Solve for x in the given fig.

Question

Find the radius of smaller circle if radius of bigger circle is 14 cm

i tried this for solving

What was the answer? Please edit your original post and show your steps. — Guy, May 12 '17 at 10:38
Hint: The circle's radius is less than $3.66$, because the area between the circle and the two lines has an area of $(28^2 - 196*\pi)/4$, or around $42.06$. — Toby Mak, May 12 '17 at 10:44

score 1 · Answer 1 · answered May 12 '17 at 11:06

1

Adding on to what @Salahamam_Fatima answered, the equation actually should be: $$14^2+14^2 = (14+x+EC)^2.$$ If we construct a line from $C$ to the other unlabelled tangent left of $C$ (call it $T_1$), and also label the tangent below $C$ $T_2$, $\angle T_1 E T_2$ is defined to be a right angle (OP said two lines were perpendicular in comments). Additionally, the two tangents are perpendicular to the centre by the tangent to a circle rule, so $ET_1CT_2$ is a square.

Following this, CE is the diagonal of a right angled triangle, so by Pythagoras's theorem, that length is $\sqrt {x^2+x^2}$ or $\sqrt {2} x$.

Now, the equation becomes: $$14^2+14^2 = (14+x+\sqrt {2} x)^2$$ = $$392 = 196 + 28\sqrt2 x + 2\sqrt2 x^2 + 3x^2$$ = $$ (3+2\sqrt2) x^2 + (28\sqrt2) x + (196) = 0$$ = $$ x = 42 - 28\sqrt2.$$

answered May 12 '17 at 11:06

Toby Mak

16,827

what do i consider the solution in the second fig right or wrong?? – vishesh das May 12 '17 at 11:10
It's wrong from what I can tell of the picture. – Toby Mak May 12 '17 at 11:11
Where was i mistaken?? It all seems right to me – vishesh das May 12 '17 at 11:12
How do you know that EC is $14(\sqrt(2)-1)(n)$? What is $n$? – Toby Mak May 12 '17 at 11:14
As EB =14{(2^1/2)-1} – vishesh das May 12 '17 at 11:17
And CB is x so EC =EB-CB=14{(2^1/2)-1}-X – vishesh das May 12 '17 at 11:18
$EB$ is $14\sqrt2- 14$, because $AB = 14$. – Toby Mak May 12 '17 at 11:19
In short, when proofwriting DO NOT skip steps. More often than not your brain will miss out a number of variable and that will affect everything after that mistake. – Toby Mak May 12 '17 at 11:20
Kk now i would make sure about it – vishesh das May 12 '17 at 11:22
But what about my ans?? – vishesh das May 12 '17 at 11:22
Nice attempt though! Sometimes one number can ruin everything1 – Toby Mak May 12 '17 at 11:22
Read my post carefully and see if you can find any mistakes. – Toby Mak May 12 '17 at 11:23
Another approach similar to what you did is to let $ED = (14\sqrt2)) - (14\sqrt2 - 14 - 2x)$, or $(AE) - (AB-BD)$ and work from there. – Toby Mak May 12 '17 at 11:24
I think Salahamam_Fatima approch is somewhat correct – vishesh das May 12 '17 at 11:41
It's very questionable — he pulls this number (${2.14}^2$) seemingly out of nowhere and skips steps. In my opinion, it's quite hard to follow what he does. – Toby Mak May 12 '17 at 11:42
@visheshdas We should probably move this discussion to chat, there's too many comments. – Toby Mak May 12 '17 at 11:43

hamam_Abdallah · Accepted Answer · 2017-05-12T14:57:00.203

0

hint

Using two Pythagoras, we get

$$14^2+14^2=(14+2x+ED)^2$$

and

$$x^2+x^2=(x+ED )^2$$

this gives $$14^2+14^2=(14+x+x\sqrt {2})^2$$

and $$x=14 (3-2\sqrt {2}) $$

edited May 12 '17 at 14:57

answered May 12 '17 at 10:45

hamam_Abdallah

62,951

You are leaving length ED in above fig – vishesh das May 12 '17 at 10:46
Yps now its corrrect b – vishesh das May 12 '17 at 10:57
But how 2 solve for x further – vishesh das May 12 '17 at 10:58
@Salahamam_Fatima Where did you get ${2.14}^2$ from? – Toby Mak May 12 '17 at 11:09
It looks too much like 2.14 squared, a better way to express this would be $(2)(14)^2$ – Toby Mak May 12 '17 at 11:50
@TobyMak just above you have$ 14^2+14^2$ – hamam_Abdallah May 12 '17 at 11:51
Never mind, it might be just my opinion but its hard to tell this if the context is not given. – Toby Mak May 12 '17 at 11:51

Solve for x in the given fig.

2 Answers2