Question

8 cm
12 om
4 om
6 cm
16 cm
B​

Answer 1

Answer:

say what's the question

Answer 2

Given- AB=16 cm and CD=14 cm are the chordsof a circle with centre at O.

OM(=6 cm)⊥AB at M and ON⊥CD at N.

To find out -

If the length of ON=

m

cm, then m=?

Solution-

We join OCand OA.

ΔOAM and ΔOCN are right ones, since OM⊥AB at M and ON⊥CD at N.

Now AM=

2

1

AB=

2

1

×16 cm =8 cm and CN=

2

1

CD=

2

1

×14 cm =7 cm since the perpendicular from the centre of a circle to a chord bisects the latter.

So, in ΔOAM, by Pythagoras theorem, we have

OA=

OM

2

+AM

2

=

6

2

+8

2

cm =10 cm and it\ is the radius of the circle.

∴OC=OA=10 cm.

Again in ΔOCN, by Pythagoras theorem, we have

ON=

OC

2

−CN

2

=

10

2

−7

2

cm =

51

cm

But given that ON=

m

.

∴ m=51cm

2