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Triangles Test - 19

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Triangles Test - 19
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  • Question 1
    1 / -0
    A ladder $$10m$$ long reaches a window $$8m$$ above the ground. Find the distance of the foot of the ladder from base of the wall.
    Solution
    Let $$AC=x$$ meters be the distance of the foot of the ladder from the base of the wall.
    $$AB=8m$$ (Height of window)
    $$BC=10m$$ (length of ladder)
    Now, $${x}^{2}+{(8)}^{2}={(10)}^{2}$$
    $$\Rightarrow$$ $${x}^{2}=100-64=36$$
    $$\Rightarrow$$ $$=6$$, i.e., $$AC=6m$$.

  • Question 2
    1 / -0
    Which figure is formed by joining the mid points of the adjacent sides of a square?
    Solution
    Let $$E,F,G,H$$ are mid points of the sides $$AB,BC,CD,DA$$ repectively.

    In $$\triangle AEH,$$
    $$\angle A=90^\circ$$                                   [$$ABCD$$ is a square.]

    So, by using Pythagoras theorem,
    $$\begin{aligned}{}HE^2&=AE^2+AH^2\\{b^2} &= {\left( {\frac{a}{2}} \right)^2} + {\left( {\frac{a}{2}} \right)^2}\\ &= 2 \times \frac{{{a^2}}}{4}\\& = \frac{{{a^2}}}{2}\\b &= \frac{a}{{\sqrt 2 }}\end{aligned}$$

    Now, in $$\triangle EHF,$$
    $$\begin{aligned}{}E{H^2} + E{F^2} &= 2 \cdot {b^2}\\& = 2 \cdot {\left( {\frac{a}{{\sqrt 2 }}} \right)^2}\\& = 2 \times \frac{{{a^2}}}{2}\\ &= {a^2}\\ &= H{F^2}\end{aligned}$$

    So, by converse of Pythagoras theorem $$\triangle EHF$$ is a right-angled triangle with right angle at $$E.$$ Similarly we can prove that there is a right angle at each vertex of quadrilateral $$EFGH$$ and each side is equal to $$\dfrac{a}{\sqrt2}.$$ Hence, quadrilateral $$EFGH$$ is a square.

  • Question 3
    1 / -0
    State which pair of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

    Solution
    In $$\triangle DEF$$,
    $$\angle D = 70^{\circ}$$
    $$\angle E = 80^{\circ}$$
    Sum of angles = 180
    $$\angle D + \angle E + \angle F = 180$$
    $$70 + 80 + \angle F = 180$$
    $$\angle F = 30^{\circ}$$

    Now, In $$\triangle DEF$$ and $$\triangle PQR$$
    $$\angle E = \angle Q = 80^{\circ}$$
    $$\angle F = \angle R = 30^{\circ}$$
    $$\angle D = \angle P$$ (third angle)
    thus, $$\triangle DEF \sim \triangle PQR$$ (AAA rule)
  • Question 4
    1 / -0
    In a $$\Delta$$ABC, if $$AB^2\, =\, BC^2\, +\, AC^2$$, then the right angle is at:
    Solution
    Given that, $$AB^2 = BC^2+AC^2$$.
    Hence, $$AB$$ is the hypotenuse          ....As per Pythagoras theorem
    Hence, angle opposite to side $$AB$$ is $$90^o$$.
    So, $$m\angle C = 90^o$$.
  • Question 5
    1 / -0
     If the two legs of a right angle triangle are equal and the square of the hypotenuse is $$100,$$ then the length of each leg is:
    Solution
    We have
    $$x^2\, +\, x^2\, =\, 100$$
    $$2x^2\, =\, 100$$
    $$x^2\, =\, 50$$
    $$x\, =\, \sqrt{50}\, =\, \sqrt{25\, \times\, 2}\, =\, 5\, \sqrt{2}$$.
  • Question 6
    1 / -0
    The sides of a triangle are given below. Check whether or not the sides form a right-angled triangle.
    $$3cm, 8cm, 6cm$$
    Solution
    Given, sides of triangle
    $$a = 3$$ cm
    $$b = 6$$ cm
    $$c = 8$$ cm

    For a triangle to be a right angle triangle, the square of largest side(hypotenuse) should be equal to the sum of squares of the other two sides.


    $$c^2 = 8^2 = 64$$
    $$a^2 + b^2 = 3^2 + 6^2 = 9 + 36 = 45$$
    Since, $$a^2 + b^2 \ne c^2$$

    Hence, the triangle is not a right-angled triangle.
  • Question 7
    1 / -0
    If triangles in figure are similar. Write the similarity criterion used by you for answering the question.

