Jump To Content

LearnHub




Circle and Triangles

Ask The Experts



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

Ask The Experts


  1. Sanchit kshirsagar saidSun, 16 Nov 2008 11:53:03 -0000 ( Link )

    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.

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

  2. wwwwwwwwwwwww saidWed, 19 Nov 2008 05:09:10 -0000 ( Link )

    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.

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

  3. asureshwaran saidWed, 04 Feb 2009 03:03:53 -0000 ( Link )

    good lesson.

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

  4. viditsa saidWed, 18 Mar 2009 12:54:06 -0000 ( Link )

    better one

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

    Post Comments

  5. deepakkumar753 saidThu, 19 Mar 2009 10:59:22 -0000 ( Link )

    basic is good.thanks

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

  6. deepakkumar753 saidThu, 19 Mar 2009 11:01:29 -0000 ( Link )

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

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

  7. pokharna saidWed, 01 Apr 2009 13:51:51 -0000 ( Link )

    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!!

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

    Post Comments

  8. gargi_l saidThu, 09 Apr 2009 13:40:58 -0000 ( Link )

    where do i get the solution for the practice problems

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

    Post Comments

  9. nickizkool saidSat, 18 Apr 2009 06:11:02 -0000 ( Link )

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

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

  10. spiyr saidMon, 27 Apr 2009 20:20:38 -0000 ( Link )

    fantastic lesson

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

    Post Comments

  11. Sureshbala saidFri, 15 May 2009 12:08:48 -0000 ( Link )

    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

    Actions
    Vote
    Current Rating
    2
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

  12. anilkumar reddy saidTue, 26 May 2009 14:41:04 -0000 ( Link )

    good

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

  13. userdce saidSat, 30 May 2009 15:58:23 -0000 ( Link )

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

    Actions
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

  14. granget saidWed, 19 Aug 2009 06:15:53 -0000 ( Link )

    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

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

  15. granget saidWed, 19 Aug 2009 06:23:32 -0000 ( Link )

    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

    Actions
    Vote
    Current Rating
    3
    Rate Up
    Rate Down
    3 Total Votes

    Post Comments

  16. granget saidWed, 19 Aug 2009 06:25:18 -0000 ( Link )

    example 3 should have better been explained with a rough diagram

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

  17. prvnyadav51 saidSat, 14 Nov 2009 10:42:52 -0000 ( Link )

    nice

    Actions
    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

    Post Comments

Your Comment
Textile is Enabled (View Reference)