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Triangles Test - 17

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Triangles Test - 17
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Two figures having the ............. shape but not necessarily the ............ size are called similar figures.
    Solution

    For two figures to be similar, their shape must be same but their sizes may be different. 

    For example, two circles of different radii, two equilateral triangles each having different lengths, etcetera.

    Hence, two figures having the $$\underline {same}$$ shape but not necessarily the $$\underline {same}$$ size are called similar figures.

  • Question 2
    1 / -0

    For two triangles, if corresponding angles are equal, then  the two triangles are similar. This is called ___ similarity.   
    Solution
    For two triangles, If corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar. This is called $$AAA$$ similarity.
    Option $$A$$ is correct.
  • Question 3
    1 / -0
    _____ congruent figures are similar but the converse is _______.
    Solution

    All congruent figures are similar but all similar figures are not congruent.

    For eg. A pair of triangles which are similar by $$A.A.A.$$ test of similarity are not congruent pairs of triangles since the definite lengths of sides are unknown.

    In $$\triangle ABC$$ and $$\triangle DEF$$, $$\angle A =\angle D= 50^o$$, $$\angle B =\angle E= 75^o$$ and $$\angle C =\angle F = 55^o$$.

    Hence, $$\triangle ABC \sim \triangle DEF$$ but they are not congruent.

  • Question 4
    1 / -0
    For two triangles, if sides of one triangle are proportional to the sides of other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is called ___ similarity.   
    Solution
    For two triangles, if sides of one triangle are proportional to the sides of other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is called $$SSS$$ similarity.   
    Option $$B$$ is correct.
  • Question 5
    1 / -0
    Complete the following sentence.
    All squares are ___________.
    Solution
    Similarity means same shape, size may be different. All squares have same shape.
    $$\therefore$$All squares $$are$$ similar.
    Congruent means same shape and size. Different squares may have same or different sizes. 
    $$\therefore$$All squares $$may$$ be congruent.
  • Question 6
    1 / -0
    If in two triangles, corresponding angles are _______ and their corresponding sides are in the ______ratio and hence the two triangles are similar.
    Solution
    Let $$ABC$$ and $$DEF$$ be two triangles 

    Here, $$\angle A=\angle D$$

              $$\angle B=\angle E$$

      and $$\angle C=\angle F$$

    That is, corresponding angles are equal

    And their corresponding sides are in the same ratio i.e., $$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}$$

    Hence the two triangles are similar. 

  • Question 7
    1 / -0
    In the figure, find $$\angle L$$.

    Solution
    From Given Figure,

    $$\cfrac { AB }{ LM } =\cfrac { 4.4 }{ 11 } =\cfrac { 2 }{ 5 } $$

    $$\cfrac { BC }{ MN } =\cfrac { 4 }{ 10 } =\cfrac { 2 }{ 5 } ;\quad \cfrac { CA }{ NL } =\cfrac { 3.6 }{ 9 } =\cfrac { 2 }{ 5 } $$

    $$\Rightarrow \cfrac { AB }{ LM } =\cfrac { BC }{ MN } =\cfrac { CA }{ NL } $$

    $$\Rightarrow \quad \triangle ABC\sim \triangle LMN$$ (SSS similarity)

    $$\Rightarrow \angle L=\angle A\\={ 180 }^{ o }-\angle B-\angle C\\={ 180 }^{ o }-{ 50 }^{ o }-{ 70 }^{ o }\\={ 60 }^{ o } \ (\text{Using Angle Sum Property})$$

    $$\therefore \angle L={ 60 }^{ o }$$
  • Question 8
    1 / -0
    If in the triangles $$ABC$$ and $$DEF$$, angle $$A$$ is equal to angle $$E$$, both are equal to $${40}^{o}$$, $$AB:ED=AC:EF$$ and angle $$F$$ is $${65}^{o}$$, then angle $$B$$ is:
    Solution
    In $$\triangle ABC$$ and $$\triangle DEF$$,
    $$\angle A = \angle E = 40^{\circ}$$
    $$\dfrac{AB}{ED} = \dfrac{AC}{EF}$$ 
    Thus, $$\triangle ABC \sim \triangle DEF$$ (SAS rule)
    or $$\angle C = \angle F = 65^{\circ}$$

    Now, In $$\triangle ABC$$,
    $$\angle A + \angle B + \angle C = 180$$ (Sum of angles of triangle)
    $$40 + 65 + \angle B = 180$$
    $$\angle B = 180 - 105$$
    $$\angle B = 75^{\circ}$$
  • Question 9
    1 / -0
    G is the centroid of $$\bigtriangleup$$ABC. GE and GF are drawn parallel to AB and AC respectively. Find A($$\bigtriangleup$$ GEF) : A($$\bigtriangleup$$ABC).

    Solution
    Given: $$\triangle ABC$$, $$GE \parallel AB$$ and $$GF \parallel AC$$
    Construction : Draw a median AD
    Since, G is the centroid.
    We know, $$\cfrac{GD}{AD} = \cfrac{1}{3}$$ (Centroid divided the median in the ratio of 2 : 1)
    Ratio of areas of the triangle is equal to the ratio of the square of medians.
    Thus, $$\cfrac{A(\triangle GEF)}{A(\triangle ABC)} = \cfrac{GD^2}{AD^2}$$
    $$\cfrac{A(\triangle GEF)}{A(\triangle ABC)} = \cfrac{1}{9}$$
  • Question 10
    1 / -0
    In $$\triangle ABC$$ and $$\triangle DEF$$, $$\angle A={50}^{o}, \angle B={70}^{o}, \angle C={60}^{o}, \angle D={60}^{o}, \angle E={70}^{o}, \angle F={50}^{o}$$, then $$\triangle ABC$$ is similar to:
    Solution
    In $$\triangle ABC$$ and $$\triangle DEF$$,
    $$\angle A = \angle F = 50^{\circ} $$
    $$\angle B = \angle E = 70^{\circ}$$
    $$\angle C = \angle D = 60^{\circ}$$
    thus, $$\triangle ABC \sim \triangle FED$$ (AAA rule)
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