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

Polygon similarity Quiz 2

mathematics

  If triangles $PQR\text{ and ABC}$ are similar to each other, then what is the measure of the angle$\angle ACB$ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4b51be63-ae4a-4842-8b27-efd23564f6fd1646091735008999649.png"/>

 Using the concept of trigonometry in a right angle triangle and determine the angle$\angle ACB$.

$\text{In triangle, PQR} \\ \text{tanR=}\dfrac{PQ}{QR} \\  \Rightarrow \text{tanR=}\dfrac{\sqrt{12}}{2} \\  \Rightarrow \text{tanR=}\dfrac{2\sqrt{3}}{2} \\  \Rightarrow \text{tanR=}\sqrt{3} \\  \Rightarrow \text{tanR=tan6}{{\text{0}}^{{}^\circ }} \\  \Rightarrow \angle \text{R=6}{{\text{0}}^{{}^\circ }} \\  \text{Corresponding angles of two triangles are same}\text{.} \\  \text{So, in triangle PQR and ABC,} \\  \Rightarrow \angle \text{PRQ=}\angle \text{ACB} \\  \Rightarrow \angle \text{ACB=6}{{\text{0}}^{{}^\circ }}$

 If area of the polygon ABCDEFGH of same side length is $\dfrac{5}{2}$  times the area of the triangle AFE, then the area of polygon PQRSTUVW which has the area equal to $\dfrac{1}{2}$  times the area of the polygon ABCDEFGH is <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ae2d8228-1b76-4033-bb22-adbe788a98025359972571677314106.png"/> $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d92282b3-dac6-49ef-857f-7cf06b8c09b08724354593941861525.png"/>

 Using the concept of the trigonometry in triangles and the area of triangle, and thus, determine the area of the polygon PQRSTUVW.

$\text{In triangle AFE,} \\ \text{tanE=}\dfrac{AF}{6} \\  \Rightarrow \text{tan 60}{}^\circ \text{=}\dfrac{AF}{6} \\  \Rightarrow \sqrt{3}\text{=}\dfrac{AF}{6} \\  \Rightarrow AF=6\sqrt{3}\text{ cm} \\  \text{Area of triangle with base ‘b’ and height ‘h’ is }\dfrac{bh}{2}. \\  \text{So, the area of triangle AFE,} \\  \Rightarrow \dfrac{\left( 6 \right)\left( 6\sqrt{3} \right)}{2}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 18\sqrt{3}\text{ c}{{\text{m}}^{2}} \\  \text{Area of polygon ABCDEFGH is,} \\  \Rightarrow \left( \dfrac{5}{2}\times 18\sqrt{3} \right)\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 45\sqrt{3}\text{ c}{{\text{m}}^{2}} \\  \text{Area of the polygon PQRSTUVW is,} \\  \Rightarrow \dfrac{45\sqrt{3}}{2}\text{ c}{{\text{m}}^{2}}$

 In a class test, a teacher drew four pairs of different shapes. Then a random girl was asked to tell which pair fails the concept of polygon similarities and the answer of the girl was correct. Then her answer was

 Using the formula for the area of circle, square and rectangle, and the perimeter of square. Also, recall the definition of polygon.

$\text{Here, Option (c) is wrong as,} \\  \text{Area of circle with radius 5 cm is,} \\  \Rightarrow \pi {{(\text{5)}}^{2}}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 2\text{5}\pi \text{ c}{{\text{m}}^{2}} \\ \text{By the definition of polygon,} \\ \text{Polygon can have only finite number of sides}. \\  \text{But, as circle is curved,}$ $\text{ so it cannot be formed from line segments}.$

 What is the perimeter of the square PQRS, if square ABCD and square PQRS are equal? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/80a8612f-95d7-4a41-9ba3-ef7edb0e7b817346916937140838712.png"/>

 Using the formula for the area of square and the perimeter of square and thus, determine the perimeter of square PQRS.

$\text{Side of square with diagonal ‘d’ is equal to }\dfrac{d}{\sqrt{2}}. \\ \text{So, the side of triangle ABCD is,} \\  \Rightarrow \dfrac{6\sqrt{2}}{\sqrt{2}}\text{ cm}  \Rightarrow 6\text{ cm} \\  \text{Perimeter of square with side ‘a’ }= 4a. \\ \text{So, the perimeter of triangle ABCD is,} \\  \Rightarrow 6\times 4\text{ cm} \Rightarrow 24\text{ cm} \\ \text{Square ABCD and PQRS are equal}$ $\text{Then, the perimeter of square PQRS is,}$ $ \\ \Rightarrow 24\text{ cm}$

 Let polygons ABCDE and PQRST are similar to each other with the same perimeter $39\text{ cm}$. Then, ${{x}^{2}}-{{y}^{2}}$ is equal to <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d70eaf7b-230f-4ab4-ae54-e74a1524dcf0636840342322248678.png"/> $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/086a795a-efcf-4185-a714-cc693ee6e1b74883554733395252230.png"/>

 The perimeters of two equal triangles are also equal and algebraic expression ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$.

$\text{Perimeter of polygon = Sum of all its sides}. \\ \text{Then, perimeter of polygon ABCDE is,} \\  \Rightarrow \left( 3x \right)+\left( 2y+1 \right)+\left( 6-x \right)+\left( 14-x \right)+\left( 13-y \right)=39 \\  \Rightarrow x+y+34=39 \\  \Rightarrow x+y=5 \\  \text{Perimeter of polygon PQRST is,} \\  \Rightarrow \left( 3x \right)+\left( 2y+1 \right)+\left( 6-x \right)+\left( 14-x \right)+\left( 17-3y \right)=39 \\  \Rightarrow x-y+38=39 \\  \Rightarrow x-y=1 \\  \text{Now,} {{x}^{2}}-{{y}^{2}} \\  \Rightarrow \left( x+y \right)\left( x-y \right) \\  \Rightarrow 5\times 1 \Rightarrow 5$