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Answer:

#"(a) "(5,-2)# is the Right Choice.

Explanation:

Let, #r gt 0# is the radius of the circle in Question.

Because the circle touches the X-Axis at #(3,0)#, its

centre C must be, #C=C(3,+-r)#.

Given that the point #P=P(1,-2)# lies on the circle, we have,

#PC^2=r^2#.

#:. (1-3)^2+(+-r+2)^2=r^2...[because," Distance Formula]"#.

#:. 4+(r^2+-4r+4)-r^2=0#.

#:. +-4r=-8 rArr r=+-2," but, as "r gt 0, r=+2#.

This means that, the centre is #C=C(3,+-2), and, r=2#.

Hence, the eqns. of the circles are,

# (x-3)^2+(y+-2)^2=2^2,#

#because P(1,-2) cancel(in) (x-3)^2+(y-2)^2=2^2#

#:., "the circle under reference is "(x-3)^2+(y+2)^2=2^2#.

Of all the given points, only #(5,-2)# satisfies the above eqn.

Hence, #"(a) "(5,-2)# is the Right Choice.

Answer:

#color(brown)(Area " " ABCD = 20.48 + 19 = 39.48 " sq. units"#

Explanation:

https://www.emathzone.com/tutorials/geometry/area-of-any-irregular-quadrilateral.html

#A (2,5), B (7,1), C (3, -4), D(-2,3)#

Using distance formula,

#bar (AB) = sqrt((7-2)^2 + (1-5)^2) ~~ 6.4#

#bar (BC) = sqrt((3-7)^2 + (-4-1)^2) ~~ 6.4#

#bar (CD) = sqrt((-2-3)^2 + (3+4)^2) ~~ 8.6#

#bar (AD) = sqrt((-2-2)^2 + (3-5)^2) ~~ 4.47#

#bar (AC) = sqrt((3-2)^2 + (-4-5)^2) ~~ 9.06#

https://www.onlinemathlearning.com/area-triangle.html

#"Area of " Delta " ABC " = sqrt(s * (s - a) * (s - b) * s - c)), " where " s = (a + b + c) / 2#

#s = (6.4 + 6.4 + 9.06) / 2 = 10.93#

#Delta_(ABC) = sqrt(10.93 * (10.93 - 6.4) * (10.93 - 6.4) * (10.93 - 9.06)) ~~ 20.48#

#"Area of " Delta " ADC " = sqrt(s * (s - a) * (s - d) * s - c)), " where " s = (a + d + c) / 2#

#s = (4.47 + 8.6 + 9.06) / 2 = 11.065#

#Delta_(ADC) = sqrt(11.065 * (11.065 - 4.47) * (11.065 - 8.6) * (11.065 - 9.06)) ~~ 19#

#"Area of quadrilateral " ABCD = " Area of " Delta_(ABC) + " Area of " Delta_(ADC)#

#color(brown)(Area " " ABCD = 20.48 + 19 = 39.48 " sq. units"#

Please solve q 19?

CW
CW
Featured 2 months ago

Answer:

#DM=3/2# units

Explanation:

enter image source here
Given #angleDAB=60^@, angleADC=120^@#,
#=> DC# // #AB#
Let #E# be the midpoint of #AD#,
#=> AE=ED=sqrt3/2#
given #M# is the midpoint of #BC, => EM# // #DC# // #AB#
#=> angleDEM=60^@#,
and #EM=(AB+DC)/2=(2*AD)/2=(2sqrt3)/2=sqrt3#
In #DeltaEDM#, as #EM=2ED, and angleDEM=60^@#,
#=> angleEMD=30^@, => angleEDM=90^@#,
#=> DeltaEDM# is right angled at #D#,
#=> DM=EMsin60=sqrt3*sqrt3/2=3/2# units

Or if you are not sure if #DeltaEDM# is a right triangle, you can also use the law of cosines to find the length #DM#,
#DM^2=ED^2+EM^2-2*ED*EM*cos60#,
#=> DM^2=(sqrt3/2)^2+sqrt3^2-2*sqrt3/2*sqrt3*1/2#
#=> DM^2=3/4+3-3/2=9/4#
#=> DM=sqrt(9/4)=3/2# units

Answer:

(d) None of the Above.

Explanation:

Clearly, the eqn. of the given line #L# is,

# L : (x,y,z)=(-1,3,3)+k(1,2,3), k in RR#.

In other words, any point on #L# is,

#(-1+k,3+2k,3+3k), k in RR#.

Also, the #XY"-plane"={(x,y,0) | x,y in RR}#.

