Featured Answers

4
Active contributors today

Answer:

Length of chord #= 9sqrt5 ~~ 20.125#

Explanation:

enter image source here

As area of a circle is given by #pir^2#, and it is #135pi#, we have #r=sqrt135=3sqrt15#

As shown in the figure, the angle #Theta# subtended by the chord at the centre is :
#Theta=(5pi)/6-pi/6=(2pi)/3#
#=> Theta/2=pi/3#
#=> AM=rsin(Theta/2)#
#=># Length of chord #AB=2*AM=2*r*sin(Theta/2)#
#=2*3sqrt15*sin(pi/3)=2*3sqrt15*sqrt3/2=9sqrt5#

Answer:

Equation of circle is #x^2+y^2-3x-3y=0#, but tangent at #(3,0)# is #y=x-3#. In fact #4x-5y=0# is a secant of the circle.

Explanation:

Let the equation of circle be #x^2+y^2+2gx+2fy+c=0#.

As it passes through #(0,0)#, #(0,3)# and #(3,0)#. Note we are having a tangent #4x-5y=0# at point #(3,0)#, hence this too is on circle. Putting these in the equation of circle, we get

Putting #x-0# and #y=0#, we get #c=0#.

Putting #x=0# and #y=3#, we get #9+6f+c=0#. But as #c=0#, #f=-3/2#.

Now putting #x=3# and #y=0#, we get #9+6g+c=0#. But as #c=0#, #g=-3/2#.

Hence, equation of circle is #x^2+y^2-3x-3y=0#, whose center is #(3/2,3/2)#.

Slope of the radius joining #(3/2,3/2)# and #(3,0)# is

#(0-3/2)/(3-3/2)=-1# and hence slope of tangent at #(3,0)# is #1#

and equation of tangent at #(3,0)# is #y=x-3# and not #4x-5y=0# ( note that it passes through #(0,0)# and hence, this cannot be a tangent, but is a secant ) as mentioned.

As is seen from figure below, the circle passes through the three points, but #4x-5y=0# is not a tangent. Tangent at #(3,0)# is #y=x-3#.

graph{(y-x+3)(x^2+y^2-3x-3y)(4x-5y)=0 [-3.63, 6.37, -0.96, 4.04]}

graph{(y+x)(x^2+y^2-3x-3y)(4x-5y)=0 [-3.63, 6.37, -0.96, 4.04]}

Answer:

The point of intersection is #(-22.5,-41)#

Explanation:

The orthocenter is where the altitudes intersect in a triangle. Technically, it's where all three altitudes meet. But we only need to find two of them - and it doesn't matter which two.

I've included two pictures - the first one is of the triangle (which is barely a triangle) and the other is a picture of the triangle, zoomed out, so we can see where the three altitudes intersect (so far outside the triangle!

geogebra.com

geogebra.com

So.... Let's find two of the altitudes. We'll need to know two things to find one altitude - one of the vertices, and the slope of the line that the altitude belongs to (we can just use the "rise over run" method for that.

Arbitrarily, let's start with the point #(5, 3)# and the line segment between #(1, 6)# and #(9,1)#. Some quick math, and the slope of that segment is #-5/8# and the negative reciprocal is #8/5#. So let's use those two pieces of information to get the equation of that altitude!

#y-y_1 = m(x-x_1)#

#y-3 = 8/5(x-5)#

#y-3 = 8/5x-8#

#y = 8/5x-5#

Sweet. Now let's do the same for a different altitude. Again, it doesn't matter which one. Let's pick the point #(9,1)# and the segment with endpoints #(5,3)# and #(1,6)#. Some counting gives us a slope of #-3/4# so the slope of the altitude is #4/3#. Let's do this!

#y-y_1=m(x-x_1)#

#y-1=4/3(x-9)#

#y-1=4/3x-12#

#y=4/3x-11#

Sweet.

Now we just need to find the point of intersection of those two altitudes. We can do that by setting them equal to each other.

#4/3x-11 = 8/5x-5#

#20/15x-165/15 = 24/15x-75/15#

#15[20/15x-165/15] = 15[24/15x-75/15]#

#20x-165=24x-75#

#-90=4x#

#x=-22.5#

So the x-coordinate is #-22.5#. Let's plug that in to one of those two equations and find the y-coordinate:

#y=8/5x-5#

#y=8/5(-22.5)-5#

#y=-36-5#

#y=-41#

So it looks like the point of intersection is #(-22.5,-41)#

drawn

The given equation of ellipse

#b^2x^2+a^2y^2+a^2b^2#

Its standard form is

#color(blue)(x^2/a^2+y^2/b^2=1)#

which has its center at origin #(0,0)# and major axis and minor axis of lengths #2a and 2b# are along x-axis and y-axis respectively.

Parametrically any point on this ellipse is represented by #(acostheta,bsintheta)#, where #theta# is the parameter

The equation of any tagent to this ellipse can be written as

#color(green)((xcostheta)/a+(ysintheta)/b=1)#

Again by the given condition base of the triangle circumscribing the ellipse is parallel to X-axis.
The tangent for #theta=270^@# represents the base of the isoscele triangle and its equation becomes

#(xcostheta)/a+(ysintheta)/b=1#

#=>(xcos270)/a+(ysin270)/b=1#

#=>0-y/b=1=>y=-b#

Now if we put #y=-b# in the equation of the tangent then we can get the abscissa of the corner points of the base #(BC)# of the triangle and this abscissa in the direction of postitive x-axis #(DC)# represents the half of the length of the base of the triangle.

