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GEOMETRY HELPP

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GEOMETRY HELPP

by arifaisal123 » Mon Nov 14, 2011 1:27 am
Q1-In the figure below, triangle ABC and XYZ are equilateral triangles. YZ is parallel to BC. BC=4, QC=1 and BP=2 unit. The perimeter of the triangle XPQ is:
A)5 B)4 C)3
D)2
E)none

Q2-in the figure BD=2DC and AD=DB. What is the value of AB?
A)root(BD^2+DC^2)
B)root3 BD
C)BC+DC
D)2BC+2
E)none

Q3-find the area, in sq cm, of a rhombus having a side measuring 20 cm and the diagonal measuring 24 cm
A) 216
B) 324
C) 384
D) 576
E) none


full explanation pls
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by neelgandham » Mon Nov 14, 2011 1:58 am
arifaisal123 wrote:Q1-In the figure below, triangle ABC and XYZ are equilateral triangles. YZ is parallel to BC. BC=4, QC=1 and BP=2 unit. The perimeter of the triangle XPQ is:
A)5 B)4 C)3
D)2
E)none
BC = 4 = BP+PQ+QC = 2+PQ+1 => PQ =1

YZ is parallel to BC and triangle ABC and XYZ are equilateral triangles,
Implies AC is || to XQ and XZ,and AB is || to XY and XP
Angle ABC = Angle XYZ = Angle XPQ = 60
Angle BCA = Angle YZX = Angle PQX = 60
Angle CAB = Angle ZXY = Angle QXP = 60

AAA theorem : The triangles XPQ and ABC are similar, i.e. XPQ is an equilateral triangle of side measure 1. and the perimeter of the triangle XPQ = 3*side = 3*1 = 3

IMO C

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by neelgandham » Mon Nov 14, 2011 2:17 am
arifaisal123 wrote: Q3-find the area, in sq cm, of a rhombus having a side measuring 20 cm and the diagonal measuring 24 cm
A) 216
B) 324
C) 384
D) 576
E) none
AD = DC = CD = BA = 20 - Please refer the attached image (Not drawn to scale).
Let BD = 24.
We know that the diagonals in a rhombus are perpendicular bisectors of each other
i.e.Angle AED = Angle AEB = Angle CEB = Angle CED = 90 degrees
BE = ED = 24/2 = 12
AE = EC = x (assumption)

Applying Pythagorean theroem on triangle AED,
AE^2 + ED^2 = AD^2
x^2 + 12^2 = 20^2
x = 16 (If you know that 3,4,5 make a Pythagorean triplet, you need not even calculate)

Area of Rhombus = 4 * Area of Triangle AED
Area of Rhombus = 4 * 0.5 * base * height
Area of Rhombus = 4 * 0.5 * AE * ED
Area of Rhombus = 4 * 0.5 * 12 * 16 = 384

IMO C
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by neelgandham » Mon Nov 14, 2011 2:37 am
arifaisal123 wrote: Q2-in the figure BD=2DC and AD=DB. What is the value of AB?
A)root(BD^2+DC^2)
B)root3 BD
C)BC+DC
D)2BC+2
E)none
Let length of the side DC be x, then length of side BD equals 2x
Applying Pythagorean theroem on triangle BCD, we get

CD^2 + BC^2 = BD^2
x^2 + BC^2 = (2x)^2
BC^2 = 4*(x^2)- x^2 = 3*x^2
BC = x * Square root(3)

AD = DB = 2x
Applying Pythagorean theroem on triangle BCA
AB^2 = BC^2 + CA^2
AB^2 = BC^2 + (CD+DA)^2
AB^2 = (x * Square root(3))^2 + (x+2x)^2 = 3*(x^2) + 9*(x^2) =12*(x^2)
AB = x*2*Square root(3)

I am leaving it here for you to substitute the values of BD,DC,BC(2x,x,x * Square root(3)) in the options to find the correct option!
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by arifaisal » Mon Nov 14, 2011 8:49 am
hey neel ..thanx a lot man
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