Free Lines and Angles Subjective Test 01 Practice Test - 7th grade
Question 1
Find the angle which is complementary to itself. [1 MARK]
SOLUTION
Solution :Given - the complementary angles are equal.
Let the angle be x. So its complement = x.
⇒x+x=90∘
⇒2x=90∘
⇒x=45∘
The angle which is complementary to itself is x=45∘.
Question 2
State at least two differences between a line and a line segment. [2 MARKS]
SOLUTION
Solution :Each difference: 1 Mark
1. A line is a set of points that can be extended in either direction whereas a line segment is a part of a line, which is bounded between two given points.
2. Length of a line segment can be measured, whereas a line can extend up to infinity and cannot be measured.
Question 3
If the number of endpoints in a line, ray and a line segment are represented by X, Y, and Z, then find the value of X + Y + Z [2 MARKS]
SOLUTION
Solution :Steps: 1 Mark
Result: 1 Mark
The number of endpoints in a line is zero.
⇒X=0
The number of endpoints in a Ray is 1
⇒Y=1
The number of endpoints in a line segment is 2.
⇒Z=2
∴X+Y+Z=3
Question 4
Find the supplementary angles corresponding to all the angles of a right angle isosceles triangle [2 MARKS]
SOLUTION
Solution :Steps: 1 Mark
Answer: 1 Mark
The angles of a right-angled isosceles triangle are:
90∘, 45∘ and 45∘.
Hence the supplementary angles are:
180∘−90∘=90∘
180∘−45∘=135∘
180∘−45∘=135∘
Question 5
(a) In the given figure, the value of x is: [2 MARKS]
(b) An angle is greater than 45∘. Its complement is ____ (>,<, = ) 45∘.
SOLUTION
Solution :Solution: 1 Mark each
(a) Sum of the angles should be equal to 360∘
110∘+60∘+40∘+x∘=360∘
210∘+x∘=360∘
⇒x∘=360∘−210∘=150∘
(b) Let A and B be two complementary angles and let A > 45∘.
A + B = 90∘
B = 90∘ − A
Therefore, B will be less than 45∘.
Question 6
(a) Prove that when two lines intersect, the vertically opposite angles are equal.
(b) Find the value of x from the given figure: [3 MARKS]
SOLUTION
Solution :(a) Proof: 2 Marks
(b) Solution: 1 Mark
(a)
We have to prove ∠1=∠3 & ∠2=∠4
∠1+∠2=180∘ [Linear pair]
⇒∠1=180∘−∠2
Again, ∠3+∠2=180∘ [Linear pair]
⇒∠3=180∘−∠2
Hence, ∠1=∠3
Similarly, we can prove that ∠2=∠4.
(b) ∠AOB = ∠DOE (vertically opposite angles)
∴∠x = 60∘
Question 7
In the given figure, CO = OD & ∠OCD=30∘. Find ∠AOB? [3 MARKS]
SOLUTION
Solution :Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Sum of the angles of a triangle is 180∘.
Since OC = OD, ∠OCD=∠ODC=30∘
[Angles opposite to equal sides are equal]
⇒∠COD=180∘−(30∘+30∘)=120∘
∠AOB=∠COD [Vertically Opposite Angles]
∴∠AOB=120∘
Question 8
(a) In the given figure, prove that LX∥MY∥NZ. [3 MARKS]
(b) If k=50∘, then find the value of ∠PSY.
SOLUTION
Solution :(a) Steps: 1 Mark
Proof: 1 Mark
(b) Solution: 1 Mark
(a) ∠XPB=∠MSP=k (Alternate Angles are equal)
⇒LX∥MY...(i).
∠PSM=∠SRN=k (Corresponding Angles)
⇒MY∥NZ....(ii).
From (i) and (ii)
LX∥MY and MY∥NZ
⇒LX∥MY∥NZ.
(b) Given k=50∘,
∠PSY=180∘−k
=180∘−50∘=130∘
Question 9
(a) In the given figure, prove that lines lm & nq are parallel to each other.
(b) Find the value of "x” if AB is parallel to CD.
[3 MARKS]
SOLUTION
Solution :(a) Steps: 1 Mark
Proof: 1 Mark
(b) Solution: 1 Mark
(a) Given, ∠LST=105∘ and ∠NTS=75∘
∠LST+∠NTS=105∘+75∘=180∘
These are co-interior angles.
Since co-interior angles are supplementary, lines are parallel.
Hence, the lines LM and NQ are parallel.
(b) Since AB is parallel to CD, corresponding angles will be equal.Therefore, x = 130°
Question 10
(a) Find out the unknown ∠BOD in the given quadrilateral ABDC in which AB∥CD. Given that AD & BC are angle bisectors of ∠BAC & ∠DCA respectively.
(b) What is the value of (x+y) in the figure below?
