SAT 2013

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Trignometry Basics - Circles and Triangles

by Suresh


Geometry has some major theorems. One should be clear about them, the ones on similarity of triangles, congruency of triangles, pythagoras, area and volume formula. Kindly refer to a text book for revising such concepts, I would recommend to go through NCERT books (from fifth standard to tenth standard). Anyway let's look at an important concept here!

The major theorems which we always need are :

Theorem 1: Pythagoras Theorem : a^2+b^2=c^2 where a,b,c are sides of a right angled triangle.

Clearly, C is the largest side, we call it hypotenuse.

The triplets of real numbers (a,b,c) which satisfy the above theorem is called pythagorean triplets. They are of real interest in all kinds of work.

Example 1: The length of one of the legs of a right triangle exceeds the length of the other leg by 10 cm but is smaller than that of the hypotenuse by 10 cm. Find the hypotenuse.

The obvious solution is (a-10)^2+a^2=(a+10)^2 ( I have jumped a step)

solving we have a^2-20a=20a =>a=40 ( a can't be zero, its side of a triangle)%

hypo is a+10=50

P.s : we have avoided the cumbersome assumption of sides as a,a+10 and a+20

Tipster clue: See this, the smallest integer Pythagorean triplet is (3,4,5) so all numbers of the form (3k,4k,5k) will be Pythagorean!

Practice Problem 1: Find the sum of the lengths of the sides of a right angled triangle if the Circumradius=15 and inradius=6

Theorem 2: Sin law

\frac {a}{\sin A}=\frac {b}{\sin B}=\frac {c}{\sin C}=2R

where a,b,c are sides opposite <A,<B and <C respectively and R is circumradius of Triangle ABC.

Very useful theorem, though we have entered the domain of trigonometry, but trigonometry, plane geometry and coordinate geometry are very important for each other to co exist.

Theorem 3: Cosine law

a^2=b^2+c^2-2bc\cos A ( the notations remain the same as Theorem 2). The theorem can be similarly used for other angles too.

Practice Problem 2: Find the angle between the diagonal of a rectangle with perimeter 2p and area (\frac {3}{16})p^2


Example 2: Find the length of the base of an isosceles triangle with area S and vertical angle A.

How do we start with this, we can off course going to need some basic geometry knowledge. let me tell you all of it. First the vertical angle of an isosceles triangle is the angle between the two equal sides( unless otherwise mentioned). The Perpendicular dropped on the unequal side from the opposite vertex, bisects the vertical angle as well as bisects the side. It means if we have a triangle ABC with AC=AB and AD perpendicular to BC then BD=BC and < BAD=< CAD.

The last thing we need is that area of a triangle is (1/2)bc \sin A or (1/2)b^2 \sin A for an isosceles triangle as b=c

now given (1/2)b^2\sin A=S\cdots (1)

Now as AD bisects the vertical angle and then use BD=b\sin (A/2)

hence BC=2BD=2b\sin (A/2)

we can put the value of b from (1) and we are done !


Example 3: In Triangle ABC, AD,BE and CF are the medians which intersect at G. ABCH is trapezium with AH=5units , and BC=10units and Area( Tr BHC)=35 Sq units. Find the ratio of Area( BDFG): Area( ABCH). ( note we have H and C on same side of B )

Here we again need to know this. The three medians divide the triangle into three triangle of equal area . Also they divide it into three quadrilaterals of equal area. So Area(BDFG)=(\frac {1}{3})Area(ABC)

Next comes, the traingles drawn on the same base and between same parallel lines have equal area. Hence Area(ABC)=Area(BHC)=35 as we know the base BC, we know the altitude D=2\frac {Area}{base}=\frac {70}{10}=7

Area-of-trapezium=(\frac {1}{2})altitude(sum-of-parallel-sides)=(\frac {1}{2})7(10+5)



so our ratio is \frac {(35/3)}{(15.7/2)}=\frac {2}{9}

Image Credit: billjacobus1

20 Comments
    Sudeepjoshi28
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    Sudeep JoshiThu, 10 Jun 2010 17:41:51 -0000

    it's tough but an intresting topic…………………………….it's usage is also vast………………………

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    mukherjipooja
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    mukherjipoojaMon, 03 May 2010 07:01:55 -0000

    i hv a question…post d explanation if anybody solves it..
    q1.a regular hexagon is circumscribed by a rectangle such that all six corners of the hexagon lie on the rectangle.what is the ratio of the area of largest possible equi. triangle that can be cut from the rectangle to the area of hexagon?
    1.3:2
    2.2:3
    3.sqrt(3):1
    4.2:sqrt(3).

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    lihosp
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    lihospThu, 18 Mar 2010 14:32:39 -0000

    hehe these questions are so simple…if these type of question gonna come in SAT paper,i can score full marks sir..so can u say these types of question gonna come in SAT Paper…

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    rubinasingh
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    Rubina SinghThu, 25 Mar 2010 11:02:47 -0000

    Hello,

    The questions mentioned here are just the examples and definitely the questions will follow the same apttern, basically this exam is to test the speed and accuracy of the candidate so you need to be good in speed and must be accurate so do it very carefully.

