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This is the second of our many quizzes on Congruence. Click on this to access more such competitive quizzes on Geometry.

Congruence Quiz 2

mathematics

  If $\Delta \text{ABC}\cong \Delta \text{PQR}$ and it follows $\angle A={{60}^{\circ}}$ and $\angle B={{70}^{\circ}}$. Then find $\angle R$?

  By applying the basic formula that sum of all angles of a triangle are 180 and then we will find the angle A which is congruent to angle R. Hence in that way we will find the angle R.

$ \text{Given }\angle A=60,\angle B=70\text{ and therefore }\angle C=180-60-70=50 \\ \text{As }\Delta \text{ABC}\cong \Delta \text{PQR so }\angle R={{50}^{\circ}} \\ $ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b3333b03-fd3d-40d5-9747-d1a13c8503757177043796591428084.png"/>

 Which of the following is not a congruence rule?

 Option (A) , (B) and (C) are rules that are really possible and the fourth option is the case where congruent triangles are possible or may not.

 $ \text{In the option }\left( D \right)\text{ we can see AAA congruence which is not possible}\text{.} \\ \text{So, it is the odd one}\text{.}$

Let us consider an equilateral triangle PQR and A is the point which is the interior point for the triangle PQR. If PA=PQ and point B is also given so that $\angle AQB=\angle AQR$ and $QB=PR$. Then find $\angle B$ ? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f835e4de-2d07-440d-9917-8258237d0bf67293030263937172114.png"/>

 First of all conceptualize the question and the extent AR and by SSS congruence rule we can find our angle B.

$ \text{Draw AR} \\ \text{PR=RQ} \\ \text{AR=AR} \\ \text{PA=AQ} \\ \text{By the SSS congruence , }\Delta \text{PAR}\cong \Delta \text{PAQ} \\ \text{Therefore, }\left| \!{\underline {\, &nbsp;&nbsp;\text{PRA} \,}} \right. =\left| \!{\underline {\, &nbsp;&nbsp;\text{QRA} \,}} \right. ={{30}^{\circ}} \\ \text{Since PQ=QB+QR} \\ \Rightarrow \left| \!{\underline {\, &nbsp;&nbsp;\text{AQB} \,}} \right. =\left| \!{\underline {\, &nbsp;&nbsp;\text{AQR} \,}} \right. \\ \Rightarrow \text{AQ=AQ} \\ \text{Therefore, }\Delta \text{QAB}\cong \Delta \text{QAR} \\ \text{Thus }\left| \!{\underline {\, &nbsp;&nbsp;\text{B} \,}} \right. ={{30}^{\circ}} $ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/149cb528-36ec-44fb-8163-9beb86cb09996178743160690946023.png"/>

In an isosceles triangle PQR of $\left| \!{\underline {\, &nbsp;&nbsp;P \,}} \right. ={{90}^{\circ}}$ and $\overrightarrow{\text{PQ}}=\overrightarrow{\text{PR}}$. If D is the interior point of the triangle PQR so that $\overrightarrow{\text{PQ}}=\overrightarrow{\text{QA}}$and$\left| \!{\underline {\, &nbsp;&nbsp;\text{PQA} \,}} \right. ={{90}^{\circ}}$. Statement: $\overrightarrow{\text{PA}}=\overrightarrow{\text{PR}}$<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/159a09cb-ee30-4715-8be7-ee44afd8bbea5569844076286850531.png"/>

 Construct $\Delta \text{P}{{\text{Q}}^{1}}\text{D}$ which is congruent to $\Delta \text{PQD}$. 

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/89234c30-cc3f-4e55-a294-f3ebb16a2d348942166037757049529.png"/> $ \text{Since PQ=DQ, }\left| \!{\underline {\, \text{PQD} \,}} \right. =\left| \!{\underline {\, \text{QDP} \,}} \right. .\text{ Then it follows that }\left| \!{\underline {\, \text{QPD} \,}} \right. ={{75}^{\circ}}\text{ since }\left| \!{\underline {\, \text{PQD} \,}} \right. ={{30}^{\circ}}. \\ \text{Since }\Delta \text{P}{{\text{Q}}^{1}}\text{D has been constructed congruent to }\Delta \text{PQD, }\left| \!{\underline {\, {{\text{Q}}^{1}}\text{PD} \,}} \right. ={{75}^{\circ}}\text{ as well}\text{.Then }\left| \!{\underline {\, \text{DPR} \,}} \right. ={{90}^{\circ}}\text{-7}{{\text{5}}^{\circ}}\text{=1}{{\text{5}}^{\circ}}\text{.} \\ \Rightarrow \left| \!{\underline {\, {{\text{Q}}^{1}}\text{PR} \,}} \right. \text{=}\left| \!{\underline {\, {{\text{Q}}^{1}}\text{PD} \,}} \right. -\left| \!{\underline {\, \text{DPR} \,}} \right. ={{60}^{0}}.\text{ Since it gives that PQ=PR and PQ=P}{{\text{Q}}^{1}}\text{ }\text{by construction}\text{.} \\ \text{It follows that P}{{\text{Q}}^{1}}=\text{PR}\text{. Therefore }\Delta \text{P}{{\text{Q}}^{1}}\text{R is equilateral}\text{.} \\ \text{Since }\left| \!{\underline {\, \text{P}{{\text{Q}}^{1}}\text{D} \,}} \right. =\left| \!{\underline {\, \text{PQD} \,}} \right. ={{30}^{\circ}},\text{ }{{\text{Q}}^{1}}\text{D must be perpendicular bisector of PR}\text{.} \\ \text{Therefore, P}{{\text{Q}}^{1}}\text{RD is a kite and PD}=\text{RD}\text{.} $

