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Answer:

#(3,6) and (1,8)#

Explanation:

enter image source here

Some of the properties of a parallelogram:
a) Opposite sides are parallel.
b) Opposite sides are congruent.
c) Opposite angles are congruent.
d) The diagonals bisect each other, #=># the two diagonals have the same midpoint.

The Midpoint Formula: The midpoint of two points, #(x_1, y_1)# and #(x_2, y2)# is the point M found by the following formula :

#M=((x_1+x_2)/2, (y_1+y_2)/2)#

First, let #XZ and WY# be the two diagonals,
#=> XZ and WY# have the same midpoint,

#=> (2+4)/2=(3+x)/2, => x=3#
#=> (7+0)/2=(1+y)/2, => y=6#

Hence, the first coordinates of vertex #Y# are #(3,6)#

enter image source here

Then, let #XW and YZ# be the two diagonals,
# => XW and YZ# have the same midpoint.

#=> (2+3)/2=(x+4)/2, => x=1#
#=> (7+1)/2=(y+0)/2, => y=8#

So the second coordinates for vertex Y are #(1,8)#

Hence, the two possible coordinates for vertex Y are #(3,6) and (1,8)#

Answer:

Area of the region outside the house that the dog can reach is #339.29# square feet

Explanation:

As the dog is tied to a corner of a hexagon, with a rope of length of #12# feet, his movement is indicated in the following figure.

enter image source here

As may be seen, he has freedom of movement in a circle of #12# feet radius but within a sector of #240^@# only. Then he reaches next corner, on either side and has a movement restricted to #6# feet in an arc of #60^@# only.

Hence the area of the region outside the house that the dog can reach is given by

#240/360xxpixx12^2+2xx60/360xxpixx6^2#

= #2/3xx144xxpi+2xx1/6xx6^2xxpi#

= #96pi+12pi#

= #108pi#

= #108xx3.1416#

= #339.29# square feet

Answer:

The equation of the circle is

#(x - -12/10)^2 + (y - 3/2)^2 = (sqrt(369)/10)^2#

Explanation:

The standard Cartesian form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

where #(x, y)# is any point on the circle, #(h,k)# is the center point and r is the radius.

Given: The tangent line #4x-5y=0# is tangent to the circle at the point #(0,0)#

Because the radius is the circle is perpendicular to the tangent, we can use the tangent line to find an equation that pass through the center. Write the tangent line in slope-intercept form:

#-5y =-4x#

#y = 4/5x#

This makes the center line:

#y = -5/4x#

Evaluate this at the point #(h,k)#:

#k = -5/4h" [2]"#

Evaluate equation [1] and the points #(0,0)# and #(0,3)#

#(0 - h)^2 + (0 - k)^2 = r^2" [3]"#
#(0 - h)^2 + (3 - k)^2 = r^2" [4]"#

Expand the squares:

#h^2 + k^2 = r^2" [5]"#
#h^2 + 9-6k+k^2 = r^2" [6]"#

Subtract equation [5] from equation [6]:

#9-6k = 0#

#k = 3/2#

Substitute into equation [2]:

#3/2 = -5/4h#

#h = -12/10#

Substitute into equation [5]:

#(3/2)^2+ (-12/10)^2 = r^2#

#(15/10)^2+ (-12/10)^2 = r^2#

#225/100+ 144/100 = r^2#

#r = sqrt(369)/10#

The equation of the circle is

#(x - -12/10)^2 + (y - 3/2)^2 = (sqrt(369)/10)^2#

Here is a graph of, the circle (red), the center point (blue), the tangent line (green) and the center line (yellow):

Desmos.com

Answer:

Area #AEF=1512 cm^2#

Explanation:

When a line from one of the vertex of a triangle, meets the opposite side, the ratio between the areas of the two triangles is same as the ratio of the two parts in which it divides the opposite side.

By corollary, in a quadrilateral whose diagonals divide it in four triangles, areas of two adjacent triangles are in the same ratio as the ratio of other two triangles.

Let #DeltaXYZ# denotes area of triangle #XYZ#.
Given #DeltaBDG : DeltaGDE = 30 : 60 = 1:2#
#=> DeltaBDC : DeltaDCE = DeltaBDC:80=1:2#
#=>DeltaBDC=40#

As #DeltaGCB:BCF=70:420=1:6#
#=> GB:BF=1:6#
#=> DeltaGAB:DeltaBAF=1:6#
let #DeltaGAB=1a, and DeltaBAF=6a#

As #DeltaEBC:DeltaCBF=120:420=2:7#
#=> EC:CF=2:7#
#=> DeltaEAC:DeltaCAF=2:7#
#=> (1a+210):(6a+420)=2:7#
#=> (1a+210)xx7=(6a+420)xx2#
#=> 1470-840=12a-7a#
#=> 5a=630#
#=> a=126#

Area of triangle #AEF=1a+6a+210+420=7a+210+420#
#=7xx126+210+420=1512 cm^2#

Answer:

see explanation.

