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

Triangle properties Quiz 2

mathematics

 An ornament has the following details. ABC is an equilateral triangle. $P, Q,$ and $R$ are the midpoint. Find the ratio of OR and ${O}'R$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3294e3ab-87e6-44b4-b524-49012207217b7377896149357096477.png"/>

  Simplify the fig. by drawing the required parts.  For two congruent triangles, every side $($median, angle bisector, altitude$)$, area, perimeter are equal.  

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f1d90277-6a09-4fd1-bbc3-d8b49a86444a8120838307928661048.png"/> $ \\ $ $\mathsf{\Delta ABC}$ is an equilateral triangle. $ \\ $ Thus, $AB=BC=CA$ $ \\ $ As, $P, Q, R$ are the midpoints of the sides, $ \\ $ Therefore, $AP=PB=BR=RC=QC=AQ=\dfrac{\text{Length of the Side of the triangle }}{\mathsf{2}} $ $ \\ $ In $\Delta PBR$, $ \\ $ $PB=BR$ and $\angle PBR=60{}^\circ$ $ \\ $ So, $\angle BPR=\angle PRB\\ \Rightarrow \angle BPR+\angle PRB+\angle PBR=180{}^\circ  \\ \Rightarrow \angle BPR+\angle BPR+60{}^\circ =180{}^\circ  \\ \Rightarrow 2\angle BPR=120{}^\circ  \\ \Rightarrow \angle BPR=60{}^\circ =\angle PRB \\ \text{As, }\angle PRB=\angle PBR=\angle BPR=60{}^\circ \\ $ Thus, $\Delta PBR$ is an equilateral triangle.$ \\ $ Similarly, it can be proved that $\Delta RQC$ is an equilateral triangle.$ \\ $ As, $PB=QC$ $ \\ $ So, $\Delta PBR$ and $\Delta RQC$ are congruent and thus both the altitude OR and O’R are equal, i.e., $OR=O'R$ $ \\ $ Therefore, the ratio is $\\ \Rightarrow \dfrac{OR}{O'R}=\dfrac{OR}{OR}=\dfrac{1}{1}$

  Find out the least degree of rotation (in anti-clockwise direction) of the bigger circle, required to get symmetry in the whole fig. , assuming that there is definitely a line of symmetry present in it. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d20e9ea0-e574-492f-a323-4d36ef64f60b9047577284146274579.png"/>

 Symmetry is the position when both sides of a line of symmetry will have every length and angle equal.  Check the fig. by rotating the bigger circular base in an anticlockwise direction.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/688f2015-faa8-4789-bccd-e8898137033a7424471054143556743.png"/> $ \\ $ From the above fig., it is evident that on rotating the middle circular base at an angle of 90 degree anticlockwise, the total ornament comes into symmetry.

 If the fig. has a line of symmetry when a bigger circle rotated anticlockwise, please signify the type of triangle BZC, when $BZ=BC$? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6c7de58a-7820-46c8-8bae-b7afaeff76453870678959615711157.png"/>

 Find out the line of symmetry.  Line of symmetry means both sides of the line will have equal measurements. After finding it, analyse the triangle $\Delta CBZ$.

 To attain symmetry, the ornament is supposed to be rotated at an angle of ${{90}^{\circ }}$ anticlockwise. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3c7083b6-b87e-4c32-9520-8b36c9f44fb72046542987721711775.png"/> $ \\ $ As, the figure is in symmetry, $CZ=BZ$ $ \\ $ It is given that, $BZ=BC$ $ \\ $ So, $CZ=BZ=BC$ $ \\ $ Thus, $\Delta CBZ$ is an equilateral triangle. 

 In an equilateral triangle ABC, from A, the angle bisector is drawn which meets BC at point P. From A, an altitude is dropped on BC at point Q. The midpoint of the side BC is marked as R. Then, find the value of: $CQ+BP+BR+BQ$

 In an equilateral triangle, median, angle bisector, altitude dropped from one vertex meets the other side at the same point. Thus, all the points are actually not different points but the same point on the line BC.

 In an equilateral triangle, median, angle bisector, altitude dropped from one vertex meets the other side at the same point. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0e220e3d-5f7e-4005-95c8-35676ebbd8ad7856589149513084578.png"/> $ \\ $ When the median, altitude, angle bisector were drawn, we find that the point P, Q, R coincide at the same point. Thus, point $P=\text{ }point\text{ }Q=\text{ }point\text{ }R\\ CQ+BQ+BR+BR \\ =BC+2BR \\ =BC+2\times \dfrac{BC}{2} \\ =BC+BC \\ =2BC \\ \text{We know that, }AB=BC=AC\\ \text{So, }2BC=2AB$

 The triangle ABC has $AB=AC\text{ and }\angle \text{A=60}{}^\circ $. P, Q are midpoints of AC and AB respectively. Find the ratio of $\dfrac{ar(\Delta PQR)}{ar(\Delta ABC)}$   $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e23ec3dd-ec26-45a8-bd04-e8a70f9de09f7984325411075198176.png"/>

  Midpoints of the triangle divide the bigger into equal triangles when joined together.  Equal triangle or congruent triangles always have the same area.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1203e1ab-be2b-485c-816f-494bad092ced5861086402056683650.png"/> $ \\ $ Simplify the figure. $ \\ \text{As, }AB=AC\text{ and }\angle \text{A=60}{}^\circ \\ \text{So,}\\ \angle B=\angle C \\ \angle A+\angle B+\angle C=180{}^\circ  \\ 60{}^\circ +\angle B+\angle B=180{}^\circ  \\ \angle B=\angle C=60{}^\circ  \\ $ Therefore, $\Delta ABC$ is an equilateral triangle. $ \\ $ P, Q, R are the midpoints of the triangle. $\\ $ So, $AP=AQ=BP=BR=CR=CQ$ $\\ $ As, $\angle C=60{}^\circ$ and $QC=CR$, so, $\Delta PBR$ is an equilateral triangle. $ \\ $ Similarly, $\Delta APQ$ and  $\Delta {QRB}$ are equilateral triangle.$ \\ $ So, $ QC=PR$ as $\Delta {PBR}$ is an equilateral triangle $\\ $ $QC=QR$ as $\Delta {QRB }$ is an equilateral triangle $ \\ $ $AP=PQ$ as $\Delta {PAQ }$ is an equilateral triangle ${PQ=PR=QR}$ $ \\ $ So, $\Delta PQR$ is also an equilateral triangle. $\\ $ Therefore, the bigger triangle is divided in 4 equal area triangles.$ \\ $ Area of $\Delta ABC=Ar.(\Delta APQ+\Delta PQR+\Delta PBR+\Delta QRC)$ $ \\ \Rightarrow Ar.(\Delta ABC)=Ar.(\Delta PQR+\Delta PQR+\Delta PQR+\Delta PQR) \\ \Rightarrow Ar.(\Delta ABC)=4\times Ar.(\Delta PQR) \\ \Rightarrow \dfrac{Ar.(\Delta PQR)}{Ar.(\Delta ABC)}=\dfrac{1}{4} $