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MATHS M.C.Q

Triangles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 274 questions.

Questions · Page 5 of 6

M.C.Q

Question 2011 Mark
In the given figure, $\text{EAD}\perp\text{BCD}.$ Ray FAC cuts ray EAD at a point A such that $\angle\text{EAF}=30^\circ.$ Also, in $\triangle\text{BAC},\angle\text{BAC}=\text{x}^\circ$ and $\angle\text{ABC}=(\text{x}+10).$ Then, value of x is:
Answer
  1. 25
    solution:
    $\angle\text{EAF}=\angle\text{CAD}$ (vertically opposite angles)
    $\Rightarrow\angle\text{CAD}=30^\circ$
    In $\triangle\text{ABD},$ by angle sum property
    $\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$
    $\Rightarrow(\text{x}+30)^\circ+(\text{x}+10)^\circ+90^\circ=180^\circ$
    $\Rightarrow2\text{x}+130^\circ=180^\circ$
    $\Rightarrow2\text{x}=50^\circ$
    $\Rightarrow\text{x}=25^\circ$
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Question 2021 Mark
In an isosceles $\triangle\text{ABC},$ if AB = AC and $\angle\text{A} = 90^\circ,$ Find $\angle\text{B}.$
Answer
  1. 45º
    Solution:
    We know that sum of all angle of a triangle is 180°
    So, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    $\angle\text{A} = 90^\circ$
    AB = AC
    $\angle\text{B} = \angle\text{C}$ (The angle opposite to equal side is also equal)
    $90^\circ + \angle\text{B} + \angle\text{C} = 180^\circ$
    $\angle\text{B} + \angle\text{C} = 180^\circ - 90^\circ$
    $\angle\text{B} + \angle\text{C} = 90^\circ ( \angle\text{B} = \angle\text{C})$
    $2\angle\text{B} = 90^\circ$
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Question 2031 Mark
$\triangle\text{ABC}\cong\triangle\text{PQR},$ then which of the following is true?
Answer
  1. CA = RP
    Solution:
    Corresponding sides are equal for two congruent triangles.
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Question 2061 Mark
In $\triangle\text{PQR},\ \angle\text{P}=60^\circ,\ \angle\text{Q}=50^\circ.$ Which side of the triangle is the longest?
Answer
  1. PQ
    Solution:
    In $\triangle\text{PQR},\ \angle\text{P}=60^\circ,\ \angle\text{Q}=50^\circ.$
    Now, by angle sum property, $\angle\text{P} +\angle\text{Q} +\angle\text{R} = 180^\circ$
    $60^\circ + 50^\circ + \angle\text{R} = 180^\circ$
    or, $ \angle\text{R} = 180^\circ - 110^\circ = 70^\circ$
    So, $\angle\text{R}$ is the largest angle and the side opposite to it, i.e, PQ will be the longest side.
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Question 2081 Mark
In $\triangle\text{ABC, }\angle\text{C}=\angle\text{A},$ BC = 4cm and AC = 5cm. then, AB =?
Answer
  1. 4cm
    Solution:
    In $\triangle\text{ABC,}$
    $\angle\text{C}=\angle\text{A}$
    $\Rightarrow\text{AB = BC}$ (sides opposite to equal angles are equal)
    $\Rightarrow\text{AB = 4cm}$
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Question 2091 Mark
In $\triangle\text{ABC}$ and $\triangle\text{DEF,}$ it is given that AB = DE and BC = EF. In order that $\triangle\text{ABC}\cong\triangle\text{DEF},$ we must have:
Answer
  1. $\angle\text{B}=\angle\text{E}$
    Solution:
    In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
    AB = DE and BC = EF
    SO, the induded angles should be equal for the triangle to be congurent by the SAS congruence criterion.
    Thus, we must have $\angle\text{B}=\angle\text{E}.$
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Question 2101 Mark
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is:
Answer
  1. 360º
    Solution:

