Acircle has radius of 18 cm. find the area of the smaller of the two regions determined by a chord with length of 18 square root of 2 cm. (hint: look for right triangles using pythagorean theorem converse) plz

Step-by-step explanation:

Let O be the radius of the circle and AB be the chord of length 18√2 cm.

Since, the radius of the circle is 18 cm,

Therefore, OA = OB = 18 cm,

Let, N be the mid point of the chord then,

AN = NB = 9√2 cm

In triangle AON,

Since, ( by the property of circle)

Similarly, in triangle BON,

⇒

Therefore, the area of sector AOB =

=

=

Area of triangle AOB =

=

=

=

=

=

Therefore, the area of smaller region by the chord AB = Area of sector AOB - Area of triangle AOB =

=

the answer would be c:   ax + b = ax + c

step-by-step explanation:

12

step-by-step explanation:

12 + 5 = 17

12x2=24

24-7=17