Triangles
Category : 10th Class
Triangles
Triangles
Snap Test
(a) \[F{{C}^{2}}\] (b) \[F{{E}^{2}}\]
(c) \[D{{C}^{2}}\] (d) \[D{{B}^{2}}\]
(e) None of these
Ans. (a)
Explanation: In \[\Delta \,ABC\], it is given that \[DE\,\,~\parallel \,\,AB\].
\[\therefore \,\,\,\,\,\frac{CD}{DA}=\frac{CE}{EB}\] …….. (i)
In \[\Delta \,CDB\], it is given that \[FE~\,\,\parallel \,\,DB\].
\[\therefore \,\,\,\,\,\frac{CF}{FD}=\frac{CE}{EB}\] …….. (ii)
From equations (i) and (ii), we get:
\[\frac{CD}{DA}=\frac{CF}{FD}\]
\[\Rightarrow \,\,\,\,\frac{DA}{CD}=\frac{FD}{CF}\Rightarrow \,\,\,\,\frac{DA}{CD}+1\,\,\,=\,\frac{FD}{CF}+1\,\,\Rightarrow \,\,\,\frac{DA+CD}{CD}=\frac{FD+CF}{CF}\]
\[\Rightarrow \,\,\,\,\,\frac{AC}{CD}=\frac{DC}{CF}\,\,\,\,\,\,\,\,\Rightarrow \,\,D{{C}^{2}}=CF\times AC\]
2. In the figure, AD is the bisector of \[\angle \,\mathbf{BAC}\]. If \[\mathbf{AB}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm}\], \[\mathbf{AC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{BC}\text{ }=\text{ }\mathbf{12}\text{ }\mathbf{cm}\], find BD and DC.
(a) 7.5 cm, 4.5 cm (b) 6.5 cm, 4.5 cm
(c) 7.5 cm, 2.5 cm (d) 7.5 cm, 3.5 cm
(e) None of these
Ans. (a)
Explanation: Let \[BD\text{ }=\text{ }x\text{ }cm\]. Then \[DC\text{ }=\text{ }BC\text{ }-\text{ }BD\text{ }=\text{ }\left( 12\text{ }-\text{ }x \right)\text{ }cm\].
In \[\Delta \,ABC\], AD is the bisector of \[\angle \,BAC\].
\[\therefore \,\,\,\,\,\frac{BD}{DC}\,\,=\,\,\frac{AB}{AC}\]
\[\Rightarrow \,\,\,\,\frac{x}{12-x}=\frac{10}{6}\,\,\,\Rightarrow \,\,6x=120\,\,-\,\,10x\]
\[\Rightarrow \,\,\,16x\,\,=\,\,120\,\,\,\Rightarrow \,\,\,x\,\,=\,\,7.5.\]
Hence, \[BD\text{ }=\text{ }x\text{ }cm\text{ }=\text{ }7.5\text{ }cm\] and \[DC\text{ }=\text{ }\left( 12\text{ }-\text{ }x \right)\text{ }cm\text{ }=\text{ }\left( 12\text{ }-\text{ }7.5 \right)\text{ }cm\text{ }=\text{ }4.5\text{ }cm\].
(a) 5.5 cm (b) 4.5 cm
(c) 6.5 cm (d) 7.5 cm
(e) None of these
Ans. (c)
Explanation: We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
\[\therefore \,\,\text{ }\Delta \,ABC\sim \Delta \,DEF\]
\[\Rightarrow \,\,\,\,\,\,\frac{Area\,(\Delta ABC)}{Area\,(\Delta DEF)}\,\,=\,\,\frac{B{{C}^{2}}}{E{{F}^{2}}}\]
\[\Rightarrow \,\,\,\,\frac{64}{169}=\left( \frac{B{{C}^{2}}}{E{{F}^{2}}} \right)\,\,\,\,\Rightarrow \,\,\,{{\left( \frac{8}{13} \right)}^{2}}={{\left( \frac{BC}{EF} \right)}^{2}}\Rightarrow \,\,\,\,\frac{BC}{EF}=\frac{8}{13}\]
\[\Rightarrow \,\,\,\,EF\,\,=\,\,\frac{13}{8}\times BC=\left( \frac{13}{8}\times 4 \right)\,cm\,\,=\,\,6.5\,cm\]
Determine AC and BC.
(a) 6.2 cm, 5 cm (b) 7.2 cm, 9 cm
(c) 5.2 cm, 9 cm (d) 6.2 cm, 8 cm
(e) None of these
Ans. (b)
Explanation: In \[\Delta \,ADE\text{ }and\,\,\Delta \,ABC\]:
\[\angle \,A\text{ }=\angle \,A\] (common)
\[\Delta \,ADE\text{ }-\Delta \,ABC\]
(corresponding angles since \[DE\,\,\parallel \,\,~BC\])
\[\Delta \,ADE\text{ }\tilde{\ }\Delta \,ABC\] (AA Similarity)
So, \[\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}\,\,\,\,\,\,\] ……. (i)
From equation (i):
\[\frac{AD}{AB}=\frac{AE}{AC}\,\,\,\Rightarrow \,\,\,\frac{2}{4.5}\,\,=\,\,\frac{3.2}{AC}\,\,\Rightarrow \,\,AC=\,\,\left( \frac{3.2\times 4.5}{2} \right)\,cm\,\,=\,\,7.2\,cm\]
Again, form equation (i):
\[\frac{AD}{AB}=\frac{DE}{BC}\,\,\,\Rightarrow \,\,\,\frac{2}{4.5}\,\,=\,\,\frac{4}{BC}\,\,\Rightarrow \,\,BC=\,\,\left( \frac{4\times 4.5}{2} \right)\,cm\,\,=\,\,9\,cm\]
(a) 24 cm (b) 20 cm
(c) 15 cm (d) 25 cm
(e) None of these
Ans. (a)
Explanation: In the following figure, Let \[\Delta \,ABC\] be an isosceles triangle having \[AB\text{ }=\text{ }AC\text{ }=\text{ }13\text{ }cm\]. Let \[AD\bot BC\] and \[AD\text{ }=\text{ }5\text{ }cm\].
In \[\Delta \,ABD\]:
\[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}},\,\,\,B{{D}^{2}}=A{{B}^{2}}-A{{D}^{2}}\]
\[\Rightarrow \,\,\,BD\,\,\,=\,\,\sqrt{A{{B}^{2}}-A{{D}^{2}}}=\,\,\sqrt{{{13}^{2}}-{{5}^{2}}}=\,\,\sqrt{169-25}\,\,=\,\,\sqrt{144}\,\,=\,\,12\,cm\]
Similarly, \[DC\text{ }=\text{ }12\text{ }cm\].
\[\therefore \] \[BC\text{ }=\text{ }BD\text{ }+\text{ }DC\text{ }=\text{ }\left( 12\text{ }+\text{ }12 \right)\text{ }cm\text{ }=\text{ }24\text{ }cm\].
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