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Triangles Test - 27

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Triangles Test - 27
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  • Question 1
    1 / -0
    In the adjacent figure, P and Q are points on the sides AB and AC respectively of a triangle ABC. PQ is parallel to BC and divides the triangle ABC into 2 parts, equal in area. The ratio of $$PA:AB=$$

    Solution
    Given that Area of the $$ \Delta $$ APQ $$ = $$  Area of PQCB
    That means Area $$ \Delta $$ ABC $$ = $$  2 Area of $$ \Delta $$ APQ
    Since PQ $$ \parallel $$ BC
    Therefore, $$ \Delta $$ APQ is similar to $$ \Delta $$ ABC
    We know that ratio of the areas of two triangles is equal to the square of ratio of their sides in case of similar triangles.
    Therefore, $$\cfrac { Area\quad of\quad \triangle APQ }{ Area\quad of\triangle ABC }$$ = $$\dfrac { { PA }^{ 2 } }{ { AB }^{ 2 } } $$
    $$ \dfrac { { PA }^{ 2 } }{ { AB }^{ 2 } } = $$$$\cfrac { Area\quad of\quad \triangle APQ }{ Area\quad of\triangle ABC }$$ $$= \dfrac { 1 }{ 2 } $$ 
    $$\dfrac{PA}{AB} $$ $$=\sqrt { \dfrac { 1 }{ 2 }  } $$
    Therefore, $$PA : AB $$ = $$ 1: $$ $$\sqrt { 2 } $$

  • Question 2
    1 / -0
    If $$\triangle ABC\sim \triangle QRP,\dfrac{Ar(ABC)}{Ar(QRP)}=\dfrac{9}{4}$$,$$AB=18\ cm$$ and $$BC=15\ cm$$; then $$PR$$ is equal to:
    Solution
    Given $$ \triangle  ABC \sim  \triangle  QRP $$
    Therefore, $$ \cfrac { Area\triangle ABC\quad  }{ Area\triangle QRP\quad }=\cfrac { { BC }^{ 2 } }{ { PR }^{ 2 } }  $$
    or $$ \cfrac { 9 }{ 4 }  = \cfrac { { 15 }^{ 2 } }{ { PR }^{ 2 } }  $$
    or $$ { PR }^{ 2 }\quad =\quad \cfrac { { 15 }^{ 2 }\quad \times \quad 4 }{ 9 }  cm$$
    Therefore, $$ { PR }=\cfrac { { 15 }\times \quad 2 }{ 3 }  = $$ $$10 cm.$$

  • Question 3
    1 / -0
    In the given figure, $$\angle YXZ=\angle XPZ,$$ then, $$\dfrac{ZX}{ZY}$$ is equal to:

    Solution
    AAA postulate for symmetry : 
    If all three corresponding interior angles of 2 triangles are the same, then the two triangles are similar. 
    In general, if two corresponding angles are equal, then the remaining angle will also be equal, as they always add up to $$180^o$$.
    In figure (i), if $$\angle A = \angle D,\angle B = \angle E,\angle C = \angle F$$ by AAA postulate we can say that  2 triangles are similar.
    In fig (2) given $$\angle YXZ = \angle XPZ$$
    We can divide the given figure (iii) into two separate triangles.
    Consider $$\triangle XYZ \text{ and } \triangle PXZ$$.
    $$\angle YXZ = \angle XPZ$$      ....(given)
    $$\angle XZY= \angle XZP$$      ...($$\because$$ they are same )
    $$\angle XYZ = \angle PXZ$$       ...($$\because$$ sum of angles = 180°)
    Hence, by AAA postulate, 
    $$\triangle XYZ \sim $$$$\triangle PXZ$$ 
    $$\therefore \dfrac{ZX}{ZY} = \dfrac{PZ}{XZ}$$     ....cpst

  • Question 4
    1 / -0
    Diagonals AC and BD of a trapezium ABCD with $$AB\parallel DC$$ intersect each other at the point O ,$$\dfrac{OA}{OC}=?$$
    Solution

    In $$\triangle AOB$$ and $$\triangle COD$$,

    $$\angle AOB = \angle COD$$          ....Vertically opposite angles

    $$\angle OAB = \angle OCD$$          ...Alternate angles since $$AB \parallel CD$$

    $$\therefore \triangle AOB \sim \triangle COD$$       ....A.A test of similarity

    $$\therefore \dfrac {OA}{OC} = \dfrac {OB}{OD} = \dfrac {AB}{CD}$$     ...C.S.S.T.

  • Question 5
    1 / -0
    In a circle with centre $$O$$, $$OD$$ $$\perp$$ chord $$AB$$. If $$BC$$ is the diameter, then which of the following option is correct:
    Solution

    Option A: $$AC \neq BC$$

    Option B: $$OD \neq BD$$

    Option C:

    In $$\triangle ABC$$ and $$\triangle ODB$$,

    $$\angle CAB = \angle ODB = 90^{0}$$       ....given
    $$\angle ABC = \angle OBD$$           ...(common angle)
    $$\angle ACB = \angle DOB$$           ...(rest of the angle)
    Hence, $$\Delta ABC \sim \Delta$$ DBO    ....AA test of similarity

    Thus, $$\dfrac{OD}{AC}= \dfrac{OB}{BC}$$    ...C.S.S.T    

    $$\dfrac{OD}{AC} = \dfrac{r}{2r}$$
    $$\therefore AC = 2 \times OD$$
    Option D: Not applicable as option C is correct.

