Class 9 Maths Exercise 11.1 Solution

To construct an angle \( 90^{\circ} \), follow the given steps:
1. Draw a ray OA
2. Take O as a centre with any radius, draw an arc DCB is that cuts OA at B .
3. With \( B \) as a centre with the same radius, mark a point \( C \) on the arc DCB.
4. With C as a centre and the same radius, mark a point D on the arc DCB.
5. Take C and D as centre, draw two arcs which intersect each other with the same radius at P .
6. Finally, the ray OP is joined which makes an angle \( 90^{\circ} \) with OP is formed.

Justification
To prove \( \angle \mathrm{POA}=90^{\circ} \)
In order to prove this, draw a dotted line from the point O to C and O to D and the angles formed are:

From the construction, it is observed that
\(\mathrm{OB}=\mathrm{BC}=\mathrm{OC}\)
Therefore, OBC is an equilateral triangle
So that, \( \angle \mathrm{BOC}=60^{\circ} \).
Similarly,
\(\mathrm{OD}=\mathrm{DC}=\mathrm{OC}\)
Therefore, DOC is an equilateral triangle
So that, \( \angle D O C=60^{\circ} \).
From SSS triangle congruence rule
\(\triangle \mathrm{OBC} \cong \mathrm{OCD}\)
So, \( \angle \mathrm{BOC}=\angle \mathrm{DOC}[ \) By C.P.C.T \( ] \)
Therefore, \( \angle \mathrm{COP}=\frac{1} {2} \angle \mathrm{DOC}=\frac{1} {2} \left(60^{\circ}\right) \).
\(\angle C O P=30^{\circ}\)
To find the \( \angle \mathrm{POA}=90^{\circ} \) :
\(\angle \mathrm{POA}=\angle \mathrm{BOC}+\angle \mathrm{COP}\)
\(\angle \mathrm{POA}=60^{\circ}+30^{\circ}\)
\(\angle \mathrm{POA}=90^{\circ}\)
Hence, justified.

1. Draw a ray OA
2. Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.
3. With \( B \) as a centre with the same radius, mark a point C on the arc DCB.
4. With C as a centre and the same radius, mark a point D on the arc DCB.
5. Take C and D as centre, draw two arcs which intersect each other with the same radius at \( P \).
6. Finally, the ray OP is joined which makes an angle \( 90^{\circ} \) with OP is formed.
7. Take B and Q as centre draw the perpendicular bisector which intersects at the point \( R \)
8. Draw a line that joins the point O and R
9. So, the angle formed \( \angle R O A=45^{\circ} \)

Justification
From the construction,
\(
\angle \mathrm{POA}=90^{\circ}
\)
From the perpendicular bisector from the point B and Q, which divides the \( \angle \) POA into two halves. So it becomes
\( \angle \mathrm{ROA}=\frac{1}{2} \angle \mathrm{POA} \)
\( \angle \mathrm{ROA}=\left(\frac{1}{2}\right) \times 90^{\circ}=45^{\circ} \)
Hence, justified

Draw a ray OA
2. Take O as a centre with any radius, draw an arc BC which cuts OA at B.
3. With B and C as centres, draw two arcs which intersect each other at the point E and the perpendicular bisector is drawn.
4. Thus, \( \angle \mathrm{EOA} \) is the required angle making \( 30^{\circ} \) with OA .

\( 22 \frac{1^{\circ}}{2} \)
1. Draw an angle \( \angle \mathrm{POA}=90^{\circ} \)
2. Take O as a centre with any radius, draw an arc BC which cuts OA at B and OP at Q
3. Now, draw the bisector from the point B and Q where it intersects at the point \( R \) such that it makes an angle \( \angle R O A=45^{\circ} \).
4. Again, \( \angle \mathrm{ROA} \) is bisected such that \( \angle \mathrm{TOA} \) is formed which makes an angle of \( 22.5^{\circ} \) with OA

\( 15^{\circ} \)
1. An angle \( \angle \mathrm{DOA}=60^{\circ} \) is drawn.
2. Take O as centre with any radius, draw an arc BC which cuts OA at B and OD at C
3. Now, draw the bisector from the point B and C where it intersects at the point E such that it makes an angle \( \angle \mathrm{EOA}=30^{\circ} \).
4. Again, \( \angle \mathrm{EOA} \) is bisected such that \( \angle \mathrm{FOA} \) is formed which makes an angle of \( 15^{\circ} \) with OA.
5. Thus, \( \angle \mathrm{FOA} \) is the required angle making \( 15^{\circ} \) with OA .

