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Answer:

#17.66 " units"#

Explanation:

To find the perimeter, you need to find the distance of each side and add them up right?

So first you need to figure out the length of each side, and you can do this by plotting the corner points and assigning vectors to each side.
When you plot the points, it should look something like this:

https://www.geogebra.org/

To find a vector #vec(PQ)# between two points #P(x_1,y_1) " and " Q(x_2,y_2)# You can use the equation #vec(PQ)=[(x_2-x_1),(y_2-y_1)] #

Hence to calculate the vectors between #A(7,5), B(1,2) " and "C(4,8)#, the below calculations can be conducted:
#vec(AB)=[(1-7),(2-5)]=[(-6),(-3)]#

#vec(BC)=[(4-1),(8-2)]=[(3),(6)]#

#vec(AC)=[(4-7),(8-5)]=[(-3),(3)]#

Then the length of each vector needs to be calculated. To calculate the length of vector #vec(MN)=[(x),(y)]#

You can use:
#"distance of " vec(MN)=|vec(MN)|=sqrt(x^2+y^2)#

Hence when we are calculating the distances of the vectors of the triangle:

#|vec(AB)|=sqrt((-6)^2+(-3)^2)=sqrt(36+9)=sqrt(45) " units"#

#|vec(BC)|=sqrt((3)^2+(6)^2)=sqrt(9+36)=sqrt(45)" units"#

#|vec(AC)|=sqrt((-3)^2+(3)^2)=sqrt(9+9)=sqrt(18)" units"#

Therefore the total perimeter is
#|vec(AB)|+|vec(BC)|+|vec(AC)|=sqrt(45)+sqrt(45)+sqrt(18)=6sqrt5+3sqrt2~~17.66 " units"#

Answer:

The coordinates are #(23.7,8.9)# and #(-8.7,-1.9)#

Explanation:

The length of side #A=sqrt((7-8)^2+(5-2)^2)=sqrt10#

Let the height of the triangle be #=h#

The area of the triangle is

#1/2*sqrt10*h=27#

The altitude of the triangle is #h=(27*2)/sqrt10=54/sqrt10#

The mid-point of #A# is #(15/2,7/2)#

The gradient of #A# is #=(2-5)/(8-7)=-3#

The gradient of the altitude is #=1/3#

The equation of the altitude is

#y-7/2=1/3(x-15/2)#

#y=1/3x-5/2+7/2=1/3x+1#

The circle with equation

#(x-15/2)^2+(y-7/2)^2=54^2/10=291.6#

The intersection of this circle with the altitude will give the third corner.

#(x-15/2)^2+(1/3x+1-7/2)^2=291.6#

#x^2-15x+225/4+1/9x^2-5/3x+25/4=291.6#

#1.11x^2-16.7x-229.1=0#

We solve this quadratic equation

#x=(16.7+-sqrt(16.7^2+4*1.11*229.1))/(2*1.11)#

#=(16.7+-36)/2.22#

#x_1=23.74#

#x_2=-8.69#

The points are #(23.7,9.65)# and #(-8.7,-1.9)#

graph{(y-1/3x-1)((x-7.5)^2+(y-3.5)^2-291.6)((x-7)^2+(y-5)^2-0.05)((x-8)^2+(y-2)^2-0.05)(y-5+3(x-7))=0 [-12, 28, -10, 10]}

Answer:

see explanation

Explanation:

1) to find length #BE#
enter image source here
Let #O# be the center of the circle.
Let #angleAOB=x#
#=> anglex=360/5=72^@#
Consider #DeltaOBE#,
As #OB=OE=r=25# cm
#=> DeltaOBE# is isosceles
#=> angleOBE=angleOEB#,
#=> M# is the mid-point of #BE#
#=> DeltaOMB and DeltaOME# are two congruent right triangles.
#=> BM=rsinx#
#=> BE=2*BM=2rsinx#
#=2*25*sin72=2*25*(sqrt(10+2sqrt5)/4)#
#=(25sqrt(10+2sqrt5))/2~~47.55# cm

2) to find length #BA# or length #BF#
enter image source here
#DeltaBOE# is isosceles
#x=360/5=72^@#
#=> y=(180-2x)/2=(180-144)/2=18^@#
#BOA# is also isosceles.
#=> angleOBA=(180-x)/2=(180-72)/2=54^@#
#=> angleFBA=z=54-18=36^@#
#=> BA=2rsin(x/2)=2*25*sin36#
#=2*25*(sqrt(10-2sqrt5))/4=(25sqrt(10-2sqrt5))/2~~29.39# cm
#DeltaBFA# is also isosceles.
#=> 2*BF*cosz=BA#
#=> BF=(BA)/(2*cosz)=(BA)/(2*cos36)#
#=1/2*(25sqrt(10-2sqrt5))/2*4/(sqrt5+1)#
#=(25sqrt(10-2sqrt5))/(sqrt5+1)~~18.16# cm

Answer:

Please see below.

