In the figure given below, the side of the square ABCD is 2 cm. E is the midpoint of AB and F is the midpoint of AD. G is a certain point on CF and 3CG＝2GF. What is the area of the shaded triangle BEG, in cm2?

Question

Sadie Allen · Accepted Answer

shaded area #=0.8 cm^2#

Given #3CG=2GF, => CG : GF = 2 : 3, => CG:CF=2:5#
#=> CG=2/5CF#

#=> CH=2/5DF=2/5xx1=0.4#

#=> HB=2-0.4=1.6#

#=> "shaded area"=1/2xxEBxxHB=1/2xx1xx1.6=0.8 cm^2#

Sadie Allen · Answer

#0.8 (cm)^2.#

Let #/_DFC=theta, and, /_GFE=/_CFE=phi.#

Clearly, we have, #CF=sqrt5, and, FE=sqrt2.#

#:.," from the right-"DeltaCDF, sintheta=2/sqrt5, costheta=1/sqrt5.#

Since, #CG:GF=2:3, CG=2/5*CF=2/5*sqrt5=2/sqrt5, GF=3/sqrt5.#

We will use, the Notation #[PQRS]# to denote the Area of a

Polygon #PQRS.#

Observe that,

#[GEB]+[BCG]+[CDF]+[FAE]+[EGF]=[ABCD]...(star)#

From Trigo., we know that, the Area of #DeltaLMN, i.e.,#

#[LMN]=1/2*LM*LN*sin/_MLN.#

#:. [BCG]=1/2*CG*CB*sin/_BCG#

#=1/2*2/sqrt5*2*sin(pi/2-/_FCD)#

#=2/sqrt5*sin/_DFC=2/sqrt5*sintheta=2/sqrt5*2/sqrt5#

#:.[BCG]=4/5..........(1).#

Clearly, #[CDF]=1, [FAE]=1/2, &, [ABCD]=4......(2).#

#[EGF]=1/2*FG*FE*sin/_GFE=1/2*3/sqrt5*sqrt2*sinphi.#

But, #/_AFE+/_GFE+/_DFC=pi#

#rArr pi/4+phi+theta=pi rArrphi=3pi/4-theta#

#rArr sinphi=sin(3pi/4-theta)=sin{pi-(pi/4+theta)}#

#=sin(pi/4+theta)#

#=1/sqrt2(costheta+sintheta)=1/sqrt2*3/sqrt5=3/sqrt10.#

#:. [EGF]=3/sqrt10*3/sqrt10=9/10..........(3).#

Finally, from #(1)-(3), and, (star),# we have,

#[GEB]=4-4/5-1-1/2-9/10=0.8" (cm)"^2#

Enjoy Maths.!

Allison Allen · Answer

#0.8 cm^2.#

As a Second Method, we will solve the Problem using the

Co-ordinate Geometry.

Let us take #A(0,0),# ray #AB# as the #+ve# direction of the

X-Axis, and, ray #AD# as the #+ve# direction of the Y-Axis.

Clearly, we have, #B(2,0), C(2,2), E(1,0), D(0,2) and F(0,1).#

Given that, #2GF=3CG, i.e., (FG)/(GC)=3/2, &, F(0,1), C(2,2).#

By the Section Formula, we get,

#G=G((3/2(2)+0)/(3/2+1),(3/2(2)+1)/(3/2+1))=G(6/5,8/5).#

Finally, with #G(6/5,8/5), E(1,0), &, B(2,0),# we have,

#"The Area of "DeltaBEG="1/2|d|, where,#

#d=|(2,0,1),(1,0,1),(6/5,8/5,1)|=2(0-8/5)-0+1(8/5-0)=-8/5.#

#"Hence, the Reqd. Area="1/2*8/5=4/5=0.8cm^2,# as before!

Enjoy Maths.!