Free Lines and Angles Subjective Test 01 Practice Test - 7th grade 

Given - the complementary angles are equal.

Let the angle be x. So its complement  = x.

$\Rightarrow x+x={90}^{\circ }$

$\Rightarrow 2x={90}^{\circ }$

$\Rightarrow x={45}^{\circ }$

The angle which is complementary to itself is $x={45}^{\circ }$.

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.

Steps: 1 Mark
Result: 1 Mark

The number of endpoints in a line is zero.

$\Rightarrow X=0$

The number of endpoints in a Ray is 1

$\Rightarrow Y=1$

The number of endpoints in a line segment is 2.

$\Rightarrow Z=2$

$\therefore X+Y+Z=3$

Steps: 1 Mark
Answer: 1 Mark

The angles of a right-angled isosceles triangle are:

${90}^{\circ }$, ${45}^{\circ }and{45}^{\circ }$.

Hence the supplementary angles are:

${180}^{\circ }-{90}^{\circ }={90}^{\circ }$

${180}^{\circ }-{45}^{\circ }={135}^{\circ }$

${180}^{\circ }-{45}^{\circ }={135}^{\circ }$

Solution: 1 Mark each

(a) Sum of the angles should be equal to ${360}^{\circ }$

${110}^{\circ }+{60}^{\circ }+{40}^{\circ }+x^{\circ }={360}^{\circ }$

${210}^{\circ }+x^{\circ }={360}^{\circ }$

$\Rightarrow x^{\circ }={360}^{\circ }-{210}^{\circ }={150}^{\circ }$
 
(b) Let A and B be two complementary angles and let A >  ${45}^{\circ }$.

A + B =  ${90}^{\circ }$

B = ${90}^{\circ }$ − A

Therefore, B will be less than ${45}^{\circ }$.

(a) Proof: 2 Marks
(b) Solution: 1 Mark


(a)
     

We have to prove $\angle 1=\angle 3\&\angle 2=\angle 4$

$\angle 1+\angle 2={180}^{\circ }$              [Linear pair]

$\Rightarrow \angle 1={180}^{\circ }-\angle 2$  

Again, $\angle 3+\angle 2={180}^{\circ }$       [Linear pair]

$\Rightarrow \angle 3={180}^{\circ }-\angle 2$

Hence, $\angle 1=\angle 3$

Similarly, we can prove that $\angle 2=\angle 4$.

(b) $\angle$AOB = $\angle$DOE    (vertically opposite angles)
     $\therefore \angle$x = 60${}^{\circ }$

Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Sum of the angles of a triangle is ${180}^{\circ }$.

Since OC = OD,  $\angle OCD=\angle ODC={30}^{\circ }$
[Angles opposite to equal sides are equal]

$\Rightarrow \angle COD={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={120}^{\circ }$

$\angle AOB=\angle COD$  [Vertically Opposite Angles]

$\therefore \angle AOB={120}^{\circ }$

(a) Steps: 1 Mark 
     Proof: 1 Mark
(b) Solution: 1 Mark

(a) $\angle XPB=\angle MSP=k$  (Alternate Angles are equal)

$\Rightarrow LX\parallel MY...\left(i\right)$.

$\angle PSM=\angle SRN=k$  (Corresponding Angles) 

 $\Rightarrow MY\parallel NZ....\left(ii\right)$.

From (i) and (ii)

$LX\parallel MYandMY\parallel NZ$

$\Rightarrow LX\parallel MY\parallel NZ$.


(b) Given  $k={50}^{\circ }$,

$\angle PSY={180}^{\circ }-k$

$={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

(a) Steps: 1 Mark
      Proof: 1 Mark
(b) Solution: 1 Mark

(a) Given, $\angle LST={105}^{\circ }$ and $\angle NTS={75}^{\circ }$

$\angle LST+\angle NTS={105}^{\circ }+{75}^{\circ }={180}^{\circ }$

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°

(a) Steps: 2 Marks
      Result: 1 Mark
(b) Solution: 1 Mark

(a) Since $AB\parallel CD$

$\Rightarrow \angle BAC+\angle DCA={180}^{\circ }$  

[Sum of Co-interior angles is equal to ${180}^{\circ }$]

$\Rightarrow \frac{\angle BAC}{2}+\frac{\angle DCA}{2}=\frac{{180}^{\circ }}{2}$

$\Rightarrow \angle OAC+\angle OCA={90}^{\circ }$   [AD & BC are angle bisectors]

In $\Delta AOC,$

$\angle AOC+\angle OAC+\angle OCA={180}^{\circ }$

 [sum of  all the angles in triangle is ${180}^{\circ }$]

$\Rightarrow \angle AOC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

$\angle AOC=\angle BOD$  [Vertically Opposite Angles]

