides 6 cm , 12 cm, and 10 cm, then the proportion between DEF and ABC is 6/3=12/6=10/5 which gives us 2/1. Therefore the ratio of corresponding sides is 2:1, so according to my theorem this would also mean that the perimeter ratio will be 2:1. Perimeter of Triangle ABC= 14cm ( 3cm +6cm + 5cm) While Perimeter of Triangle DEF= 28 cm ( 6cm+12cm+ 10cm). Perimeter of Triangle DEF -----------------------------------Perimeter of Triangle ABC 28 cm / 14 cm = 2/1 This theorem proves to us that in a similar triangle, the ratio between their corresponding sides is the same ratio between its perimeter. Therefore if the ratio between corresponding sides is z:x then the ratio between their perimeter will also be z:x. Examples ABC ~ DEF, so according to the theorem, this means that: Perimeter of ABC Perimeter of DEF = AB DE 24 inches / x (perimeter of DEF) = 6/9 6x = 216 ABC ~ DEF x= 36 In this figure, GHI ~ JKL, therefore one can prove that the ratio between the corresponding sides is the same as the ratio between the figures’ perimeter. Perimeter of GHI Perimeter of JKL = HG KJ 24/36 = 6/9 Ratio simplifies to 2/3=2/3. Well, now we have reached in the areas’ section. We will first try to find the relationship between the areas of triangle (right angled) and corresponding sides. AREAS: GHI ~ JKL. Corresponding Sides ratio = ( GH/JK or any other can be chosen i.e. GI/JL or HI/KL) Ratio= 6/9 simplified 2/3. Ratio of area = Area of GHI divided by AREA of KJL, GHI =( ½ (6)(8) = 24) KJL= ( ½ (12) (9)= 54 ) Area of GHI = 24 Area of JKL 54 = 4 9 Corresponding Sides = HG/ KJ= 6/9 = 2/3 4/9 and 2/3 . Relationship (2/3)2 ----- Therefore, relationship of the Area of Similar Triangle is such that, if ratio of corresponding sides are a:b then relationship between area of the triangle will be a2 : b2 Examples: 4cm2 36cm2 6 cm y cm Now, 4/36 is the ratio between the area of the two triangle. a2 : B2 this means we have to square root. By square rooting 4/36 it is simplified to 2/6. ( 2 x 2) / ( 6 x 6). Ratio is 2/6 therefore solution will be as follow: 2/6 = 6/y 2y = 36 Y= 18 cm. 2 Eg 2: The areas of two similar triangles are 45cm and 80 cm Find the perimeter of each triangle. 2. In the figure below, the first triangle has an area of 45cm square while the second triangle has an area of 80 The sum of their perimeters is 35cm. cm square. First of all we’ll try to find the ratio between these triangles. Soln: 45cm2 80cm 2 1st Part: 45/80 if simplified by 5, then it results 9/16: therefore, by square rooting, ratio will be ¾ (3:4) 2nd Part: If we take sides of the triangles as X, as they are corresponding. Then the equation will be as follows, 3x + 4x = 35, 7x= 35 Therefore x=5. Perimeter of triangle 1 will be 3x5= 15cm Perimeter of triangle 2 willobviously be 20cm, as they add up to 35 cm, but even then to prove it we can take 4 x (X) which is 5cm, resulting to 20cm. ------------------------------------------------------------------------------------------------------------------------------------- VOLUMES OF SIMILAR FIGURES When three dimensional figures are similar, the shapes are the same but the sizes are often different. A line has one dimension, and the scale factor is used once, An area has two dimensions and thus the scale factor is used twice. Lastly volume has three dimensions and the scale factor is used thrice. :- this means, in a one dimension figure ratio of corresponding side a:b is corresponding to perimeter a:b a:b = a:b :- In a two dimension figure, ratio of corresponding side a : b is corresponding to area of figure a2 : b2 :- In a three dimension figure, ratio of corresponding side a : b is corresponding to area of figure a3 :b3. However, when considering the surface area, if ratio of corresponding sides is “a” then surface area will be “a” square. 3cm 6cm 6cm 12 12cm 6cm 3cm 6cm 5cm 10cm In the following shape we can find that the ratio between the corresponding sides first, and then we can find the surface area of the Cubes, and then we can work out the ratio. Soln: 3cm/6cm = 5cm/10cm= 6cm/12cm Therefore, ratio between sides is 1:2 . Surface area of figure one is: Surface area of figure two is: 2(LW) + 2(LH) + 2 ( HW) 2(LW) + 2(LH)+ 2(HW) = 2 (15) + 2(30) + 2(18) = 2(60) + 2 (120) + 2(72) = 126cm square. = 504 cm2 126/504= By simplifying this, ¼, if we square root one divided by 4 or ( 0.25) answer will be 1/2 . This Proves that the surface area of a figure corresponds with its sides. If ratio of sides are a:b then ratio of surface area will be a2 : b2 Now using the same figure as above, we will calculate the volume. Volume of a cuboid is L x W x H( length x width x height). However, as we have already got the ratios between their sides that is 1:2. Soln: If we compare the two volumes, in ratio. It gives us 90/720 this is 1/8. While our ratio between corresponding sides is ½. Therefore, the relationship between the ratio of corresponding sides and its volume is as follows: If ratio of corresponding sides is a:b then ratio between volume will be a3 : b3 = 3cm x 6cm x 5cm is equal to 90 cubic centimetres. = 6cm x 12cm x 10cm is equal to 720 cubic centimetres. Questions: Two similar jugs have heights of 4cm and 6cm respectively. If the capacity of the smaller jug is 50cm3. Find the capacity of the larger jug. Soln. First of all, we have to find the ratio between jug A and jug B a:b = 4 : 6 which is simplified to 2 : 3 In this question we can use cross multiplication, so thus, if ratio is 2:3 then 50cm/x. 2x = 150cm, X=75 cm3 Question 2: Two similar spheres are made of the same material having weights of 32kg and 108kg respectively. If the radius of the larger sphere is 9cm, find the radis ouf the smaller sphere. A:b = a3 : b3 , therefore 32/108= 8/27, if we cube root this, our answer will be 2/3. If the larger sphere is 9 cm, that mean 2/3 x 9 = x Therefore x= 6 cm.