Question

Find the distance of a chord of length 16cm from the centre of a circle of radius 12cm. Answer with screenshot of the working note

Answer 1

αиѕωєя

• The length of the chord = 16cm

• The radius of the circle = 10cm

• When a perpendicular is drawn on the chord such that the chord is bisecting into two. Then the length of the chord will be halved, that is it becomes 8cm.

Using the Pythagorean theorem,

$OA^2 = OC^2 + AC^2 \\ </h3><h3>10^2 = OC^2 + 8^2 \\ </h3><h3>100 = OC^2 + 64 \\ </h3><h3>OC^2 = 36 \\ </h3><h3>OC = 6cm$

• Therefore, the distance of the chord from the centre of the circle is 6cm.

$hope \: its \: help \: u$

Answer 2

ANSWER

______________

The length of the chord = 16cm

The radius of the circle = 10cm

When a perpendicular is drawn on the chord such that the chord is bisecting into two. Then the length of the chord will be halved, that is it becomes 8cm.

Using the Pythagorean theorem,

[ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2

$\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 </u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2</u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 </u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC </u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2</u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2 +64</u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2 +64OC </u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2 +64OC 2</u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2 +64OC 2 =36</u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2 +64OC 2 =36OC=6cm</u></p><p><u>[tex]\begin{gathered}OA^2 = OC^2 + AC^2 \\ 10^2 = OC^2 + 8^2 \\ 100 = OC^2 + 64 \\ \\ OC^2 = 36 \\ OC = 6cm\end{gathered} OA 2 =OC 2 +AC 2 10 2 =OC 2 +8 2 100=OC 2 +64OC 2 =36OC=6cm$

Therefore, the distance of the chord from the centre of the circle is 6cm.

HOPE ITS HELP U