3 Mark Question - Maths STD 9 Questions - Vidyadip

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Question 13 Marks

If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.

Answer

It is given that the six spokes are equally spaced, thus, two adjacent spokes subtend equal angle at the centre of the wheel. Let that angle measures x°. Also, The six spokes form a complete angle, that is 360°. Therefore, x + x + x + x + x + x = 360° 6x = 360°$\text{x}=\frac{360^\circ}{6}$
x = 60° Hence, the measure of the angle between two adjacent spokes measures 60°.

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Question 23 Marks

Prove that the straight lines perpendicular to the same straight line are parallel to one another.

Answer

Let AB and CD be drawn perpendicular to the Line MN$\angle\text{ABD}=90^\circ$ ... (i) [AB is perpendicular to MN]
$\angle\text{CON}=90^\circ$ ... (ii) [CD is perpendicular to MN ]
Now,$\angle\text{ABD}=\angle\text{CDN}=90^\circ$[From (i) and (ii)]
Therefore, AB || CD, Since corresponding angles are equal.

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Question 33 Marks

In figure, arms BA and BC of $\angle\text{ABC}$ are respectively parallel to arms ED and EF of $\angle\text{DEF}.$ Prove that $\angle\text{ABC}=\angle\text{DEF}.$

Answer

Given AB || DE and BC || EF To Prove:$\angle\text{ABC}=\angle\text{DEF}$
Construction: Produce BC to x such that it intersects DE at M. Proof: Since AB || DE and BX is the transversal ABC = DMX ...(1) [Corresponding angle] Also, BX || EF and DE is the transversal DMX = DEF ...(2) [Corresponding angles] From (1) and (2)$\angle\text{ABC}=\angle\text{DEF}$

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Question 43 Marks

How many pairs of adjacent angles, in all, can you name in figure 10.36.

Answer

Pairs of adjacent angles are:$\angle\text{EOC},\angle\text{DOC}$
$\angle\text{EOD},\angle\text{DOB}$
$\angle\text{DOC},\angle\text{COB}$
$\angle\text{EOD},\angle\text{DOA}$
$\angle\text{DOC},\angle\text{COA}$
$\angle\text{BOC},\angle\text{BOA}$
$\angle\text{BOA},\angle\text{BOD}$
$\angle\text{BOA},\angle\text{BOE}$
$\angle\text{EOC},\angle\text{COA}$
$\angle\text{EOC},\angle\text{COB}$
Hence, 10 pair of adjacent angles.

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Question 53 Marks

In figure, AB || CD || EF and GH || KL. Find $\angle\text{HKL}.$

Answer

Produce LK to meet GF at N. Now, alternative angles are equal$\angle\text{CHG}=\angle\text{HGN}=60^\circ$
$\angle\text{HGN}=\angle\text{KNF}=60^\circ$ [Corresponding angles]
Hence,$\angle\text{KNG}=180-60=120$
$\Rightarrow\ \angle\text{GNK}=\angle\text{AKL}=120^\circ$ [Corresponding angles]
$\angle\text{AKH}=\angle\text{KHD}=25^\circ$ [alternative angles]
Therefore,$\angle\text{HKL}=\angle\text{AKH}+\angle\text{AKL}=25+120=145^\circ.$

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Question 63 Marks

In figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that $\angle\text{ABP}+\angle\text{CDP}=\angle\text{DPB}.$

Answer

Given that AB || CD Let EF be the parallel line to AB and CD which passes through P. It can be seen from the figure. Alternative angles are equal$\angle\text{ABP}=\angle\text{BPF}$
Alternative angles are equal$\angle\text{CDP}=\angle\text{DPF}$
$\angle\text{ABP}+\angle\text{CDP}=\angle\text{BPF}+\angle\text{DPF}$
$\angle\text{ABP}+\angle\text{CDP}=\angle\text{DPB}$
Hence proved AB parallel to CD, P is any point To prove: $\angle\text{ABP}+\angle\text{BPD}+\angle\text{CDP}=360^\circ$
Construction: Draw EF || AB passing through P.

Proof: Since AB || EF and AB || CD, Therefore EF || CD [Lines parallel to the same line are parallel to each other]$\angle\text{ABP}+\angle\text{EPB}=180^\circ$[Sum of co-interior angles is 180]
$\angle\text{EPD}+\angle\text{COP}=180^\circ\dots(1)$ [Sum of co-interior angles is 180]
$\angle\text{EPD}+\angle\text{CDP}=180^\circ\dots(2)$
By adding (1) end (2)$\angle\text{ABP}+\angle\text{EPB}+\angle\text{EPD}+\angle\text{CDP}=(180+180)^\circ$
$\angle\text{ABP}+\angle\text{EPB}+\angle\text{COP}=360^\circ$

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Question 73 Marks

In figure, $\angle{1}=60^\circ$ and $\angle{2}=\Big(\frac{2}{3}\Big)^\text{rd}$ of a right angle. Prove that l∥m.

