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#### Question 1:

Write the correct answer in each of the following:
In the figure, if AB ∥ CD ∥ EF, PQ ∥ RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(A) 85°
(B) 135°
(C) 145°
(D) 110°

As $\mathrm{PQ}\parallel \mathrm{RS}$, produce PQ to M.

$\therefore 60°=\angle 1+25°\phantom{\rule{0ex}{0ex}}⇒\angle 1=35°$

Also,

Since, $\mathrm{PM}\parallel \mathrm{RS}$.
$\therefore \angle 2+\angle \mathrm{SRM}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{SRM}=180°-60°=120°$

Now,

Applying angle sum property in $△$MQR,
$\angle 1+\angle 3+\angle \mathrm{QMR}=180°\phantom{\rule{0ex}{0ex}}⇒35°+\angle 3+120°=180°\phantom{\rule{0ex}{0ex}}⇒\angle 3=25°$

Finally,
$\begin{array}{rcl}\angle \mathrm{QRS}& =& \angle \mathrm{ARQ}+\angle \mathrm{ARS}\\ & =& 25°+120°\\ & =& 145°\end{array}$

Hence, the correct answer is option C.

#### Question 2:

Write the correct answer in each of the following:
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(A) an isosceles triangle
(B) an obtuse triangle
(C) an equilateral triangle
(D) a right triangle

Take a $△\mathrm{ABC}$ with the angles being

Given that,

Thus, the triangle is a right angled triangle.
Hence, the correct answer is option D.

#### Question 3:

Write the correct answer in each of the following:
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is

(A) $37\frac{1°}{2}$

(B) $52\frac{1°}{2}$

(C) $72\frac{1°}{2}$

(D) 75°

Consider each of the two interior opposite angle be $x$.
An exterior angle of the given triangle is 105°.
The exterior angle of a triangle is equal to the sum of the two interior opposite angles.

$\therefore 105°=x+x\phantom{\rule{0ex}{0ex}}⇒2x=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}×105°=52\frac{1°}{2}$
So, each angle is 52.5°.
Hence, the correct answer is option B.

#### Question 4:

Write the correct answer in each of the following:
The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is
(A) an acute angled triangle
(B) an obtuse angled triangle
(C) a right triangle
(D) an isosceles triangle

By angle sum property of triangles, the sum of the angles of the triangle is 180°.
Thus, let the angle of triangle be
$⇒5x+3x+7x=180°\phantom{\rule{0ex}{0ex}}⇒15x=180°⇒x=\frac{180°}{15}\phantom{\rule{0ex}{0ex}}⇒x=12°$

∴ The angle measures of the triangle are:

As, the measure of each angle of triangle is less than 90°, so the angles of the triangle are acute angles.
Hence, the correct answer is option A.

#### Question 5:

Write the correct answer in each of the following:
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be
(A) 50°
(B) 65°
(C) 145°
(D) 155°

Consider $△\mathrm{PQR}$.
Given that, $\angle \mathrm{P}=130°$.

Now, OQ and OR are the bisectors of the angles Q and R.
$\begin{array}{rcl}\angle \mathrm{QOR}& =& 180°-\frac{1}{2}\left(\angle \mathrm{PQR}+\angle \mathrm{PRQ}\right)\\ & =& 180°-\frac{1}{2}×50°\\ & =& 180°-25°\\ & =& 155°\end{array}$
Hence, the correct answer is option D.

#### Question 6:

Write the correct answer in each of the following:
In the figure, POQ is a line. The value of x is

(A) 20°
(B) 25°
(C) 30°
(D) 35°

Given that,

Thus, the value of $x$ is 20°.
Hence, the correct answer is option A.

