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This is one of our many quizzes on Triangle similarity. Click on this to access more such competitive quizzes on Geometry.

Triangle similarity Quiz 2

mathematics

 In the given figure, if $AB=5cm,BC=12cm,DC=4cm$ and $\angle ABC=\angle CDE={{90}^{\circ }}$, then find $DE$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/616b9e27-a654-4a55-8344-52299810cd682456858047058870416.png"/>

 From the figure choose the triangles which are similar and then identify the relation between the sides. Using this, find the value of $DE$.

  From the figure , $\angle ABC=\angle CDE={{90}^{\circ }}$ and $C$ is the common vertex. $ \\ $ As per AA similarity criteria,  $ \\ \Delta ABC\sim \Delta EDC \\ \therefore \dfrac{AB}{ED}=\dfrac{BC}{DC}=\dfrac{AC}{EC} \\ $ Given , $AB=5cm,BC=12cm,DC=4cm$ $ \\ \dfrac{5}{ED}=\dfrac{12}{4} \\ \Rightarrow ED=\dfrac{5}{3}$

 In the given figure , $BC=12cm,AD=6cm,AC=13cm$$,\angle BED=\angle DFB=\angle FBE={{90}^{\circ }}$and $ED=x$. Find the value of $x$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7be52a2d-7a53-45a6-a4a6-e45065638c971634368133644487212.png"/>

 Identify the triangles which are similar and then use properties of similar triangles to find the required value of $x$.

  It is given that $\angle BED=\angle DFB=\angle FBE$ $ \\ $ $A$ is the common vertex for $\Delta ABC$ and $\Delta AED$ $ \\ $ As per AA similarity criteria,  $ \\ \Delta ABC\sim \Delta AED \\ \therefore \dfrac{AB}{AE}=\dfrac{BC}{ED}=\dfrac{AC}{AD} \\ \Rightarrow \dfrac{12}{x}=\dfrac{13}{6} \\ \Rightarrow x=\dfrac{72}{13}$

  In $\Delta PQR$ $PR=5.5cm,PQ=3.5cm,QR=4cm$ $,\angle Q={{80}^{\circ }},\angle P={{60}^{\circ }}$ and  In $\Delta ABC$ , $AC=11cm,AB=7cm,BC=8cm$ , find $\angle B$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1650f25e-2cf9-4e6d-a591-a188869ef5264865724964012188687.png"/>

 From the figure , find the similar triangles and by using the property of similar triangles then find the required angle.

From the figure, $ \\ \dfrac{PQ}{AB}=\dfrac{3.5}{7}=\dfrac{1}{2} \\ \dfrac{QR}{BC}=\dfrac{4}{8}=\dfrac{1}{2} \\ \dfrac{PR}{AC}=\dfrac{5.5}{11}=\dfrac{1}{2} \\ $ Hence , $\Delta PQR\sim \Delta ABC$ $ \\ $ By similarity property , $\angle Q=\angle B$ $ \\ $ Given that , $\angle Q={{80}^{\circ }} \\ \therefore \angle B={{80}^{\circ }}$

 In the given figure, $AB=10cm,AC=5cm,BD=a,CE=4cm$, then find the roots of the equation ${{x}^{2}}-ax+12=0$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/69bddd7a-b247-45f8-b47a-16a161cc58248407096889123566188.png"/>

 From the figure , find the value of $a$ using similarity of triangles and then find the roots of the equations after substituting the value of $a$.

 From the figure, $ \\ $ $\angle B=\angle C$ and $A$ is the common vertex. $ \\ $ As per AA similarity criteria,  $ \\ \Delta ABD \sim \Delta ACE \\ \Rightarrow \dfrac{AB}{AC}=\dfrac{BD}{CE}=\dfrac{AD}{AE} \\ $ Given that , $AB=10cm,AC=5cm,BD=a, CE=4cm$ $ \\ $ Hence, we get   $\dfrac{10}{5}=\dfrac{a}{4} \\ \Rightarrow a=8 \\ $ Equation obtained is ${{x}^{2}}-8x+12=0$ $ \\ $ This can be written as $\left( x-6 \right)\left( x-2 \right)=0$ $ \\ $ Hence, the roots are $x=6$ and $x=2$

 $ABCD$ is a rectangle with $PQ=6,OQ=2,OP=4,RS=2,OS=x$  and $OR=y$. Find $\dfrac{y}{x}$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c9eeae34-47ae-41c6-be16-3118d3909ace3561080008563461606.png"/>

 Use the property of a rectangle to find the similar triangles and then by using the property of similar triangles find the value of $x$ and $y$.

 As  $ABCD$ is a rectangle , $AB$ is parallel to $CD$. $ \\ $ Considering $SQ$ as transversal , $\angle RSO=\angle OQP$ since they are alternate interior angles. $ \\ $ Considering $RP$ as transversal, $\angle ORS=\angle OPQ$ since they are alternate interior angles. $ \\ $ Also , $\angle ROS=\angle POQ$ as they are vertically opposite angles. $ \\ $ Hence , $\Delta POQ\sim \Delta ROS$ $ \\ \therefore \dfrac{PO}{RO}=\dfrac{OQ}{OS}=\dfrac{PQ}{RS} \\ \Rightarrow \dfrac{4}{y}=\dfrac{6}{2}\Rightarrow y=\dfrac{4}{3}$ and $\dfrac{2}{x}=\dfrac{6}{2}\Rightarrow x=\dfrac{2}{3} \\ $ Hence , $\dfrac{y}{x}=2$