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hariharakarthi
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Triangle Problem 01 Sep 2009, 20:06
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solve the attached Traingle problem and explain in detail,

regards,
hhk



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gmate2010
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Triangle Problem 01 Sep 2009, 20:35
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Let height of triangle PDC = h
base is a = side of a

given, area of triangle = 2 \(a^2\)
=> \(\frac{1}{2}*a*h = a^2\)
=> h = 2a

Let E and F are the points of intersection of square and triangle..
then, EF || to CD

and triangle PEF will be similar to triangle PDC
from simlar triangle property,
EF/DC = height of PEF/h

=> \(EF = a*\frac{(h-a)}{h}\)
=> EF = a*a/2a = a/2

area of PEF = 1/2 * EF * a
= 1/2 *a/2 *a
= \(\frac{a^2}{4}\)

ans - 1/4
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defoue
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Triangle Problem 03 Sep 2009, 08:09
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Any alternate way of solving it?
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meenal8284
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Triangle Problem 03 Sep 2009, 11:40
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solving by similarity of triangles this is how i do it..:

let side of square be equal to 4

area of square is 16
area of triangle is 32
correspondingly area ht of the trianlge PDC is 16

Base of the smaller triangle is 3/4 of side (derived from similarity of triangles)
this gives me that the ht of the smaller tri is total ht - side = 16-4 = 12
area of the triangle is 18

that of square is 16

so the answer is 9/8
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CasperMonday
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Triangle Problem 03 Sep 2009, 12:23
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\(\frac{1}{2}PH*DC=DC^2\)
\(PH=2DC\)

\(PO=PH-OH\)
since ABCD is a squre, \(OH=DC\)
hence, \(PO=2DC-DC=DC\)

triangulars PEF and PDC are similar, hence
\(\frac{MN}{DC}=\frac{PO}{PH}\)

\(\frac{MN}{DC}=\frac{DC}{2DC}\)

from here, \(MN=\frac{DC}{2}\)

\(Area of PMN=\frac{1}{2}*PO*MN\)

\(Area of PMN=\frac{DC^2}{4}\)

\(\frac{Area of PMN}{Area of ABCD}=\frac{DC^2}{4}:DC^2=\frac{1}{4}\)
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Triangle Problem 03 Sep 2009, 15:56
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mnpqxyzt
The answer is 9/8

gmate2010
Let height of triangle PDC = h
base is a = side of a

given, area of triangle = 2 \(a^2\)
=> \(\frac{1}{2}*a*h = 2*a^2\)
=> h = 4a

Let E and F are the points of intersection of square and triangle..
then, EF || to CD

and triangle PEF will be similar to triangle PDC
from simlar triangle property,
EF/DC = height of PEF/h

=> \(EF = a*\frac{(h-a)}{h}\)
=> EF = a*3a/4a = 3a/4

area of PEF = 1/2 * EF * a
= 1/2 *3a/4 *3a
= \(\frac{9a^2}{8}\)

ans - 9/8
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thanks 4 correcting me...
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gurpreet07
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Triangle Problem 05 Sep 2009, 12:18
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gmate2010
mnpqxyzt
The answer is 9/8

gmate2010
Let height of triangle PDC = h
base is a = side of a

given, area of triangle = 2 \(a^2\)
=> \(\frac{1}{2}*a*h = 2*a^2\)
=> h = 4a

Let E and F are the points of intersection of square and triangle..
then, EF || to CD

and triangle PEF will be similar to triangle PDC
from simlar triangle property,
EF/DC = height of PEF/h

=> \(EF = a*\frac{(h-a)}{h}\)
=> EF = a*3a/4a = 3a/4

area of PEF = 1/2 * EF * a
= 1/2 *3a/4 *3a
= \(\frac{9a^2}{8}\)

ans - 9/8
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Can you please explain me this part
EF/DC = height of PEF/h

=> \(EF = a*\frac{(h-a)}{h}\)
=> EF = a*3a/4a = 3a/4

area of PEF = 1/2 * EF * a
= 1/2 *3a/4 *3a
= \(\frac{9a^2}{8}\)
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thb
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Triangle Problem 06 Sep 2009, 01:15
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CasperMonday


\(\frac{1}{2}PH*DC=DC^2\)
\(PH=2DC\)

\(PO=PH-OH\)
since ABCD is a squre, \(OH=DC\)
hence, \(PO=2DC-DC=DC\)

triangulars PEF and PDC are similar, hence
\(\frac{MN}{DC}=\frac{PO}{PH}\)

\(\frac{MN}{DC}=\frac{DC}{2DC}\)

from here, \(MN=\frac{DC}{2}\)

\(Area of PMN=\frac{1}{2}*PO*MN\)

\(Area of PMN=\frac{DC^2}{4}\)

\(\frac{Area of PMN}{Area of ABCD}=\frac{DC^2}{4}:DC^2=\frac{1}{4}\)
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Your approach was good, except that you misread that the triangle is twice the size of the square.
\(\frac{1}{2}DC*PH=2DC^2\) => \(PH=4DC\)



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