Question

Three semi circles are drawn outside a triangle so that the sides of the triangle are the diameters of the semi circles. If the triangle is a right angled triangle and the semi circles have areas 18pie, 32pie and 50pie, determine the are of the triangle.

Answer 1

48?

Answer 2

how did you get that?

Answer 3

wait ill explain

Answer 4

thanks

Answer 5

See,we know that area of one circle is pi r^2
yes or no

Answer 6

yes or no?

Answer 7

yes

Answer 8

So, diameters of the 3 circles is the same as the 3 sides
and pi r^2 can be written as (d/2)^2
yes or no?

Answer 9

yes

Answer 10

can you do the equation? please

Answer 11

so calculate the diameter for all 3 circles
I shall work it out for one circle
\[18\pi=\pi \times d ^{2}/4\]
so pi is cancelled
\[18 \times 4=d ^{2}\]
\[d=\sqrt{72}\]
that is 6sqrt2

Answer 12

work out for remaining 2 and tell me what u get
be sure to get it as something sqrt 2

Answer 13

are u doing it ?

Answer 14

anyways after finding all 3 sides or diameters find semi-perimeter
semi perimeter is perimeter by 2
then use heron's formula to find area heron's formula is
\[\sqrt{s(s-a)(s-b)(s-c)}\]
where s is semi-perimeter

Answer 15

ok did u understand
do u want me to explain some part?

Answer 16

is there another formula I can use?

Answer 17

none that i know of
sry

Answer 18

ok

Answer 19

ok so u understand everything rite?

Answer 20

yes but how do I get 48?

Answer 21

find all 3 sides
they should be \[6\sqrt{2},8\sqrt{2},10\sqrt{2}\]
s=\[24\sqrt2/2=12\sqrt2\]
area=\[\sqrt{12\sqrt2\times 12\sqrt2-6\sqrt2 \times 12\sqrt2- 8\sqrt2 \times 12\sqrt2-10\sqrt2}\]
that is
\[\sqrt{12\sqrt2 \times 6\sqrt2 \times 4\sqrt2 \times 2\sqrt2}\]
that is
sqrt2 squared *24squared
that is 24*2=48

Answer 22

thanks a lot ur the best
also my hero