mohon bantuannya ya guys

1) jumlah sudut segitiga ABC = 180°
    ∠ABC + ∠BAC + ∠ACB      = 180°
         40° +     80° + ∠ACB     = 180°
                                ∠ACB     = 180° - 40° - 80°
                                               =  60°
    sudut berpelurus = 180°
    ∠ACB + ∠ACD    = 180°
        60°  + ∠ACD   = 180°
                   ∠ACD  = 180° - 60°
                               = 120°

2) sudut berpelurur = 180°
    ∠CBA + ∠ABD    = 180°
        4x   + 6x + 10  = 180
               10 x + 10  = 180
                       10 x  = 180 - 10
                            x  = 170 /10
                                = 17°
    ∠CBA = 4x
               = 4 ( 17° )
               = 68°
    jumlah sudut segitiga ABC = 180°
    ∠CAB + ∠CBA + ∠ACB      = 180°
       54°  +  68°    +  ∠ACB      = 180°
                                ∠ACB      = 180 - 54 - 68
                                                = 58°
3) Luas lingkaran = π r²
                  462    = 22/7 x r²
                       r²   = 462 x 7/22
                            = 147
                       r    = √147
                            = 12,12 cm
    keliling daerah yang diarsir = 3/4 keliling lingkaran + 2 x jari jari
                                                = 3/4 ( 2 π r ) + 2 x 12,12
                                                = 3/4 ( 2 x 22/7 x 12,12) + 24,24
                                                = 81,44 cm

4) pada belah ketupat berlaku ketentuan sudut yang berhadapan sama besar maka
    ∠BCD  =  ∠DOB
                = 150°