Answer:
b=2.5
Step-by-step explanation:
The given line has equation: [tex]y=b-x[/tex]
The x-intercept is (b,0)
The y-intercept is (0,b)
The area of triangle QOP [tex]=\frac{1}{2} |OQ|*|OP|=[/tex]
[tex]=\frac{1}{2}b*b[/tex]
[tex]=\frac{1}{2}b^2[/tex]
The area of triangle QRS [tex]=\frac{1}{2} |RQ|*|RS|=[/tex]
[tex]=\frac{1}{2}(4-b)*(b-4)[/tex]
The area of triangle QRS and QOP are in the ratio 9:25
We must negate the area under the x-axis.
[tex]\implies \frac{- \frac{1}{2}(4-b)(b-4)}{\frac{1}{2}b^2} =\frac{9}{25}[/tex]
[tex]\implies \frac{-(4-b)(b-4)}{b^2} =\frac{9}{25}[/tex]
[tex]\implies {-(4-b)(b-4) =\frac{9}{25}b^2[/tex]
[tex]\implies b^2-8b+16=\frac{9}{25}b^2[/tex]
[tex]\implies 25b^2-200b+400=9b^2[/tex]
[tex]\implies 25b^2-9b^2-200b+400=0[/tex]
[tex]\implies 16b^2-200b+400=0[/tex]
[tex]\implies 2b^2-25b+50=0[/tex]
[tex](2b-5)(b-10)=0[/tex]
[tex]b=2.5\:or\:b=10[/tex]
But 0<b<4
Therefore b=2.5
VERIFY
The area of triangle QOP
[tex]=\frac{1}{2}*2.5*2.5=3.125[/tex]
The area of triangle QRS
[tex]=|\frac{1}{2}(4-2.5)*(2.5-4)|=|-1.125|=1.125[/tex]
We must take absolute value because this area is below the x-axis.
[tex]\frac{1.125}{3.125} =\frac{9}{25}[/tex]
