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This is one of our many quizzes on Congruent triangle proofs. Click on this to access more such competitive quizzes on Geometry.

Congruent triangle proofs Quiz 2

mathematics

 In a viva, the examiner gives four options $($mentioned below$)$ to Ram, and asks which of the following options states that triangles are congruent? What Ram should answer from the options to answer the question, correctly?

 For two triangles to be congruent, recall the fact that the same area, same perimeter and same corresponding angles is applicable on similar triangles which can be and cannot be congruent.

 If the perimeter, the area and the corresponding angles of two triangles are equal, then it doesn’t fully imply that those two triangles are congruent. $ \\ $ For the above statement to be true, their sides must be equal to. But, if the ratio of the corresponding sides of two triangles is $1:1$, then this implies that those two triangles are congruent.

 If triangles $\Delta \text{ABC}$  and $\Delta \text{PQR}$ are congruent, the what is the value of $x$? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/536e8ede-f397-4f60-a19b-6d46796ad7bf2215560250766402296.png"/>

 Recall the fact that, perimeters of two congruent triangles are always the same.

  Triangles $\Delta{ABC}$ and $\Delta {PQR}$ are congruent, then by SSS criterion for congruence of triangles, corresponding sides of the triangle will be the same. $ \\ $ Also, their perimeter will be the same. $ \\ \Rightarrow \left( 2x+1 \right)+\left( 2x-3 \right)+\left( x+1 \right)=\left( 3x-4 \right)+(x+2)+(2x-4) \\  \Rightarrow 5x-1=6x-6 \\  \Rightarrow x=5 \\ $ 

 If $y-x=6$  and for values of $x$ and $y$, triangles $\Delta \text{ABC}$  and $\Delta \text{PQR}$ are congruent, then, value of ${{x}^{2}}-{{y}^{2}}$? $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/33a8d677-bf0a-4ac8-8c44-3971f11863a1353078663638403985.png"/>

 Corresponding angles of congruent triangles are always the same. Using fact that, sum of all angles of triangle is equal to $\text{180}{}^\circ $ evaluate value of $x+y$and then determine ${{x}^{2}}-{{y}^{2}}$.

  For, $x$ and $y$, $\Delta ABC$ and $\Delta PQR$ are congruent and $y-x=6$ $ \\ $ Then, corresponding angles of the triangle will be same, $ \\  \Rightarrow \angle RPQ=\angle CAB \\  \Rightarrow \angle RPQ=5y+10 \\  \text{and, } \\  \Rightarrow \angle QRP=\angle ACB \\  \Rightarrow \angle QRP=5x+10 \\ $ Sum of all angles in triangle is equal to $180{}^\circ$ $ \\ $ In $\Delta PQR$, $\angle PQR=90{}^\circ$  $ \\  \Rightarrow \angle QRP+\angle PQR+\angle RPQ=180{}^\circ  \\  \Rightarrow \left( 5x+10 \right){}^\circ +\left( 5y+10 \right){}^\circ +90{}^\circ =180{}^\circ  \\  \Rightarrow \left( 5x+5y+20 \right){}^\circ =90{}^\circ  \\  \Rightarrow 5(x+y){}^\circ =70{}^\circ  \\  \Rightarrow x+y\text{=14} \\  \text{Now, }{{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)=-\left( x+y \right)\left( y-x \right) \\  \Rightarrow -\left( 14 \right)\left( 6 \right) \\  \Rightarrow -84 \\ $ 

 From the given options, choose the correct pair which is congruent to each other?

 Recall the SSS criterion for congruence of a triangle and the formula for area of equilateral triangle.