$\Rightarrow \angle BOD={90}^{\circ }$

(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${}^{\circ }$, the sum of the two interior angles is also 120${}^{\circ }$.
Thus, (x+y) = 120${}^{\circ }$

Steps: 3 Marks
Result: 1 Mark

Given  $DE\parallel BC,\angle CBA={40}^{\circ }$

$\Rightarrow \angle ADE=\angle CBA={40}^{\circ }$   [corresponding angle]

Since, $\Delta DEF$ is equilateral

$\Rightarrow \angle DEF={60}^{\circ }$

 $\angle FEC={50}^{\circ }$   (Given)
$\angle AED={180}^{\circ }-({50}^{\circ }+{60}^{\circ })$     (Angles on a straight line)
                     = ${70}^{\circ }$
Consider $\Delta ADE$ , 

$\angle AED={70}^{\circ }$ & $\angle ADE={40}^{\circ }$ 

$\angle ADE+\angle AED+\angle DAE={180}^{\circ }$  [Angle sum property]

$\Rightarrow \angle DAE={180}^{\circ }-({70}^{\circ }+{40}^{\circ })={70}^{\circ }$.

$\therefore \angle BAC=\angle DAE={70}^{\circ }$.

(a) Solution: 2 Marks
(b) Each part: 1 Mark

(a) $\Rightarrow$ Corresponding Angles

$\angle 1=\angle 5,\angle 3=\angle 7,\angle 2=\angle 6,\angle 4=\angle 8$;

$\Rightarrow$ Vertically Opposite Angles:

 $\angle 1=\angle 4,\angle 2=\angle 3,\angle 5=\angle 8,\angle 6=\angle 7$;

$\Rightarrow$ Alternate Interior Angles:

$\angle 3=\angle 6,\angle 4=\angle 5$;

$\Rightarrow$ Alternate Exterior Angles:

$\angle 1=\angle 8,\angle 2=\angle 7$;

$\Rightarrow$ Co-interior Angles:

$\angle 4,\angle 6,\angle 3,\angle 5$.



(b) (i) $x={60}^{\circ }$ (Alternate Angles)

     (ii) $x={120}^{\circ }$ (Corresponding Angles)

(a) Steps: 1 Mark
     Result: 1 Mark
(b) Steps: 1 Mark
     Result: 1 Mark
 
(a) ${40}^{\circ }+4x+3x={180}^{\circ }$  [since POQ is a straight line]

$7x={180}^{\circ }-{40}^{\circ }$

$x=\frac{140}{7}={20}^{\circ }$

Hence, the value of 'x' is

$x={20}^{\circ }$


(b) Given, x = ${30}^{\circ }$,

So, $2x={60}^{\circ }$

Also, $2x+3y={180}^{\circ }$  (linear pair)

Replacing x in the above equation,
${60}^{\circ }+3y={180}^{\circ }$

$3y={120}^{\circ }$

$y={40}^{\circ }$

The value of $6y-3x={150}^{\circ }$

Each part: 1 Mark

Two angles are complementary when they add up to ${90}^{\circ }$ 

Two angles are supplementary when they add up to ${180}^{\circ }$

(i) ${120}^{\circ }+{50}^{\circ }={170}^{\circ }$

Angles are neither complementary nor supplementary.

(ii) ${60}^{\circ }+{120}^{\circ }={180}^{\circ }$

Angles are supplementary.

(iii) ${39}^{\circ }+{61}^{\circ }={100}^{\circ }$

Angles are neither complementary nor supplementary.

(iv) ${65}^{\circ }+{25}^{\circ }={90}^{\circ }$

Angles are complementary.

(a) Steps: 1 Mark
      Angles: 1 Mark
(b) Steps: 1 Mark
      Result: 1 Mark

(a) Given $\angle 1={45}^{\circ }$

$\Rightarrow \angle 3=\angle 1={45}^{\circ }$    [Vertically opposite angle]

$\angle 2+\angle 3={180}^{\circ }$   [Linear pair]

$\angle 2+\angle {45}^{\circ }={180}^{\circ }$   

$\angle 2={180}^{\circ }-\angle {45}^{\circ }$  

$\Rightarrow \angle 2={135}^{\circ }$

$\Rightarrow \angle 2=\angle 4$   [Vertically opposite angle]

$\Rightarrow \angle 4=\angle 2={135}^{\circ }$


(b) Let the angle be $x^{\circ }$

$\therefore$ its supplementary angle = $(180-x{)}^{\circ }$

Given that $(180-x{)}^{\circ }$ is the complementary angle of ${70}^{\circ }$

$\therefore (180-x)+70=90$

$\Rightarrow x=180-20$

$\Rightarrow x={160}^{\circ }$
The required angle is ${160}^{\circ }$.