    Solution
    In $$\triangle MNL$$, $$MN = 3$$ and $$ML = 5$$ 
    Thus, $$\frac{MN}{ML} = \frac{3}{5}$$

    In $$\triangle PQR$$, $$PQ = 6$$ and $$QR = 10$$
    Thus, $$\frac{PQ}{QR} = \frac{6}{10} = \frac{3}{5}$$

    In $$\triangle MNL$$ and $$\triangle PQR$$
    $$\frac{MN}{ML} = \frac{PQ}{QR} = \frac{3}{5}$$
    $$\angle M = \angle Q = 70^{\circ}$$
    Thus, $$\triangle MNL \sim \triangle QPR$$ (SAS rule)
  • Question 8
    1 / -0
    Sides of two similar triangles are in the ratio $$4:9$$. areas of these triangles are in the ratio
    Solution
    Given, $$\dfrac{S_1}{S_2} = \dfrac{4}{9}$$
    For Similar triangles,
    $$\dfrac{A_1}{A_2} = \dfrac{(S_1)^2}{(S_2)^2}$$
    $$\dfrac{A_1}{A_2} = \dfrac{4^2}{9^2}$$
    $$\dfrac{A_1}{A_2} = \dfrac{16}{81}$$ 
  • Question 9
    1 / -0
    In the figure $$D\;$$and$$\;E$$ are the mid-points of sides $$AB\;$$and$$\;AC$$ respectively of $$\Delta ABC$$. Find $$\angle EDB.$$

    Solution
    In $$\triangle ABC$$
    $$\Rightarrow$$  $$\angle A+\angle B+\angle C=180^o$$
    $$\Rightarrow$$  $$60^o+\angle B+50^o=180^o$$
    $$\Rightarrow$$  $$110^o+\angle B=180^o$$
    $$\Rightarrow$$  $$\angle B=70^o$$
    $$\therefore$$  $$\angle ABC=70^o$$          ----- ( 1 )

    In $$\triangle ADE$$ and $$\triangle ABC$$,
    $$\Rightarrow$$  $$AB=2AD$$                     [ $$D$$ is mid-point of $$AB$$ ]
    $$\therefore$$  $$\dfrac{AD}{AB}=\dfrac{1}{2}$$     ---- ( 2 )
    $$\Rightarrow$$  $$AC=2AE$$                  [ $$E$$ is a mid-point of side $$AC$$ ]
    $$\therefore$$  $$\dfrac{AE}{AC}=\dfrac{1}{2}$$      ---- ( 3 )

    $$\Rightarrow$$  $$\dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{1}{2}$$           [ From ( 1 ) and ( 2 ) ]
    $$\Rightarrow$$  $$\angle A=\angle A$$           [ Common angle ]
    $$\therefore$$  $$\triangle ADE\sim \triangle ABC$$       [ By $$SAS$$ similarity rule ]

    $$\Rightarrow$$  $$\angle ADE=\angle ABC$$        [ Corresponding angles of similar triangle are equal ]
    $$\therefore$$  $$\angle ADE=70^o$$          [ From ( 1 ) ]

    $$\Rightarrow$$  $$\angle ADE+\angle EDB=180^o$$           [ Linear pair ]

    $$\Rightarrow$$  $$70^o+\angle EDB=180^o$$

    $$\Rightarrow$$  $$\angle EDB=110^o$$.

  • Question 10
    1 / -0
    If the ratio of the corresponding sides of the two similar triangles is 2 : 3, then the ratio of their corresponding altitudes is
    Solution
     If the two triangle are similar, the ratio of corresponding sides is same as the ratio of their altitudes.
    So, $$2:3$$ is required ratio.
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