Math, asked by chethanBC6841, 1 year ago

A circular table is pushed into the corner of a rectangular room in such a way that it just touches two adjacent walls. An ant crawling on the table rim (on the minor arc between the two points of contact) observes that it is 3 inches from one wall and 6 inches from the other. Find the diameter (in inches) of the table.

Answers

Answered by knjroopa

Step-by-step explanation:

Given A circular table is pushed into the corner of a rectangular room in such a way that it just touches two adjacent walls. An ant crawling on the table rim (on the minor arc between the two points of contact) observes that it is 3 inches from one wall and 6 inches from the other. Find the diameter (in inches) of the table.

So let walls be x and y axis. The circle is a tangent to x and y axis. Let the centre be (m,n). Let x axis tangent point be (m,0) and y axis tangent point be (0,n).
Since radii of circle are constant m = n
There is a point on circle which is 6 inches away from one axis and 3 inches away from other axis. So it is equal to (6,3)
By using distance formula we get d = √(m2 – n1) ^2 + (m2 – n1)^2
(m – 3)^2 + (n – 6)^2 = √(m – m)^2 + (n – 0)^2
(m – 3)^2 + (m -6)^2 = m^2
So m^2 – 6m + 9 + m^2 – 12 m + 36 = m^2
So m^2 – 18 m + 45 = 0
So m^2 – 15 m – 3m + 45 = 0
So m(m – 15) – 3(m – 15) = 0
Or m – 15 = 0, m – 3 = 0
Or m = 15,3
The points could not exist between m and n tangent points and so we have the radius r = 15 inches.
Now diameter = 2 x radius
= 2 x 15
= 30 inches.

Reference link will be

https://brainly.in/question/9063527

Answered by Surbhitjangra

Answer:

(r-1)^2 + (r-8)^2 = r^2

solving the equation we get the final equation as

r^2 + 18r + 45 = 0

which in fact gives us two solutions which are

x= 3

x=15

hence radius cannot be 3 hence it will be 15 thus

d= 2r

d= 30

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