Answer

Given:$\angle{1}=60^\circ$ and $\angle{2}=\Big(\frac{2}{3}\Big)^\text{rd}$ of a right angle
To Prove: Parallel Drawn to m Proof:$\angle{1}=60^\circ$
$\angle{2}=\Big(\frac{2}{3}\Big)\times90=60$
Since$\angle{1}=\angle{1}=60^\circ$
Therefore, Parallel to m as pair of corresponding angles are equal.

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Question 83 Marks

Define supplementary angles.

Answer

supplementary Angles: Two angles, the sum of whose measures is 180°, are called suppplementary angles.
Thus,
angles $\angle\text{BAC}$ and $\angle\text{DAC}$ are aupplementary angles.
If x + y = 180°

Example 1: Angles of measure 50° and 130° are supplementary angles, because
50° + 130° = 180°
Example 2: Angles of measure 60° and 120° are supplementary angles, because
60° + 120° = 180°

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Question 93 Marks

In figure, state Which lines are parallel and why.

Answer

Vertically opposite angles are equal$\Rightarrow\ \angle\text{EOC}=\angle\text{DOK}=100^\circ$
$\Rightarrow\ \angle\text{DOK}=\angle\text{ACO}=100^\circ$
Here two lines EK and CA cut by a third line and the corresponding angles to it are equal Therefore, EK || AC.

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Question 103 Marks

Given $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10,$ Find the value of x for which POQ will be a line. (Figure 10.41).

Answer

For the case that POR is a line$\angle\text{POR}$ and $\angle\text{QOR}$ are linear parts
$\angle\text{POR}+\angle\text{QOR}=180^\circ$
Also, given that,$\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10$
2x + 10 + 3x = 180 5x + 10 = 180 5x = 180 - 10 5x = 170 x = 34 Hence the value of x is 34°

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Question 113 Marks

In figure, AB || CD and P is any point shown in the figure. Prove that: $\angle\text{ABP}+\angle\text{BPD}+\angle\text{CDP}=360^\circ$

Answer

The given figure is as follows:

It is give that AB || CD Let us draw a line XY passing through point P and parallel to AB and CD. We have XY || CD, thus, $\angle\text{CDP}$ and $\angle{2}$ are consecutive interior angles. Therefore,$\angle{2}+\angle\text{CDP}=180^\circ\dots(\text{i})$
Similarly, we have XY || AB, thus, $\angle\text{ABP}$ and $\angle{1}$ are consecutive interior angles. Therefore,$\angle{1}+\angle\text{ABP}=180^\circ\dots(\text{ii})$
On adding equation (i) and (ii), we get:$\angle{2}+\angle\text{CDP}+\angle{1}+\angle\text{ABP}=180^\circ+180^\circ$
$(\angle{2}+\angle{1})+\angle\text{CDP}+\angle\text{ABP}=360^\circ$
$\angle\text{ABP}+\angle\text{BPD}+\angle\text{CDP}=360^\circ$
Hence proved.

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Question 123 Marks

In figure, OA and OB are opposite rays:
If x = 25°, what is the value of y?

Answer

Given that, x = 25 Since $\angle\text{AOC}$ and $\angle\text{BOC}$ form a linear pair$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
Given that $\angle\text{AOC}=2\text{y}+5$ and $\angle\text{BOC}=3\text{x}$
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
(2y + 5) + 3x = 180 (2y + 5) + 3(25) = 180 2y + 5 + 75 = 180 2y + 80 = 180 2y = 180 - 80 = 100$\text{y}=\frac{100}{2}$
y = 50

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Question 133 Marks

If figure, if AB || CD and CD || EF, find $\angle\text{ACE}.$

Answer

Since EF || CD
Therefore, EFC + ECD = 180 [Co-interior angles are supplementary]
⇒ ECD = 180 - 130 = 50
Also BA || CD
⇒ BAC = ACD = 70 [alternative angles]
But, ACE + ECD = 70
⇒ ACE = 70 - 50 = 20

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Question 143 Marks

In figure POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle\text{ROS}=\frac{1}{2}(\angle\text{QOS}-\angle\text{POS}).$

