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Triangles Test - 23

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Triangles Test - 23
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  • Question 1
    1 / -0
    Which postulate can be used to prove the triangles are similar?

    Solution
    From the figure, triangle $$DEF$$ and triangle $$IGH$$ are similar by SAS postulate as $$\angle {F} = \angle {H}$$
    Therefore, the given triangles are similar by SAS similarity postulate.
  • Question 2
    1 / -0
    If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
    Solution
    If the corresponding angles are equal then the triangles are similar by $$AAA$$ similarity criteria.

  • Question 3
    1 / -0
    In $$\displaystyle \Delta ABC\sim \Delta DEF$$ and their areas are $$\displaystyle { 36cm }^{ 2 }$$ and $$\displaystyle { 64cm }^{ 2 }$$ respectively.If side AB=3 cm. Find DE.
    Solution
    In $$\displaystyle \Delta ABC\sim \Delta DEF$$
    $$\displaystyle \frac { ar.\left( \Delta ABC \right)  }{ ar.\left( \Delta DEF \right)  } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } $$
    $$\displaystyle \frac { 36 }{ 64 } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } $$
    $$\displaystyle \frac { 6 }{ 8 } =\frac { 3 }{ DE } $$
    $$\displaystyle DE=\frac { 8\times 3 }{ 6 } =\frac { 24 }{ 6 } =4cm$$
    Therefore, D is the correct answer.
  • Question 4
    1 / -0
    Which of the following postulate can be used to prove $$\Delta {PQR}$$ and $$\Delta {ABC}$$ are similar?

    Solution
    From the figure, $$\Delta {PQR}$$ and $$\Delta {ABC}$$ are similar by SAS postulate as $$\angle {Q} = \angle {B}$$
    Therefore, the given triangles are similar by SAS similarity postulate.
  • Question 5
    1 / -0
    Basic proportionality theorem  is also known as
    Solution
    Basic proportionality theorem is also known as Thales Theorem. Thales was a famous Greek mathematician who gave an important truth relating two equiangular triangles.
    Therefore, $$B$$ is the correct answer.
  • Question 6
    1 / -0
    If one shape becomes another using a resize, then the shapes are __________. 
    Solution
    Resizing leads to change in scale factor and if the scale factor remains equal, then the figures will be similar to each other.
  • Question 7
    1 / -0
    The areas of two similar triangles are $$121 cm^2$$ and $$81 cm^2$$ respectively. Find the ratio of their corresponding heights.
    Solution
    Given the areas of two similar triangles are $$121$$ sq. cm and $$81$$ sq. cm
    We know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding heights.
    The ratio of area of triangles $$= \dfrac{121}{81}=\dfrac{(11)^{2}}{(9)^{2}} $$ $$=$$  $$\dfrac{(h_1)^2}{(h_2)^2}$$
    Then ratio of heights of triangle $$=\sqrt{\left [ \dfrac{11}{9} \right ]^{2}}=\dfrac{11}{9}$$
  • Question 8
    1 / -0
    Which of the following can be used to prove $$\Delta {ABC}$$ and $$\Delta {DEF}$$ are similar?

    Solution
    From the figure, $$\Delta {ABC}$$ and $$\Delta {DEF}$$ are similar by SSS postulate because three sides of one triangle are congruent to three sides of another triangle.
    Therefore, the given triangles are similar by SSS similarity postulate.
  • Question 9
    1 / -0
    If the pictured triangles are similar, what reason can be given?

    Solution
    From the figure, triangles are similar by SSS postulate because three sides of one triangle are congruent to three sides of another triangle.
    Therefore, the given triangles are similar by SSS similarity postulate.
  • Question 10
    1 / -0
    In the given figure, find the value of x, if AB||CD.

    Solution
    In given figure AB II CD 
    And line $$DO$$ and $$CO$$ cut side $$AB$$ and $$CD$$ at $$D,A,C$$ and $$B$$
    $$\angle OCD=\angle OBA$$ .......  (congruent angle )
    $$\angle ODC=\angle OAB$$ ...... (congruent angle )
    $$\angle COD=AOB$$ ....... (Common angle )
    $$\Delta COD $$ and $$\Delta BOA$$ are similar triangle BY AAA
    Then $$\dfrac{DO}{AO}=\dfrac{CD}{AB}$$
    $$\Rightarrow \dfrac{6+x}{x}=\dfrac{5.49}{1.83}$$
    $$\Rightarrow 6+x=3x$$
    $$\Rightarrow 3x-x=6$$
    $$\Rightarrow x=3$$
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