#:." A point "P in L nn XY"-plane" rArr P in L & P in XY"-plane"#.

# rArr 3+3k=0, or, k=-1#.

#:. P =P(-1+k,3+2k,3+3k)#

#=P(-1-1,3-2,3-3)=P(-2,1,0)#.

Hence, the right choice is (d) None of the Above.

Answer:

Continuation.

Explanation:

The common-volume surface (CVS) is segmented by #2^N#

equal segments, each reaching the common chord ( axis of

symmetry ) through planar sides, like segments of a peeled orange.

To get a typical segment, rotate the LHS circle in the planar-section

graph ( in my 1st part answer ) about the chord, through #pi/2^(N-1)#

rad. The RHS part of the circle generates a segment for the inner

surface, and correspondingly, the larger LHS part forms the

opposite segment, for the outer surface.

Let us study the limit, as #N to oo#. The CVS #to# ( Rugby ball

shaped ) prolate spheroid of semi-axes

a = b = 1/2 and c = #sqrt 3 /2#. Its volume is

#4/3 pi (1/2)(1/2)(sqrt3 /2) = pi sqrt 3 / 6# cu.

The outer surface #to# an Earth-like oblate spheroid of semi-

axes a = b = 1.5 and c = 1, with conical dimples at the poles that

are #(1 - sqrt 3/2)# unit deep.

The volume enclosed, in the limit, is nearly

#4/3 pi (3/2)(3/2)(1) - 2 (1/3pi(1/2)(1/2)(1-sqrt3/2) )# cu

#= pi (17/6 + sqrt3/12)# cu.

Now, the elusive volume of the common-to #2^N# conjoined

spheres is expressed as a double integral

V = #2^N# ( volume of a typical segment)

#= 2^(N + 2) int int sqrt( 3/4 - rho cos theta - rho^2) rho d theta d rho#,

with limits

#rho# from 0 to #1/2 (sqrt( cos^2theta + 3 ) - cos theta )# and

#theta# from 0 to #pi/2^N#

I had used cylindrical polar coordinates #( rho, theta, z )# ,

referred to the common chord as z-axis and its center as origin.

Choosing the center of the LHS sphere as origin, this becomes

#4/3 2^N int (1- cos^2 alpha sec^2theta)^(3/2) d theta#, with

#theta# from #0 to alpha = pi/2^N- sin^(-1)(1/2 sin (pi /2^N))#.

This is indeed a Gordian knot. So, I look for another method that

leads to a closed form solution. The planar section z = 0 for 8 ( N =

3 ) spheres appears below. The graph reveals most of the aspects

in the description.
graph{((x-0.5)^2+y^2-1)((x+0.5)^2+y^2-1)((y-0.5)^2+x^2-1)((y+0.5)^2+x^2-1)((x-0.3536)^2+(y-0.3536)^2-1)((x+0.3536)^2+(y-0.3536)^2-1)((y+0.3536)^2+(x+0.3536)^2-1)((y+0.3536)^2+(x-0.3536)^2-1)=0[-3 3 -1.5 1.5]}
Now, the common space has just disappeared at z = #+- sqrt3/2#. This height is #1 +- sqrt3/2# above the Table.
graph{((x-0.5)^2+y^2-.25)((x+0.5)^2+y^2-.25)((y-0.5)^2+x^2-.25)((y+0.5)^2+x^2-.25)((x-0.3536)^2+(y-0.3536)^2-.25)((x+0.3536)^2+(y-0.3536)^2-.25)((y+0.3536)^2+(x+0.3536)^2-.25)((y+0.3536)^2+(x-0.3536)^2-.25)=0[-3 3 -1.5 1.5]}
The graphs are on uniform scale.

Answer:

Hence, the orthocentre of #triangle ABC# is #C(6,3)#

Explanation:

Let , #triangle ABC # ,be the triangle with corners at

#A(2,3) ,B(6,1) and C(6,3) # .

We take, #AB=c, BC=a and CA=b#

So,

#c^2=(2-6)^2+(3-1)^2=16+4=20#

#a^2=(6-6)^2+(1-3)^2=0+4=4#

#b^2=(2-6)^2+(3-3)^2=16+0=16#

It is clear that, #a^2+b^2=4+16=20=c^2#

# i.e. color(red)(c^2=a^2+b^2=>mangleC=pi/2#

Hence, #bar(AB)# is the hypotenuse.

#:.triangle ABC # is the right angled triangle.

#:.#The orthocenter coindes with #C#

Hence, the orthocentre of #triangle ABC# is #C(6,3)#

Please see the graph:

enter image source here

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