Answer

Given that OR perpendicular$\therefore\angle\text{POR}=90^\circ$
$\angle\text{POS}+\angle\text{SOR}=90$ $[\therefore\angle\text{POR}=\angle\text{POS}+\angle\text{SOR}]$
$\angle\text{ROS}=90^\circ-\angle\text{POS}\dots(1)$
$\angle\text{QOR}=90$ $(\because\text{OR}\perp\text{PQ})$
$\angle\text{QOS}-\angle\text{ROS}=90^\circ$
$\angle\text{ROS}=\angle\text{QOS}-90^\circ\dots(2)$
By adding (1) and (2) equations, we get:$\therefore\ \angle\text{ROS}=\angle\text{QOS}-\angle\text{POS}$
$\angle\text{ROS}=\frac{1}{2}(\angle\text{QOS}-\angle\text{POS})$

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Question 153 Marks

In figure, p is a transversal to lines m and n, $\angle{2}=120^\circ$and $\angle{5}=60^\circ.$Prove that m || n.

Answer

Given that$\angle{2}=120^\circ$ and $\angle{5}=60^\circ.$
To prove,$\angle{2}+\angle{1}=180^\circ$ [Linear pair]
$120+\angle{1}=180^\circ$
$\angle{1}=180-120$
$\angle{1}=60^\circ$
Since $\angle{1}=\angle{5}=60^\circ$ Therefore, m || n. [As pair of corresponding angles are equal]

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Question 163 Marks

In figure transversal l intersects two lines m and n, $\angle{4}=110^\circ$ and $\angle{7}=65^\circ.$ Is m || n?

Answer

Given:$\angle{4}=110^\circ$and $\angle{7}=65^\circ.$
To find: is m || n Here,$\angle{7}=\angle{5}=65^\circ$ [Vertically opposite angle]
Now,$\angle{4}+\angle{5}=110+65=175^\circ$
Therefore, m is not parallel to n as the pair of co interior angles is not supplementary.

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Question 173 Marks

If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.

Answer

Let the measure of the angle be x°.
Its complement will be (90° - x°) and its supplement will be (180° - x°).
Supplement of thrice of the angle = (180° - 3x°)
According to the given information:
(90° - x°) = (180° - 3x°)
3x - x = 180 - 90
2x = 90
x = 45
Thus, the measure of the angle is 45°.
The measure of the angle is 45°.

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Question 183 Marks

In figure, PQ || AB and PR || BC. If $\angle\text{QPR}=102^\circ,$ determine $\angle\text{ABC}.$ Give reasons.

Answer

AB is produce to meet PR at K Since PQ || AB$\angle\text{QPR}=\angle\text{BKR}=102^\circ$[corresponding angles]
Since PR ∥ BC$\angle\text{RKB}+\angle\text{CBK}=180^\circ$[Since Corresponding angles are supplementary]
$\angle\text{CKB}=180-102=78$
$\therefore\ \angle\text{CKB}=78^\circ$

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Question 193 Marks

AB, CD and EF are three concurrent lines passing throught the point O such that OF bisects $\angle\text{BOD}.$ If $\angle\text{BOF}=35^\circ,$ find $\angle\text{BOC}$ and $\angle\text{AOD}.$

Answer

$\angle\text{BOF}=35^\circ$
$\therefore\angle\text{BOD}=2\angle\text{BOF}=70^\circ$ $[\because$ OF bisects $\angle\text{BOD}]$
$\angle\text{BOD}=\angle\text{AOC}=70^\circ$ [vertically opposite angles]
Now,$\angle\text{BOC}+\angle\text{AOC}=180^\circ$ [linear pair]
$\Rightarrow\ \angle\text{BOC}+70^\circ=180^\circ$
$\Rightarrow\ \angle\text{BOC}=110^\circ$
$\therefore\ \angle\text{AOD}=\angle\text{BOC}=110^\circ$ [vertically opposite angles]

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Question 203 Marks

In figure if l || m, n∥p and $\angle\text{1}=85^\circ,$ find $\angle{2}.$

Answer

Corresponding angles are equal$\Rightarrow\ \angle{1}=\angle{3}=85^\circ$
By using the property of co-interior angles are supplementary$\angle{2}+\angle{3}=180^\circ$
$\angle{2}+55=180^\circ$
$\angle{2}=180-85$
$\angle{2}=95^\circ$

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Question 213 Marks

What value of y would make AOB a line in the below figure, If $\angle\text{AOC}=4\text{y}$ and $\angle\text{BOC}=(6\text{y}+30)?$

Answer

Since, $\angle\text{AOC}$ and $\angle\text{BOC}$ are linear pairs$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
6y + 30 + 4y = 180 10y + 30 = 180 10y = 180 - 30 10y = 150$\text{y}=\frac{150}{10}$
y = 15 Hence value of y that will make AOB a line is 15°.

