In a circle of radius 10 cm, with centre O, PQ and PR are two chords each of length 12 cm. PO intersects chord QR at the point S. The length of OS is:
Answers
Answer:
2.8
Step-by-step explanation:
firstly we find the area of quadrilateral
then value of its diagonal with help of area
in the end we apply pythagoras theorem.
in Which book you have seen this question??
Answer:
The length of OS is 2.8 cm
Step-by-step explanation:
In a circle of radius 10 cm, with centre O, PQ and PR are two chords each of length 12 cm.
PO intersects chord QR at the point S.
Please see the attachment for figure and name of figure.
In ΔPRS, ∠PSR=90°
PR² = PS² + SR²
12² = (10-y)² + x² -------------- (1)
In ΔRSO, ∠RSO=90°
RO² = OS² + SR²
10² = y² + x² -------------- (2)
Subtract eq(1) - eq(2)
12² - 10² = (10-y)² - y² + x² - x²
(12+10)(12-10) = (10-y+y)(10-y-y) [∴ a²-b²=(a+b)(a-b)]
(22)(2) = (10)(10-2y)
44 = 100-20y
20y = 66
y = 2.8 cm
Hence, The length of OS is 2.8 cm