Explanation:

Let us consider the following figure, where we have the median #AM# and #AP_|_BC#.

enter image source here

Now in #DeltaABM#, we have

#AB^2=AP^2+PB^2=AP^2+(BM-PM)^2#

= #color(red)(AP^2)+BM^2+color(red)(PM^2)-2BMxxPM#

or #AB^2=color(red)(AM^2)+BM^2-2BMxxPM# .....................(1)

and in #DeltaACM#, we have

#AC^2=AP^2+PC^2=AP^2+(MC+PM)^2#

= #color(red)(AP^2)+MC^2+color(red)(PM^2)-2MCxxPM#

or #AC^2=color(red)(AM^2)+MC^2-2MCxxPM#

But as #BM=MC#, we can write it as

#AC^2=color(red)(AM^2)+BM^2-2BMxxPM# .....................(2)

Adding (1) and (2), we get

#AB^2+AC^2=2AM^2+2BM^2# and as #BM=(BC)/2#

#AB^2+AC^2=2AM^2+(2BC^2)/4#

or #AB^2+AC^2=2AM^2+1/2BC^2#

This is Apollonius theorem.

Answer:

area #=3sqrt3 " units"^2#

Explanation:

enter image source here
Circle 1 : #x^2+y^2+6x=0, =># center #C_1=(-3,0) , r_1=3#,
Circle 2: #x^2+y^2-2x=0, =># center #C_2=(1,0), r_2=1#
As the two circles touch each other externally, they have 3 common tangents.
Obviously, Y-axis #(x=0)# is one of the common tangents.
Let #y=mx+c# be the equation of the other tangents.
Circle 1 : # x^2+(mx+c)^2+6x=0#
#=> x^2+m^2x^2+2mcx+c^2+6x=0#
#=> (1+m^2)x^2+2(mc+3)x+c^2=0#
As the line #y=mx+c# is tangent to the circle, it touches the circle at one point, so set the discriminant #b^2-4ac# equal to #0#,
#=> (2(mc+3))^2-4(1+m^2)c^2=0#
#=> 4(mc+3)^2-4(1+m^2)c^2=0#
#=> (mc+3)^2-(1+m^2)c^2=0#
#=> m^2c^2+6mc+9-c^2-m^2c^2=0#
#=> 6mc=c^2-9# ----------- #color(red)(EQ(1))#
Similarly, Circle 2 : #x^2+(mx+c)^2-2x=0#
#=> (1+m^2)x^2+2(mc-1)x+c^2=0#
similarly, set the discriminant #b^2-4ac=0#,
#=> 2mc=-c^2+1# ------------#color(red)(EQ(2))#
Solving #EQ(1) and EQ(2)# gives #c=+-sqrt3#
a) when #c=sqrt3, => m=-1/sqrt3#
b) when #c=-sqrt3, => m=1/sqrt3#
#=># equations of the two tangents :
#y=-1/sqrt3x+sqrt3, => sqrt3y=-x+3#, and,
#y=1/sqrt3x-sqrt3, => sqrt3y=x-3#
Setting #x=0, sqrt3y=-x+3, and sqrt3y=x-3# equal to one another, we get :
#P=(3,0), Q=(0,sqrt3) and R=(0,-sqrt3)#
As #PQ=PR=QR=sqrt12=2sqrt3#,
#=> DeltaPQR# is an equilateral triangle.
#=># area of #DeltaPQR=sqrt3/4*(2sqrt3)^2=3sqrt3 " units"^2#

drawn
Given that

  • the equation of parabola #y=x^2#
  • the equation of a straight line #y=2x-17#

Let the corner points #A(t_1,t_1^2) and D(t_2,t_2^2) # of the square #ABCD# are on the parabola and other two corner points #BandD# are on the given line #y=2x-17#

So slope of AD #=(t_1^2-t_2^2)/(t_1-t_2)=2-> "the slope of the given line"# , since side AD is parallel to the given line.

Hence we have #t_1+t_2=2#

The length of side #AD=sqrt((t_1-t_2)^2+(t_1^2-t_2^2)^2)#

#=(t_1-t_2)sqrt(1+(t_1+t_2)^2)#

#=(t_1-t_2)sqrt(1+2^2)=sqrt5(t_1-t_2)=2sqrt5(t_1-1)#

Now length of the perpendicular from #A# to the line

#AB=(t_1^2-2t_1+17)/sqrt(1^2+2^2)=(t_1^2-2t_1+17)/sqrt5#

#ABCD# being a square

We have #AB=AD#

#=>(t_1^2-2t_1+17)/sqrt5=2sqrt5(t_1-1)#

#=>t_1^2-2t_1+17-10(t_1-1)=0#

#=>t_1^2-12t_1+27=0#

#=>t_1^2-9t_1-3t_1+27=0#

#=>t_1(t_1-9)-3(t_1-9)=0#

#=>(t_1-9)((t_1-3)=0#

Hence we get

#t_1=3 or 9#

When #t_1= 3#, the area of the square will be minimum

So

#"AREA"_"min"= AD^2# at #t_1=3#

#=(2sqrt5(t_1-1))^2#

#=20xx(3-1)^2#

#=80#squnit

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