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

Similarity Quiz 2

mathematics

 A figure is constructed to be painted for the next art exhibition as shown. In which order are the two triangles similar?  $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/de2113c6-6480-48c8-9537-645c3240bb591255070023812010515.png"/>

 The ordering of the points of triangles is equal while finding similarity between triangles. BC is a common edge between triangles.

  $\text{In triangles ABC and DBC } \\  \text{AB=BD} \\  \angle BAC=\angle BDC={{53}^{\circ }} \\  \text{As BC is common in both the triangles} \\  BC=CB \\  \angle ADB=\angle EDC \\  \text{By SAS property the above triangles are similar} \\ \text{So }\vartriangle \text{ABC}\approx \vartriangle \text{DBC} $

 Riya wants to make wings for her next fancy dress completion such that one wing is smaller than the other. She maps out the diagram as follows. What would be the length of the smaller wing? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/03cab953-5abc-4ac5-870c-a3332b4e68571501662455830923345.png"/>

 The ordering of the points of triangles is equal while finding similarity between triangles. If two triangles are similar, their lengths are in the same ratio. 

  $   \text{In triangles ABP and XZP } \\   \text{Since AB  }\!\!|\!\!\text{   }\!\!|\!\!\text{  XZ} \\   \text{By Alternate angle theorem} \\   \angle BAP=\angle PZX \\   \text{By opposite angles} \\   \angle APB=\angle XPZ \\   \text{By AA property the above triangles are similar} \\   \text{So }\vartriangle \text{ABP}\approx \vartriangle XZP \\   \text{Since the triangles are similar the lengths will be in same ratio} \\   \Rightarrow \dfrac{AB}{XZ}=\dfrac{8}{4}=2 \\   \Rightarrow \dfrac{AP}{PZ}=\dfrac{6.5}{PZ}=2 \\   \Rightarrow PY=3.25 \\   \text{Similarly} \\   \Rightarrow \dfrac{BP}{PX}=\dfrac{7}{PX}=2 \\   \Rightarrow PX=4.5 \\ $

 Ram and Shyam both make separate triangles such that their corresponding angles are equal. If the area of the ram's triangle is one fourth of Shyam’s. What is the ratio of the lengths of Ram’s Triangle to Shayam’s? 

 If all angles of a triangle are of the same measurement, they are similar triangles. If two triangles are similar, their lengths are in the same ratio. 

$   \text{Here the angles are equal which means that the triangles are similar} \\   \text{Since the Areas are in ratio 1:4 that} \\   \text{let }{{\text{L}}_{1}}\text{ and }{{\text{L}}_{2}}\text{ be the lengths of two triangles} \\   \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{(1)}{(4)} \\   \Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\sqrt{\dfrac{1}{4}} \\   \Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{1}{2} \\ $

 A triangle is made such that two sides are equal, another triangle is constructed such that it has the angles as the first triangle. The areas of these triangles are found out to be in a ratio of 16:100. What do you think their heights relate to?

 If all angles of a triangle are of the same measurement, they are similar triangles. If two triangles are similar, their lengths are in the same ratio.

 $   \text{Here the angles are equal which means that the triangles are similar} \\   \text{Since the Areas are in ratio 16:100 that} \\   \text{let }{{\text{L}}_{1}}\text{ and }{{\text{L}}_{2}}\text{ be the heights of two triangles} \\   \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{(16)}{(100)} \\   \Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\sqrt{\dfrac{16}{100}} \\   \Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{4}{10}=\dfrac{2}{5} \\ $

 A person makes an isosceles triangular two tier cake such that each level is twice the previous in terms of height and is similar to each other. Find the total area of the top to be covered on the top side, if the area of the smallest cake is 16 sq cm. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d993810b-e808-495c-a1ca-9f56287de2a94139483718171628332.png"/>

 $   \text{let }{{\text{H}}_{1}}\text{ and }{{\text{H}}_{2}}\text{ be the heights of two triangles} \\   \dfrac{{{H}_{1}}}{{{H}_{2}}}=\dfrac{(1)}{(2)} \\   \text{let }{{\text{A}}_{1}}\text{ and }{{\text{A}}_{2}}\text{ be the Areas of two triangle} \\   \Rightarrow {{\left( \dfrac{{{H}_{1}}}{{{H}_{2}}} \right)}^{2}}=\dfrac{{{A}_{1}}}{{{A}_{2}}}={{\left( \dfrac{1}{2} \right)}^{2}}=\left( \dfrac{1}{4} \right) \\   \text{We know that the area of smaller cake is 16 c}{{\text{m}}^{2}} \\   \text{Therefore} \\   \Rightarrow \dfrac{16}{{{A}_{2}}}=\dfrac{1}{4} \\   \Rightarrow {{A}_{2}}=64c{{m}^{2}} \\   \text{ so total area to cover is 64+16=80c}{{\text{m}}^{2}} \\ $