 Let us consider a triangle $\Delta \text{PQR}$. Such that $\text{PA}$ is perpendicular to $\text{QR}$at $\text{B}$ so that $AR=4$as it is shown in the figure. If $\angle PQR={{45}^{\circ}}$. Then find the area of triangle PQR?

 Construct right angled triangles $\Delta \text{PQ}{{\text{B}}_{1}}$ and $\Delta \text{PQ}{{\text{B}}_{2}}$ in which both are congruent to $\Delta \text{PQB}$. Meet the ${{\text{B}}_{1}}\text{P}$ and ${{\text{B}}_{2}}\text{P}$ at ${{\text{Q}}_{1}}$ and then by doing some calculations we will get area of triangle $\Delta \text{PQR}$.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/244d6d6a-0d4d-4d44-9812-b09d99b7b11c8138886923301237847.png"/> $ \\ \text{Construct right angled triangle }\Delta \text{PQ}{{\text{B}}_{1}}\text{ is congruent to }\Delta \text{PQB}\text{. } \\ \text{In like manner, construct a right angled triangle }\Delta \text{PQ}{{\text{B}}_{2}}\text{ is congruent to }\Delta \text{PQB}\text{.} \\ \text{We then }{{\text{B}}_{1}}\text{P and }{{\text{B}}_{2}}\text{P to meet }{{\text{Q}}_{1}}. \\ \text{Since }\Delta \text{PQ}{{\text{B}}_{1}}\cong \Delta \text{PQB and }\Delta \text{PQ}{{\text{B}}_{2}}\cong \Delta \text{RQB, }\angle \text{PQ}{{\text{B}}_{1}}+\angle \text{RQ}{{\text{B}}_{2}}=\angle \text{PQB}+\angle \text{RQB}={{45}^{\circ}}. \\ \text{Therefore, }\angle {{\text{B}}_{1}}\text{Q}{{\text{B}}_{2}}={{90}^{\circ}}. \\ \text{Since }\angle {{\text{B}}_{1}}=\angle {{\text{B}}_{2}}={{90}^{\circ}}\text{ and the sum of the angles of quadrilateral Q}{{\text{B}}_{1}}{{\text{Q}}_{1}}{{\text{B}}_{2}}\text{ is 36}{{\text{0}}^{\circ}}, \\ \text{it follows }\angle {{\text{Q}}_{1}}={{90}^{\circ}}\text{ and that is a rectangle}\text{.} \\ \text{Since QB}=\text{Q}{{\text{B}}_{1}}=\text{Q}{{\text{B}}_{2}},\text{ the rectangle is also a square}\text{. Therefore QB}=\text{Q}{{\text{B}}_{1}}=\text{Q}{{\text{B}}_{2}}\text{ as well}\text{.} \\ \text{Now P}{{\text{B}}_{1}}=\text{PB}=2\text{ so that }{{\text{Q}}_{1}}\text{P}=\text{Q}{{\text{B}}_{1}}-\text{P}{{\text{B}}_{1}}=\text{QB}-2.\text{ Similarly, }{{\text{Q}}_{1}}\text{C}=\text{QB-3} \\ \text{Since }\Delta \text{P}{{\text{Q}}_{1}}\text{C is a right triangle, }{{\text{Q}}_{1}}{{\text{P}}^{2}}+{{\text{Q}}_{1}}{{\text{R}}^{2}}=\text{P}{{\text{R}}^{2}}={{\left( \text{PB+BR} \right)}^{2}}=25 \\ \text{Therefore} \\ \text{The area of }\Delta \text{PQR is }\dfrac{1}{2}\left( 5 \right)\left( 6 \right)=15 $