    In $\triangle\text{ABC},$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    Now, $\angle\text{FAB}=180^\circ-\angle\text{A}\dots(1)$
    $\angle\text{DCA}=180^\circ-\angle\text{C}\dots(2)$
    $\angle\text{EBC}=180^\circ-\angle\text{B}\dots(3)$
    Adding equation (1), (2) and (3)
    $\angle\text{FAB}+\angle\text{DCA}+\angle\text{EBC}=180^\circ-\angle\text{A}+180^\circ-\angle\text{C}+180^\circ-\angle\text{B}$
    $=540^\circ-(\angle\text{A}+\angle\text{B}+\angle\text{C})$
    $=540^\circ-180^\circ$
    ⇒ Sum of all exterior angles = 360º
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Question 2111 Mark
For any $\triangle\text{ABC},$ AB + BC is always:
Answer
  1. Greater than AC
    Soluton:
    Sum of any two sides is greater than third side.
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Question 2131 Mark
In Fig. if $\text {DC}\parallel\text{DC}$ and $\text{AB}=\text{AC},$ the value of $\angle\text{ABD}$ is:
Answer
  1. 110º
    Solution:

    If $\text{AE}\parallel\text{DC}$ and AC is transversal,
    then $\angle\text{FAC}=70^\circ$ (opposite angles)
    Also $\angle\text{FAC}-=\angle\text{ACB}=70^\circ$ (Alternate angles)
    Since $\text{AB}=\text{AC},\triangle\text{ABC}$ is isosceles.
    So $\angle\text{ABC}=\angle\text{ACB}$
    $\Rightarrow\angle\text{ABC}=70^\circ$
    Now $\angle\text{ABD}=180^\circ-\angle\text{ABC}=180^\circ-70^\circ=100^\circ$
    Hence, correct option is (b).
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Question 2141 Mark
In a $\triangle\text{ABC},$ if $\angle\text{A}-\angle\text{B}=42^\circ$ and $\angle\text{B}-\angle\text{C}=21^\circ$ then $\angle\text{B}=?$
Answer
  1. 53º
    Solution:
    $\angle\text{A}-\angle\text{B}=42^\circ$
    $\Rightarrow\angle\text{A}=\angle\text{B}+42^\circ$
    $\angle\text{B}-\angle\text{C}=21^\circ$
    $\Rightarrow\angle\text{C}=\angle\text{B}-21^\circ$
    In $\triangle\text{ABC},$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{B}+42^\circ+\angle\text{B}+\angle\text{B}-21^\circ=180^\circ$
    $\Rightarrow3\angle\text{B}=159$
    $\Rightarrow\angle\text{B}=53^\circ$
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Question 2151 Mark
Answer
  1. 230º
    Solution:
    In $\triangle\text{ACO}$
    $\angle\text{ACO} + \angle\text{COA} + \angle\text{OAC} = 180^\circ$
    Now, $\angle\text{OAC} = 180^\circ$
    $⇒\ 80^\circ + 40^\circ + 180^\circ - \text{x}^\circ= 180^\circ$
    $⇒\ \text{x}^\circ = 120^\circ$
    $ \angle\text{BOD} = \angle\text{COA} = 40^\circ$ (Opposite angles)
    $\angle\text{BDO} = 70^\circ$
    In $\triangle\text{OBD}$
    $\angle\text{OBD} = 180^\circ - 40^\circ - 70^\circ = 70^\circ$
    Also, $\text{y}^\circ = 180^\circ - \angle\text{OBD} = 180^\circ - 700^\circ = 110^\circ$
    $⇒\ \text{x}^\circ + \text{y}^\circ = 120^\circ + 110^\circ = 230^\circ$
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Question 2161 Mark
In a triangle, an exterior angle at a vertex is 95º and its one of the interior opposite angle is 55º, then the measure of the other interior angle is:
Answer
  1. 40º
    Solution:
    Let the other interior opposite angle be xº.
    Then, we have xº + 55º = 95º
    ⇒ xº = 95º - 55º = 40º
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Question 2181 Mark
In $\triangle\text{RST}$ what is the value of x?
Answer
  1. 100º
    Solution:

