Better to break down the problem in 2 dimension where the circle replaces the sphere and a right angle isosceles triangle replaces the cone.
The largest circle fits into the triangle
so the radius R is perpendicular to the sides of the the triangle
R² = (h-R)² - (l-r)²
= h²+R²-2hR - l²-r²+2lr
2hR = h²-l² -r²+2lr (l²-h² =r²)
= 2lr - 2 r²
R = r(l-r)/h
Option D

kstv wrote:Better to break down the problem in 2 dimension where the circle replaces the sphere and (1) a right angle isosceles triangle replaces the cone.
The largest circle fits into the triangle
so the radius R is perpendicular to the sides of the the triangle
R² = (h-R)² - (l-r)²
= h²+R²-2hR - l²-r²+2lr
2hR = h²-l² -r²+2lr (l²-h² =r²)
= 2lr - 2 r²
(2) R = r(l-r)/h
Option D

(1) Why?

(2) How [spoiler]D[/spoiler]?

sanju09 wrote:Which of the following expression gives the radius R of the largest sphere that can be carved out of a right cone of radius r, upright height h, and slant height l?
(A) R = r l / (r + h)
(B) R = l h / (r + h)
(C) R = l h / (r + l)
(D) R = r h / (r + l)
(E) R = r h / (h + l)

buddy,

is this really a GMAT question? :roll: : finding it very tough!

anyway, have to resort to logical-guessing, as can't use calculus ....

in a right angled triangle, l can be considered the hypotenuse with base as r/2 and perpendicular as h.

so, among the options, the contender will be any option which has l in the numerator. this gives A, B, and C.

now, in C, l comes in the sum in the denominator, which will reduce the overall figure . So, left with A & B.

notice that , l/(r+h) is same in both.

So, assuming h > r, B has to be the correct option

(trying to make value of fraction bigger by increasing the numerator and decreasing the denominator).

Please give the original solution and OA. This is killing me !!!

harshavardhanc wrote:

sanju09 wrote:Which of the following expression gives the radius R of the largest sphere that can be carved out of a right cone of radius r, upright height h, and slant height l?
(A) R = r l / (r + h)
(B) R = l h / (r + h)
(C) R = l h / (r + l)
(D) R = r h / (r + l)
(E) R = r h / (h + l)

buddy,

is this really a GMAT question? :roll: : finding it very tough!

anyway, have to resort to logical-guessing, as can't use calculus ....

in a right angled triangle, l can be considered the hypotenuse with base as r/2 and perpendicular as h.

so, among the options, the contender will be any option which has l in the numerator. this gives A, B, and C.

now, in C, l comes in the sum in the denominator, which will reduce the overall figure . So, left with A & B.

notice that , l/(r+h) is same in both.

So, assuming h > r, B has to be the correct option

(trying to make value of fraction bigger by increasing the numerator and decreasing the denominator).

Please give the original solution and OA. This is killing me !!!

I neither have the OA nor the OE to this question, but if this is killing somebody then I cannot let it happen...

I am poor at drawing on BTG, so please serve yourselves...

Draw an isosceles triangle ABC, with the legs AB = AC = l and the base BC = 2 r. Drop altitude AD on BC and read AD = h. The incircle to the triangle ABC will have its center O on the altitude AD. Let the incircle touch AC in P and BC in D such that OP = OD = R.

Note that ∆OPA ~ ∆CDA, hence

OP/CD = AO/AC

Or R/r = (h - R)/l

Or l R = r h - r R

Or R (r + l) = r h

Or [spoiler]R = r h / (r + l)

D[/spoiler]

sanju09 wrote:

harshavardhanc wrote:

sanju09 wrote:Which of the following expression gives the radius R of the largest sphere that can be carved out of a right cone of radius r, upright height h, and slant height l?
(A) R = r l / (r + h)
(B) R = l h / (r + h)
(C) R = l h / (r + l)
(D) R = r h / (r + l)
(E) R = r h / (h + l)

buddy,

is this really a GMAT question? :roll: : finding it very tough!

anyway, have to resort to logical-guessing, as can't use calculus ....

in a right angled triangle, l can be considered the hypotenuse with base as r/2 and perpendicular as h.

so, among the options, the contender will be any option which has l in the numerator. this gives A, B, and C.

now, in C, l comes in the sum in the denominator, which will reduce the overall figure . So, left with A & B.

notice that , l/(r+h) is same in both.

