Two chords AB and CD of length 10cm and 24cm respectively of a circle are parallel to each other and are on opposite side of its centre.If the distance between ABand CD is 17cm, find the radius of the circle.
Answers
now
a line drawn which passes through center and cut AB at T and CD at R
for triangle OTA
OT=x
TA=5
OA=r ( where r is the radius of circle
because OTA is right angle triangle
so,
r^2=5^2+x^2 =================(1)
again
for triangle ORC is right angle triangle
OR=17-x
CR=12
OC=r
hence,
r^2=(17-x)^2+12^2 ==============(2)
now equation (1) and (2)
x^2+5^2=(17-x)^2+12^2
x^2-(17-x)^2=144-25=119
(x-17+x)(x+17-x)=119
2x-17=7
x=12 put this equation (1)
now
r^2=5^2+(12)^2=169
r=13cm
hence radius of circle is 13cm
Answer:
radius is 13 cm
Step-by-step explanation:
Given two chords AB and CD of lengths 24 cm and 10 cm respectively of a circle are parallel . if the chord lie on the same side of the center and the distance between them is 7 cm. we have to find the length of diameter.
As we know the line passing through center on the chord perpendicular bisect the chord.
Let OF=x gives OE=7+x
Hence, ΔOED and ΔOFB both are right angled triangle.
By Pythagoras theorem
In ΔOED, OD^{2}=OE^{2}+ED^{2}OD2=OE2+ED2
⇒ r^{2}=(7+x)^{2}+5^{2}r2=(7+x)2+52
In ΔOFB, OB^{2}=OF^{2}+FB^{2}OB2=OF2+FB2
⇒ r^{2}=x^{2}+12^{2}r2=x2+122
From above two equations,
(7+x)^{2}+5^{2}=x^{2}+12^{2}(7+x)2+52=x2+122
⇒ (49+x^2+14x)+25=x^{2}+144(49+x2+14x)+25=x2+144
⇒ 14x=7014x=70 ⇒ x=5
∴ r^{2}=5^{2}+12^{2}=25+144=169r2=52+122=25+144=169
⇒ r=13
diameter is 26 cm
I hope it will help you out