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MATHS M.C.Q

Lines and Triangles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 28 questions.

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Question 11 Mark
If one of the angles of a triangle is 130º then the angle between the bisectors of the other two angles can be:
  1. 50º
  2. 65º
  3. 90º
  4. 155º
Answer
  1. 155º
Solution:

Let $\angle\text{A}=130^\circ$
In $\triangle\text{ABC},$ by angle sum property,
$\angle\text{B}+\angle\text{C}+\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}+130^\circ=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=50^\circ$
$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=25^\circ$
Now, in $\triangle\text{BOC},$
$\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
$\Rightarrow25^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=155^\circ$
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Question 21 Mark
In the adjoining figure, what is the value of y?
  1. 36
  2. 54
  3. 63
  4. 72
Answer
  1. 54
Solution:
AOB is a straight line.
$\therefore$ xº + yº 90º = 180º
⇒ x + y = 90 .....(i)
Since the angles around a point sum up to 360º,
⇒ xº + 90º + yº + 72º + 3xº = 360º
⇒ 4x + y = 198 .....(ii)
Subtracting (i) from (ii), we get
3x = 108 ⇒ x = 36º
Substituting in (i), we get
y = 54º
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Question 31 Mark
An angle is one fifth of its supplement. The measure of the angle is:
  1. 15º
  2. 30º
  3. 75º
  4. 150º
Answer
  1. 30º
Solution:
Let the measure of the angle be xº.
So, its supplement = (180º - x)
According to the given condition,
$\text{x}=\frac{1}{5}(180^\circ-\text{x)}$
$\Rightarrow5\text{x}=180-\text{x}$
$\Rightarrow6\text{x}=180$
$\Rightarrow\text{x}=30^\circ$
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Question 41 Mark
The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is:
  1. 120º
  2. 100º
  3. 80º
  4. 60º
Answer
  1. 80º
Solution:
By angle sum property,
2x + 3x + 4x = 180° 
⇒ 9x = 180° 
⇒ x = 20° 
Hence, largest angle = 4x = 4(20°) = 80° 
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Question 51 Mark
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:
  1. An isosceles triangle.
  2. An obtuse triangle.
  3. An equilateral triangle.
  4. A right triangle.
Answer
  1. A right triangle.
Solution:
In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.
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Question 61 Mark
In the given figure, AOB is a straight line. If $\angle\text{AOC}=(3\text{x} +10)^\circ$ and $\angle\text{BOC}=(4\text{x}-26)^\circ,$ then $\angle\text{BOC}=?$
  1. 96º
  2. 86º
  3. 76º
  4. 106º
Answer
  1. 86º
Solution:
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow(3\text{x}+10)+(4\text{x}-26)=180^\circ$
$\Rightarrow7\text{x}-16=180^\circ$
$\Rightarrow7\text{x}=196$
$\Rightarrow\text{x}=28$
So, $\angle\text{BOC}=4 \text{x}-26=4(28)-26=86^\circ$
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Question 71 Mark
In the given figure, AB || CD. If $\angle\text{BAO}=60^\circ$ and $\angle\text{OCD}=110^\circ$ then $\angle\text{AOC}=?$
  1. 70º
  2. 60º
  3. 50º
  4. 40º
Answer
  1. 50º
Solution:
Let $\angle\text{AOC}=\text{x}^\circ$
Draw YOZ || CD || AB.

Now, YO || AB and OA is the transversal.
$\Rightarrow\angle\text{YOA}=\angle\text{OAB}=60^\circ$ (alternate angles)
Again, OZ || CD and OC is the transversal.
$\Rightarrow\angle\text{COZ}+\angle\text{OCD}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{COZ}+110^\circ=180^\circ$
$\Rightarrow\angle\text{COZ}=70^\circ$
Now, $\angle\text{YOZ}=180^\circ$ (straight angle)
$\Rightarrow\angle\text{YOA}+\angle\text{AOC}+\angle\text{COZ}=180^\circ$
$\Rightarrow60^\circ+\text{x}+70^\circ=1806^\circ$
$\Rightarrow\text{x}=50^\circ$
$\Rightarrow\angle\text{AOC}=50^\circ$
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Question 81 Mark
An exterior angle of a triangle is 110º and its two interior opposite angles are equal. Each of these equal angles is:
  1. $70^\circ$
  2. $55^\circ$
  3. $35^\circ$
  4. $27\frac{1^\circ}{2}$
Answer
  1. $55^\circ$
Solution:
Let each interior opposite angle be x.
Then, x + x = 110° (Exterior angle property of a triangle)
⇒ 2x = 110°
⇒ x = 55°
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Question 91 Mark
The angles of a triangle are in the ration 3 : 5 : 7. The triangle is:
  1. Acute-angled.
  2. Obtuse-angled.
  3. Right-angled.
  4. An isosceles triangle.
Answer
  1. Acute-angled.
Solution:
Let the angles measure (3x)º, (5x)º and (7x)º.
Then,
3x + 5x + 7x = 180º
⇒ 15x = 180º
⇒ x = 12º
Therefore, the angles are 3(12)º = 36º, 5(12)º = 60º and 7(12)º = 84º.
Hence, the triangle is acute-angled.
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Question 101 Mark
If two angles are complements of each other, then each angle is:
  1. An acute angle.
  2. An obtuse angle.
  3. A right angle.
  4. A reflex angle.
Answer
  1. An acute angle.
Solution:
If two angles are complements of each other, that is, the sum of their measures is 90º, then each angle is an acute angle.
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Question 111 Mark
In the given figure, AOB is a straight line. The value of x is:
  1. 12
  2. 15
  3. 20
  4. 25
Answer
  1. 15
Solution:
AOB is a straight line.
$\Rightarrow\angle\text{AOB}=180^\circ$
⇒ 60° + 5x° + 3x° = 180° 
⇒ 60° + 8x° = 180° 
⇒ 8x° = 120° 
⇒ x = 15° 
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Question 121 Mark
In the given figure, $\angle\text{OAB}=110^\circ$ and $\angle\text{BCD}=130^\circ$ then $\angle\text{ABC}$ is equal to:
  1. 40º
  2. 50º
  3. 60º
  4. 70º
Answer
  1. 60º
Solution:
Through B draw YBZ || OA || CD.