Explanation:

1.
Area of a cricle #A=pir^2#, where #r# is the radius of the circle.
#=> 16pi=pir^2, => r=sqrt16=4# m
circumference of a circle #c=2*pi*r#
#=> c=2*pi*4=8pi# m

2.
enter image source here
Let #r_1# be the radius of the inscribed circle and #s# be the side of the square.
As shown in the figure, the inscribed circle has a diameter #d=s#
#=> r_1=d/2=s/2=4/2=2# cm
Let #r_2# be the radius of the circumsribed circle,
#r_2^2=r_1^2+r_1^2=2r_1^2#
#=> r_2=sqrt2r_1=2sqrt2# cm

3.
enter image source here
The largest circle that can fit insides a square of edge of #4# in. has a diameter #d=4# in.
Area of circle #A_c=pir^2=pi(d/2)^2=(pid^2)/4=(16pi)/4=4pi# sq. in.
Let #A_s# be the area of the square, #=> A_s=4^2=16# sq. in.
Area of the materail wasted = shaded area = #A_s-A_c#
#=16-4pi=4(4-pi)~~3.434# sq. in.

4.
enter image source here
Area of washer (one face) = shaded area #= pir_1^2-pir_2^2#
given #r_1=1/2, and r_2=1/4#
#=pi(r_1^2-r_2^2)=pi((1/2)^2-(1/4)^2)=3/16pi# sq. in.

5.
enter image source here
An equilateral triangle is a triangle in which all 3 sides are equal and all 3 angles are each #60^@#
As two tangents to a circle from an external point are equal,
#=> BD=BE#
As #D and E# are points of tangency and #BO# is a common side,
#=> DeltaODB and DeltaOEB# are congruent.
#=> BO# bisects #angleDBE#, which is #60^@#
#=> angleOBD=angleOBE=30^@#
#BE=1/2BC=6/2=3#
#BOcos30=BE=3#
#BO=3/cos30=3/(sqrt3/2)=6/sqrt3=(6sqrt3)/3=2sqrt3#
#OE=BOsin30=2sqrt3xx1/2=sqrt3#
#BO# is the radius of the circumsribed circle and #OE# the radius of the inscribed circle.
annulus area = shaded area #= pi(BO)^2-pi(OE)^2#
#=pi(2sqrt3)^2-pi(sqrt3)^2#
#=12pi-3pi=9pi " cm"^2#

drawn
Let the equation of the circle be

#x^2+y^2=a^2.............[1]#,

where #a# is the radius and #O(0,0)#is the center of the circle.

Let # A(acostheta,asintheta)# be any point on the circle parametrically and #theta # is the parameter.

So the equation of the tangent on the circle at #A# will be given by

#x xxacostheta+yxxasintheta=a^2#

#or,x costheta+ysintheta=a#

So the equations of set of chords parallel to that tangent can be written as

#x costheta+ysintheta=c..........[2]#,

where c is another parameter which changes from chord to chord.

Multiplying {1} by #cos^2theta # we get

#x^2cos^2theta+y^2cos^2theta=a^2cos^2theta.............[3]#

From [2] and [3] we can write

#(c-ysintheta)^2+y^2cos^2theta=a^2cos^2theta#

#=>y^2-2csintheta y+(c^2-a^2cos^2theta)=0#

If #y_1 and y_2# are two roots of this quadratic equation then

#y_1+y_2=2csintheta#

If #(h,k)# represents the coordinates of the mid point of the chord then #k=(y_1+y_2)/2=csintheta#

Now #(h,k)# is on [2} so we get

#h costheta+ksintheta=c..........[4]#

#=>h costheta+csinthetaxxsintheta=c#

#=>h costheta=c(1-sin^2theta)#

#=>h costheta=c cos^2theta#

#=>h =c costheta#
So

#k/h=tantheta#

Hence #y=xtantheta # will represent the locus of the mid point of parallel chords of a particular tangent for particular value of # theta#.

Now equation of the diameter of the circle passing through #O(0,0) and P(acostheta,asintheta)# is given by

#(y-0)/(x-0)=(asintheta-0)/(acostheta-0)#

#=>y=xtantheta#, which is also the locus of the mid point of the chords parallel to the tangent.

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