Question

In a circle of radius 17 cm, two parallel chords of length 16 cm and 30 cm respectively are
drawn on the same side of the centre. Find the distance between them....​

Answer 1

Answer:

Step-by-step explanation:

here , XY is the perpendicular distance

now perpendicular from center will bisect the chord

AX=15

AO = 17...(radius)

apply pythagoras to find  

OX=8cm

similarly in triangle OCY

OY=15 cm

total XY = 23 cm

Answer 2

Given :

• Radius of circle = 17 cm
• Length of one chord = 16 cm
• Other chord = 30 cm

To find :

• Distance between the chords

According to the question,

$\sf : \implies{(OD) {}^{2} = (OF) {}^{2} + (FD) {}^{2} }$

$\sf : \implies{(17) {}^{2} = (OF) {}^{2} + (15) {}^{2} }$

$\sf : \implies {289 = (OF) {}^{2} + 225}$

$\sf : \implies{289 - 225 =( OF) {}^{2} }$

$\sf : \implies{64 = (OF) {}^{2} }$

$\sf : \implies{ \sqrt{64} = OF}$

${ \underline{ \boxed{ : \implies{ \sf \red{8 \: cm = OF}}}}} \: \bigstar$

Now,

$\sf : \implies{(OB) {}^{2} = (EB) {}^{2} + (OE) {}^{2} }$

$\sf : \implies{(17) {}^{2} = ( {8)}^{2} +(OE) {}^{2} }$

$\sf : \implies{289 = 64 + ( {OE)}^{2} }$

$\sf : \implies{225 = {(OE)}^{2} }$

$\sf : \implies{225 = ( {OE)}^{2} }$

$\sf : \implies{ \sqrt{225} = OE}$

${ \underline{ \boxed{ \sf \red{ : \implies{15 \: cm = OE}}}}} \: \bigstar$

So,

➞ Distance between the chords = 15 cm + 8 cm = 23 cm

$\therefore { \underline{ \sf{ \: \: \: So, the \: distance \: between \: the \: chords \: is \: { \textsf{ \textbf{23 cm.}}} }}}$