#### Question 7:

Write the correct answer in each of the following:
In the figure, if OP ∥ RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR is equal to

(A) 40°
(B) 50°
(C) 60°
(D) 70°

In the figure given, producing OP, intersects RQ at X.
Since OP RS and RX is the transversal.
So,

Now, RQ is a line segment.
So,

is an exterior angle.
[exterior angle = sum of 2 opposite interior angle]
$⇒110°=50°+\angle \mathrm{PQX}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{PQX}=110°-50°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{PQX}=60°$

Hence, the correct answer is option C.

#### Question 8:

Write the correct answer in each of the following:
Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
(A) 60°
(B) 40°
(C) 80°
(D) 20°

The ratio of angles of a triangle is 2 : 4 : 3.
Let the angles of the triangle be

Thus, the smallest angle of a triangle is 40°.
Hence, the correct answer is option B.

#### Question 1:

For what value of x + y In the Figure will ABC be a line? Justify your answer.

In the given figure,  are two adjacent angles.
For ABC to be a straight line, the sum of two adjacent angles  must be 180°.

#### Question 2:

Can a triangle have all angles less than 60°? Give reason for your answer.

A triangle cannot have all angles less than 60°. This is because the sum of the angles of the triangle will be less than 180°, which does not satisfy the angle sum property of triangles.

#### Question 3:

Can a triangle have two obtuse angles? Give reason for your answer.

According to angle sum property of triangles, the sum of the three angles of a triangle is 180°.

Thus, a triangle cannot have 2 obtuse-angles because the sum of all the angles cannot exceed 180°.

#### Question 4:

How many triangles can be drawn having its angles as 45°, 64° and 72°? Give reason for your answer.

As, $45°+64°+72°=181°\ne 180°$.
We cannot draw any triangle having its angle as 45°, 64° and 72° because the sum of the angles cannot be 181°.

#### Question 5:

How many triangles can be drawn having its angles as 53°, 64° and 63°? Give reason for your answer.

As, sum of the angles = $53°+64°+63°=180°.$
Thus, we can draw infinitely many triangles, whose sum of the angles of every triangle, having angles as 53°, 64° and 63° is 180°.

#### Question 6:

In the figure, find the value of x for which the lines l and m are parallel.

If a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary.

Here, the two given lines $l$ and $m$ are parallel.
Now, the angles  are consecutive interior angles on the same side of the transversal.

#### Question 7:

Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.

No, each of these angles will be a right angle only when they form a linear pair. i.e. when the non-common arms of the given two adjacent angle are two opposite rays.

#### Question 8:

If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give reason for your answer.

If two line intersect each other at a point, then four angles are formed. If one of these four angles is a right angle, then each of other three angle will be a right angle by linear pair axiom.

#### Question 9:

In the figure, which of the two lines are parallel and why?

For figure (1), a transversal intersects two lines such that the sum of interior angles on the same side of the transversal is
$132°+48°=180°$
Thus, the lines $l$ and $m$ are parallel.

For figure (2), a transversal intersects two lines such that the sum of interior angles on the same side of the transversal is
$73°+106°=179°$
Therefore, the line $p$ and $q$ are not parallel.

#### Question 10:

Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.

When 2 lines $l$ and $m$ are perpendicular to the same line $n$, each of the two corresponding angles formed by these lines $l$ and $m$ with the line $n$ equal to 90°.
Hence, $l$ and $m$ are parallel to each other.

#### Question 1:

In the figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that the points A, O and B are collinear.

In the figure, OD ⏊ OE, OD and OE are the bisector of  respectively.

As, OD and DE bisects  respectively.

On adding equations (1) and (2),

So, $\angle \mathrm{AOC}+\angle \mathrm{COB}$ are forming linear pair i.e. AOB is a straight line.
Hence, point A, O and B are collinear.

#### Question 2:

In the figure, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

From the figure,

But, these are corresponding angles. Hence, $m$ and $n$ are parallel.

#### Question 3:

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m. Show that AP ∥ BQ.

As, $l\parallel m$ and $t$ is the transversal.

