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Circle and Triangles

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

Sanchit kshirsagar
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Sanchit kshirsagar said:

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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  • Posted about 1 month ago.
wwwwwwwwwwwww
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wwwwwwwwwwwww said:

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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  • Posted about 1 month ago.
wwwwwwwwwwwww
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wwwwwwwwwwwww said in response to:
Sanchit kshirsagar
Sanchit kshirsagar’s post:
Citation Body

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.

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 quadrilaterals 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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  • Posted about 1 month ago.
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