Question

Two concentric circles are of radii 25 cm and 24 cm. The length of the chord of the larger circle which touches the smaller circle is?​

Answer 1

Step-by-step explanation:

AB = 14 cm

plz mark as brainliest

Answer 2

The length of the chord of the larger circle which touches the smaller circle is 14 cm.

Given:

Two concentric circles of radii 25 cm and 24 cm

To Find:

The length of the chord of the larger circle which touches the smaller circle

Solution:

Let QS be the chord of the larger circle which touches the smaller circle at P. Let us join OP.

We have

OP = 24 cm

OQ= 25 cm

OP ⊥ QS and since perpendicular from the center bisects the chord, OP bisects QS.

Therefore QS = 2QP = 2PS

In right-angled triangle Δ OQP

$QP^{2}$ = $OQ^{2}$ - $OP^{2}$

$QP^{2}$= $25^{2}$ - $24^{2}$ = $49^{2}$

QP = √49 = 7 cm

Therefore QS = 2*QP= 14 cm.

Hence, the length of the chord of the larger circle which touches the smaller circle is 14 cm.

#SPJ2