Question

a chord of a circle divides the circle into two parts such that the squares inscribed in the two parts have areas 16 and 144 units. the radius of the circle

Answer 1

A chord of a circle divides the circle into two parts such that the squares inscribed in the two parts have areas 16 and 144 unit as shown in figure.
Let radius of Circle is r
And distance between centre of circle to chord = x [ see attachment]
so, distance between centre to side of big square {e.g., AB} = 12 - x

Now, ∆ABC is right angled triangle .
so, AC² = AB² + BC²
Here, AC = r
AB = 12 - x
BC = half of side length = 12/2 = 6 [ because radius is the perpendicular bisector on chord ]
r² = (12 - x)² + 6² ----------(1)

Similarly,
∆ADE is also a right angled triangle ,
so, AD² = AE² + DE²
r² = (x + 4)² + 2² ----------(2)

Now, from equations (1) and (2),
(12 - x)² + 6² = (x + 4)² + 2²
⇒144 + x² - 24x + 36 = x² + 16 + 8x + 4
⇒ 180 -24x = 20 + 8x
⇒ 160 = 32x
⇒ x = 5

Now, radius , r = √{(x +4)² + 2²}
r = √{9² + 2²} = √85

Answer 2

Hello Dear.

Here is your answer---


As per as the Question, Diagram is shown in the attachment.

Please Kindly Refers to the Attachment for the Question.

Let the side of the square be r cm.

Area of the Square ABCD = 144 cm²
                             (Side)² = 144
                               Side = √144
                              Side = 12 cm.

Thus, AB = BC = CD = AD = 12 cm.

Area of the Square PQRS = 16 cm²
                              (Side)² = 16
                               Side = √16
                               Side = 4 cm.

Thus, PQ = QR = RS = PS = 4 cm.

Now, We know, OL and OK are Perpendicular on BC and PS.

Also,
PK = KS = 4/2  [The perpendicular from the centre to the chord Bisects it]                     = 2 cm                             
                     
Similarly, BL = LC = 12/2
                      = 6 cm

OP = OB = r cm    [Since Line from the Centre to any part of the circle is                                                        Radius]
Let OK be x cm.
From the attachment, KL = 4 cm + 12 cm.
                                         = 16 cm.
Thus, OL = KL - OK
               = (16 - x) cm.

In Δ OPK,
  OP² = OK² + KP²            [By Pythagoras Theorem]
   r² = x² + (2)²
  r² = x² + 4    ---------------------------eq(i)

Similarly,

In ΔOBL,
OB² = OL² + BL²
 r² = (16 - x) + (6)
 r² = 256 + x² - 32x + 36               [(a - b)² = a² + b² - 2ab] 
 r² = 292 + x² - 32x    ----------------------------------eq(i)

From eq(i) and eq(ii)

 x² + 4 = 292 + x² - 32x
    32 x = 288
         x = 388/32
         x = 9 cm.

Putting the Value of x = 9 cm in eq(i),

r² = x² + 4
r² = (9)² + 4
r² = 81 + 4
r² = 85
r = √85 cm.
or r = 9.22 cm

Thus, the Radius of the Circle is √85 cm or 9.22 cm.


Hope it helps.

Have a Marvelous Day.