## Concept 3 Circle and Triangles ( Part 1)

*Geometry as a section wa spretty popular in CAT till 2004, consiting of 1/3rd of the problems and people used to think if they could handle it well they are clear with quant cutoff. And really that used to be the case. Cat has changed the trend reducing geometry every year since and last year in CAT 2007, there was not a single problem from geometry. But, we can for sure be ready for nice geometry problems, so that if they come, we are up and ready for it.*

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 Quantum Cat By Arihant Publishers. Lets roll then !

**The major theorems which we always need are** :

**Theorem 3.1 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 3.1 *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 3.1 FInd the sum of the lengths of the sides of a right angled triangle if the Circumradius=15 and inradius=6**

**Theorem 3.2 Sin law **

**a/Sin A=b/SinB =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 3.3 Cosine law**

**a^2=b^2+c^2-2bcCos A ***( the notations remain the same as Theorem 3.2). The theorem can be similarly used for other angles too.*

**Practice Problem 3.2 Find the angle between the diagonal of a rectangle with perimeter 2p and area (3/16)p^2**

*Example 3.2 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 offcourse 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)bcsinA or (1/2)b^2Sin A for an isosceles triangle as b=c

now given (1/2) b^2sin A=S…….(1)

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

hence BC=2BD=2bSin(A/2)

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

**Practice Problem 3.3 Find the largest angle of a triangle in which the altitude and the median drawn from the same vertex divide the angle at the vertex into three equal parts**

Lets do a more involved example. This came in IMS SimCat 9. Nice and easy problem, but it might scare you for a moment if you look at the figure they drew. So I am not giving it 🙂

*Example 3.3 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)= (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= 2Area/base=70/10=7

Area of trapezium =(1/2)altitude( sum of paralle sides)= (1/2)7(10+5)

so our ratio is (35/3)/(15.7/2)=2/9

We are done 🙂

Thanks implex looking forward for more such concepts ./.. keep posting dude

shivamSeptember 17, 2008 at 10:42 am

Please explain the answer to practice problems 3.2 and 3.3

tusharApril 30, 2014 at 6:20 pm