Question

PQRS is a common diameter of three circles. The area of the middle circle is the average of the area of the other two. If PQ = 2 and RS = 1 then the length QR

Answer 1

Answer:

hy ur answer is

Step-by-step explanation:

Given

PQRS is the common diameter of all the three circles

PQ = 2 , RS = 1 , QR = ?

Area of circle which has diameter PR = Average of other circles =

= Ar(Circle no.1) + Ar(Circle no.2) /2

♦ We know that:

Area of Circle = πr²

Middle circle's (Circle no.2)

→ Diameter = PR = PQ + QR = 2 + QR

→ Radius = Diameter/2 = 2 + QR/2

First circle's (Circle no.1)

→ Diameter = PQ = 2

→ Radius = 2/2 = 1

→ Area = π(1)² = π = 3.14

Third circle's (Circle no. 3)

→ Diameter = PS = PQ + QR + RS = 2 + QR + 1

PS = 3 + QS

→ Radius = 3 + QR/2

Ar(Circle no.2) = Ar(circle no. 1) + Ar(circle no. 3)/2

★ π(2 + QR/2)² = 3.14 + π(3 + QS/2)²

★ π[(2 + QR)²/2²] = 3.14 + π[(3 + QS)²/2²)]

Using

(a + b)² = a² + 2ab + b²

★ π(4 + 4QR + QR²/4) - π(9 + 6QR + QR²/4) = 3.14

★ π(4 + 4QR + QR² - (9 + 6QR + QR²)/4) = 3.14

★ π( 4 + 4QR + QR² - 9 - 6QR - QR²/4) = 3.14

★ -5 - 2QR/4 = 3.14/3.14

★ - 5 - 2QR = 4

★ - 2QR = 4 + 5

★ QR = 9/-2

★ QR = - 4.5

Hence, distance cannot be negative so, QR = 4.5

Hence, QR = 4.5

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