Question

The line $y = b-x$ with $0 < b < 4$ intersects the $y$-axis at $P$ and the line $x=4$ at $S$. If the ratio of the area of triangle $QRS$ to the area of triangle $QOP$ is 9:25, what is the value of $b$? Express the answer as a decimal to the nearest tenth.

Answer 1

Answer:

b=2.5

Step-by-step explanation:

The given line has equation: $y=b-x$

The x-intercept is (b,0)

The y-intercept is (0,b)

The area of triangle QOP $=\frac{1}{2} |OQ|*|OP|=$

$=\frac{1}{2}b*b$

$=\frac{1}{2}b^2$

The area of triangle QRS $=\frac{1}{2} |RQ|*|RS|=$

$=\frac{1}{2}(4-b)*(b-4)$

The area of triangle QRS and QOP are in the ratio 9:25

We must negate the area under the x-axis.

$\implies \frac{- \frac{1}{2}(4-b)(b-4)}{\frac{1}{2}b^2} =\frac{9}{25}$

$\implies \frac{-(4-b)(b-4)}{b^2} =\frac{9}{25}$

$\implies {-(4-b)(b-4) =\frac{9}{25}b^2$

$\implies b^2-8b+16=\frac{9}{25}b^2$

$\implies 25b^2-200b+400=9b^2$

$\implies 25b^2-9b^2-200b+400=0$

$\implies 16b^2-200b+400=0$

$\implies 2b^2-25b+50=0$

$(2b-5)(b-10)=0$

$b=2.5\:or\:b=10$

But 0<b<4

Therefore b=2.5

VERIFY

The area of triangle QOP

$=\frac{1}{2}*2.5*2.5=3.125$

The area of triangle QRS

$=|\frac{1}{2}(4-2.5)*(2.5-4)|=|-1.125|=1.125$

We must take absolute value because this area is below the x-axis.

$\frac{1.125}{3.125} =\frac{9}{25}$

Answer 2

Answer:

b=2.5

Step-by-step explanation:

The line y=b-x intersects the x-axis at the point where 0 = b-x, or x=b. So, we seek the x-coordinate of point Q.

Since the y-axis is parallel to the line x = 4, we see that angle QSR = angle QPO. Also QOP = QRS = 90 degrees. Thus triangle QOP is similiar to triangle QRS, so [QRS]/[QOP] =(QR/QO)^2, which means we have (QR/QO)^2/9/25, so QR/QO= 3/5. Since QR + QO= 4, we have 3/5(QO + QO) = 4, and QO =4 * 5/8= 5/2. Therefore, the x-coordinate of Q is 5/2 = 2.5