1. Two circles of radii 5 cm and 3 cm intersect at two points and the

1.	1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance betweentheir centres is 4 cm. Find the length of the common chord,2. If two equal chords of a circle intersect within the circle, prove that the segments ofone chord are equal to corresponding segments of the other chord,3. If two equal chords of a circle intersect within the circle, prove that the linejoining the point of intersection to the centre makes equal angles with the who so ever will solve this question will get 50 thanksbut you also have to give me back by following and y'all if you want you thanks my answers also!!!!!!!!
Answer» Given parameters are: OP = 5cm OS = 4CM and PS = 3cm Also, PQ = 2PR Now, suppose RS = x. The diagram for the same is shown below. Ncert solutions class 9 chapter 10-10 Consider the ΔPOR, OP2 = OR2+PR2 ⇒ 52 = (4-x)2+PR2 ⇒ 25 = 16+x2-8x+PR2 ∴ PR2 = 9-x2+8x — (i) Now consider ΔPRS, PS2 = PR2+RS2 ⇒ 32 = PR2+x2 ∴ PR2 = 9-x2 — (ii) By equating EQUATION (i) and equation (ii) we get, 9 -x2+8x = 9-x2 ⇒ 8x = 0 ⇒ x = 0 Now, put the VALUE of x in equation (i) PR2 = 9-0 ⇒ PR = 3cm ∴ The length of the cord i.e. PQ = 2PR So, PQ = 2×3 = 6cm Pls give one ques at a time.

1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance betweentheir centres is 4 cm. Find the length of the common chord,2. If two equal chords of a circle intersect within the circle, prove that the segments ofone chord are equal to corresponding segments of the other chord,3. If two equal chords of a circle intersect within the circle, prove that the linejoining the point of intersection to the centre makes equal angles with the who so ever will solve this question will get 50 thanksbut you also have to give me back by following and y'all if you want you thanks my answers also!!!!!!!!

Answer»

Given parameters are:

OP = 5cm

OS = 4CM and

PS = 3cm

Also, PQ = 2PR

Now, suppose RS = x. The diagram for the same is shown below.

Ncert solutions class 9 chapter 10-10

Consider the ΔPOR,

OP2 = OR2+PR2

⇒ 52 = (4-x)2+PR2

⇒ 25 = 16+x2-8x+PR2

∴ PR2 = 9-x2+8x — (i)

Now consider ΔPRS,

PS2 = PR2+RS2

⇒ 32 = PR2+x2

∴ PR2 = 9-x2 — (ii)

By equating EQUATION (i) and equation (ii) we get,

9 -x2+8x = 9-x2

⇒ 8x = 0

⇒ x = 0

Now, put the VALUE of x in equation (i)

PR2 = 9-0

⇒ PR = 3cm

∴ The length of the cord i.e. PQ = 2PR

So, PQ = 2×3 = 6cm