Free Lines and Angles Subjective Test 02 Practice Test - 7th grade 

Two angles are supplementary if their sum is equal to ${180}^{\circ }$.

Each Part: 1 Mark

Sum of complementary angle is equal to ${90}^{\circ }$.

Therefore, the complement of:

(a)  ${65}^{\circ }$
      $\Rightarrow {90}^{\circ }-{65}^{\circ }={25}^{\circ }$

(b) ${32}^{\circ }$
      $\Rightarrow {90}^{\circ }-{32}^{\circ }={58}^{\circ }$

Solution: 1 Mark each

(a) Let the angle be x.

therefore its supplement  =  x  (as the supplementary angles are equal)

Sum of supplementary angles is  ${180}^{\circ }$

$x+x={180}^{\circ }$

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

(b) Let the angle be x.
Its supplement = 2x.
Sum of supplementary angles is  ${180}^{\circ }$
$\therefore x+2x={180}^{\circ }$
$\Rightarrow 3x={180}^{\circ }$
$\Rightarrow x={60}^{\circ }$

$\therefore$ The angle which is double its supplement = 2x = ${120}^{\circ }$

Each part: 1 Mark

(a) By adding consecutive triangular numbers, we get square numbers.
e.g. 1 + 3 = 4, 6 + 10 = 16, etc.

(b)  $\angle F={90}^{\circ }-{28}^{\circ }$  (Sum of complementary angles is 90${}^{\circ }$)
              $={62}^{\circ }$

Proof: 2 Marks

$\angle 1+\angle 2+\angle 3+\angle 4={360}^0$  [Angles about a point]

$\angle 4=\angle 8$    [Pair of corresponding angles]
 
and $\angle 2=\angle 6$   [Pair of corresponding angles]

$\Rightarrow \angle 1+\angle 3+\angle 8+\angle 6={360}^0$

Acute Angle: 1 Mark
Right Angle: 1 Mark
Obtuse Angle: 1 Mark

Angles are classified based on their measure as:

(i) Acute Angle: An angle that is > 0 and < 90${}^{\circ }$ is called an acute angle.

(ii) Right Angle: An angle that is exactly equal to ${90}^{\circ }$, is called a right angle.

(iii) Obtuse Angle: An angle that is > ${90}^{\circ }$ and < ${180}^{\circ }$, is called an obtuse angle.

(a) Conditions: 2 Marks
(b) Explanation: 1 Mark

(a) Two angles are said to be adjacent if they have:   

      i) Common vertex

     ii) Common arm

    iii) Non-common arms are on either side of the common arm

(b) If  $\angle 1$ decreases then, $\angle 2$ will increase by the same measure, so that both the angles are still supplementary.

(a) Answer: 1 Mark
     Reason: 1 Mark
(b) Solution: 1 Mark

(a) A triangle can have only one obtuse angle.

If we have more than one obtuse angle in a triangle, then the sum of two (obtuse) angles will exceed ${180}^{\circ }$. 

Since the sum of all angles of a triangle is ${180}^{\circ }$. 

$\Rightarrow$ A triangle cannot have more than one obtuse angle.



(b)

$\angle BCD={180}^{\circ }$ (Straight angle)
$\therefore \angle ACB=\angle BCD-\angle ACD=(180-160{)}^{\circ }={20}^{\circ }$
$\Rightarrow \angle BAC={180}^{\circ }-(\angle ABC+\angle ACB)=180-(30+20)=180-50=130$
$\therefore \angle BAC={130}^{\circ }$

(a) Solution: 1 Mark
(b) Application of concept: 1 Mark
     Correct Answer: 1 Mark

(a) There are 4 points of intersection between the parallel lines & transversal.

(b)  In the above figure since the two lines are parallel and cut by a transversal, using the property of corresponding angles:

=> 2x - 60${}^{\circ }$ = x

=> x - 60${}^{\circ }$ = 0

=> x = 60${}^{\circ }$

Each part: 2 Marks

(a) Given, $a\parallel b$ and $l\parallel m$

Mark an angle 1 in the figure.

$\angle 1={80}^{\circ }$   [Corresponding angles]

$\angle 1+\angle x={180}^{\circ }$ [Since $l\parallel m,\angle 1and\angle x$ are cointerior angles]

$\Rightarrow \angle x={180}^{\circ }-{80}^{\circ }$

$\Rightarrow \angle x={100}^{\circ }$


(b) $\angle 5=5x+35$          (Alternate angles)
 ${45}^{\circ }+\angle 5={180}^{\circ }$     (Linear pair)
 $\angle 5={180}^{\circ }-{45}^{\circ }$
                $={135}^{\circ }$

$\therefore 5x+35=135$
$\Rightarrow 5x=135-35=100$
$\Rightarrow x=\frac{100}{5}=20$

(a) Correct answer: 1 Mark 
(b) Each angle: 1 Mark


(a)

The total number of line segments is 8. The line segments are AF, AB, BF, FD, FC, CD, AC and BD.

