In the diagram shown, ΔABC is isosceles with AB = CB. Circle M is inscribed in ΔABC such that it is tangent at points D, E, and F. If the length of BF is twice the length of CF and the perimeter of ΔABC is 32 inches, then determine the length of side BC in inches.

BC = 12 inches

Step-by-step explanation:

As the triangle is isosceles, we have that AE = CF and EB = FB

If we call AE by x, we have that EB = 2x, FB = 2x and CF = x

As E and D are points tangent to the circle, we have that AE = AD = x

As D and F are points tangent to the circle, we have that CD = CF = x

So if the perimeter is 32, we have that:

AE + EB + BF + FC + CD + DA = 32

x + 2x + 2x + x + x + x = 32

8x = 32

x = 4

The length of BC is equal BF + FC = 2x + x = 3x = 12 inches

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

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