In the figure above the ratio of the area of WXZ to the area of WYZ is 7:2. If XY = 21, what is the length of segment WY?

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Because ∆XZW and WYZ is the same height and
Because ∆XZW and WYZ is the same height and the are of ∆WXZ to the the WYZ
[tex]so \:( \frac{1}{2}xw \times h)( \frac{1}{2} w)=7:2[/tex]
Because xy=21
[tex]so \: WY=2 \frac{1}{2}(7 + 2x2) \\ = 21 \frac{1}{2}p \times 2 \\ =\small\color{blue}{{{\boxed{\tt\red{} \:\:\:\:\:\:\:\:\:\: WY=\frac{14}{3}\:\:\:\: }}}}[/tex]
The length of the line segment WY is [tex]\frac{14}{3}[/tex] unit.
The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle.
Area of triangle = (1/2)base × height
According to the given question.
The ratio of the area of triangle WXZ to the area of triangle WYZ is 7:2.
Since, the height WZ for both the triangles WXZ and WYZ is same.
Let, WZ = h
Therefore, the ratio of the area of the triangles is given by
[tex]\frac{\frac{1}{2}WX(h) }{\frac{1}{2}(WY)(h) } =\frac{7}{2}[/tex]
⇒ [tex]\frac{WX}{WY} =\frac{7}{2}[/tex]
⇒ [tex]WX = \frac{7}{2} WY..(i)[/tex]
Since, in the given figure
[tex]XY = WY + WX[/tex]
⇒ [tex]21 = WY + \frac{7}{2} WY[/tex] (from i)
⇒ [tex]21 = \frac{9}{2}WY[/tex]
⇒ [tex]WY = \frac{14}{3}[/tex]
Hence, the length of the line segment WY is [tex]\frac{14}{3}[/tex] unit.
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