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Triangles Test - 18

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Triangles Test - 18
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  • Question 1
    1 / -0
    If in two triangles $$ABC$$ and $$PQR$$,
    $$\cfrac{AB}{QR}=\cfrac{BC}{PR}=\cfrac{CA}{PQ}$$, then
    Solution
    Given,
    $$\dfrac{AB}{QR}=\dfrac{BC}{PR}=\dfrac{CA}{PQ}$$
    Therefore,
    $$\triangle PQR\sim \triangle CAB$$  --------(By SSS Congruency)

  • Question 2
    1 / -0
    In $$\triangle ABC$$, $$D$$ and $$E$$ are midpoints of $$AB$$ and $$AC$$ respectively. Find the ratio of the areas of $$\triangle ADE$$ and $$\triangle ABC$$.

    Solution
    Given: $$D$$ and $$E$$ are mid points of $$AB$$ and $$AC$$
    By mid point theorem, $$DE \parallel BC$$
    In $$\triangle ADE$$ and $$\triangle ABC$$
    $$\angle DAE = \angle BAC$$ (Common)
    $$\angle ADE = \angle ABC$$ (Corresponding angles)
    $$\angle AED = \angle ACB$$ (Corresponding angles)
    Thus, $$\triangle ABC \sim \triangle ADE$$ ($$AAA$$ rule)
    Hence, $$\dfrac{Area(\triangle ADE)}{Area(\triangle ABC)} = \dfrac{AD^2}{AB^2}$$ (Similar triangle property)
    $$\dfrac{Area(\triangle ADE)}{Area(\triangle ABC)}= \dfrac{AD^2}{(2 AD)^2}$$
    $$\dfrac{Area(\triangle ADE)}{Area(\triangle ABC)} = 1 : 4$$
  • Question 3
    1 / -0
    In triangles $$ABC$$ and $$DEF$$, $$\angle B=\angle E, \angle F=\angle C$$ and $$AB=3DE$$. Then, the two triangles are:
    Solution
    In $$\triangle ABC$$ and $$\triangle DEF$$
    $$\angle B = \angle E$$ (given)
    $$\angle C = \angle F$$ (Given)
    $$\angle A = \angle D$$ (third angle) 
    Thus, $$\triangle ABC \sim \triangle DEF$$ (AAA rule)
    Given, $$AB = 3 DE$$ Since, $$AB \ne DE$$
    The two triangles are similar but not congruent.
  • Question 4
    1 / -0
    The areas of two similar triangles are $$12$$ $${cm}^{2}$$ and $$48$$ $${cm}^{2}$$. If the height of the smaller one is $$2.1$$ $$cm$$, then the corresponding height of the bigger one is:
    Solution

    Areas of two similar triangles are $$12 cm^2$$ and $$48 cm^2$$
    For similar triangles the ratio of areas is equal to the ratio of square of corresponding heights
    Hence, $$\dfrac{A_1}{A_2} = \dfrac{(h_1)^2}{(h_2)^2}$$

    $$\dfrac{12}{48} = \dfrac{(2.1)^2}{(h_2)^2}$$

    $$(h_2)^2= 4 \times (2.1)^2$$

    $$h_2 = 2 \times 2.1$$

    $$h_2 = 4.2 cm$$

  • Question 5
    1 / -0
    State which pair of triangles in figure are similar.  Write the similarity criterion used by you for answering the question.

    Solution
    In $$\triangle ABC$$ and $$\triangle PQR$$,
    $$\angle A = \angle P = 60^{\circ}$$
    $$\angle B = \angle Q = 80^{\circ}$$
    $$\angle C = \angle R = 40^{\circ}$$
    Thus, $$\triangle ABC \sim \triangle PQR$$ (AAA rule)
  • Question 6
    1 / -0
    If in two triangles $$DEF$$ and $$PQR$$, $$\angle D=\angle Q$$ and $$\angle R=\angle E$$, then which of the following is not true?
    Solution
    If in $$\triangle DEF$$ and $$\triangle QRP$$
    $$\angle D=\angle Q , \angle R=\angle E$$
    Then by AA similarity we have,
    $$\triangle DEF\sim \triangle QRP$$
    $$\dfrac{DE}{QR}=\dfrac{EF}{RP}=\dfrac{DF}{PQ}$$
    Thus, is not correct.

  • Question 7
    1 / -0
    In the given figure, $$\angle ABC={90}^{o}$$ and $$BD\bot AC$$. If $$AB=5.7cm, BD=3.8cm$$ and $$CD=5.4cm$$, Find the length of $$BC$$.

    Solution
    In $$\triangle BDC$$,
    $$BD^2 + DC^2 = BC^2$$
    $$(3.8)^2 + (5.4)^2 = BC^2$$
    $$43.6 = BC^2$$
    $$BC = 6.6$$ cm
  • Question 8
    1 / -0
    In $$\triangle PQR$$, $$LM\parallel QR$$ and $$PM:MR=3:4$$. Calculate$$\cfrac{PL}{PQ}$$ and then $$\cfrac{LM}{QR}$$

    Solution
    Given, $$LM \parallel QR$$, $$PM : MR = 3 : 4$$
    $$\dfrac{PM}{MR} = \dfrac{3}{4}$$
    $$\dfrac{PM + MR} {MR} = \dfrac{3+ 4}{4}$$
    $$\dfrac{PR}{MR} = \dfrac{7}{4}$$

    Now, In $$\triangle PLM$$ and $$\triangle PQR$$
    $$\angle LPM = \angle QPR$$ (Common angle)
    $$\angle PLM = \angle PQR$$ (Corresponding angles)
    $$\angle PML = \angle PRQ$$ (Corresponding angles)
    Thus, $$\triangle PLM \sim \triangle PQR$$ (AAA rule)
    Hence, $$\dfrac{PL}{PQ} = \dfrac{LM}{QR} = \dfrac{PM}{PR}$$ (Corresponding sides)
    $$\dfrac{PL}{PQ} = \dfrac{LM}{QR} = \dfrac{3}{7}$$

  • Question 9
    1 / -0
    Fill in the blank
    All .............. triangles are similar
    Solution
    For equilateral triangles, each side is equal. Hence the ratio between two sides of different triangles will be constant. Hence, all equilateral triangles are similar.
  • Question 10
    1 / -0
    Fill in the blank
    All squares are ............... (similar, congruent)
    Solution
    All sides of a square are constant. Hence, the ratio of sides will be constant for a pair of squares. Thus, all squares are similar.
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