  From option $(d)$, Triangle with all sides equal is called equilateral triangle $ \\ $ Thus, triangle will have all sides equal to $4 cm$ $ \\ $ Area of an equilateral triangle with side $a$ is $\dfrac{\sqrt{3}}{4}{{a}^{2}}$ $ \\ $ Then, side of an equilateral triangle with area $4\sqrt{3}\text{ c}{{\text{m}}^{2}}$ is, $ \\ \Rightarrow \dfrac{\sqrt{3}}{4}{{a}^{2}}=4\sqrt{3}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow \sqrt{3}{{a}^{2}}=16\sqrt{3}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow {{a}^{2}}=16\text{ c}{{\text{m}}^{2}} \\  \Rightarrow a=4\text{ cm} \\ $ Then, by SSS criterion for congruence of triangle $

 Let, the length of the side of a square is $x$  cm, E and F is the midpoint of the sides AB and DC respectively. If the triangles, $\Delta \text{AOE}$  and $\Delta \text{COF}$ are congruent then, what is the value of $\text{ar}\left( \Delta \text{ABC} \right)-\text{ar}\left( \text{EBCF} \right)$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5b1815d2-2e1a-45a5-bb4e-cd7fc24d3d643649167023280678725.png"/>

 Recall the formula for evaluating area of right angled triangle and area of rectangle, and using figure, obtain the value of $\text{ar}\left( \Delta \text{ABC} \right)-\text{ar}\left( \text{EBCF} \right)$

  $E$ and $F$ is the midpoint of sides $AB$ and $CD$ respectively $ \\ $ Side of square is $x$cm $ \\ $ Then, $\text{EB = FC = }\dfrac{x}{2}\text{ cm}$ $ \\ $ And, $\text{EF = BC = }x\text{ cm}$ $ \\ $ Area of rectangle whose length is $l$ and breadth is $b$ is equal to $l \cdot b$ $ \\ $ Then, area of rectangle EBCF is, $ \\ \Rightarrow x\cdot \dfrac{x}{2}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow \dfrac{{{x}^{2}}}{2}\text{ c}{{\text{m}}^{2}} \\  \text{Now, OF =}\dfrac{x}{2}\text{ cm and FC =}\dfrac{x}{2}\text{ cm} \\ $ Area of right-angled triangle whose base is $b$ and height is $h$, is $\dfrac{bh}{2}$ $ \\ $ Then, area of right angled triangle OFC is, $ \\  \Rightarrow \dfrac{\left( \dfrac{x}{2} \right)\cdot \left( \dfrac{x}{2} \right)}{2}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow \dfrac{{{x}^{2}}}{8}\text{ c}{{\text{m}}^{2}} \\ $ As, $\Delta AOE$ and $\Delta COF$ are congruent, then, $ar\left( \Delta \text{AOE} \right)\text{ = ar}\left( \text{ }\Delta \text{OFC} \right) \\  \Rightarrow \text{ar}\left( \text{ }\Delta \text{OFC} \right)=\dfrac{{{x}^{2}}}{8}\text{ c}{{\text{m}}^{2}} \\ \text{So, } \Rightarrow \text{ar}\left( \Delta \text{ABC} \right)=\text{ar}\left( \Delta \text{AOE} \right)+\text{ar}\left( \text{EBOC} \right) \\  \Rightarrow \text{ar}\left( \Delta \text{ABC} \right)=\text{ar}\left( \Delta \text{AOE} \right)+\left[ \text{ar}\left( \text{EBCF} \right)-\text{ar}\left( \Delta \text{OFC} \right) \right] \\  \Rightarrow \text{ar}\left( \Delta \text{ABC} \right)-\text{ar}\left( \text{EBCF} \right)=\text{ar}\left( \Delta \text{AOE} \right)-\text{ar}\left( \Delta \text{OFC} \right) \\  \Rightarrow \text{ar}\left( \Delta \text{ABC} \right)-\text{ar}\left( \text{EBCF} \right)=\left( \dfrac{{{x}^{2}}}{8}-\dfrac{{{x}^{2}}}{8} \right)\text{ c}{{\text{m}}^{2}} \\  \Rightarrow \text{ar}\left( \Delta \text{ABC} \right)-\text{ar}\left( \text{EBCF} \right)=0 \\ $ 