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Triangle : gmat prep

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by thephoenix » Mon Apr 05, 2010 8:16 am
find out the three sides which comes to be PQ=5 PR=5 and QR=5v2
its an isoceles triangle
area=1/2 * b* h
h will bisect base(isoceles tri property)
base=5 sqrt2
so h=[5^2-(5/2 *v2)^2]^(1/2)
=5/2(sqrt2)

area=1/2 * 5(sqrt 2)* 5/2 *( sqrt 2)=25/2=12.5
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by smar83 » Mon Apr 05, 2010 11:11 am
Try using Heron's formula.

Heron's formula is used to find area of a triangle when you know all three sides.
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by bhumika.k.shah » Mon Apr 05, 2010 12:13 pm
huh????????????????/ :shock:
smar83 wrote:Try using Heron's formula.

Heron's formula is used to find area of a triangle when you know all three sides.
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by student22 » Mon Apr 05, 2010 12:36 pm
I just googled that formula:


woah...I never knew about this. This is so useful!!!
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by smar83 » Mon Apr 05, 2010 12:40 pm
yeah... generally ppl having engineering background know about it.

And that's how I know about it too...
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by student22 » Mon Apr 05, 2010 1:00 pm
any other random (secret) formulas that you know of?
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by smar83 » Mon Apr 05, 2010 2:17 pm
unfortunately not.
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by akahuja143 » Mon Apr 05, 2010 2:55 pm
I wakely remember Heron's formula back in a day we use it 7th or 8th grade

But anyways here is the formula sqrt(s(s-a)(s-b)(s-c)) where =s =a+b+c/2 a,b,care sides of the triangles
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by [email protected] » Mon Apr 05, 2010 3:16 pm
two ways:
1. find the total area of rectangle ( 7 * 4) = 28 unit sqaure then subtract individual areas of small triangles (1/2 * 1*7) (1/2 * 4*3) (1/2 * 3 * 4) which is 12.5

2. Find the sides of 3 sides of traingle which are 5 5 and 5 root 2 if we have 3 sides as a ,a, a root 2 it means that it is a right angled triangle with hypotenuse as a root 2 . so area is 1/2 * base * height so 1/2 * 5 * 5 which is 12.5

Thanks
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by LMK27 » Tue Apr 06, 2010 7:29 pm
How do you know the sides are 5, 5, and 5v2?
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by thephoenix » Tue Apr 06, 2010 7:32 pm
LMK27 wrote:How do you know the sides are 5, 5, and 5v2?
section formula dist b/n point(x1,y1) and( x2,y2) is

dist=sqrt[(x1-x2)^2+(y1-y2)^2]
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by gmat.cracker24 » Wed Apr 07, 2010 4:31 am
Going by Hero's formula would take atleast 1 min...since it involves sqrt... however earlier suggested method of finding area saves time:
Area of triangle=Area of sqaure -area of 3 triangles
=7X4 -1/2( 7X1 + 4X3 + 3X4 )
=12.5
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by akhpad » Thu Apr 08, 2010 2:38 am
Lets try this method, which I studied long time back.


Image


This is 2 by 2 matrix

This method is valid for polygon as well.
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by [email protected] » Thu Apr 08, 2010 8:36 am
easier way
consider RX is perpendicular on x-axis
then

OQRX a trapezoid, and

areaPQR=area (OQRX)-area(OPQ)-area(RPX)
=(OQ+RX)/2*OX - 1/2 *OP*OQ - 1/2 * PX* RX
=(3+4)/2*7 - 1/2 * 4*3 - 1/2 *3*4
=24.5 - 6 - 6
=12. 5
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