$⇒\frac{1}{2}\angle \mathrm{MAB}=\frac{1}{2}\angle \mathrm{ABN}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{PAB}=\angle \mathrm{ABQ}$

Now,  are alternate angles.
Hence, AP ∥ BQ.

#### Question 4:

In the figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l m.

In the given figure, it is given that AP ∥ BQ.

Now, AP is the bisector of  is the bisector of $\angle \mathrm{ABN}$.

As AP ∥ BQ,

As, these are alternate angles.
Hence, $l\parallel m$.

#### Question 5:

In the figure, BA ∥ ED and BC  EF. Show that ∠ABC = ∠DEF.

Produce DE to intersect BC at P.
$\mathrm{EF}\parallel \mathrm{BC}$ and DP is the transversal.

Now, $\mathrm{AB}\parallel \mathrm{DP}$ and BC is the transversal.

From (1) and (2),
$\angle \mathrm{ABC}=\angle \mathrm{DEF}$
Hence, proved.

#### Question 6:

In the Figure BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°

Produce ED to meet BC at point P.

Now,
is the transversal.

The sum of co-interior angles on the same side of the transversal is 180°.

Again,  is transversal.

From (1) and (2),
$\angle \mathrm{DEF}+\angle \mathrm{ABC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ABC}+\angle \mathrm{DEF}=180°$
Hence, proved.

#### Question 7:

In the figure, DE  QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB.

Given that, $\mathrm{DE}\parallel \mathrm{QR}$ and line $n$ is the transversal line.

The sum of two co-interior angles on the same side of the transversal is 180°.

$⇒\angle \mathrm{PAB}+\angle \mathrm{PBA}=90°$        [∵ AP is the bisector of $\angle \mathrm{EAB}$ and BP is the bisector of $\angle \mathrm{RBA}$]

Now, in $△$APB,
$\angle \mathrm{APB}=180°-\left(\angle \mathrm{PAB}+\angle \mathrm{PBA}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{APB}=180°-90°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{APB}=90°$
Hence, $\angle \mathrm{APB}=90°.$

#### Question 8:

The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.

Given that, the ratio of angles 2 : 3 : 4.

Let angles of triangle be .

Thus,
$\angle \mathrm{A}=2×10=40°\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}=3×20=60°\phantom{\rule{0ex}{0ex}}\angle \mathrm{C}=4×20=80°$

Hence, the angles of the $△$ABC are .

#### Question 9:

A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL = ∠ACB.

In $∆\mathrm{ABC}$

Substituting the value from equation $\overline{)4}$ and $\overline{)5}$ in equation $\overline{)3}$, we get.
$180°-\angle \mathrm{ACB}=180°-\angle \mathrm{BAL}⇒\angle \mathrm{ACB}=\angle \mathrm{BAL}.$
Hence, proved.

#### Question 10:

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Two lines $p$ and $n$ are respectively perpendicular to two parallel lines $l$ and  $m$, i.e.,
Show that: $p$ is parallel to $n$.

As $n\perp m$,
so,

Again, $p\perp l$,
so,

But $l$ is parallel to $m$, so

From (1) and (2),

As, these are corresponding angles.
Hence, $p\parallel n$

#### Question 1:

If two lines intersect, prove that the vertically opposite angles are equal.

Given that, the two lines AB and CD intersect at point O.

To prove:
$\left(\mathrm{i}\right)\angle \mathrm{AOC}=\angle \mathrm{BOD}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\angle \mathrm{AOD}=\angle \mathrm{BOC}$

Proof:

Similarly, ray OD stands line AB.

From (1) and (2),
$\angle \mathrm{AOC}+\angle \mathrm{AOD}=\angle \mathrm{AOD}+\angle \mathrm{BOD}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{AOC}=\angle \mathrm{BOD}$

Hence, proved.

(ii) Since, ray OD stands on line AB.

Similarly, ray OB stands on line CD.