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Question 223 Marks

How many pairs of adjacent angles are formed when two lines intersect in a point?

Answer

Let us draw the following diagram showing two lines AB and CD intersecting at a point O.

We have the following pair of adjacent angles, so formed:$\angle\text{AOC}$ and $\angle\text{BOC}$
$\angle\text{AOC}$ and $\angle\text{AOD}$
$\angle\text{BOD}$ and $\angle\text{BOC}$
$\angle\text{BOD}$ and $\angle\text{AOD}$
Hence, in total four pair of adjacent angles are formed.

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Question 233 Marks

An angle is equal to five times its complement. Determine its measure.

Answer

Let the complement of the required angle measures x°. Therefore, the required angle becomes 5x°. Since, the angles are complementary. Thus, their sum must be equal to 90°. Or we can say that : x + 5x = 90° 6x = 90°$\text{x}=\frac{90^\circ}{6}$
x = 15° Hence, the required angle becomes: 5x = 5(15°) = 75°.

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Question 243 Marks

An angle is equal to its supplement. Determine its measure.

Answer

Let the supplement of the angle be x°. According the given statement, the required angle is equal to its supplement, therefore, the required angle becomes x°. Sine both the angles are supplementary, therefore, their sum must be equal to 180°. Or we can say that: x + x = 180° 2x = 180°$\text{x}=\frac{180^\circ}{2}$
x = 90° Hence, the required angle measures 90°.

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Question 253 Marks

Two lines AB and CD intersect at O. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$ find the measure of $\angle\text{AOC},\angle\text{COB},\angle\text{BOD}$ and $\angle\text{DOA}.$

Answer

Given:$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ$
To find:$\angle\text{AOC},\angle\text{COB},\angle\text{BOD},\angle\text{DOA}$
Here,$\angle\text{AOC},\angle\text{COB},\angle\text{BOD}=270^\circ$[Complete angle]
⇒ 270 + AOD = 360 ⇒ AOD = 360 - 270 ⇒ AOD = 90 Now, AOD + BOD = 180 [Linear pair] 90 + BOD = 180 ⇒ BOD = 180 - 90 ⇒ BOD = 90 AOD = BOC = 90 [Vertically opposite angles] BOD = AOC = 90 [Vertically opposite angles]

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Question 263 Marks

In figure, l, m and n are parallel lines intersected by transversal p at X. Y and Z respectively. Find $\angle\text{1},\angle\text{2}$ and $\angle\text{3}.$

Answer

From the given figure:$\angle\text{3}+\angle\text{myz}=180^\circ$ [Linear pair]
$\Rightarrow\ \angle{3}=180-120$
$\Rightarrow\ \angle3=60^\circ$
Now line I parallel to m$\angle{1}=\angle{3}$ [Corresponding angles]
$\angle{1}=60^\circ$
Also m parallel to n$\Rightarrow\ \angle{2}=120^\circ$ [Alternative interior angle]
Hence,$\angle{1}=\angle{3}=60^\circ$
$\angle{2}=120^\circ$

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Question 273 Marks

In figure, lines AB and CD intersect at O. If $\angle\text{AOC}+\angle\text{BOE}=70^\circ$ and $\angle\text{BOD}=40^\circ,$ find $\angle\text{BOE}$ and reflex $\angle\text{COE}.$

Answer

In the figure, $\angle\text{AOC},\angle\text{BOE}$ and $\angle\text{COE} $ from a linear pair. Thus,

$\angle\text{AOC}+\angle\text{BOE}+\angle\text{COE}=180^\circ$
It is given that $\angle\text{AOC}+\angle\text{BOE}=70^\circ,$ on substituting this value, we get:$70^\circ+\angle\text{COE}=180^\circ$
$\angle\text{COE}=180^\circ-70^\circ$
$\angle\text{COE}=110^\circ$
Thus, reflex $\angle\text{COE}=360^\circ-110^\circ$ Therefore, reflex $\angle\text{COE}=250^\circ.$

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Question 283 Marks

In figure, show that AB || EF.

Answer

Produce EF to intersect AC at K. Now,$\angle\text{DCE}+\angle\text{CEF}=35+145=180^\circ$
Therefore, EF || CD ... (1) [Since Sum of Co-interior angles is 180] Now,$\angle\text{BAC}=\angle\text{ACD}=57^\circ$
⇒ BA || EF ... (2) [Alternative angles are equal] From (1) and (2) AB || EF [Since, Lines parallel to the same line are parallel to each other] Hence proved.

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