Rectangle in a Triangle

Let $x=BD$ and $y=BF$ .

By similarity of $\triangle EDC$ and $\triangle ABC$ , we have $\dfrac{y}{20-x}=\dfrac{10}{20}\implies x=2(10-y)$ .

We know that the area of $BDEF$ is less than or equal $32$ and $y>5$ .

The area of the rectangle $BDEF$ is $A=xy=2y(10-y)$ .

$2y(10-y)\leq 32$

$y^2-10y+16\geq 0$

$y^2-10y+25\geq 9$

$(y-5)^2\geq 3^2$

$\left| y-5\right| \geq 3$

Since $y>5$ , we have $y-5>0\implies \left| y-5\right| =y-5$ .

$y-5\geq 3$

$\boxed{y\geq 8}$

Therefore, the minimum value of $BF$ is $8$ .

The question says , that BF>5 and the area was > or =32 so it had to be 8 as none of the given options except 8 was a factor of 32.

let BF = y so FA = 10-y , let BD = x hence DC = 20-x triangles ECD and AEF are similar and right angled. Hence applying cosine ratio for respective angles we obtain y/(20-x) = 10-y/(x) , and solving this equation we obtain x+2y = 20. substituting the given values and checking for the rectangle area <= 32 we are left with 8 and 9 for y. The least among the given is 8 hence it is the answer

as EF is parallel to BC SO,AF/FB=AE/EC Also , triangle AFE and triangle ABC are similar(angle AFE=angle ABC, angle AEF=angle ACB,) SO,AF/AB=AE/AC=FE/BC Or, AR/10=CE/10 root 5=FE/20 Or, EF=2AF --------------------------------------------------------------------------------------------(1) Area of rectangle =FE x BF =less or equal 32 but ,2AFxBF=less or equal 32 Or, AFxBF = less or equal 16 As BF> 5 If we put BF=6 , get AF=4 , so the area BDEF=24 .>16 (can't possible) lly, put BF=7 , get AF=3 again the area BDEF=21>16(can't possible) lly, put BF=8 , get AF =2, we get the area BDEF=16(possible) lly, put BF=9 ,GET AF= 1,we get the area BDEF=9(also possible) so , from this solution we see the required value of BF must be greater than or equal 8

Area of Rectangle = BF x FE And, we know that, 10>BF>5. And rectangle is possible with value of BF ranging from 6 to 9. Putting Values of BF from 6 to 9, given area 32 is only divisible by 8. So, BF has a value of 8. Just use little common sense.

Solve by the option method.......

Since CED & EAF are similar So

CD/EF = ED/ AF

(20-BD)/BD = FB/(10-FB)

Putting Each Value Of FB from options we get different values of BD

Area = BD x FB is Less Than Or Equal to When FB = 9 or 8

So The Answer Is The Least Taht Is 8

If DE=BF = x then AF = 10-x; Also AEF is similar to ABC so EF = 2 (10-x) . So area of rectangle BDEF = 2 (10-x) x <=32; this meand 20x-2x^2 <= 32 => 2x^2-20x+32 >= 0 i.e. 2 (x-8)*(x-2) >= 0 x>=8 or x=< 2 , since x=< 2 is not acceptable (one of the other conditions is x > 5) x= 8 is minimum possible value that will satisfy all constraints.

You can use Hit and trial method. Best for beginners, at least. find the area of triangle ABC(1/2 X 20 X 10=100) and then, find the area of the rectangle BDEF( with dimensions such that its area <32) as: for example, first take BF=6, then take the value of EF= 5 as these values satisfy the area of BDEF to be <32(take the highest value of both these which satisfy ar(BDEF)<32 as 100 is a big value) then take BF= 7, then, BF=8 and so on..

now find the area of triangle AFE and triangle EDC with the formula 1/2 X b X h in AFE b=breadth of the rectangle BDEF h= 10 - taken value of AF( in BDEF) in EDC h= taken value of BF in BDEF, b= 20 - breadth of BDEF

now add ar(AFE), ar(EDC), ar(BDEF) while changing the value of BF and subsequently of FE and so on and when it equals 100, that's your answer.