[4 MARKS]
SOLUTION
Solution :(a) Steps: 2 Marks
Result: 1 Mark
(b) Solution: 1 Mark
(a) Since AB∥CD
⇒∠BAC+∠DCA=180∘
[Sum of Co-interior angles is equal to 180∘]
⇒∠BAC2+∠DCA2=180∘2
⇒∠OAC+∠OCA=90∘ [AD & BC are angle bisectors]
In ΔAOC,
∠AOC+∠OAC+∠OCA=180∘
[sum of all the angles in triangle is 180∘]
⇒∠AOC=180∘−90∘=90∘
∠AOC=∠BOD [Vertically Opposite Angles]
⇒∠BOD=90∘
(b) The value of an exterior angle of a triangle is equal to the sum of the opposite interior angles.As the exterior angle value is given to be 120∘, the sum of the two interior angles is also 120∘.
Thus, (x+y) = 120∘
Question 11
In the given figure, find the value of ∠BAC. ΔDEF is equilateral, ∠CBA=40∘, ∠FEC=50∘ and DE is parallel to BC. [4 MARKS]
SOLUTION
Solution :Steps: 3 Marks
Result: 1 Mark
Given DE∥BC,∠CBA=40∘
⇒∠ADE=∠CBA=40∘ [corresponding angle]
Since, ΔDEF is equilateral
⇒∠DEF=60∘∠FEC=50∘ (Given)
∠AED=180∘−(50∘+60∘) (Angles on a straight line)
= 70∘
Consider ΔADE ,
∠AED=70∘ & ∠ADE=40∘
∠ADE+∠AED+∠DAE=180∘ [Angle sum property]
⇒∠DAE=180∘−(70∘+40∘)=70∘.
∴∠BAC=∠DAE=70∘.
Question 12
(a) Lines AB & CD are parallel and cut by a transversal t. List the alternate, corresponding, co-interior and vertically opposite angles.
(b) (i)
l and m are parallel lines and t is the transversal. Find x
(ii)
l and m are parallel lines and t is the transversal. Find x.
[4 MARKS]
SOLUTION
Solution :(a) Solution: 2 Marks
(b) Each part: 1 Mark
(a) ⇒ Corresponding Angles
∠1=∠5,∠3=∠7,∠2=∠6,∠4=∠8;
⇒ Vertically Opposite Angles:
∠1=∠4,∠2=∠3,∠5=∠8,∠6=∠7;
⇒ Alternate Interior Angles:
∠3=∠6,∠4=∠5;
⇒ Alternate Exterior Angles:
∠1=∠8,∠2=∠7;
⇒ Co-interior Angles:
∠4,∠6,∠3,∠5.
(b) (i) x=60∘ (Alternate Angles)
(ii) x=120∘ (Corresponding Angles)
Question 13
(a) From the given figure find the value of 'x'. [4 MARKS]
(b) In the figure below, x = 30∘. The value of (6y – 3x) is ___.
SOLUTION
Solution :(a) Steps: 1 Mark
Result: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a) 40∘+4x+3x=180∘ [since POQ is a straight line]
7x=180∘−40∘
x=1407=20∘
Hence, the value of 'x' is
x=20∘
(b) Given, x = 30∘,So, 2x=60∘
Also, 2x+3y=180∘ (linear pair)
Replacing x in the above equation,
60∘+3y=180∘3y=120∘
y=40∘
The value of 6y−3x=150∘
Question 14
Classify the following pair of angles as complementary and supplementary. [4 MARKS]
(i) 120∘,50∘
(ii) 60∘,120∘
(iii) 39∘,61∘
(iv) 65∘,25∘
SOLUTION
Solution :Each part: 1 Mark
Two angles are complementary when they add up to 90∘
Two angles are supplementary when they add up to 180∘
(i) 120∘+50∘=170∘
Angles are neither complementary nor supplementary.
(ii) 60∘+120∘=180∘
Angles are supplementary.
(iii) 39∘+61∘=100∘
Angles are neither complementary nor supplementary.
(iv) 65∘+25∘=90∘
Angles are complementary.
Question 15
(a) Find the value of angles 2,3,4 if ∠1 is 45. [4 MARKS]
(b) The supplement of which angle is the complement of 70∘?
SOLUTION
Solution :(a) Steps: 1 Mark
Angles: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a) Given ∠1=45∘
⇒∠3=∠1=45∘ [Vertically opposite angle]
∠2+∠3=180∘ [Linear pair]
∠2+∠45∘=180∘
∠2=180∘−∠45∘
⇒∠2=135∘
⇒∠2=∠4 [Vertically opposite angle]
⇒∠4=∠2=135∘
(b) Let the angle be x∘
∴ its supplementary angle = (180−x)∘
Given that (180−x)∘ is the complementary angle of 70∘
∴(180−x)+70=90
⇒x=180−20
⇒x=160∘
The required angle is 160∘.