Question

Please solve question c . I'm having a test tomorrow please help me . Answer as fast as possible

Answer 1

Answer:

Step-by-step explanation:

angle DOC =90= angle AOB

FINDING OC

OC=AC/2

    FINDING AC

     AC^2 =28^2 +28^2

     AC^2 = 1568

     AC= 39.6

OC= 39.6/2

OC= 19.8

FINDING AREA OF ONE SECTOR DOCY / AOBX

((22/7)(19.8^2))/4

308 cm^2

AREA OF ONE SEGMENT DYC / ABX= AREA OF SECTOR- AREA OF TRIANGLE DOC / AOB

FINDING AREA OF TRIANGLE DOC/AOB

AREA= (BASE*HEIGHT)/2

AREA= 28*14/2

AREA= 196 cm^2

AREA OF ONE SEGMENT DYC / ABX= AREA OF SECTOR- AREA OF TRIANGLE DOC / AOB

segment area= 308-196

==112 cm^2

1 segment area==112

2 segment area==112*2

===224cm^2