Question

two concentric of radii 15 cm and 12cm are drawn .find the length of chord of larger circle which touches the smaller circle​

Answer 1

Given :

• Two concentric of radii 15 cm and 12cm are drawn .

To Find :

• The length of chord of larger circle which touches the smaller circle.

Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(1.2,0)(1.121,1.121)(0,1.2)\qbezier(1.2,0)(1.121,-1.121)(0,-1.2)\qbezier(0,-1.2)(-1.121,-1.121)(-1.2,0)\qbezier( -1.2,0)(-1.121,1.121)(0,1.2)\put(2, - 1.2){\line(-1,0){4}}\put(0, - 1.2){\line(0,1){1.2}}\put( - 0.2, - 1.5){M}\put(0,0){O}\qbezier(0,0)(0,0)(2, - 1.2)\put( - 2.3, - 1.5){A}\put(2, - 1.5){B}\put( - 1.1, - 0.5){ \sf 12 cm}\put(0.5, - 0.3){ \sf 15 cm}\end{picture}$

Solution :

• O is the centre of circle.
• AB is the chord of circle with radius 15 cm.
• and AB touches smaller circle at a point M. Hence, AB is the tangent to the circle with radius 12 cm.

Since, Radius ⏊ Tangent

Therefore,∠OMB = 90°

• Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

In ∆ OMB, by Using Pythagoras theorem we get :-

᠉ (BM)² = (OB)² - (OM)²

᠉ (BM)² = (15)² - (12)²

᠉ (BM)² = 225 - 144

᠉ (BM)² = 81

᠉ BM = 9 cm

Since,∆AOM ≅ ∆BOM

⛬ AM = BM (By CPCT rule)

So, AB = 2BM

⛬ AB = 2(9) = 18 cm

Therefore,The length of chord of larger circle which touches the smaller circle is 18 cm

Answer 2

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: Question :- \: \star}}}}}$

Two concentric of radii 15 cm and 12cm are drawn .find the length of chord of larger circle which touches the smaller circle.

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

$\underline{\boxed{\textsf{l = {\textbf{18 cm}}}}} \qquad\qquad \bigg\lgroup\bold{length\ of \ the \ Chord} \bigg\rgroup$

$\Large{\underline{\underline{\bf{GiVen:-}}}}$

➹ The radii of 2 circles = 15 cm and 12cm

$\Large{\underline{\underline{\bf{Find:-}}}}$

✦ Find the length of chord of larger circle which touches the smaller circle .

$\Large{\underline{\underline{\bf{Solution:-}}}}$

So we consider CB a tangent of the small circle and chord to the larger circle.

➹ Two circles of radii 15 cm & 12 cm with common centre.

➹ Let A be the point of contact.

We have to do Construction

So we Join OC and OB.

The radius is perpendicular to the tangent { property applied }

In ∆OPB ∠OAB = 90°

OA = 15cm

OB = 12cm

Simply applying Pythagoras theorem

( Hy ) ² = ( per ) ² + ( base ) ²

➹ (12)² = (15)² + (AB)²

➹ 144 = 225 + AB²

✦ AB² = 225 - 144

AB² = 81

AB = √81

↠ AB = 9 cm

The perpendicular from the centre of circle to a chord,

will bisects :--

↠ CB = 2 × AB

↠ CB = 2 × 9 = 18 cm

Therefore :- )

$\large{\boxed{\bold{Length \:of\:chord\:=\:l\:=\:18\:cm}}}$

What is Tangent ?

A tangent to a circle is defined as a line that passes through exactly one point on a circle, and is perpendicular to a line passing through the center of the circle. A line that is tangent to more than one circle is referred to as a common tangent of both circles.

Define Pythagoras Theorem ?

Pythagoras' theorem states that for all right-angled triangles, 'The square on the hypotenuse is equal to the sum of the squares on the other two sides'. The hypotenuse is the longest side and it's always opposite the right angle. In this triangle a² = b² + c² and angle is a right angle.

Can we apply the Pythagoras Theorem for any triangle?

No, this theorem is applicable only for the right-angled triangle.