Question

suppose AB, CD, EF, GH, are four equally inclined chords of a circle concurrent at an Interior point P of the circle dividing the circle into 8 parts. show that the sum of the areas of the sector like regions HOA, CPE, GPF, DPF is equal to the sum of the areas of the remaining regions APC, RPG, BPD, FPH.

the one solving will be marked brainliest
and the person would be considered best mathematician....​

Answer 1

Answer:

area of sector =Ф/360*Пr^2

hpa+cpe+gpb+dpf  (ange is 45)

=Ф/360*П(r^2)

the remaining sector

=Ф/360*Пr^2

Step-by-step explanation:

Answer 2

Answer:

Solution:

$\bf {= area \: of \: sector = \frac{( ° )}{360°} \pi {r}^{2} }$

$\bf {= hpa + cpe + gpd + dpf }$

$\bf{ = \frac{( ° )}{360°} \pi {r}^{2} }$

Note.... ( ° ) is in Degrees.

Hope it will be helpful :)...✍️