    In $\triangle\text{RST}$
    $\angle\text{R}+\angle\text{S}+\angle\text{T}=180^\circ$
    $\Rightarrow2\text{a}^\circ+\text{x}^\circ+2\text{b}^\circ=180^\circ$
    $\Rightarrow\text{x}^\circ=180^\circ-2(\text{a}+\text{b})^\circ\dots(1)$
    Now in $\triangle\text{ROT}$
    $\angle\text{ORT}+\angle\text{ROT}+\angle\text{OTR}=180^\circ$
    $\Rightarrow\text{a}^\circ+140^\circ+\text{b}^\circ=180^\circ$
    $\Rightarrow(\text{a}+\text{b})^\circ=180^\circ-140^\circ=40^\circ\dots(2)$
    From (1) and (2)
    $\text{x}^\circ=180^\circ-2(40^\circ)$
    $\Rightarrow\text{x}=100^\circ$
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Question 2191 Mark
If O is any point in the interior of $\triangle\text{ABC},$ then, which of the following is true?
Answer
  1. $(\text{OA + }\text{OB} + \text{OC})\ >\ \frac{1}{2}(\text{AB + }\text{BC} + \text{CA})$
    Solution:
    Sum of any two sides of a triangle is greater than the third side.
    Join O with all sides of the triangle, we have
    OA + OB > AB ...(i)
    OA + OC > AC ...(ii)
    and, OB + OC > BC ...(iii)
    Adding the three inequalities i.e. (i) + (ii) + (iii), we get
    2(OA + OB + OC) > AB + BC + AC
    i.e. $(\text{OA + }\text{OB} + \text{OC})\ >\ \frac{1}{2}(\text{AB + }\text{BC} + \text{CA})$
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Question 2201 Mark
The side BC of $\triangle\text{ABC}$ is produced to a point D. The bisector of $\angle\text{A}$ meets side in L. If $\angle\text{ABC} = 30^\circ$ and $\angle\text{ACD} = 115^\circ$ then $\angle\text{ALC} =$
Answer
  1. $72\frac{1^\circ}{2}$
    Solution:

    $\angle\text{C} = 180^\circ - \angle\text{ACD} = 180^\circ - 115^\circ = 65^\circ$
    In $\triangle\text{ABC}$
    $\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
    $⇒\angle\text{A} = 180^\circ - 30^\circ - 65^\circ$
    $⇒\angle\text{A} = 85^\circ$
    Now in $\triangle\text{ALC}$
    $\angle\text{ALC} + \angle\text{LAC} + \angle\text{C} = 180^\circ$
    $⇒\angle\text{ALC} = 180^\circ - \angle\text{LAC} - \angle\text{C}$
    $=180^\circ-\frac{\angle\text{C}}{2}-\angle\text{C}$
    $=180^\circ-\frac{85^\circ}{2}-65^\circ$
    $=\frac{145^\circ}{2}$
    $=72\frac{1^\circ}{2}$
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Question 2211 Mark
Answer
  1. 120º
    Solution:

    In $\triangle\text{ABD}$
    $\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$
    $\Rightarrow55^\circ+\angle\text{DBA}+25^\circ=180^\circ$
    $\Rightarrow\angle\text{DBA}=180^\circ-55^\circ-25^\circ$
    $=180^\circ-80^\circ$
    $\Rightarrow\angle\text{DBA}=100^\circ$
    So, $\angle\text{DBC}=180^\circ-\angle\text{DBA}$
    $=180^\circ-100^\circ$
    $\Rightarrow\angle\text{DBC}=80^\circ$
    Now, in $\triangle\text{EBC}$
    $\angle\text{E}+\angle\text{EBC}+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{E}+80^\circ+40^\circ=180^\circ$ $\big(\angle\text{DBC}=\angle\text{EBC}\big)$
    $\Rightarrow\angle\text{E}=180^\circ-120^\circ=60^\circ$
    Also, $\text{x}=180^\circ-\angle\text{E}=180^\circ-60^\circ$
    $\Rightarrow\text{x}=120^\circ$
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Question 2221 Mark
Answer
  1. $\frac{3}{2}\text{X}^\circ$
    Solution:
    From Figure, $\angle\text{DOC} = 180^\circ - \angle\text{AOD}$ (Both are Supplementary)
    $⇒\ \angle\text{DOC} = 180^\circ−3\text{y}^\circ$
    Also, $\angle\text{ACB} = 180^\circ- \angle\text{A} - \angle\text{B}$
    $⇒\angle\text{ACB} = 180^\circ - \text{x}^\circ−2\text{x}^\circ = 180^\circ - 3\text{x}^\circ$
    And $ \angle\text{ACD} = 180^\circ - \angle\text{ACB}$
    $= 180^\circ - (180^\circ- 3\text{x}^\circ)$
    $⇒\angle\text{ACD}=3\text{x}^\circ$
    Now, in $\triangle\text{OCD}$
    $\angle\text{DOC} + \angle\text{OCD} + \angle\text{D} = 180^\circ$
    $180^\circ − 3\text{y}^\circ + 3\text{x}^\circ + \text{y}^\circ = 180^\circ\ [\angle\text{OCD} = \angle\text{ACD}]$
    $⇒2\text{y}^\circ=3\text{x}^\circ$
    $\Rightarrow\ \text{Y}=\frac{3}{2}\text{X}^\circ$
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Question 2231 Mark
In Fig. AB and CD are parallel lines and transversal EF intersect them at P and Q respectively. If $\angle\text{APR}=25^\circ,\angle\text{RQC}=30^\circ$ and $\angle\text{CQF}=65^\circ,$ then:
Answer
  1. x = 55º, y = 40º
    Solution:

    $\angle\text{OQP}=180^\circ-\angle\text{OQF}$
    $=180^\circ-(30^\circ+65^\circ)$
    $\Rightarrow\angle\text{OQP}=85^\circ\dots(1)$
    $\angle\text{APQ}=\angle\text{CQF}$ (Corresponding angles)
    $\Rightarrow25^\circ+\text{y}^\circ=65^\circ$
    $\Rightarrow\text{y}^\circ=65^\circ-25^\circ$
    $\Rightarrow\text{y}^\circ=40^\circ$
    Now in $\triangle\text{OPQ}$
    $\angle\text{O}+\angle\text{OPQ}+\angle\text{PQO}=180^\circ$
    $\Rightarrow\text{x}^\circ+40^\circ+85^\circ=180^\circ$
    $\text{x}^\circ=180^\circ-85^\circ-40^\circ=55^\circ$
    $\Rightarrow\text{x}^\circ=55^\circ,\text{y}=40^\circ$
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Question 2241 Mark
In triangles ABC and PQR, AB = AC, $\angle\text{C} = \angle\text{P}$ and $\angle\text{B} = \angle\text{Q}.$ The two triangles are:
Answer
  1. Isosceles but not congruent.
    Solution:
    Given: $\triangle\text{ABC}$ and $\triangle\text{PQR}, \ \text{AB = AC}, \ \angle\text{C} = \angle\text{P}$ and $\angle\text{B} = \angle\text{Q}.$

    $\text{AB = AC}$
    $\Rightarrow\ \angle\text{B} = \angle\text{C}$ (opposite angles to equal sides are equal)
    Hence, $\triangle\text{ABC}$ is an isosceles triangle.
    $\angle\text{C} = \angle\text{P}$ and $\angle\text{B} = \angle\text{Q}$ (given)
    $⇒\angle\text{P} = \angle\text{Q}\ \ (∴\angle\text{B} = \angle\text{C})$
    $⇒ \text{QR} = \text{PR}$ (opposite sides to equal angles are equal)
    Hence, $\triangle\text{PQR}$ is an isosceles triangle.
    So, the two triangles are isosceles but not congruent.
    As AAA is not the criterion for a triangle to be congruent.
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Question 2251 Mark
In the given figure, two rays BD and CE intersect at a point A. The side BC of $\triangle\text{ABC}$ have been produced on both sides to points F and G respectively. If $\angle\text{ABF}=\text{x}^\circ,\angle\text{ACG}=\text{y}^\circ$ and $\angle\text{DAE}=\text{z}^\circ$ then z =?
Answer
  1. x + y - 180
    Solution:
    $\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (linear pair)
    $\Rightarrow\text{x}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\angle\text{ABC}=180^\circ-\text{x}$
    $\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (linear pair)
    $\Rightarrow\text{y}+\angle\text{ACB}=180^\circ$
    $\Rightarrow\angle\text{ACB}=180^\circ-\text{y}$
    In $\triangle\text{ABC},$ by angle sum property
    $\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$
    $\Rightarrow(180^\circ-\text{x})+(180^\circ-\text{y})+\angle\text{BAC}=180^\circ$
    $\Rightarrow\angle\text{BAC}-\text{x}-\text{y}+180^\circ=0$
    $\Rightarrow\angle\text{BAC}=\text{x}+\text{y}-180^\circ$
    Now, $\angle\text{EAD}=\angle\text{BAC}$ (vertically opposite angles)
    $\Rightarrow\text{z}=\text{x}+\text{y}-180^\circ$
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Question 2261 Mark
Answer
  1. $\frac{3}{2}\text{x}^\circ$
    Solution:

    From figure,
    $\angle\text{DOC}=180^\circ-\angle\text{AOD}$ (Both are Supplementary)
    $\Rightarrow\angle\text{DOC}=180^\circ-3\text{y}^\circ$
    Also, $\angle\text{ACB}=180^\circ-\angle\text{A}-\angle\text{B}$
    $\Rightarrow\angle\text{ACB}=180^\circ-\text{x}^\circ-2\text{x}^\circ=180^\circ-3\text{x}^\circ$
    And $\angle\text{ACD}=180^\circ-\angle\text{ACB}$
    $=180^\circ-(180^\circ-3\text{x}^\circ)$
    $\Rightarrow\angle\text{ACD}=3\text{x}^\circ$
    Now, in $\triangle\text{OCD}$
    $\angle\text{DOC}+\angle\text{OCD}+\angle\text{D}=180^\circ$
    $180^\circ-3\text{y}^\circ+3\text{x}^\circ+\text{y}^\circ=180^\circ$ $\big[\angle\text{OCD}=\angle\text{ACD}\big]$
    $\Rightarrow2\text{y}^\circ=3\text{x}^\circ$
    $\Rightarrow\text{y}=\frac{3}{2}\text{x}^\circ$
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Question 2271 Mark
In fig, in $\triangle\text{ABC},$ $AB = AC,$ then the value of $x$ is:
Answer
Triangle $ABC$ is an iscosceles triangle and hence in the triangle other two angles are $50$ and $50.$
Therefore, $X = 180 - 50 = 130$
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Question 2281 Mark
In Fig. $\text{AB}\perp\text{BE}$ and $\text{FE}\perp\text{BE},$ If $\text{BC}=\text{DE}$ and $\text{AB}=\text{EF},$ then $\triangle\text{ABD}$ is congruent to:
Answer
  1. $\triangle\text{FEC}$
    Solution:

    AB = EF
    BC = DE
    BC + CD = DE + CD (adding CD both sides)
    BC + CD = BD, DE + CD = CE
    So BD = CE
    Now Consider $\triangle\text{ABD},$ & $\triangle\text{FEC}$
    $\text{AB}=\text{FE}$
    $\text{BD}=\text{EC}$
    $\angle\text{ABD}=\angle\text{FEC}=90^\circ$
    So $\triangle\text{ABD}\cong\triangle\text{FEC}$ by SAS creterion.
    Hence, correct option is (d).
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Question 2301 Mark
In the given figure, BO and CO are bisectors of $\angle\text{B}$ and $\angle\text{C}$ respectively. If $\angle\text{A}=50^\circ$ then $\angle\text{BOC}=?$
Answer
  1. 115º
    Solution:
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle sum property)
    $\Rightarrow50^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{B}+\angle\text{C}=130^\circ$
    $\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=65^\circ$
    In $\triangle\text{OBC},$
    $\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$ (Angle sum property)
    $\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
    $\Rightarrow65^\circ+\angle\text{BOC}=180^\circ$
    $\Rightarrow\angle\text{BOC}=115^\circ$
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Question 2311 Mark
An angle is 14º more than its complement. Find its measure.
Answer
  1. 52º
    Solution:
    Let the angle = x
    Its complement = 90º - x
    According to the question, x is 14° more than its complement,
    ⇒ x = (90° - x) + 14°
    ⇒ x + x = 104°
    ⇒ 2x = 104°
    $\Rightarrow\ \text{X}=\frac{104^\circ}{2}=52^\circ$
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Question 2331 Mark
In a triangle, an exterior angle at a vertex is 95º and its one of the interior opposite angle is 55º, then the measure of the other interior angle is:
Answer
  1. 40º
    Solution:
    Let the other interior opposite angle be xº.
    Then, we have
    xº + 55º = 95º
    ⇒ xº = 95º - 55º = 40º
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Question 2341 Mark
In the adjoining figure, O is Mid–point of AB. If $\angle\text{ACO} = \angle\text{BDO},$ then $\angle\text{OAC}$ is equal to:
Answer
  1. $\angle\text{OBD}$
    Solution:
    In $\triangle\text{OAC}$ and $\triangle\text{OBD},$
    AO = OB As O is the mid-point of AB
    $\triangle\text{AOC}=\ \triangle\text{BOD}$ (Vertically Opposite Angles)
    $\triangle\text{AOC}=\ \triangle\text{BDO}$ (Given)
    $\therefore\ \triangle\text{OAC}=\ \triangle\text{OBD}$ (AAS Axiom)
    $\therefore\ \triangle\text{OAC}=\ \triangle\text{OBD}$ (C.P.C.T.)
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Question 2351 Mark
If the measure of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?
Answer
  1. 45º
    Solution:
    The measures of angles of a triangle are in ratio 3 : 4 : 5.
    Let the angles be 3x, 4x, and 5x.
    in any triangle, sum of all angles = 180º
    ⇒ 3x + 4x + 5x = 180º
    ⇒ 12x = 180º
    ⇒ x = 15º
    So, smallest angle = 3 × 15º = 45º
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Question 2361 Mark
It is given that $\triangle\text{ABC} ≅\triangle\text{FDE}$ and AB = 5cm, $\angle\text{B} = 40^\circ$ and $\angle\text{A} = 80^\circ$ Then which of the following is true?
Answer
  1. $\text{DF} = 5\text{cm}, \ \angle\text{E} = 60^\circ$
    Solution:
    Given that: In $\triangle\text{ABC}, \ \text{AB} = 5\text{cm},\ \angle\text{B} = 40^\circ$ and $\angle\text{A} = 80^\circ$
    Using angles sum property of triangle, we have
    $\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
    $⇒ 80^\circ + 40^\circ + \angle\text{C} = 180^\circ$
    $⇒ 120^\circ + \angle\text{C} = 180^\circ$ [$\therefore\ \angle\text{B} = 40^\circ$ and $\angle\text{A} = 80^\circ$]
    $⇒ \angle\text{C} = 180^\circ – 120^\circ$
    $⇒ \angle\text{C} = 60^\circ$
    It is given that $\triangle\text{ABC} ≅\triangle\text{FDE},$ so we have
    AB = FD, BC = DE and $\text{AC}=\text{FE}\ \&\ \angle\text{A} = \angle\text{F}, \ \angle\text{B} = \angle\text{D}$ and $\angle\text{C} =
    \angle\text{E}$
    ⇒ AB = FD = 5cm and $\angle\text{C} = \angle\text{E} = 60^\circ$
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Question 2371 Mark
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is:
Answer
  1. 360º
    Solution:

    In $\triangle\text{ABC}$
    $\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
    Now, $\angle\text{FAB} = 180^\circ- \angle\text{A}\ \cdots(\text{i})$
    $\angle\text{DCA} = 180° - \angle\text{C}\ \cdots(\text{ii})$
    $\angle\text{EBC} = 180^\circ - \angle\text{B}\ \cdots(\text{iii})$
    Adding equations (i), (ii) and (iii)
    $\angle\text{FAB} + \angle\text{DCA} + \angle\text{EBC} = 180^\circ \\- \angle\text{A} + 180^\circ - \angle\text{C} + 180^\circ - \angle\text{B}$
    $= 540^\circ - (\angle\text{A} + \angle\text{B} + \angle\text{C})$
    $= 540^\circ - 180^\circ$
    ⇒ Sum of All exterior angles = 360°
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Question 2391 Mark
In the following, write the correct answer.
Which of the following is not a criterion for congruence of triangles?
Answer
  1. SSA
    Solution:
    We know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
    Also, criterion for congruence of triangles are SAS (Side-Angle-Side), ASA (Angle-Side- Angle), SSS (Side-Side-Side) and RHS (right angle-hypotenuse-side).
    So, SSA is not a criterion for congruence of triangles.
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Question 2401 Mark
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, $\angle\text{BAD}=$
Answer
  1. 55°
    Solution:

    If AD is the median, then D is the mid-point of BC.
    $\text{BD}=\text{DC}$
    So consider $\triangle\text{ADB}$ and $\triangle\text{ADC}$
    $\text{AD}=\text{AD}$ (common)
    $\text{DB}=\text{DC}$
    $\text{BA}=\text{CA}$
    So by SSS, $\triangle\text{ADB}\cong\triangle\text{ADC}$
    Now $\angle\text{B}=\angle\text{C}=35^\circ$
    $\Rightarrow\angle\text{BAD}=\angle\text{DAC}$
    So in $\triangle\text{ABC},$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    $\Rightarrow2\angle\text{BAD}+35^\circ+35^\circ=180^\circ$
    $\Rightarrow2\angle\text{BAD}=110^\circ$
    $\Rightarrow\angle\text{BAD}=55^\circ$
    Hence, correct option is (a).
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Question 2411 Mark
In the following, write the correct answer.
In $\triangle\text{ABC}$ if BC = AB and $\angle\text{B}=80^{\circ}$ then is equal to:
Answer
  1. 50º

    Solution: In $\triangle\text{ABC}$ we have BC = AB
    But $\angle\text{B}=80^{\circ}$ $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
    $=\angle\text{A}+80^{\circ}\angle\text{A}=180^{\circ}$ $=2\angle\text{A}=100^{\circ}$
    $=\angle\text{A}=100^{\circ}\div2=50^{\circ}$
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Question 2431 Mark
If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is:
Answer
  1. 135º
    Solution:

    In $\triangle\text{ABC},$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{A}+90^\circ+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{A}+\angle\text{C}=90^\circ\dots(1)$
    Now, in $\triangle\text{AOC},$
    $\angle\text{COA}+\angle\text{OAC}+\angle\text{OCA}=180^\circ$
    $\Rightarrow\angle\text{COA}+\frac{\angle\text{A}}{2}+\frac{\angle\text{C}}{2}=180^\circ$ {AO and CO bisects angle (\angle\text{A}$ and $\angle\text{C}$}
    $\Rightarrow\angle\text{COA}=180-\Big(\frac{\angle\text{A}+\angle\text{C}}{2}\Big)$
    $=180^\circ-\Big(\frac{90^\circ}{2}\Big)$ {From (1)}
    $=100^\circ-45^\circ$
    $=135^\circ$
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Question 2441 Mark
In the adjoining figure, $\text{AB}\bot\text{BE}$ and $\text{FE}\bot\text{BE}.$ If $AB = FE$ and $BC = DE,$ then.
Answer
Given:
$AB = FE, BC = ED,$
$\text{AB}\bot\text{BE}$ and $\text{FE}\bot\text{BE}.$
To Prove: $AD = FC$
Proof: In $\triangle\text{ABD}$ and $\triangle\text{FEC},$
$AB = FE ...(1) ($Given$)$
$\triangle\text{ABD}=\triangle\text{FEC} ...(2)$
Each $= 90^\circ$
$BC = ED ($Given$)$
$\Rightarrow BC + CD = ED + DC$
$\Rightarrow BD = EC ...(3)$
In view of $(1), (2)$ and $(3),$
$\triangle\text{ABD}\cong\triangle\text{FEC}$ using $\text { SAS}$ congruence rule.
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Question 2451 Mark
Answer
  1. 30º
    Solution:

    $\angle\text{PBC}=\angle\text{QCD}$ (Corresponding angles, OP || CQ and BC is transverse)
    $\Rightarrow\angle\text{PBC}=70^\circ$
    Now, $\angle\text{PBA}+\angle\text{ABC}+\angle\text{PBC}$
    $\Rightarrow20^\circ+\angle\text{ABC}=70^\circ$
    $\Rightarrow\angle\text{ABC}=50^\circ$
    In $\triangle\text{ABC},$
    $\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ\dots(1)$
    Now, $\angle\text{ABC}=\angle\text{BAC}=50^\circ$ $($isosceles $\triangle)$
    And, $\angle\text{ACB}=180^\circ-(70^\circ+\text{x})$
    From (1)
    $50^\circ+50^\circ+180^\circ-(70^\circ+\text{x})=180^\circ$
    $\Rightarrow\text{x}=30^\circ$
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Question 2461 Mark
In the given figure, AD is the median, then $\angle\text{BAD}$ is:
Answer
  1. 50º
    Solution:
    In the given $\angle\text{ABC},\ \text{AB}\ =\ \text{AC}$
    Hence, $\angle\text{B} = \angle\text{C} = 40^\circ$
    Now, by angle sum property,
    $\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
    $\Rightarrow\ \angle\text{A} + 40^\circ + 40^\circ = 180^\circ$
    $\Rightarrow\ \angle\text{A} = 100^\circ$
    Now, in $\triangle\text{BAD}$ and $\triangle\text{DAC}.$
    AB = AC (given)
    BD = DC (D is the mid-point of BC)
    $\angle\text{ABD} = \angle\text{ACD} = 40^\circ$
    $\angle\text{A}= \frac{1}{2} × 100^\circ = 50^\circ $
    Hence by SAS, $\triangle\text{BAD}$ and $\triangle\text{DAC}$ are congruent.
    This means, $\angle\text{BAD} = \angle\text{CAD} = \frac{1}{2}$ Therefore, $\triangle\text{BAD}=50^\circ.$
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Question 2471 Mark
In the adjoining figure, $\text{ABCD}$ is a quadrilateral in which $BN$ and $DM$ are drawn perpendiculars to $AC$ such that $BN = DM$. If $OB = 4\ cm$. then $BD$ is:
Answer
In Triangle $DMO$ and triangle $BNO,$
$BN = DM$ and $\angle\text{DMO} = \angle\text{BNO} (90^\circ)$
$\angle\text{DMO} = \angle\text{BNO}$
Therefore, Triangle $\text{DMO}$ and triangle $\text{BNO}$ are congruent by $\text{AAS}$ criteria
Therefore, $OB = OD ($by $\text{CPCT})$
So $OD = 4\ cm, BD = OD + OB = 4 + 4 = 8\ cm$
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Question 2481 Mark
In $\triangle\text{ABC, }\angle\text{A}=40^{\circ}$ and $\angle\text{B}=60^{\circ}.$ Then the longest side of $\triangle\text{ABC}$ is:
Answer
  1. AB
    Solution:

    In $\triangle\text{ABC},$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ ...(Using Angle Sum Property)
    $\Rightarrow40^{\circ}+60^{\circ}+\angle\text{C}=180^{\circ}$
    $\Rightarrow\angle\text{C}=80^{\circ}$
    We know that, the greater angle has the longest side opposite to it.
    Since $\angle\text{A}<\angle\text{B}<\angle\text{C, }\text{BC < AC < AB}.$
    So, the longest side is AB.
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Question 2491 Mark
In $\triangle\text{ABC},$ if $\angle\text{B} = 30^\circ$ and $\angle\text{C} = 70^\circ,$ then which of the following is the longest side?
Answer
  1. BC
    Solution:
    Since the sum of all sides of a triangle is 180°.
    So, $\angle\text{C}=70^\circ,\ \angle\text{B}=70^\circ,\ \angle\text{A}=80^\circ,$
    We have a theorem which states that the side opposite to the greatest angle is the longest.
    So, the side opposite to angle A is the longest.
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Question 2501 Mark
In the adjoining figure, $\triangle\text{ABC}\cong\triangle\text{ADC}.$ If $\angle\text{BAC} = 30^\circ$ and $\angle\text{ABC} = 100^\circ$ then $\angle\text{ACD}$ is equal to:
Answer
In triangle ABC, $\angle\text{BAC} = 30^\circ$ and $\angle\text{ABC} = 100^\circ ($Given$)$
$\angle\text{BAC} + \angle\text{ABC} + \angle\text{BCA} = 180^\circ$
$\angle\text{BAC} = 50^\circ$
Also $\angle\text{ACD} = 50^\circ($Since, $\triangle\text{ABC}\cong\triangle\text{ADC})$
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M.C.Q - Page 5 - MATHS STD 9 Questions - Vidyadip