Question

There are two concentric circles with radii 12 cm and 15 cm. Find the length of the chord of the larger circle which touches the smaller circle

Answer 1

Answer:

$\huge{\boxed{\boxed{\sf{CHORD=18CM}}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION}}}}}$

>>BIG CIRCLE RADIUS(R1)= 15CM

>>SMALL CIRCLE RADIUS(R2)=12CM

>>Draw two concentric circles and draw a chord inside big circle which touches the outer end of small circle.

>>Such that the point where small circle touches the chord bisect the chord in two equal parts.

>>and draw a pependicular line on that point to the center of small circle.

>>and also join a line from the center of circle to the one end of the chord such that it will be the hypotenuse. then it form a right-angled triangle.

□ hypotenuse =15cm

□ perpendicular=12cm

□ base=?

>> first we find the length of base then doubled it bcz we draw a perpendicular line on that chord which divides the chord in two equal parts.

>> According to question:

$</u><u>I</u><u>N</u><u> \: right - angled \: triangle\\ = ){h}^{2} = {p}^{2} + {b}^{2} \\ = ) {15}^{2} = {12}^{2} + {b}^{2} \\ = )225 - 144 = {b}^{2} \\ = ) {b}^{2} = 81 \\ = )b = \sqrt{81} \\ = ) b = 9 \\ chord = 2 \times 9 \\ \: \: \: \: \: \: \: \: \: \: \: \: = 18cm$

$\huge{\boxed{\boxed{\sf{CHORD=18CM}}}}$

Answer 2

>>BIG CIRCLE RADIUS(R1)= 15CM