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Answer:

(a) Perimeter of the triangle ABC

#P = (a + b + c) = 80.68 + 72 + 76.34 color(blue)(= 229.02)#

Explanation:

enter image source here

(a)
#b_1 / b_2 = (a_1 + a_2) / (c_1 + c_2)# Eqn (1)

#a_1 / a_2 = (c_1 + c_2) / (b_1 + b_2)# Eqn (2)

#c_1 / c_2 = (b_1 + b_2) / (a_1 + a_2) # Eqn (3)

Multiplying Eqn (1) by (3),

#(b_1/ b_2) * (c_1/c_2) = (cancel(a_1 + a_2) / (c_1 + c_2)) / ( (b_1 + b_2) / cancel(a_1 + a_2))#

#(c_2 * b_1) / (c_1 * b_2) = (b_1 + b_2) / (c_1 + c_2)#

#(c_2 * b_1) * (c_1 + c_2) = (b_1 + b_2) * (c_1 * b_2)#

#c_2^2 b_1 + c_2 c_1 B-1 = b_1 b_2 c_1 + b_2^2 c_1#

Substituting values of #b_1, b_2, c_1# in the above equation,

#32c_2^2 + 32*36*c_2 = 32*40*36 + 40^2 * 36#

#cancel(32)^1c_2^2 + cancel(1152)^36 c_2 = cancel(46080)^1280 + cancel(57600)^1800#

#c_2^2 + 36c_2 - 3080 = 0#

#c_2 = (-36 +- sqrt(36^2 + 4*3080))/2 = (-36 + 116..69)/2 = 40.34# rounded to two decimals and leaving the negative value.

Considering Equation (3),

#c_1 / c_2 = (b_1 + b_2) / (a_1 + a_2)#

#36 / 40.34 = (40 + 32) / a# where #a = (a_1 + a_2)#

#a = (72 * 40.34) / 36 = 80.68#

Considering Eqn (2),

#(a - a_2) / a_2 = (c_1 + c_2) / (b_1 + b_2) = 76.34 / 72#

#80.68 - a_2 = (76.34/72) * a_2

#148.34a_2 = 80.68 * 72#

#a_2 = (80.68 * 72) / 148.34 = 39.16#

#a_1 = 80.68 - a_2 = 80.68 - 39.16 = 41.52#

Perimeter of the triangle ABC

#P = (a + b + c) = 80.68 + 72 + 76.34 color(blue)(= 229.02)#

Similarly, we can find P for Question (b)

Answer:

Length of a leg #vec(AC) = vec(BC) = color(green)(21.21)#

Explanation:

enter image source here

In triangle ABC in the above figure' hypotenuse vec(AB) = 30"

#hatB, hatA# are equal and #hatC = 90^0#

#:. hatB = hatA = (180 - 90)/2 = 45^0#

Using trigonometric functions,

#vec(AC) = vec(AB) * sin B = 30 * sin(45) = 30 * (1/sqrt2) = 21.21"#

#vec(BC) = vec(AB) = 21.21"#

Answer:

Please see below.

Explanation:

I did not use the midpoint at all.

Using the equation of a circle:

#x^2-2hx+h^2+y^2-2ky+k^2=r^2#

Because it passes through #(0,0)#, we find the #r^2=h^2+k^2#.

So, #x^2-2hx+y^2-2ky=0#

Using the point #(4,-8)#, we get

#16-8h+64+16k=0#

So, #h=2k+10#

Every circle that contains the points #(0,0)# and #(4,-8)# has equation:

#(x-h)^2+(y-k)^2 = h^2+k^2# where #h=2k+10#

Answer:

Hence, the ratio in which the point divides the line is
#4:-3=-4:3#

Explanation:

We need to find the point of intersection of the lines
#5x+4y=4# and
#(y-5)/(x-4)=(-1-y)/(7-x)#

Simplifying the second line

#(y-5)(7-x)=(-1-y)(x-4)#

#(y-5)(x-7)-(y+1)(x-4)=0#

Expanding
#xy-7y-5x+35-(xy-4y+x-4)=0#

#xy-7y-5x+35-xy+4y-x+4=0#

Rearanging and simplifying
#-6x-3y+39=0#
Dividing by 13
#-2x-y+13=0#

or

#2x+y=13#

The equations are:

#5x+4y=4#-------(1)
#2x+y=13#-------(2)

#detA=5xx1-2xx4=5-8=-3#
#detx=4xx1-13xx4=4-52=-48#
#dety=5xx13-2xx4=65-8=57#

#x=detx/detA=-48/-3=16#
#y=dety/detA=57/-3=-19#

#(x,y)-=(16,-19)#

Check

#5x+4y=5xx16+4xx(-19)=80-76=4#

#2x+y=2xx16+(-19)=32-19=23#

Verified.

Hence, the intersection point is #P-=(16,-19)#
One end of the line is
Other end of the line is #B-=(7,-1)#
Arranging in the form of #APB#
#(x_A,y_A)-=(4,5)#
#(x_P,y_P)-=(16,-19)#
#(x_B,y_B)-=(7,-1)#
P divides AB in the ratio
#(x_P-x_A)/(x_B-x_P)=(16-4)/(7-16)=12/-9=4/-3#along x axis
#(y_P-y_A)/(y_B-y_P)=(-19-5)/(-1-(-19))=(-24)/18=(-4)/3 #along y axis
Check:
Both are same
Justifying the coordinate for intersection point

Hence, the ratio in which the point divides the line is
#4:-3=-4:3#

Answer:

#"arc DF "=pi~~3.14" feet to 2 dec. places"#

Explanation:

#"the length of the arc is calculated using"#

#• " length of arc "="circumference "xx"fraction of circle"#

#color(white)(xxxxxxxxxxxx)=2pirxx60/360#

#color(white)(xxxxxxxxxxxx)=6pixxcancel(60)^1/cancel(360)^6#

#color(white)(xxxxxxxxxxxx)=(cancel(6) pi)/cancel(6)=pi~~3.14#

Answer:

#80^@,60^@" and "40^@#

Explanation:

#"sum the parts of the ratio "#

#rArr4+3+2=9" parts"#

#• " the sum of the 3 angles in a triangle "=180^@#

#rArr180^@/9=20^@larrcolor(blue)"1 part"#

#rArr4" parts "=4xx20^@=80^@#

#rArr3" parts "=3xx20^@=60^@#

#rArr2" parts "=2xx20^@=40^@#

#"the 3 angles in the triangle are"#

#80^@,60^@" and "40^@#

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