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by stubbornp » Sun Oct 12, 2008 10:54 am
Image


18. In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?
(1) The area of triangular region ABD is equal to the area of triangular region DBC.
(2) BD┴AC and AD = DC
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Source: — Data Sufficiency |
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by nitin86 » Sun Oct 12, 2008 7:43 pm
IMO B

I think because for a isoceles triangle, Median is perpendicular to the base and that's what B is stating
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by stubbornp » Sun Oct 12, 2008 9:03 pm
nitin86 wrote:IMO B

I think because for a isoceles triangle, Median is perpendicular to the base and that's what B is stating
but isn't this the same situation for equilateral triangle??
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by arorag » Mon Oct 13, 2008 6:15 am
ANs should be D
From I if area is equal and both have common height.
So AD=DC,
therefore AB=AC
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by raju232007 » Tue Oct 14, 2008 6:04 am
@ stubbornp An equilateral triangle is a special case of isosceles triangle wherein all the sides & angles are equal...it is no where mentioned in this question that all the 3 sides or 3 angles are equal..


statement 2 is sufficient because BD perpendicular to AC and AD=DC indicates that BD is the perpendicular bisector of base AC and so BD is the altitude..

If a line drawn from the vertex bisects the base into 2 equal halves then the triangle must be isosceles...

The OA should be B...what is the source of this question?Do let me know if i am wrong...
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by stubbornp » Tue Oct 14, 2008 7:00 pm
raju232007 wrote:@ stubbornp An equilateral triangle is a special case of isosceles triangle wherein all the sides & angles are equal...it is no where mentioned in this question that all the 3 sides or 3 angles are equal..


statement 2 is sufficient because BD perpendicular to AC and AD=DC indicates that BD is the perpendicular bisector of base AC and so BD is the altitude..

If a line drawn from the vertex bisects the base into 2 equal halves then the triangle must be isosceles...

The OA should be B...what is the source of this question?Do let me know if i am wrong...

answer is ofcourse B...source is 1000DS....

But i am unable to digest...that equilateral is a special case of isosceles...Any commemts??
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by sudhir3127 » Wed Oct 15, 2008 4:28 am
stubbornp wrote:
raju232007 wrote:@ stubbornp An equilateral triangle is a special case of isosceles triangle wherein all the sides & angles are equal...it is no where mentioned in this question that all the 3 sides or 3 angles are equal..


statement 2 is sufficient because BD perpendicular to AC and AD=DC indicates that BD is the perpendicular bisector of base AC and so BD is the altitude..

If a line drawn from the vertex bisects the base into 2 equal halves then the triangle must be isosceles...

The OA should be B...what is the source of this question?Do let me know if i am wrong...

answer is ofcourse B...source is 1000DS....

But i am unable to digest...that equilateral is a special case of isosceles...Any commemts??
Triangles can be classified according to the number of their equal sides.

1., A triangle with 3 equal sides is called equilateral,

2. A triangle with 2 equal sides are isosceles and

3. A triangle with no equal sides is called scalene.

we can therefore notice that an equilateral triangle is also isosceles

Another way of looking is

1.An isosceles triangle has at least two congruent sides.

2.An equilateral triangle has three congruent sides.

So, an equilateral triangle is a special case of isosceles triangles.

Since the equilateral triangle has three congruent sides, it satisfies the conditions of isosceles triangle.

So, equilateral triangles are always isosceles triangles

i can only make this much ....hope its clear...
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by stubbornp » Wed Oct 15, 2008 1:25 pm
Clear.....tnx
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by sumithshah » Thu Oct 16, 2008 6:39 am
Can anyone explain why D is not the answer?
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by mental » Wed Nov 12, 2008 3:48 am
sumit A is not sufficient
as the areas can be equal even whrn the triangle is not isoceles

area can be defined as 1/2(base * height)

following the diagram for triangle ABC

the base being AC...
lets say the height H be the height of B from AC
draw a perpendicular from B on AC extended.
Clearly it will fall outside the triangle ABC
so the area will be 1/2(AC * H)

lets consider the two triangles ABD and DBC
they share the common vertex B above line ADC and hence they have the same height H

so their areas will be 1/2(AD * H) and 1/2(DC*H)

if the areas are equal then AD = DC.....one side equal
but we have no information on other two sides and any of the angles

in B
there is a angle and two sides..........making both triangles similar

hence answer is B
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