QR is a chord of the circle with centre at O and POR is a diamter of it. OD is perpendicular to QR. If OD= 4 cm , then the length of PQ is

Options

A. 4 cm
B. 2 cm
C. 8 cm
D. None of these

Answer

C

Explanation

`OD bot QR :. angle ODR`= 1 right angle
Again POR is a diameter of the circle with centre at O.
`:. angle PQR` is an angle is semicircle.
`:. angle PQR` =1 right angle.
`:. OD||PQ and angle PRQ` is common to both `Delta PQR and Delta ODR`
`:. Delta PQR and Delta ODR` are similar triangle.
`:. (OD)/(PQ)=(OR)/(PR) or, (4)/(PQ)=(OR)/(2QR) [ :. PR= 2OR] or, (4)/(PQ)=(1)/(2) ork, PQ=8`
`:.` the length of PQ= 8 cm `:.` (c) is correct.

Aliter : POR is a diamter of the circle with center at `O . angle PQR` is an angle in semicircle.
`:. PQR=` right angle or `90^(@)`
`:. Delta PQR` is a right- angled triangle of which PR is the hypotenuse.
`:. PR^(2)=PQ^(2)+Qr^(2)....(1)` [ by Pythagoras' theorem]
Again `OD bot QR`, where is a chor of the circle.
`:.D` is the mid-point of `QR, :. DR=(1)/(2)QR`
`:. Delta ODR` is a right-angled triange of which OR is the hypotenus.
`:. OR^(2)=OD^(2)+DR^(2)`
` or, ((1)/(2)PR)^(2)=4^(2) +((1)/(2)QR)^(2)[ :. QR=(1)/(2) PR and DR and OD =4 cm]`
` or, (PR^(2))/(4)=16+(QR^(2))/(4) or, PR^(2)=64+QR^(2)......(2)`
Now, from (1) and (2) we get,
`PQ^(2)+QR^(2)=64+QR^(2)`
`or, PQ^(2)=64 or, PQ sqrt(64)=8`
`:.` the length of PQ =8 cm.
`rArr (c)` is correct.