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Question: In the rectangle \(ABCD\), what is the area of \(\Delta BEF\) in terms of \(x,y,z\) ? ![](https://...

In the rectangle ABCDABCD, what is the area of ΔBEF\Delta BEF in terms of x,y,zx,y,z ?

Explanation

Solution

First, we will find the length of all the sides and then compute the area of all triangles and the rectangle ABCDABCD , by solving all these equations and areas we will get the area of ΔBEF\Delta BEF in terms of x,y,zx,y,z .

Complete answer:
Let us assume the length of sides AB=a,BC=b,AF=c,CE=dAB = a,BC = b,AF = c,CE = d respectively.

Area of triangle = 12×base×height\dfrac{1}{2} \times base \times height
Consider ΔABF\Delta ABF,
\Rightarrow Area of ΔABF\Delta ABF = yy …. (Given)
12×a×c=y\dfrac{1}{2} \times a \times c = y
c=2yac = \dfrac{{2y}}{a}
Consider ΔBCE\Delta BCE,
Area of ΔBCE\Delta BCE = xx …. (Given)
12×b×d=x\dfrac{1}{2} \times b \times d = x
d=2xbd = \dfrac{{2x}}{b}
side BC=side AD=bside{\text{ }}BC = side{\text{ }}AD = b …. (Opposite side of rectangle are congruent)
side AB=side CD=aside{\text{ }}AB = side{\text{ }}CD = a …. (Opposite side of rectangle are congruent)
\Rightarrow sideAD=sideAF+sideFDsideAD = sideAF + sideFD …. (AFDA - F - D)
b=c+sideFDb = c + sideFD
sideFD=bc=b2ya\therefore sideFD = b - c = b - \dfrac{{2y}}{a}
\Rightarrow side  CD=side  CE+side  EDside\;CD = side\;CE + side\;ED …. (CEDC - E - D)
a=d+sideEDa = d + sideED
sideED=ad=a2xb\therefore sideED = a - d = a - \dfrac{{2x}}{b}
Consider ΔDEF\Delta DEF,
Area of ΔDEF\Delta DEF = zz …. (Given)
\Rightarrow 12×(b2ya)×(a2xb)=z\dfrac{1}{2} \times \left( {b - \dfrac{{2y}}{a}} \right) \times \left( {a - \dfrac{{2x}}{b}} \right) = z
\Rightarrow ab2x2y+4xyab=2zab - 2x - 2y + \dfrac{{4xy}}{{ab}} = 2z
\Rightarrow ab2x2y2z+4xyab=0ab - 2x - 2y - 2z + \dfrac{{4xy}}{{ab}} = 0
Multiplying both sides by abab we get,
\Rightarrow (ab)22(ab)(x+y+z)+4xy=0{(ab)^2} - 2(ab)(x + y + z) + 4xy = 0
This equation is a quadratic equation of form ax2+bx+c=0a{x^2} + bx + c = 0 and the roots of the equation are given by x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Here, x=ab,a=1,b=2(x+y+z),c=4xyx = ab,a = 1,b = - 2(x + y + z),c = 4xy
ab=(2(x+y+z))±(2(x+y+z))24(1)(4xy)2(1)\therefore ab = \dfrac{{ - ( - 2(x + y + z)) \pm \sqrt {{{( - 2(x + y + z))}^2} - 4(1)(4xy)} }}{{2(1)}}
ab=2(x+y+z)±4(x+y+z)216xy2\therefore ab = \dfrac{{2(x + y + z) \pm \sqrt {4{{(x + y + z)}^2} - 16xy} }}{2}
ab=2(x+y+z)±4[(x+y+z)24xy]2\therefore ab = \dfrac{{2(x + y + z) \pm \sqrt {4[{{(x + y + z)}^2} - 4xy]} }}{2}
Taking 4 common and inside the radical symbol and taking it out the radical sign we have,
ab=2(x+y+z)±2(x+y+z)24xy2\therefore ab = \dfrac{{2(x + y + z) \pm 2\sqrt {{{(x + y + z)}^2} - 4xy} }}{2}
Taking 2 common in the numerator, we have:
ab=2[(x+y+z)±(x+y+z)24xy]2\therefore ab = \dfrac{{2[(x + y + z) \pm \sqrt {{{(x + y + z)}^2} - 4xy} ]}}{2}
cancelling 2, then we have,
ab=(x+y+z)±(x+y+z)24xy\therefore ab = (x + y + z) \pm \sqrt {{{(x + y + z)}^2} - 4xy}
The area of rectangle ABCDABCD = base×height=a×b=abbase \times height = a \times b = ab
Substituting the value of abab which we got by solving the quadratic equation, we get,
Area of rectangle ABCDABCD = base×height=a×b=ab=(x+y+z)±(x+y+z)24xybase \times height = a \times b = ab = (x + y + z) \pm \sqrt {{{(x + y + z)}^2} - 4xy}
Consider the area of ΔBEF=ω\Delta BEF = \omega ,
Therefore, we can say
Area of rectangle ABCDABCD = area of ΔABF\Delta ABF+ area of ΔBCE\Delta BCE + area of ΔDEF\Delta DEF+ area of ΔBEF\Delta BEF
Substituting all the values we got into this equation we get,
\Rightarrow ab=x+y+z+ωab = x + y + z + \omega
(x+y+z)±(x+y+z)24xy=(x+y+z)+w\therefore (x + y + z) \pm \sqrt {{{(x + y + z)}^2} - 4xy} = (x + y + z) + w
Since, abab is the area of entire rectangle, it is larger than (x+y+z)(x + y + z) and hence we can reject the negative radical and consider only positive radical,
\Rightarrow w=(x+y+z)24xyw = \sqrt {{{(x + y + z)}^2} - 4xy}
\Rightarrow ΔBEF=ω=(x+y+z)24xy\Delta BEF = \omega = \sqrt {{{(x + y + z)}^2} - 4xy}
Therefore, the area of ΔBEF\Delta BEF in terms of x,y,zx,y,z is (x+y+z)24xy\sqrt {{{(x + y + z)}^2} - 4xy} .

Note:
Consider the sides of the rectangle and triangle as some constants and then form equations and find the solution of those equations. The area of a triangle inside a rectangle always has an area less than that of the rectangle.