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MATHS M.C.Q

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

Questions · Page 6 of 6

M.C.Q

Question 2511 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, $\angle\text{C} = 70^\circ,\ \angle\text{B} = 30^\circ,\ \angle\text{A} = 80^\circ.$
    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 2521 Mark
The cost of turfing a triangular field at the rate of $Rs. 45$ per $100\ m^2$ is $Rs. 900.$ If the double the base of the triangle is $5$ times its height, then its height is:
Answer
Cost of turfing a triangilar field at the rate of $Rs. 45$ per $100 = Rs. 900$
$\frac{\text{Arae}\times45}{100}=900$
$\Rightarrow$ Area $= 2000\  sq.cm$
According to question,
$2 \times$ Base $= 5 \times$ Height
$\Rightarrow\ \text{Base}=\frac{\text{Height}\times5}{2}$
Area of a triangle $= 2000\  sq.cm$
$\Rightarrow\ \frac{1}{2}\times\text{Base}\times\text{Height}=2000$
$\Rightarrow\ \frac{1}{2}\times\frac{\text{Height}\times5}{2}\times\text{Height}=2000$
$\Rightarrow\ \text{(Height)}^2=1600$
$\Rightarrow\ \text{Height}=40\ \text{cm}$
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Question 2531 Mark
In the adjoining figure, $\triangle\text{ABG}\cong\triangle\text{ADG}.$ If $\angle\text{BAC} = 30^\circ$ and $\angle\text{ABC} = 100^\circ$ then $\angle\text{ACD}$ is equal to:
Answer
  1. 50º
    Solution:
    In $\triangle\text{ABC}, \triangle\text{BAC} = 30^\circ$ and $\triangle\text{ABC}= 100^\circ$ (Given)
    $\angle\text{BAC} + \angle\text{ABC}+ \angle\text{BCA} = 180^\circ$
    $\angle\text{BCA}= 50^\circ$
    Also $\angle\text{ACD}= 50^\circ$ (Since, $\triangle\text{ABC}\cong\triangle\text{ADC}$).
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Question 2541 Mark
If all the three angles of a triangle are equal, then each one of them is equal to:
Answer
  1. 60º
    Solution:
    Let the measure of each angle be xº
    Now, the sum of all angles of any triangle is 180°
    Thus, xº + xº + xº = 180º
    i.e. 3xº = 180º
    i.e. xº = 60º
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Question 2551 Mark
Two sides of a triangle are of length 4cm and 2.5cm. The length of the third side of the triangle cannot be.
Answer
  1. 6.5cm
    Solution:
    Length of the greatest side of a triangle must be less than the sum of the other two sides.
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Question 2561 Mark
In the adjoining fig. $AB = AC$. If $\angle\text{C} = 50^\circ,$ then the value of $x$ and $y$ are:
Answer
In triangle$ ABC, AB = AC,$ hence their opposite angles will be equal.
$\Rightarrow\angle\text{B}=\angle\text{C}= 50^\circ$
$\Rightarrow y = 50^\circ$
Now, by angle sum property,
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
or$, x + 50^\circ + 50^\circ = 180^\circ$
or$, x + 100^\circ = 180^\circ$
$\Rightarrow x = 80^\circ$
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Question 2571 Mark
Line segments AB and CD intersect at O such that AC || DB. If $\angle\text{CAB}=45^\circ$ and $\angle\text{CDB}=55^\circ,$ then $\angle\text{BOD}=$
Answer
  1. 80º
    Solution:

    AC || BD
    And, AB is transverse to these parallal lines
    So $\angle\text{CAB}=\angle\text{ABD}$ (Alternate angles)
    $\Rightarrow\angle\text{ABD}=45^\circ$
    Now In $\triangle\text{BOD}$
    $\angle\text{BOD}+\angle\text{ODB}+\angle\text{DBA}=180^\circ$
    $\angle\text{DBA}=\angle\text{ABD}=45^\circ,\angle\text{ODB}=55^\circ$
    So $\angle\text{BOD}=180^\circ-45^\circ-55^\circ$
    $=80^\circ$
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Question 2581 Mark
In a $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}$ then A : B : C =?
Answer
  1. 4 : 3 : 2
    Solution:
    Given that $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}=\text{k}.$
    $\Rightarrow\angle\text{A}=\frac{\text{k}}{3},\angle\text{B}=\frac{\text{k}}{4}\text{ and}\ \angle\text{C}=\frac{\text{k}}{6}$
    $\Rightarrow\text{A}:\text{B}:\text{C}=\frac{\text{k}}{3}:\frac{\text{k}}{4}:\frac{\text{k}}{6}$
    $\Rightarrow\text{A}:\text{B}:\text{C}=\frac{1}{3}:\frac{1}{4}:\frac{1}{3}$
    The LCM of 3, 4 and 6 is 12.
    Multiply by 12 throughout.
    ⇒ A : B : C = 4 : 3 : 2
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Question 2611 Mark
In figure, for which value of x is $11||12?$
Answer
  1. 47
    Solution:
    Let if $11||12$ and AB is transverse to it
    Then,
    $\angle\text{PBA}$ should be equal to $\angle\text{BAS}$ (Alternate angles)
    So if $11||12,$ then $\angle\text{BAS}=70^\circ$
    $⇒\angle\text{BAC} = 78^\circ - 35^\circ = 43^\circ\ ...\ (\text{i})$
    Now, in $\angle\text{ABC}$
    $\text{x}^\circ + \angle\text{C} + \angle\text{BAC} = 180^\circ$
    $⇒ \text{x}^\circ + 90^\circ + 43^\circ = 180^\circ$
    $⇒ \text{x}^\circ = 180^\circ - 90^\circ - 43^\circ = 47^\circ$
    $⇒ \text{x}^\circ = 47^\circ$
    So if $\text{x}^\circ = 47^\circ$ then $11||12$
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Question 2621 Mark
In the following, write the correct answer.
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$ if AB = AC, $\angle\text{A}=\angle\text{D}.$ The two triangles are:
Answer
  1. AC = DE
    Solution:
    In $\triangle\text{ABC},$
    AB = DF
    $\angle\text{A}=\angle\text{D}$
    We know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle.
    AC = DE.
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Question 2641 Mark
If triangle PQR is right angled at Q, then
Answer
  1. PR > PQ
    Solution:
    Then the hypotenuse should be always greater than the remaining two sides.
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Question 2651 Mark
In fig, $ AC = BC $and $\angle\text{ACY} = 140^\circ.$ Find $X$ and $Y:$
​​
Answer
The two equal angles are $70$ since angle
$C = 180 - 140$
$ = 40X$
$Y = 180 - 70$
$= 110$
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Question 2661 Mark
In the given figure, the sides CB and BA of $\triangle\text{ABC}$ have been produced to D and E respectively such that $\angle\text{ABD}=110^\circ$ and $\angle\text{CAF}=135^\circ,$ Then $\angle\text{ACB} = ?$
Answer
  1. 65º
    Solution:
    We can find $\angle\text{CBA}$ as follows:
    Given that $\angle\text{EBA}=110^\circ$
    $\angle\text{CBA} = 180 - 110$ ... (linear pair)
    $=70^\circ$
    Given $\angle\text{CAD}=135^\circ$
    So, $\angle\text{CAB} = 180 - 135$ ... (linear pair)
    $=45^\circ$
    So, $\angle\text{ACB} = 180 - (70 + 45)$ ... (angle sum property of triangle)
    $=65^\circ$
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Question 2671 Mark
In $\triangle\text{ABC},$ if $\angle\text{A}=100^\circ,\text{AD}$ bisects $\angle\text{A}$ and $\text{AD}\perp\text{BC}.$ Then, $\angle\text{B}=$
Answer
  1. 40º
    Solution:

    $\text{AD}\perp\text{BC}$ and AD bisects $\angle\text{A}.$
    $\Rightarrow\angle\text{BAD}=\angle\text{CAD}=50^\circ$
    In Right $\triangle\text{ADB}$
    $\angle\text{BAD}=50^\circ,\angle\text{ADB}=90^\circ$
    Also sum of all interior angles = 180º
    $\Rightarrow\angle\text{BAD}+\angle\text{ADB}+\angle\text{B}=180^\circ$
    $\Rightarrow\angle\text{B}=180^\circ-50^\circ-90^\circ$
    $\Rightarrow\angle\text{B}=40^\circ$
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Question 2681 Mark
PQR is a right-angled triangle in which $\angle\text{P} = 90^\circ$ and PQ = PR. What is the value of $\angle\text{Q}$ and $\angle\text{R}.$
Answer
  1. 45º, 45º
    Solution:
    $\angle\text{P} = 90^\circ$
    Since, PQ = PR
    $\angle\text{Q} = \angle\text{R}$
    So, $\angle\text{Q} = \angle\text{R}=\frac{180^\circ\ -\ \angle\text{P}}{2}=\frac{180^\circ-90^\circ}{2}=\frac{90^\circ}{2}=45^\circ$
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Question 2691 Mark
In $\triangle\text{PQR},\ \angle\text{R} = \angle\text{P} $ and QR = 4cm and PR = 5cm. Then the length of PQ is:
Answer
  1. 4cm
    Solution:
    In a triangle, if two of its angles are equal then the sides opposite to equal angles are also equal.
    In $\triangle\text{PQR},\ \angle\text{R} = \angle\text{P}$
    ⇒ QR (side opposite to $ \angle\text{P}$) = PQ (side opposite to $ \angle\text{R}$)
    Given that, QR = 4cm
    ⇒ PQ = 4cm
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Question 2701 Mark
In Figure, if $\text{EC}\ ||\text{ AB}, \angle\text{ECD} = 70^\circ$ and $\angle\text{BDO} = 20^\circ,$ then $\angle\text{OBD}$ is:
Answer
  1. 50°
    Solution:
    $\text{EC}\ ||\text{ AB}$ and CD is transverse to it.
    Now $\angle\text{ECD} = \angle\text{AOD} = 70^\circ$ (Corresponding angles)
    In $\angle\text{OBD}$
    $\angle\text{OBD} + \angle\text{BOD} + \angle\text{ODB} = 180^\circ$
    $\angle\text{BOD} = 180^\circ - \angle\text{AOD} = 180^\circ - 70^\circ = 110^\circ$
    $\angle\text{ODB} = 20^\circ$ (Given)
    So, $\angle\text{OBD} = 180^\circ - \angle\text{BOD} - \angle\text{ODB}$
    $= 180^\circ - 110^\circ - 20^\circ$
    $= 50^\circ$
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Question 2711 Mark
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, $\angle\text{BAD} =$
Answer
  1. 55º
    Solution:
    It is given that $\angle\text{B} = 35^\circ, \text{AB} = \text{AC}$ and Ad is the median of BC
    We know that in isosceles triangle the median from he vertex to the unequal side divides it into two equal part at right angle.
    Therefore,
    $\angle\text{ADB} = 90^\circ$
    $\angle\text{B} = \angle\text{ADB} + \angle\text{A}= 180^\circ$ (Property of triangle)
    $35^\circ + 90^\circ + \angle\text{A} = 180^\circ$
    $\angle\text{A} = 180^\circ - 125^\circ$
    $\angle\text{A} = 55^\circ$
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Question 2721 Mark
In the adjoining figure, AB = FC, EF = BD and $\angle\text{AFE} = \angle\text{CBD}.$ Then the rule by which $\triangle\text{AFE}\cong\triangle\text{CBD}.$
Answer
  1. SAS
    Solution:
    ASA
    In $\triangle\text{DBC}$ and $\triangle\text{AEF},$ we have
    AB = FC (given) by adding BF on both sides
    AF = CB
    $\triangle\text{AEF}=\ \triangle\text{CBD}$ (given)
    EF = BD (given)
    Hence, $\triangle\text{AFE}\cong\triangle\text{CBD}$ by SAS as the corresponding sides and their included angles are equal.
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Question 2731 Mark
Which of the following is not a criterion for congruence of triangles?
Answer
  1. SSA
    Solution:
    The criteria for congruence of triangles are SSS(Side-Side-Side), SAS(Side-Angle-Side), ASA(Angle-Side- Angle) and RHS(Right angle-Hypotenuse-Side).
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Question 2741 Mark
Answer
  1. 7cm
    Solution:

    Consider $\triangle\text{ AZD}$ and $\triangle\text{AXB}$
    $\text{AZ}=\text{AX}=2\text{cm}$ (AXYZ is a square)
    $\angle\text{AZD}=\angle\text{AXB}=90^\circ$
    $\text{AD}=\text{AB}$ (ABCD is a square)
    So by RHS creterion, $\triangle\text{AZD}\cong\triangle\text{AXB}$
    $\Rightarrow\text{ZD}=\text{XB}$
    Now, $\text{ZD}=\text{ZY}+\text{DY}$
    =2cm + 3cm (ZY = AZ = 2cm)
    =5cm
    ⇒ XB = 5cm
    ⇒ BY = YX + XB = 2cm + 5cm = 7cm
    Hence, correct option is (c)
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