Now, OA || YB and AB is the transversal.
$\Rightarrow\angle\text{OAB}+\angle\text{YBA}=180^\circ$ (interior angles are supplementary)
$\Rightarrow\angle\text{YBA}=70^\circ$
Also, CD || BZ and BC is the transversal.
$\Rightarrow\angle\text{DCB}+\angle\text{CBZ}=180^\circ$ (interior angles are supplementary)
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Question 131 Mark
In the given figure, AB || CD. If $\angle\text{CAB}=180^\circ$ and $\angle\text{EFC}=25^\circ$ then $\angle\text{CEF}=?$
  1. 65º
  2. 55º
  3. 45º
  4. 75º
Answer
  1. 55º
Solution:
Since AB || CD,
$\Rightarrow\angle\text{ACE}=\angle\text{BAC}=80^\circ$
In $\triangle\text{CEF},$
$\angle\text{ACE}=\angle\text{CEF}+\angle\text{CFE}$ (Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow80^\circ=\angle\text{CEF}+25^\circ$
$\Rightarrow\angle\text{CEF}=55^\circ$
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Question 141 Mark
The measure of an angle is five times its comlement. The angle measure.
  1. 25º
  2. 35º
  3. 65º
  4. 75º
Answer
  1. 75º
Solution:
Let the measure of the angle be xº,
So, its complement = (90 - x)º
According to the given condition,
x = 5(90 - x)
⇒ x = 450 - 5x
⇒ 6x = 450
⇒ x = 75º
So, the angle measures 75º.
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Question 151 Mark
In the given figure, straight lines AB and CD interect at O.If $\angle\text{AOC}=\phi,\angle\text{BOC}=\theta$ and $\theta=3\phi,$
then $\phi=?$
  1. 30º
  2. 40º
  3. 45º
  4. 60º
Answer
  1. 45º
Solution:
$\angle\text{AOD}=\angle\text{COB}=\theta$
$\angle\text{AOC}=\angle\text{BOD}=\phi$
Since the sum of the measures of the angles around a point is 360º,
$\angle\text{AOD}+\angle\text{COB}+\angle\text{AOC}+\angle\text{BOD}=360^\circ$
$\Rightarrow\theta+\theta+\phi+\phi=360$
$\Rightarrow2(\theta+\phi)=360$
$\Rightarrow\theta+\phi=180$
Given that $\theta=3\phi.$
So, $3\phi+\phi=180$
$\Rightarrow4\phi=180$
$\Rightarrow\phi=45$
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Question 161 Mark
In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y =?
  1. 60º
  2. 80º
  3. 48º
  4. 72º
Answer
  1. 60º
Solution:
The ratio of the angles is given to be 4 : 5 : 6.
So, let the measure of the angles be 4m, 5m and 6m.
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow4\text{m}+5\text{m}+6\text{m}=180^\circ$
$\Rightarrow15\text{m}=180$
$\Rightarrow\text{m}=12$
So, y = 5m = 5(12) = 60º.
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Question 171 Mark
Two complementary angles are such that twich the measure of the one is equal to three times the measure of the other. The larger of the two measure.
  1. 72º
  2. 54º
  3. 63º
  4. 36º
Answer
  1. 54º
Solution:
Let the measure of each angle be xº and (90 - x)º.
According to the given condition,
2x = 3(90 - x)
⇒ 2x = 270 - 3x
⇒ 5x = 270
⇒ x = 54º
So, (90 - x)º = (90 - 54)º = 36º
So, the larger of the two angles is 54º.
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Question 181 Mark
In the given figure, AOB is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$
  1. 40º
  2. 60º
  3. 80º
  4. 50º
Answer
  1. 80º
Solution:
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$
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Question 191 Mark
In the given figure, straight lines AB and CD intersect at O. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$
  1. 65º
  2. 115º
  3. 110º
  4. 125º
Answer
  1. 115º
Solution:
$\angle\text{AOC}+\angle\text{BOD}=1306^\circ$ (given)
But $\angle\text{AOC}=\angle\text{BOD}$ (Vartically Opposite angles)
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Since COD is a straight line,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOD}=115^\circ$
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Question 201 Mark
In the given figure AB || CD and CD || EF. If y : z = 3 : 7 then x = ?
  1. 108º
  2. 126º
  3. 162º
  4. 63º
Answer
  1. 126º
Solution:
AB || CD
x + y = 180º and y = p (Vertically opposite angles)
Also, CD || EF and t is the transversal.
$\therefore$ p + z = 180º
⇒ y + z = 180º $(\because\text{p}=\text{y})$
$\therefore$ x + y = y + z
⇒ x = z
But y : z = 3 : 7
$\therefore\text{y}=\Big(180\times\frac{3}{10}\Big)=54^\circ$ and $\text{z}=\Big(180\times\frac{7}{10}\Big)=126^\circ$
x = 126º $(\because\text{x}=\text{z})$
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Question 211 Mark
In the figure, AB || CD. If $\angle\text{APQ}=70 ^\circ$ and $\angle\text{PRD}=120^\circ$ then $\angle\text{QPR}=?$
  1. 55º
  2. 60º
  3. 40º
  4. 35º
Answer
  1. 55º
Solution:

Since AB || CD,
$\angle\text{APQ}=\angle\text{PQR}=70^\circ$ (Alternate angles)
$\Rightarrow\angle\text{PRD}=\angle\text{PQR}+\angle\text{QPR}$ (Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow120^\circ=70^\circ+\angle\text{QPR}$
$\Rightarrow\angle\text{QPR}=506^\circ$
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Question 221 Mark
In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If $\angle\text{PQR}=108^\circ$ then $\angle\text{AQP}=?$
  1. 72º
  2. 18º
  3. 36º
  4. 54º
Answer
  1. 36º
Solution:
We know that, angle of incidance = angle reflection.
that is, $\angle\text{AQP}=\angle\text{BQR}$
Since AOB is a straight line,
$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$
$\Rightarrow\angle\text{AQP}+\angle\text{AQP}+\angle\text{PQR}=180^\circ$
$\Rightarrow2\angle\text{AQP}=72$
$\Rightarrow \angle\text{AQP}=36^\circ$
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Question 231 Mark
In the given figure, $\angle\text{OAB}=75^\circ, \angle\text{OBA}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ then, $\angle\text{ODC}=?$
  1. 20º
  2. 25º
  3. 30º
  4. 35º
Answer
  1. 30º
Solution:
In $\triangle\text{OAB},$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property)
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ$ (Vertivcally opposite angles)
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
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Question 241 Mark
In the given figure, AOB is a straight line. If $\angle\text{AOC}=4\text{x}^\circ$ and $\angle\text{BOC}=5\text{x}^\circ$ then $\angle\text{AOC}=?$
  1. 40º
  2. 60º
  3. 80º
  4. 100º
Answer
  1. 80º
Solution:
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow4\text{x}+5\text{x}=180^\circ$
$\Rightarrow9\text{x}=180^\circ$
$\Rightarrow\text{x}=206^\circ$
So, $\angle\text{AOC}=4\text{x}=4(20)=80^\circ$
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Question 251 Mark
Which of the following statements is false?
  1. Through a given point, only one straight line can be drawn.
  2. Through two given points, it is possible to draw one and only one straight line.
  3. Two straight lines can intersect only at one point.
  4. A line segment can be produced to any desired length.
Answer
  1. Through a given point, only one straight line can be drawn.
Solution:
Option (a) is false, since through a given point we can draw an infinite number of straight lines.
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MCQ 261 Mark
An angle which measures more than $180^\circ$ but less than $360^\circ,$ is called.
 
  • A
    An acute angle.
     
  • B
    An obuse angle.
     
  • C
    A straight angle.
     
  • ✓
    A reflex angle.
Answer
Correct option: D.
A reflex angle.
An angle which measures more than $180^\circ$ but less than $360^\circ$ is called a reflex angle.
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Question 271 Mark
In the given figure, AB || CD. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ$ then $\angle\text{AEB}=?$
  1. 50º
  2. 60º
  3. 70º
  4. 55º
Answer
  1. 70º
Solution:
Let $\angle\text{AEB}=\text{x}^\circ$
Now, AB || CD and BC is the transversal
$\therefore\angle\text{ABE}=\angle\text{BCD}=60^\circ$ (Alterante angles)
In $\triangle\text{ABE},$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{ABE}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+\text{x}^\circ+60^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$
$\therefore\angle\text{AEB}=70^\circ$
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Question 281 Mark
In the given figure, AB || CD. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$
  1. 130º
  2. 150º
  3. 80º
  4. 100º
Answer
  1. 130º
Solution:
Construction: Through O, draw OE || AB || CD
$\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
$\Rightarrow\angle\text{EOA}=80^\circ$
So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}=80^\circ-30^\circ=50^\circ$
Since CD || EO
$\angle\text{OCD}+\angle\text{EOC}=1806^\circ$
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ$
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