6. Let ABCD be a square with side length 100. A circle with centre C and radius CD is drawn.
Another circle of radius r, lying inside ABCD, is drawn to touch this circle externally and
such that the circle also touches AB and AD. If r = m+nk, where m, n are integers andm+n/k is a prime number, find the value of
Answers
Correct question.
Let ABCD be a square with side length 100. A circle with centre C and radius CD is drawn another circle of radius r, lying inside ABCD, is drawn to touch this circle externally and such that the circle also touches AB and AD.If r = m + n√k where m, n are integers and k is prime numbers. find the value of (m + n)/(k).
EXPLANATION.
Let ABCD be a square with side length 100.
A circle with Centre "C" and radius "CD" is drawn.
Another circle of radius "r" lying inside ABCD is drawn to touch this circle externally and such that the circle also touches AB and AD.
If r = m + n√k where m, n are integers and k is prime numbers.
As we know that,
Pythagoras Theorem.
⇒ H² = P² + B².
Using this theorem in this question, we get.
In Δ ABC.
⇒ (AC)² = (BC)² + (AB)².
⇒ (AC)² = (100)² + (100)².
⇒ (AC)² = 10000 + 10000.
⇒ (AC)² = 20000.
⇒ (AC) = 100√2.
Circle with radius "r".
Inside a circle we need to find length of AO.
In Δ AOY.
⇒ (AO)² = (AY)² + (OY)².
⇒ (AO)² = (r)² + (r)².
⇒ (AO)² = 2r².
⇒ (AO) = r√2.
Now, we need to find radius "r".
We can write expression as,
⇒ AC = AO + OZ + ZC.
⇒ 100√2 = r√2 + r + 100.
⇒ 100√2 - 100 = r√2 + r.
⇒ 100(√2 - 1) = r(√2 + 1).
⇒ r = 100[(√2 - 1)/(√2 + 1)].
Rationalize the denominator of the expression, we get.
⇒ r = 100[(√2 - 1)/(√2 + 1) x (√2 - 1)/(√2 - 1)].
⇒ r = 100[(√2 - 1)²/(2 - 1)].
⇒ r = 100(√2 - 1)².
⇒ r = 100[(√2)² + (1)² - 2(√2)(1)].
⇒ r = 100[2 + 1 - 2√2).
⇒ r = 100[3 - 2√2].
⇒ r = 300 - 200√2.
⇒ r = m + n√k [given].
⇒ 300 - 200√2 = m + n√k.
⇒ m = 300 and n = - 200 and k = 2.
To find : The value of (m + n)/(k).
⇒ [300 + (-200)]/(2).
⇒ (300 - 200)/(2).
⇒ (100/2).
⇒ 50.
∴ The value of (m + n)/(k) is equal to 50.