#=>(xcostheta)/a+(-bsintheta)/b=1#
#=>x=(a(1+sintheta))/costheta#

Hence #color(red)(1/2"Base"=(a(1+sintheta))/costheta)#

It is obvious from the symmetry of the figure that ordinate of the point of intersection of the tangent with y-axis i.e. #x=0# is the distance of the top vertex from the center #(0,0)#

and then this ordinate

#OA= b/(sintheta)#

So height of the triangle

#color(magenta)("Height"=b+b/sintheta)#

#=(b(1+sintheta))/sintheta#

Hence area

#color(red)(S=1/2"Base"xx"Height)#

#=>S=(a(1+sintheta))/costhetaxx(b(1+sintheta))/sintheta#

#=>S=(a(1+sintheta))/costhetaxx(b(1+sintheta))/sintheta#

#=>S=ab(secthetacsctheta+2sectheta+tantheta)#

Now imposing the condition of minimization i.e.#(dS)/(d theta)=0# we get

So

#d/(d theta)((ab(secthetacsctheta+2sectheta+tantheta)))=0#

#=>d/(d theta)(secthetacsctheta+2sectheta+tantheta)=0#

#=>cscthetasecthetatantheta-secthetacscthetacottheta+2secthetatantheta+sec^2theta=0#

#=>sec^2theta-1/sin^2theta+2sintheta/cos^2theta+sec^2theta=0#

#=>2sec^2theta-1/sin^2theta+2sintheta/cos^2theta=0#

multiplying bothsides by #sin^2thetacos^2theta#

#=>2sin^2theta-cos^2theta+2sin^3theta=0#

#=>2sin^2theta-1+sin^2theta+2sin^3theta=0#

#=>2sin^3theta+3sin^2theta-1=0#

#=>2sin^3theta-sin^2theta+4sin^2theta-1=0#

#=>sin^2theta(2sintheta-1)+(2sintheta-1)(2sintheta+1)=0#

#=>(2sintheta-1)(sin^2theta+2sintheta+1)=0#

#=>(2sintheta-1)(sintheta+1)^2=0#

Hence #sintheta=1/2#

#sintheta=-1->"not possible"# as the area becomes zero then.

Inserting the value #sintheta=1/2# in the equation of Height and equation of Base we get

#"Height"=b+b/sintheta=b+b/(1/2)=3b#

#1/2"Base"=(a(1+sintheta))/costheta#

#=>"Base"=(2a(1+sintheta))/sqrt(1-sin^2theta)#

#=>"Base"=(2a(1+1/2))/sqrt(1-1/4)=(3a)/(sqrt3/2)=2asqrt3#

drawn

Given

#x-y=4....[1]# as one side of the triangle.

and

#x^2+4xy+y^2# ,the equation of a pair of straight line representing other two sides of the triangle.

Now

#x^2+4xy+y^2=0#

#=>x^2+2*x*2y+(2y)^2-(sqrt3y)^2=0#

#=>(x+2y)^2-(sqrt3y)^2=0#

#=>(x+2y+sqrt3y)(x+2y-sqrt3y)=0#

So

#x+(2+sqrt3)y=0......[2]#

and

#x+(2-sqrt3)y=0......[3]# are the equations of other two sides of the triangle.

Now

slope of straight line [1]

#m_1=1#

slope of straight line [2]

#m_2=-1/(2+sqrt3)=-(2-sqrt3)/((2+sqrt3)(2-sqrt3))#

#=-(2-sqrt3)/(4-3)=-(2-sqrt3)#

slope of straight line [3]

#m_3=-1/(2-sqrt3)=-(2+sqrt3)/((2+sqrt3)(2-sqrt3))#

#=-(2+sqrt3)/(4-3)=-(2+sqrt3)#

Angle between line [1] and [2]

#alpha=tan^-1((m_1-m_2)/(1+m_1xxm_2))#

#=tan^-1((1+2-sqrt3)/(1-(2-sqrt3)))#

#=tan^-1((3-sqrt3)/(sqrt3-1))#

#=tan^-1(sqrt3)=pi/3#

Angle between line [1] and [3]

#beta=tan^-1((m_3-m_1)/(1+m_1xxm_3))#

#=tan^-1((-(2+sqrt3)-1)/(1-(2+sqrt3)))#

#=tan^-1((-(3+sqrt3))/(-(sqrt3+1)))#

#=tan^-1(sqrt3)=pi/3#

So the two angles of the triangle are same and each equals to #pi/3#.

Hence the triangle is an equilateral triangle.

Answer:

shaded area #=6*(3sqrt3-pi) ~~ 12.33 " cm"^2#

Explanation:

enter image source here
See the figure above.

Green area #=#area of sector #DAF# - yellow area

As #CF and DF# are the radius of the quadrants,
#=> CF=DF=BC=CD=6#
#=> DeltaDFC# is equilateral.
#=> angleCDF=60^@#
#=> angleADF=30^@#
#=> EF=6sin60=6*sqrt3/2=3sqrt3#

Yellow area = area of sector #CDF- #area #DeltaCDF#
#= pi*6^2*60/360-1/2*3sqrt3*6#
#= 6pi-9sqrt3#

Green area = #=#area of sector #DAF# - yellow area
#=pi*6^2*30/360-(6pi-9sqrt3)#
#= 3pi-(6pi-9sqrt3)#
#= 9sqrt3-3pi#

Hence, the shaded area #A_s# in your figure #= 2xx# green area
#=> A_s=2*(9sqrt3-3pi)#
#=18sqrt3-6pi=6(3sqrt3-pi)~~12.33 " cm" ^2#

View more
Questions
Ask a question Filters
Loading...
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions