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Triangles Test - 26

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Triangles Test - 26
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  • Question 1
    1 / -0
    The areas of two similar triangles are $$16\ \text{cm}^2$$ and $$36\ \text{cm}^2$$ respectively. If the altitude of the first triangle is $$3\ \text{cm}$$, then the corresponding altitude of the other triangle is:
    Solution
    Let $${A}_{1}$$ and $${A}_{2}$$ be the areas of the similar triangles and $$s_1$$ and $$s_2$$ be the altitudes of these triangles respectively.
    Then, using the property that the ratio of the areas of similar triangles is equals to square of the ratio of the corresponding sides or altitudes.
    i.e. $$\dfrac{{A}_{1}}{{A}_{2}}=\dfrac{{s}_{1}^{2}}{{s}_{2}^{2}}$$

    Substituting the values of $$A_1, A_2$$ and $$s_1$$ in the above equation:

    $$\dfrac{16}{36}=\dfrac{{\left(3\right)}^{2}}{{s}_{2}^{2}}$$ 

    $${s}_{2}^{2}=\dfrac{36\times 9}{16}$$

    $${s}_{2}=\dfrac{6\times 3}{4}$$
         $$=4.5 \ \text{cm}$$  

  • Question 2
    1 / -0
    Pythagoras was a student of:
    Solution
    Pythagoras (572 BC) was a student of Thales. Pythagoras and his group discovered many geometric properties and developed the theory of geometry to a great extent. This process continued till 300 BC. At that lime. Euclid. a teacher of mathematics at Alexandria in Egypt. collected all the known work and arranged it in his famous treatise. 
    Hence. (a) is the correct answer.
  • Question 3
    1 / -0
    Area of a right-angled triangle is $$30 \,cm^2$$. If its smallest side $$5\, cm$$, then its hypotenuse is.
    Solution
    Option (b) is correct.
    Given, area of right angled triangle $$= 30 \,cm^2$$
    Smallest side $$= 5\, cm$$
    We know that, Area & triangle $$PQR =\dfrac {1}{2} \times  base \times  height$$
    $$ \Rightarrow  30 =\dfrac {1}{2} \times 5 \times  H$$

    $$ \Rightarrow  H =\dfrac {30\times 2}{5}$$

    $$ \Rightarrow  Height  =12\,cm$$
    According to the Pythagoras theorem, $$ \Rightarrow  (Hypotenuse)^2 =(Perpendicular)^2 + (Base)^2$$
    $$ \Rightarrow  (Hypotenuse)^2 = 144+ 25$$ 
    $$ \Rightarrow (Hypotenuse)^2 = 169$$ 
    $$ \Rightarrow Hypotenuse =13\, cm$$
  • Question 4
    1 / -0
    In Fig triangle MNO is a right-angled triangle. Its legs are $$6\, cm$$ and $$8\, cm$$ long. Length of perpendicular NP on the side MO is.

    Solution
    Option (s) is correct.
    Given, triangle MNO is a right $$-$$
    So,by pythagoras theorem we have
    $$ \Rightarrow MO^2 = MN^2  +NO^2$$
    $$ \Rightarrow MO^2 = 6^2 +8^2$$
    $$\Rightarrow MO= \sqrt {100}\, cm$$
    $$ \Rightarrow MO = 10\,cm $$
    Area of triangle MNO 
    $$\Rightarrow \dfrac {1}{2} \times MN  \times NO  $$

    $$\Rightarrow \dfrac {1}{2} \times  6 \times 8 = \frac {1}{2} 10 \times NP$$

    $$\Rightarrow NP =\dfrac {24}{5} $$

    $$\Rightarrow  NP = 4.8\, cm$$
  • Question 5
    1 / -0
    Thales belongs to the country:  
    Solution
    Thales belongs to the country Greece. The Greeks were interested in establishing the truth of the statements they discovered using deductive reasoning. Thales, a Greeks mathematician, is credited with giving the first known proof.

    Hence, option $$C$$ is correct.
  • Question 6
    1 / -0
    In $$ \triangle ABC\sim \triangle DEF$$,  BC $$ = $$ 4 cm, EF $$ =$$ 5 cm and area($$\triangle $$ABC)$$ = $$ 80 $$cm^2$$, the area($$\triangle$$ DEF) is:
    Solution
    Given $$\triangle ABC\sim \triangle DEF$$
    In two similar triangles, the ratio of their areas is the square of the ratio of their sides
    $$\Rightarrow \dfrac { ar(ABC) }{ ar(DEF) } ={ \left( \dfrac { BC }{ EF }  \right)  }^{ 2 }\\ \Rightarrow \dfrac { 80 }{ ar(DEF) } ={ \left( \dfrac { 4 }{ 5 }  \right)  }^{ 2 }\\ \Rightarrow \dfrac { 80 }{ ar(DEF) } =\dfrac { 16 }{ 25 } \\ \Rightarrow ar(DEF)=125{ cm }^{ 2 }$$
  • Question 7
    1 / -0
    Fill in the blank:

    The ratio of the areas of two similar triangles is equal to the _______ of the ratio of their corresponding sides.
    Solution

    Given : $$\triangle ABD \sim \triangle EFH$$

    $$\therefore \dfrac {AB}{EF} = \dfrac {BD}{FH} = \dfrac {AD}{EH}$$      ...C.S.S.T      ...(1)

    Now, $$A(\triangle ABD) = \dfrac 12 \times BD \times AD$$

    And, $$A(\triangle EFH) = \dfrac 12 \times FH \times EH$$

    $$\therefore \dfrac {A(\triangle ABD)}{A(\triangle EFH)} = \dfrac {\frac 12 \times BD \times AD}{\frac 12 \times FH \times EH}$$

    $$\therefore \dfrac {A(\triangle ABD)}{A(\triangle EFH)} = \dfrac {BD}{FH} \times \dfrac {AD}{EH}$$

    Hence, $$\dfrac {A(\triangle ABD)}{A(\triangle EFH)} = \dfrac {BD}{FH} \times \dfrac {BD}{FH}$$          ....From (1)

    $$\therefore \dfrac {A(\triangle ABD)}{A(\triangle EFH)} = \left(\dfrac {BD}{FH}\right)^2$$

  • Question 8
    1 / -0
    State True or False-
    E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. $$\triangle ABE$$ is similar to $$\triangle CFB$$:
    Solution

    Let $$ \angle EBA = x$$ and $$\angle EAB =y. $$
    Then, $$\angle EAB =\angle FCB $$ (Opposite angles of parallelogram)

    Also, $$\angle {EBA} = \angle {CFB}$$ ($$ \because CF||AB$$ )

    $$ \Rightarrow \angle AEB =\angle CBF$$

    Therefore by AAA Theorem of similarity,

    $$\Delta BCF$$ and $$\Delta BAE$$ are similar.

  • Question 9
    1 / -0
    If in $$\Delta ABC$$ and $$\Delta DEF, \dfrac{AB}{DE}=\dfrac{BC}{FD},$$ then they will be similar if :
    Solution
    In $$\triangle ABC$$ and $$\triangle DEF$$
    Given $$\dfrac { AB }{ DE } =\dfrac { BC }{ FD } $$
    Now included angle between side $$AB$$ and $$BC$$ of $$\triangle ABC$$ is $$\angle B$$.
    included angle between side $$DE$$ and $$FD$$ of $$\triangle DEF$$ is $$\angle D$$
    So for triangles to be similar $$\angle B=\angle D$$
    Option $$C$$ is correct.
  • Question 10
    1 / -0
    In $$\Delta ABC \sim  \Delta PQR$$, $$M$$ is the midpoint of $$BC$$ and $$N$$ is the midpoint of $$QR$$. The area of $$\Delta ABC =$$ $$100$$ sq. cm and the area of $$\Delta PQR =$$ $$144$$ sq. cm. If $$AM = 4$$ cm, then $$PN$$ is:
    Solution
    $$\dfrac { ar(\triangle ABC) }{ ar(\triangle PQR) } =\dfrac { 100 }{ 144 } $$
    If triangles are similar then ratio of their areas is equal to ratio of square of their corresponding sides
    $$\dfrac { AB^{ 2 } }{ PQ^{ 2 } } =\dfrac { 100 }{ 144 } \\ \dfrac { AB }{ PQ } =\dfrac { 10 }{ 12 } $$
    $$AM$$ and $$PN$$ are medians
    Therefore, $$ \dfrac { AM }{ PN } =\dfrac { AB }{ PQ } $$
    $$\Rightarrow  \dfrac { 4 }{ PN } =\dfrac { 10 }{ 12 } \\ \Rightarrow PM=4.8 cm$$
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