From (3) and (4),
$\angle \mathrm{AOD}+\angle \mathrm{BOD}=\angle \mathrm{DOB}+\angle \mathrm{BOC}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{AOD}=\angle \mathrm{BOC}$

Hence, proved.

#### Question 2:

Bisectors of interior ∠B and exterior ∠ACD of a $△$ABC intersect at the point T. Prove that $\angle \mathrm{BTC}=\frac{1}{2}\angle \mathrm{BAC}$.

For a triangle $△$ABC, produce BC to D and the bisectors of $\angle \mathrm{ABC}$ and $\angle \mathrm{ACD}$ meet at point T.

Now,

Also,

In $△$ABC, $\angle \mathrm{ACD}$ is an exterior angle.
$\therefore \angle \mathrm{ACD}=\angle \mathrm{ABC}+\angle \mathrm{CAB}$

Again, in $△$BTC,

From (1) and (2),

Hence, proved.

#### Question 3:

A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.

Consider AB and CD to be the two parallel lines and are intersected by transversal GH at P and Q.
Now, EP and FQ be the bisectors of corresponding angles $\angle \mathrm{APG}$ and $\angle \mathrm{CQP}.$

Since, $\mathrm{AB}\parallel \mathrm{CD}$.

As, the corresponding angle made by the lines EP and FQ and the transversal line GH are $\angle \mathrm{EPG}$ and $\angle \mathrm{FQP}$ and they are equal.
Hence, $\mathrm{EP}\parallel \mathrm{FQ}.$

#### Question 4:

Prove that through a given point, we can draw only one perpendicular to a given line.

Consider three lines  A point P formed by intersection of the lines $m$ and $n$.

The lines $m$ and $n$ are intersecting at P and are perpendicular to line l.

i.e., the two lines $m$ and $n$ coincide.

Hence, only one perpendicular line can be drawn through a given point on a given line.

#### Question 5:

Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.

Consider two lines $l$ and $m$ to be intersecting.
Also, consider another two lines $n$ and $p$ perpendicular to $m$ and $l$ respectively.

Assume that the lines $n$ and $p$ are not intersecting. This means they will be parallel to each other.

Since, $n$ and $p$ are perpendicular to $m$ and $l$ respectively.

But from (1), $l\parallel m$ which shows contradiction.
Thus, our assumption was wrong.
Hence, the lines $n$ and $p$ intersect at a point D.

#### Question 6:

Prove that a triangle must have atleast two acute angles.

Consider $△$ABC. By angle sum property of triangles,

Case 1: When two angles are 90°.
Let ∠B and ∠C be 90° in measure.

Thus, from (1),
$\angle \mathrm{A}+90°+90°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}+180°=180°$
$⇒\angle \mathrm{A}=0$
Thus, triangle is not possible.

Case 2: When two angles are obtuse.
Let ∠B and ∠C be obtuse angles.

Thus, from (1),

Thus, triangle is not possible.

Case 3: When one angle is 90°.
Let ∠B = 90°.

Therefore,
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}+\angle \mathrm{C}+90°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}+\angle \mathrm{C}=90°$
Thus, sum of the other two angles are 90°.
Hence, both angles are acute.

Case 4: When two angles are acute.
Then, the sum of two angles < 180°, then the third angle may be an acute or obtuse angle.
Thus, a triangle should have at least 2 acute angles.

#### Question 7:

In the figure, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that $\angle \mathrm{APM}=\frac{1}{2}\left(\angle \mathrm{Q}-\angle \mathrm{R}\right)$.

As, PA is bisector of $\angle \mathrm{QPR}$.

In $△$PQM, by angle sum property of triangles,

Again, in $△$PMR, by angle sum property of triangles,

Subtracting (3) from (2),
$\angle \mathrm{Q}-\angle \mathrm{R}=\left(90°-\angle \mathrm{QPM}\right)-\left(90°-\angle \mathrm{RPM}\right)$

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