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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles1 Mark
Question
Match the following columns.
 
Column $I$
 
Column $II$
$(a)$
In a given $\triangle\text{ABC},\text{DE }||\text{ BC}$ and $\frac{\text{AD}}{\text{DB}}=\frac{3}{5}.$ If $AC = 5.6\ cm$ then $\text{AE} = ....cm.$
 $(p)$ $6$
$(b)$
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $2\text{AB = 3DE}$ and $\text{BC} = 6\ cm$ then $\text{EF} = ....cm.$
 $(q)$ $4$
$(c)$
If $\triangle\text{ABC}\sim\triangle\text{PQR}$ such that $\text{ar}(\triangle\text{ABC}):\text{ar}(\triangle\text{PQR})=9:16$ and $\text{BC} = 4.5\ cm$ then $\text{QR} ...cm.$
 $(r)$ $3$
$(d)$
In the given figure, $\ce{AB \| CD}$ and $\ce{OA = (2x + 4)cm, OB = (9x - 21)cm, OC = (2x - 1)cm}$ and $\text{OD} = 3cm.$ Then $x = ?$
 $(s)$ $2.1$
The correct answer is:
$a............$
$b............$
$c............$
$d............$
✓

Answer

 
Column $I$
 
Column $II$
$(a)$
In a given $\triangle\text{ABC},\text{DE }||\text{ BC}$ and $\frac{\text{AD}}{\text{DB}}=\frac{3}{5}.$ If $\text{AC} = 5.6\ cm$ then $\text{AE} = ....cm.$
 $(s)$ $2.1$
$(b)$
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{2AB = 3DE}$ and $\text{BC} = 6\ cm$ then $\text{EF} = ....cm.$
 $(q)$ $4$
$(c)$
If $\triangle\text{ABC}\sim\triangle\text{PQR}$ such that $\text{ar}(\triangle\text{ABC}):\text{ar}(\triangle\text{PQR})=9:16$ and $\text{BC} = 4.5\ cm$ then $\text{QR} ...cm.$
 $(p)$ $6$
$(d)$
In the given figure, $\ce{AB\| CD}$ and $\text{OA} = (2x + 4)cm, \text{OB} = (9x - 21)cm, \text{OC} = (2x - 1)cm$ and $\text{OD} = 3cm.$ Then $x = ?$
 $(s)$ $3$

In a given $\triangle\text{ABC},\text{DE }||\text{ BC}$
By the Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{3}{5}=\frac{\text{AE}}{5.6-\text{AE}}$
$\Rightarrow\text{AE}=2.1\text{cm}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{3}{2}=\frac{\text{6}}{\text{EF}}$
$\Rightarrow\text{EF}=4\text{cm}$
$\triangle\text{ABC}\sim\triangle\text{PQR}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\Rightarrow\frac{9}{16}=\frac{\text{4.5}^2}{\text{QR}^2}$
$\Rightarrow\text{QR}^2=36$
$\Rightarrow\text{QR}=6\text{cm}$
Since $\ce{AB \| CD}$, the quadrilateral is a trapzium.
We know that,
diagonals of a trapzium divede each propertionally.
$\Rightarrow\frac{\text{OA}}{\text{OB}}=\frac{\text{OC}}{\text{OC}}$
$\Rightarrow\frac{2\text{x}+4}{9\text{x}-21}=\frac{2\text{x}-1}{3}$
$\Rightarrow6\text{x}+12=18\text{x}^2-51\text{x}+21=0$
$\Rightarrow18\text{x}^2-51\text{x}+9=0$
$\Rightarrow\text{x}=3=0 $ or $\text{x}=\frac{1}{6}$
If $\text{x}=\frac{1}{6},$ then $\text{OC}=2\text{x}-1=2\Big(\frac{1}{6}\Big)-1<0$
This is not possible since lenght cannot be nagetive.

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