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interesting ;)

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by mehrasa » Mon Sep 19, 2011 8:55 am
If the altitude of a triangle increases by 6% and the base of the triangle increases by 8%, by what percent will the area of the triangle increase?

A) 6%

B) 8%

C) 7.24%

D) 14.48%

E) 48%
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by cans » Mon Sep 19, 2011 9:03 am
a = bh/2
a' = 1.08*1.06*bh/2 = 1.1448a
thus % increase =( 1.1448a - a)/a * 100 = 14.48%
IMO D
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by knight247 » Mon Sep 19, 2011 9:21 am
Area=Base*Height/2=0.5*Base*Height

Increased Base=Base+6(Base)/100=1.06Base
Increased Height=Height+8(Height)/100=1.08Height

New Area=(0.5)(1.06)*(1.08)*(Base)*(Height)=(0.5)(1.1448)*(Base)*(Height)

Percentage Increase=[(0.5)(1.1448)*(Base)*(Height)-(0.5)*(Base)*(Height)]/(0.5)*(Base)*(Height)]*100=14.48

Hence D
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by Abhishek009 » Mon Sep 19, 2011 9:30 am
mehrasa wrote:If the altitude of a triangle increases by 6% and the base of the triangle increases by 8%, by what percent will the area of the triangle increase?

A) 6%

B) 8%

C) 7.24%

D) 14.48%

E) 48%
Plug in some values of Base and height...


Let initially the dimensions are as follows :


Height : 10

Base : 20


So area of the triangle is 1/2(10)(20) => 100

Now height increases 6% and the base of the triangle increases by 8% , so now the dimensions are as follows :


Height : 10.6

Base : 21.6


Now Area is : 1/2(10.6)(21.6) => 114.8


So % change in Area is 114.8 - 100 => 14.8%
Abhishek
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