So, assuming h > r, B has to be the correct option

(trying to make value of fraction bigger by increasing the numerator and decreasing the denominator).

Please give the original solution and OA. This is killing me !!!

I neither have the OA nor the OE to this question, but if this is killing somebody then I cannot let it happen...

I am poor at drawing on BTG, so please serve yourselves...

Draw an isosceles triangle ABC, with the legs AB = AC = l and the base BC = 2 r. Drop altitude AD on BC and read AD = h. The incircle to the triangle ABC will have its center O on the altitude AD. Let the incircle touch AC in P and BC in D such that OP = OD = R.

Note that ∆OPA ~ ∆CDA, hence

OP/CD = AO/AC

Or R/r = (h - R)/l

Or l R = r h - r R

Or R (r + l) = r h

Or [spoiler]R = r h / (r + l)

D[/spoiler]

Got saved....I think


Buddy, I think this is just one part of the question where you find the radius of a sphere inscribed in a cone.

But, I would really appreciate if you could prove the second part which says that this WILL be the largest sphere that can be carved out of the cone.

still...... knock knock knocking on heaven's door ...........

still...... knock knock knocking on heaven's door ...........

Knocking is impossible as heaven has no doors to it, believe me, I have a visitors' visa for the heavens' too...

dear dangerous, incircle happens to be the largest possible circle in a triangle, isn't that OK?

sanju09 wrote:

still...... knock knock knocking on heaven's door ...........

Knocking is impossible as heaven has no doors to it, believe me, I have a visitors' visa for the heavens' too...

dear dangerous, incircle happens to be the largest possible circle in a triangle, isn't that OK?

was it written somewhere in the last post.... ? oooohhh.... I guess I must have missed it.......

Yes, incircle is the largest circle that can be inscribed in a circle. Your assumption that I know this concept is absolutely correct ! I am privileged that you just wrote that post for me.

My Bad! I thought that other members on this forum would read this explanation too and would require explanation for any additional concepts involved in solving this question!

harshavardhanc wrote:

sanju09 wrote:

still...... knock knock knocking on heaven's door ...........

Knocking is impossible as heaven has no doors to it, believe me, I have a visitors' visa for the heavens' too...

dear dangerous, incircle happens to be the largest possible circle in a triangle, isn't that OK?

was it written somewhere in the last post.... ? oooohhh.... I guess I must have missed it.......

Yes, incircle is the largest circle that can be inscribed in a circle. Your assumption that I know this concept is absolutely correct ! I am privileged that you just wrote that post for me.

My Bad! I thought that other members on this forum would read this explanation too and would require explanation for any additional concepts involved in solving this question!

why inscribe circle in circle, man?

sanju09 wrote:

harshavardhanc wrote:

sanju09 wrote:

still...... knock knock knocking on heaven's door ...........

Knocking is impossible as heaven has no doors to it, believe me, I have a visitors' visa for the heavens' too...

dear dangerous, incircle happens to be the largest possible circle in a triangle, isn't that OK?

was it written somewhere in the last post.... ? oooohhh.... I guess I must have missed it.......

Yes, incircle is the largest circle that can be inscribed in a circle. Your assumption that I know this concept is absolutely correct ! I am privileged that you just wrote that post for me.

My Bad! I thought that other members on this forum would read this explanation too and would require explanation for any additional concepts involved in solving this question!

why inscribe circle in circle, man?

ha ha! was a typo! wanted it to be a triangle, but wrote circle. thanks for correcting me !

I guess my sentence-construction is becoming like your drawing on BTG

sanju09 wrote: I am poor at drawing on BTG, so please serve yourselves…