From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle?

Since, the tangent of any point of a line is perpendicular to the radius through the point of contact.


Since, the tangent of any point of a line is perpendicular to the rad

Therefore, ∠OPO =90°
It is given that PQ =15 cm
and    OQ = 17 cm
In right ΔOPQ
OQ2 = OP2 + PQ2
[Using Pythagoras theorem]
⇒ (17)2 = OP2 + (15)2
OP2 = (17)– (15)2
= 289 – 225 = 64 ⇒ OP = 8 cm
Hence, radius of the circle = 8 cm.

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ABCD is a quadrilateral such that ∠D = 90°. A cirlce C (O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, Find r.

So,    BP = BQ
(Tangents from external point B)
But    BP = 27 cm
⇒    BQ = 27 cm
It is given that BC = 38 cm
⇒    BQ + CQ = 38
⇒    27 + CQ = 38
⇒    CQ = 11 cm
⇒ CQ = CR (Tagents from an external point C)
But    CQ = 11 cm
⇒    CR = 11 cm


So,    BP = BQ(Tangents from external point B)But    BP = 27 cm

It is given that : CD = 25 cm
⇒    CR + DR = 25
⇒    11 + DR = 25
⇒    DR = 14 cm
Since, tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠ORD = ∠OSD = 90°
It is given that
∠D = 90°
Now, in quadrilateral ORDS,
∠ORD = ∠OSD = ∠RDS = ∠ROS = 90°
and    OR = OS [radii of circle]
Therefore, ORDS is a square
So,    OR = DR = 14 cm
Hence    r = 14 cm.

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A line intersecting a circle in two points is called a ___________.
  • Diameter
  • Chord   

  • Secant
  • Secant

C.

Secant
118 Views

In the given figure, ABC is a right-angled triangle with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle.


It is given that ABC is a right angle triangle with AB = 6 cm and AC = 8 cm and a circle with centre O has been inscribed.
Using Pythagoras theorem, we get
BC2 = AC2 + AB2
= (8)2 + (6)2
= 64 + 36 = 100
⇒ BC = 10 cm

Now comma space area space of space increment space ABC space equals space 1 half straight x space AB space straight x space AC
rightwards double arrow space space space space space equals space 1 half straight x space 6 space straight x space 8 space equals space 24 space cm squared space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
Area space of space increment space ABC comma space equals space ar space left parenthesis increment AOB right parenthesis plus ar left parenthesis increment BOC right parenthesis space

because Tangents at any point of a circle is perpendicular to the radius through the point of contact 

So comma space space OP space perpendicular space AB comma space OQ space perpendicular space BC comma space OR space perpendicular space AC

equals space 1 half cross times space AB space cross times space OP space plus 1 half cross times space AC space cross times space OR space plus space 1 half space cross times space BC space cross times space OQ
equals space 1 half cross times space 6 space cross times space straight r space plus space 1 half space cross times space 8 space cross times space straight r space plus 1 half space cross times space 10 space cross times space straight r space

=  3r + 4r + 5r
= 12r                                              ...(ii)

Comparing (i) and (ii), we get
24 = 12r ⇒ r = 2 cm
Method – II:
In quadrilateral APOR,
∵ ∠OPA = ∠ORA = 90°
∠PAR = 90°
⇒ ∠OPA = ∠ORA = ∠PAR = ∠POR = 90° ....(i)
and AP = AR (ii) (length of tangents drawn from an external point are equal)
Using result (i) and (ii), we get
APOR is a square
Therefore,
OR = AR = r [Sides of square] and OR = AP = r    [Sides of square]
Now, BP = AB – AP = 6 – r and, CR = AC – AR = 8 – r
Since tangents from an external point are equal
CR = CQ = 8 – r and    BP = BQ = 6 – r
Now, In ΔABC,
BC2 = AC2 + AB2
⇒ (CQ + BQ)= (8)2 + (6)2
⇒ (8 – r + 6 – r)2 = 64 + 36
⇒ (14 – 2r)2 = 100
⇒ (14 – 2r)2 = (10)2
⇒ 14 – 2r = 10
⇒ –2r = –4
⇒    r = 2
Hence radius of circle (r) = 2 cm.



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The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.


Given : Two concentric circles C1 and Cof radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C1) and BD is a tangent to the smaller circle (C2).
∴ The tangent at any point of a circle is perpendicular to the radius through the point of contact.

therefore space space OD space perpendicular space BD

Now, in right triangle BOD,  we have

OB2 = OD2 = BD2                [ Using Pythagoras theorem]

rightwards double arrow   (13)= (8)+ BD

rightwards double arrow   169 = 64 + BD2 
rightwards double arrow   BD2 = 169 - 64 
rightwards double arrow space  BD2 = 105

rightwards double arrow space space space BD space equals space square root of 105 space cm
Since, perpendicular drawn from the centre to the chord bisects the chord.

rightwards double arrow space space space space space BD space equals space DE space equals space square root of 105 space cm
rightwards double arrow space space space space BE space equals space 2 BD
rightwards double arrow space space space space space space space space space equals space 2 square root of 105 space cm

Now comma space in space increment space BOD space and space increment space BAE comma
angle BDo space equals space angle BEA space equals space 90 degree

(∵ ∠BEA = 90°, angle in semicircle is right angle] and, ∠OBD = ∠ABE     (common)

Therefore, using AAS similar condition ΔBOD ~ ΔBAE

rightwards double arrow space space BD over BE equals OD over AE

[Proportional sides of two similar triangles]

rightwards double arrow space space fraction numerator square root of 105 over denominator 2 square root of 105 end fraction equals 8 over AE
rightwards double arrow space space space 1 half equals 8 over AE
rightwards double arrow space space space space AE space equals space 16 space cm

Now, in right triangle ADE, we have

AD2 = AE2 + DE

rightwards double arrow space space AD squared space equals space 16 squared space plus space left parenthesis square root of 105 right parenthesis squared
                = 256 + 105
                = 361

rightwards double arrow     AD  = 19   cm.

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