#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# two parallel chords are drawn on the same side of the centre of circle of radius r. it is found that they subtend an angle of 2A and A at the centre of the circle. Then find the perpendicular distance between two chords

Sujit Kumar
111 Points
4 years ago
$y=r(Cos\frac{A}{2}-CosA)$Sorry the last Step is wrong in the above solution. The answer is
Let The perpendicular distance from center to side opposite to angle A be x
and the perpendicular distance from center to side opposite to angle 2A be x-y
Required to find:- The value of y

Applying Trigonometory,

On triangle with angle A
$Cos\frac{A}{2}= \frac{x}{r}=>x=r(Cos\frac{A}{2})$__________(1)

On triangle with angle 2A
$CosA= \frac{x-y}{r}=>x-y=r(CosA)$__________(2)

Subtracting Equation (2) from (1)

$y=r(Cos\frac{A}{2}-CosA)$
$Ans: \ The \ perpendicular \ distance \ between \ the \ two \ parallel \ chords \ is \ r(Cos\frac{A}{2}-CosA)$