two concentric circles of radii 15cm,12cm are drawn.find the lenght of chord of larger circle whic touches the smaller circle
Answers
Answer:
Step-by-step explanation:
\huge{\boxed{\boxed{\underline{\sf{SOLUTION}}}}}
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:
\begin{gathered}IN \: 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\end{gathered}
INright−angledtriangle
=)h
2
=p
2
+b
2
=)15
2
=12
2
+b
2
=)225−144=b
2
=)b
2
=81
=)b=
81
=)b=9
chord=2×9
=18cm
\huge{\boxed{\boxed{\sf{CHORD=18CM}}}}
CHORD=18CM