  • Question 6
    1 / -0
    In the above figure, find the length of $$AD$$

    Solution
    In the $$\Delta ABC$$, 
    Given $$AB = 6 \,\mathrm{cm}$$ and $$BC= 8\,\mathrm{cm}$$
    and $$\angle B = 90 ^\circ$$
    So we can apply Pythagoras theorem 
    $${AB}^2+{BC}^2={AC}^2$$    ...(1)

    $$\Rightarrow 36+64={AC}^2$$
    $$\Rightarrow {AC}^2=10^2$$

    In the $$\Delta ACD$$,
    Given $$CD = 24\,\mathrm{cm}$$
     $$\angle C= 90^\circ$$ 
    Applying Pythagoras theorem we get,
    $${AC}^2+{CD}^2={AD}^2$$
    $$\Rightarrow 10^2 + 24^2 = {AD}^2$$
    $$\Rightarrow {AD}^2=26^2$$
    $$\therefore AD=26$$

    Hence, option C is the correct answer.
  • Question 7
    1 / -0
    It is given that $$\triangle ABC\sim \triangle PQR$$ with $$\dfrac{BC}{QR}=\dfrac{1}{3}$$. Then, $$\dfrac{A(\triangle PRQ)}{A(\triangle BCA)}$$ is equal to:
    Solution

    Given $$ \triangle ABC \sim  \triangle  PQR$$ 

    and $$ \dfrac { BC }{ QR } = \dfrac { 1 }{ 3 } $$ or $$ \dfrac { QR }{ BC } =\dfrac { 3 }{ 1 }  $$.

    To find the ratio of the areas of the given triangles.

    Since the triangles are similar, the ratio of their areas is equal to the ratio of the square of their corresponding sides.

    Therefore, $$ \dfrac { Area\quad of\quad the\quad \triangle \quad PQR }{ Area\quad of\quad the\quad \triangle \quad ABC }=\dfrac { { QR }^{ 2 } }{ { BC }^{ 2 } } =\dfrac { { 3 }^{ 2 } }{ 1 } =\dfrac { 9 }{ 1 }=9 $$


  • Question 8
    1 / -0
    The line segments AB and BC  are perpendicular to each  other. If $$AB = 4$$ cm and $$BC = 3$$  cm, then radius of the circle  passing through A, B, C  will be
    Solution
    If AB and BC are perpendicular to each other and a circle passes through A,B,C all points then the line joining A and C will be the diameter of the circle.
    therefore, using Pythagoras theorem,
    $$AB^2 + BC^2 = AC^2$$
    $$4^2 + 3^2 = AC^2$$
    AC = 5 cm
    Hence, radius of circle = 2.5 cm

  • Question 9
    1 / -0
    The area of a right angled isosceles triangle, whose hypotenuse is equal to $$270$$ m, is:
    Solution
    Let $$\triangle ABC$$ be right angled with $$ \angle ABC = { 90 }^{ \circ  } $$ 
    and  $$AB = BC = a$$  and  hypotenuse  $$AC =  270 \text{ m}$$
    Using Pythagoras Theorem, we get
    $$ { AC }^{ 2}={ a }^{ 2 }+{ a }^{ 2 }= { 2a }^{ 2 } $$
    $$ { 2a }^{ 2 } = { 270 }^{ 2 } $$
    $$ { a }^{ 2 }=\dfrac { 270 \times 270 }{ 2 }\text { m }^{ 2 } $$
    Now, area of the $$ \triangle  ABC = \dfrac { 1 }{ 2 } \times a \times  a = \dfrac { 1 }{ 2 } { a }^{ 2 } $$
    $$ = \dfrac { 1 }{ 2 } \times \dfrac { 270 \times  270  }{ 2 }\text{ m}^{ 2 } $$
    $$ = 18,225\text { m}^{ 2 } $$
    Hence, option B is correct.

  • Question 10
    1 / -0
    The intercepts made by 3 parallel lines on a transverse line $$(l_1)$$ are in the ratio 1 : 1. A second transverse line $$(l_2)$$ making angle of $$30^{\circ}$$ with $$l_1$$ is drawn. The corresponding intercepts on $$l_2$$ are in the ratio:
    Solution
    The Triangle Proportionality Theorem states that if a line is parallel to one side of a triangle and it intersects the other two sides, then it divides those sides proportionally.

    On extending the tranversals $$l1$$ and $$l2$$, they join at $$A$$. join the other two ends with line $$BC$$ , forming a triangle $$- \ \triangle ABC$$

    $$\therefore \dfrac{AB}{BC}=\dfrac{PQ}{QR}$$

    Given
    $$ \dfrac{AB}{BC}=\dfrac{1}{1}$$

    $$\therefore \dfrac{PQ}{QR}=\dfrac{1}{1}$$ . 


    This can also be solved as below:

    If three parallel lines are intercepted by 2 transversals, in such a way that the two intercepts on one traversal are equal, then the intercepts formed by the other transversal are also equal.

    Given
    $$ \dfrac{AB}{BC}=\dfrac{1}{1} \ \implies \ AB=BC$$

    $$\therefore \dfrac{PQ}{QR}=\dfrac{1}{1}$$ 

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