    Here are some books which you can refer for its preparation:

    The Official SAT Study Guide: 2nd Edition by College Board (Editor)

    Answers and Explanations by Peter Tanguay

    The Ultimate SAT Supplement by Erik Klass

    Up Your Score 2009-2010: The Underground Guide to the SAT

    The Full Potential SAT Audio Program by Bara Sapir

    The Ultimate SAT Tutorial: The Easiest and Most Effective Way to Raise Your Score by Erik Klass

    SAT Practice: The New Verbal Section by K. Titchenell

    Try to solve more and more sample papers for a good practice. Don't forget that if you can score full marks in this then you can apply for various scholarships and can also apply to top class Universities.

    Start preparing for it and do well in exams.

    Good Luck!!!!

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    prvnyadav51
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    praveen yadavSat, 14 Nov 2009 10:42:52 -0000

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    granget
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    crsvaidehiWed, 19 Aug 2009 06:25:18 -0000

    example 3 should have better been explained with a rough diagram

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    Sureshbala
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    SureshThu, 20 Aug 2009 10:48:45 -0000

    Hi,

    I thought it is easy to understand the solution with a diagram. Anyway, will try to be more lucid in the coming lessons.

    Regards

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    granget
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    crsvaidehiWed, 19 Aug 2009 06:23:32 -0000

    for practice problem 2,on solving the given 2 conditions I got it as 3a=b or a=3b.
    so wat next…..
    I couldn't get it.
    plz post the answer for it

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    Sureshbala
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    SureshThu, 20 Aug 2009 10:50:50 -0000

    Hi, Apply the Cosine rule and you will get the answer quite comfortably.

    Regards

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    granget
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    crsvaidehiWed, 19 Aug 2009 06:15:53 -0000

    Solution to problem 1 :
    R=15
    r=6
    let a,b,c be the sides of the triangle.
    c be the hypotenuse.
    a+b=?
    as R is the circumradius,the circumdiameter i.e,c=30
    a²+b²=c²----(1)
    so a=18,b=24
    which is the only condition satisfying (1)
    so a+b=42

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    userdce
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    user dceSat, 30 May 2009 15:58:23 -0000

    area of triangle = 1/2(base*altitude)
    ie. altitude D= 2(Area/Base)

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    anilkumar reddy
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    anilkumarreddyTue, 26 May 2009 14:41:04 -0000

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    Sureshbala
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    SureshFri, 15 May 2009 12:08:48 -0000

    Folks, seems that there is some problem with the format of this lesson. I will try to fix this soon and also will try to come up with answers for the practice problems as well.

    Regards

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    amahapatra
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    amahapatraMon, 27 Jul 2009 13:35:58 -0000

    Please add answers to the practice problem also

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    spiyr
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    suman sourabhMon, 27 Apr 2009 20:20:38 -0000

    fantastic lesson

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    nickizkool
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    Nick WatsonSat, 18 Apr 2009 06:11:02 -0000

    i didnt understand the last question…… i mean where's the point H. thats y i m not able 2 form the trapezium…….

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    gargi_l
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    gargi_lThu, 09 Apr 2009 13:40:58 -0000

    where do i get the solution for the practice problems

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    pokharna
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    pokharnaWed, 01 Apr 2009 13:51:51 -0000
    i was considering that my level for quant was suffic. but now i must rethink

    i took a few mins to solve these with my weak application skills
    great work!!

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    deepakkumar753
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    deepak agarwalThu, 19 Mar 2009 11:01:29 -0000

    , the formula you used for Altitude calculation D= 2(Area/Base) is new to me

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    deepakkumar753
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    deepak agarwalThu, 19 Mar 2009 10:59:22 -0000

    basic is good.thanks

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    viditsa
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    viditsa handooWed, 18 Mar 2009 12:54:06 -0000

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    asureshwaran
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    asureshwaranWed, 04 Feb 2009 03:03:53 -0000

    good lesson.

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    wwwwwwwwwwwww
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    wwwwwwwwwwwwwWed, 19 Nov 2008 05:09:10 -0000

    Dude u said that u studied maths in grade school and also in engg, but it seems u didnt study it well. medians of triangle make 3 qudrilaterals of equal area, this concept is there in the high school course of CBSE board. and the formula Altitude = 2(Area/Base) is new to you ? u r kidding right ???

    Area of a triangle = 1/2(Base* Altitude)

    So, (Base * Altitude) = 2 * Area…….. understood ???

    We can also say, Altitude = 2* (Area/Base)….. Clear ???

    U r a funny guy…. U dont seem to be a maths scholar.

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    Sanchit kshirsagar
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    Sanchit kshirsagarSun, 16 Nov 2008 11:53:03 -0000

    I have studied Maths in grade school or in Engg as well, but to be really true, no one ever mentioned that medians of triangle make 3 qudrilaterals of equal area. Also, the formula you used for Altitude calculation D= 2(Area/Base) is new to me. Any ref book you would like to suggest that will only talk about such clues and relationships between geometrical figures.

    Seriously this one was interesting topic. I will recommend, GMAT test takers to go through this session.

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Sureshbala
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About: Worked for more than 6 years in renowned corporate institutes as their core faculty/lead content developer for C.A.T,G.R.E, G.M.A.T and Campus Recruitment Training Programs.

Last Updated At Dec 07, 2012
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