(b)  (i) $\angle x$ and $\angle {55}^{\circ }$ are vertically opposite,

$\angle x$ = ${55}^{\circ }$

$\angle x$ + $\angle z$ = ${180}^{\circ }$

${55}^{\circ }$ + $\angle y$ = ${180}^{\circ }$

$\angle y$ = ${180}^{\circ }$  –  ${55}^{\circ }$  =  ${125}^{\circ }$

$\angle y$ = $\angle z$  (Vertically Opposite angles)

$\angle z$ = ${125}^{\circ }$


(ii) $\angle z$ = ${40}^{\circ }$ (Vertically opposite angles)

$\angle y$ + $\angle z$ = ${180}^{\circ }$

$\angle y$ = ${180}^{\circ }$ – ${40}^{\circ }$ =  ${140}^{\circ }$

${40}^{\circ }$ + $\angle x$ + ${25}^{\circ }$ = ${180}^{\circ }$

${65}^{\circ }$ + $\angle x$  =  ${180}^{\circ }$

$\angle x$ = ${180}^{\circ }$ – ${65}^{\circ }$  =  ${115}^{\circ }$

Steps: 2 Marks
Each angle: 1 Mark

Since ABCD is a rectangle, $AB\parallel CD$

$\angle EDC=\angle EIJ={50}^{\circ }$ (Corresponding Angles).

Since ABCD is rectangle, $\angle ADC={90}^{\circ }$

$\Rightarrow \angle ADG+\angle GDC={90}^{\circ }$

 $\angle ADG={90}^{\circ }-{50}^{\circ }={40}^{\circ }$.

$\angle EJI={180}^{\circ }-{105}^{\circ }={75}^{\circ }$ (Linear pair);

$\angle IEJ={180}^{\circ }-({75}^{\circ }+{50}^{\circ })$ (Angle sum property)

$\Rightarrow \angle IEJ={55}^{\circ }$.

Steps: 3 Marks
Proof: 1 Mark

 $\Delta AEC$ and $\Delta DBF$ are equilateral, therefore angle made at each vertex is $={60}^{\circ }$

Consider $\Delta AMN$,

$\angle MAN={60}^{\circ }$

$\angle AEC=\angle AMN={60}^{\circ }$  (Corresponding Angles as $FB\parallel EC$)

Similarly $\angle ACE=\angle ANM={60}^{\circ }$   (Corresponding Angles).

Hence all angles of $\Delta AMN$ are equal to ${60}^{\circ }$ and is an equilateral triangle.

Consider $\Delta BNO,\angle FBD={60}^{\circ }$ 

$\angle BNO=\angle ANM={60}^{\circ }$  (vertically Opposite Angles)

$\angle BON={60}^{\circ }$  (Angle sum property of triangle).

Hence $\Delta BNO$ is also equilateral.
Similarly, we can prove it for the remaining triangles

(a) Steps: 1 Mark
     Proof: 1 Mark
(b) Steps: 1 Mark
     Correct Answer: 1 Mark

 $\angle 1=\angle 2$   (given)  
They are also corresponding angles. 

$\Rightarrow l\parallel m...\left(i\right)$  (If corresponding angles are equal, then the lines are parallel)

Also, $l\parallel BC....\left(ii\right)$  

From (i) and (ii):

$m\parallel l\parallel BC$.

$\Rightarrow m\parallel BC$.


(b)  In $\Delta BDA$, x + x + y = ${180}^{\circ }$ (by angle sum property)
similarly, in $\Delta BDC$, x + x + $\angle BDC$ = ${180}^{\circ }$
Comparing the two equations, we have $\angle BDC$ = y.
Also, $\angle BDC$ + $\angle BDA$ = ${180}^{\circ }$ (linear pair)
or, y + y = ${180}^{\circ }$
or, y = ${90}^{\circ }$

Application of theorems: 1 Mark
Steps: 1 Mark
Correct answers: 2 Marks

Given p $\parallel$ q 

${125}^{\circ }+e={180}^{\circ }$     [Linear pair]

$\therefore e={180}^{\circ }-{125}^{\circ }={55}^{\circ }\left(i\right)$

$e=f={55}^{\circ }$    [Vertically opposite angles]

$a=f={55}^{\circ }$     [Alternate interior angles] 

$a+b={180}^{\circ }$     [Linear pair]

$\Rightarrow {55}^{\circ }+b={180}^{\circ }$  [From eq. (i)]

$\Rightarrow b={180}^{\circ }-{55}^{\circ }={125}^{\circ }$

$a=c={55}^{\circ },b=d={125}^{\circ }$  [Vertically opposite angles]

$\Rightarrow a={55}^{\circ },b={125}^{\circ }$

$c={55}^{\circ },d={125}^{\circ },e={55}^{\circ },f={55}^{\circ }$.