CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

ravishah17

Applying similar triangles after drawing the figure,

we get

8/2=x+8/x-8

simplifying we get
x=40/3 which is the required answer

sudeepdeb

ans) 40/3.

applying similar triangles.

srikar2097 · 07:26 PM

Hi Rajiv_hts and slam.
My method might be flawed, but I'm having trouble with your method. Let me try to explain -

In the attached diagram, if you see the triangle created by the circle1. It is a right triangle with hypotenuse 8cm, one of the sides 2cm (height). So we get the third side (base) as (60)^1/2. But here's my problem. Since from the centre of circle1 to the point of contact is 8cm, the base of this triangle should be > 8cm (I'm sure you'll picture it when I say that we have a sector of a circle here, with centre as cirlce1 and radius 8cm).

But in a right triangle the hypotenuse is > than the other 2 sides. Do you see the contradiction here? Such a triangle is not possible. Please explain this...

P.S: Thank you slam for the diagram

hibernal

Quote:

Originally Posted by srikar2097

Hi Rajiv_hts and slam.
My method might be flawed, but I'm having trouble with your method. Let me try to explain -

In the attached diagram, if you see the triangle created by the circle1. It is a right triangle with hypotenuse 8cm, one of the sides 2cm (height). So we get the third side (base) as (60)^1/2. But here's my problem. Since from the centre of circle1 to the point of contact is 8cm, the base of this triangle should be > 8cm (I'm sure you'll picture it when I say that we have a sector of a circle here, with centre as cirlce1 and radius 8cm).

But in a right triangle the hypotenuse is > than the other 2 sides. Do you see the contradiction here? Such a triangle is not possible. Please explain this...

P.S: Thank you slam for the diagram

I wanted to clarify 1 thing ...
When you consider the triangle ABC(refer to attached diag), you cannot have a sector A-B-C . The sector can only be A-B-D.
I think the triangle that you are talking about is ABD(not ABC), which is right angled at B and AD as the hyp(here, the side is >8 cm) . (So, these are 2 different triangles)
I hope this explains your doubt.

srikar2097

Thanks hibernate for replying.
Me earlier explanation was regarding triangle ABC only, not ABD. But according to what you say, ABD would be the sector, then that would make BC extended downwards a secant. Isn't it. It'e mentioned that we have a 'point of contact' i.e. a single point. so it has to be a tangent to both the circles. Doesn't your explantion fail? Are you clear or do you still need more explanation?

My earlier doubt still stands...

srikar2097

Quote:

Originally Posted by hibernal

I wanted to clarify 1 thing ...
When you consider the triangle ABC(refer to attached diag), you cannot have a sector A-B-C . The sector can only be A-B-D.
I think the triangle that you are talking about is ABD(not ABC), which is right angled at B and AD as the hyp(here, the side is >8 cm) . (So, these are 2 different triangles)
I hope this explains your doubt.

Thanks hibernate for replying.

Me earlier explanation was regarding triangle ABC only, not ABD. But according to what you say, ABD would be the sector, then that would make BC extended downwards a secant. Isn't it. It'e mentioned that we have a 'point of contact' i.e. a single point. so it has to be a tangent to both the circles. Doesn't your explantion fail? Are you clear or do you still need more explanation? My earlier doubt still stands...

slam

Quote:

Originally Posted by srikar2097

Hi Rajiv_hts and slam.
My method might be flawed, but I'm having trouble with your method. Let me try to explain -

In the attached diagram, if you see the triangle created by the circle1. It is a right triangle with hypotenuse 8cm, one of the sides 2cm (height). So we get the third side (base) as (60)^1/2. But here's my problem. Since from the centre of circle1 to the point of contact is 8cm, the base of this triangle should be > 8cm (I'm sure you'll picture it when I say that we have a sector of a circle here, with centre as cirlce1 and radius 8cm).

But in a right triangle the hypotenuse is > than the other 2 sides. Do you see the contradiction here? Such a triangle is not possible. Please explain this...

P.S: Thank you slam for the diagram

Well, the third vertex of the triangle would lie inside the circle (that has radius

in that case.
To explain it better, one of the vertices is the centre of the circle, the second is the point of contact. We have drawn a perpendicular to the ground through the point of contact, and a parallel to the ground through the centre, to get the intersection point as the third vertex. It's not a sector of the circle like you've mentioned it to be.
Hope that makes it clear.

----------
slam.

srikar2097

Quote:

Originally Posted by slam

Well, the third vertex of the triangle would lie inside the circle (that has radius

in that case.
To explain it better, one of the vertices is the centre of the circle, the second is the point of contact. We have drawn a perpendicular to the ground through the point of contact, and a parallel to the ground through the centre, to get the intersection point as the third vertex. It's not a sector of the circle like you've mentioned it to be.
Hope that makes it clear.

----------
slam.

So in other words we are actually constructing the third side (of the triangle) so as to exploit the similarity of triangles properties? Is that right?

slam

Quote:

Originally Posted by srikar2097

So in other words we are actually constructing the third side (of the triangle) so as to exploit the similarity of triangles properties? Is that right?

To be honest, I constructed it because the height of the pt of contact was given (10). But if that makes yous you understand the method, yes I constructed to 'exploit' similarity rules.

----------
slam.

srikar2097

Quote:

Originally Posted by slam

To be honest, I constructed it because the height of the pt of contact was given (10). But if that makes yous you understand the method, yes I constructed to 'exploit' similarity rules.

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slam.

Got it! Thanks. I think I really got stuck in treating it as a sector that...