\( 75^{\circ} \)
1. A ray OA is drawn.
2. With O as centre draw an arc of any radius and intersect at the point B on the ray OA .
3. With \( B \) as centre draw an \( \operatorname{arc} \mathrm{C} \) and C as centre draw an \( \operatorname{arc} \mathrm{D} \).
4. With D and C as centre draw an arc, that intersect at the point P .
5. Join the points O and P
6. The point that arc intersect the ray OP is taken as Q .
7. With Q and C as centre draw an arc, that intersect at the point R .
8. Join the points O and R
9. Thus, \( \angle \mathrm{AOE} \) is the required angle making \( 75^{\circ} \) with OA .

\( 105^{\circ} \)
1. A ray OA is drawn.
2. With \( O \) as centre draw an arc of any radius and intersect at the point \( B \) on the ray OA .
3. With \( B \) as centre draw an \( \operatorname{arc} \mathrm{C} \) and C as centre draw an \( \operatorname{arc} \mathrm{D} \).
4. With \( D \) and \( C \) as centre draw an arc, that intersect at the point \( P \).
5. Join the points \( O \) and \( P \)
6. The point that arc intersect the ray OP is taken as Q .
7. With Q and Q as centre draw an arc, that intersect at the point R .
8. Join the points O and R
9. Thus, \( \angle \mathrm{AOR} \) is the required angle making \( 105^{\circ} \) with OA.

\( 135^{\circ} \)
1. Draw a line \(AOA^{\prime}\)
2. Draw an arc of any radius that cuts the line \(AOA^{\prime}\) at the point \( B \) and \(B^{\prime}\)
3. With \( B \) as centre, draw an arc of same radius at the point \( C \).
4. With C as centre, draw an arc of same radius at the point D
5. With D and C as centre, draw an arc that intersect at the point O
6. Join OP
7. The point that arc intersect the ray OP is taken as Q and it forms an angle \( 90^{\circ} \)
8. With \( \mathrm{B}^{\prime} \) and Q as centre, draw an arc that intersects at the point R
9. Thus, \( \angle \mathrm{AOR} \) is the required angle making \( 135^{\circ} \) with OA .

1. Let draw a line segment \( A B=4 \mathrm{~cm} \).
2. With \( A \) and \( B \) as centres, draw two arcs on the line segment \( A B \) and note the point as D and E .
3. With D and E as centres, draw the arcs that cuts the previous arc respectively that forms an angle of \( 60^{\circ} \) each.
4. Now, draw the lines from A and B that are extended to meet each other at the point \( C \).
5. Therefore, ABC is the required triangle.

Justification:
From construction, it is observed that
\(
\mathrm{AB}=4 \mathrm{~cm}, \angle \mathrm{A}=60^{\circ} \text { and } \angle \mathrm{B}=60^{\circ}
\)
We know that, the sum of the interior angles of a triangle is equal to \( 180^{\circ} \)
\(
\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}
\)
Substitute the values
\(\Rightarrow 60^{\circ}+60^{\circ}+\angle \mathrm{C}=180^{\circ}\)
\(\Rightarrow 120^{\circ}+\angle \mathrm{C}=180^{\circ}\)
\(\Rightarrow \angle \mathrm{C}=60^{\circ}\)
While measuring the sides, we get
\( \mathrm{BC}=\mathrm{CA}=4 \mathrm{~cm} \) (Sides opposite to equal angles are equal)
\(\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=4 \mathrm{~cm}\)
\(\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ}\)
Hence, justified.