we have to find the ratio of the area of triangle ABC to the area of sector BDCA .

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=9:4

Step-by-step explanation:

The relationship between area and side of triangle is

= (S1/S2)^2= (A1/A2)

=(3/2)^2=(A1/A2)

=9/4=(A1/A2)

=A1:A2=9:4




Ratio of areas =144:81

∴ ratio of sides =12:1

Let the longest side be x. Then

36:x=12:9 or x=1236×9=27cm



The sides of the regular hexagon ABCDEF can be posed as a. If so, the area of ABCDEF should be 6 times the are of the interior angle in the hexagon, considering there are 6 equilateral triangle that can fit in this regular hexagon,

Area = 6( a * a * sin 60 ) / 2,

Area = ( About ) 2.6 sq units

Now applying cosine for triangle ABC -

AC^2 = AB^2 + BC^2 – ( 2*AB*BC*cos 120 ),

a^2 + a^2 – ( 2a^2 * ( - 0.5 ) ) = a^2 + a^2+a^2 =3a^2,

AC = a√3

The area of ABC should thus be the following -

( a√3 * a√3 * sin 60 )/2 = 1.299038106 sq units

As you can see, the area of ABC is half the area of ABCDEF, thus the ratio of the area of ABC to ABCDEF is 1 : 2



Answer from: kaffolter25

We have to find the" ratio of the area of sector ABC to the area of sector DBE".

Now,

the general formula for the area of sector is

Area of sector= 1/2 r²θ

where r is the radius and θ is the central angle in radian.

180°= π rad

1° = π/180 rad

For sector ABC, area= 1/2 (2r)²(β°)

= 1/2 *4r²*(π/180 β)

= 2r²(π/180 β)

For sector DBE, area= 1/2 (r)²(3β°)

= 1/2 *r²*3(π/180 β)

= 3/2 r²(π/180 β)

Now ratio,

Area of sector ABC/Area of sector DBE =

*

= 4/3


Answer from: neariah24
C) 2/3Make a comparison, and do that by dividing the length if its a comparison of length or areas if its comparison of area and so on..To find the length, use the distance formula.Length of BC = 10Length of BD = 1510/15 = 2/3
Answer from: foreignev
Ratio = A_ABC / A_BDCA = r^2/2 / 3/4(π)r^2 = 4 / 6π = 2 / 3π = 0.21 

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