NCERT Solutions Class 9 Maths Chapter 11 – Constructions

Exercise 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Construction Procedure:
Follow these steps to make a 90-degree angle:
Make an OA ray
Draw an arc DCB that cuts OA at B, using O as the center with any radius.
Draw a point C on the arc DCB with B as the center and the same radius.
Draw a point D on the arc DCB with C as the center and the same radius as C.
Starting at C and D, draw two arcs that cross at P and have the same radius.
Finally, the ray OP is linked, resulting in a 90° angle with OP.

Justification
To prove \angle POA\ =\ 90°
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
OB\ =\ BC\ =\ OC
Therefore, OBC is an equilateral triangle
So that, \angle BOC\ =\ 60°.
Similarly,
OD\ =\ DC\ =\ OC
Therefore, DOC is an equilateral triangle
So that, \angle DOC\ =\ 60°.
Now, From SSS triangle congruence rule
\triangle OBC\ \cong\ OCD
So, \angle BOC\ =\ \angle DOC [By C.P.C.T]
Therefore, \angle COP\ =\frac{1}{2}\ \angle DOC\ =\frac{1}{2}\ (60°). \angle COP\ =\ 30°
To find the \angle POA\ =\ 90°: \angle POA\ =\ \angle BOC+\angle COP \angle POA\ =\ 60°+30° \angle POA\ =\ 90°
Hence, justified.

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Construction Procedure:
Make an OA ray
Draw an arc DCB that cuts OA at B, using O as the center with any radius.
Draw a point C on the arc DCB with B as the center and the same radius.
Draw a point D on the arc DCB with C as the center and the same radius as C.
Starting at C and D, draw two arcs that cross at P and have the same radius.
Finally, the ray OP is linked, resulting in a 90° angle with OP.
Using B and Q as a starting point, construct a perpendicular bisector that intersects at R.
Draw a line connecting points O and R.
As a result, the angle created was \angle ROA\ =\ 45°.

Justification
From the above construction,
\angle POA\ =\ 90°
From the perpendicular bisector from the point B and Q, which divides the ∠POA into two halves. So it becomes
\angle ROA\ =\frac{1}{2}\angle POA \angle ROA\ =\ \left(\frac{1}{2}\right)\times90° = 45°
Hence, justified

Question 3.
Construct the angles of the following measurements:
30°
22{\frac{1}{2}}^0
15°
Solution:
30^0
Construct Procedure:
Draw an OA ray
Draw an arc BC that cuts OA at B, using O as the center with any radius.
Draw two arcs that overlap at point E, and the perpendicular bisector is created, using B and C as centers.
As a result, EOA is the required angle to make 30 degrees with OA.

22{\frac{1}{2}}^0
Construction Procedure:

Draw a 90-degree angle (\angle POA).
Draw an arc BC that cuts\ OA at\ B and OP at Q, using\ O as the center with any radius.
Now draw the bisector between points B and Q, intersecting at point R, and making an angle of \angle ROA\ =\ 45°.
∠ROA is bisected once more, resulting in ∠TOA, which forms a 22.5° angle with OA.

{15}^0
Construction Procedure:
A 60° DOA angle is drawn.
Draw an arc BC that cuts OA\ at B and OD at C, with\ O as the center and any radius.
Now draw the bisector between points B and C, intersecting at point E, and making an angle of [Katex] \angle EOA\ =\ 30°.[/katex]
\angle EOA is bisected once more, resulting in \angle FOA, which forms a 15° angle with OA.
As a result, \angle FOA is the needed angle, which is 15° with OA.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
75° (ii) 105° (iii) 135°
Solution:
75^0
Construction Procedure:
The OA ray is drawn.
Draw an arc of any radius with O as the centre and intersect it at point B on the ray OA.
Draw an arc C with B as the centre and an arc D with\ C\ as the centre.
Draw an arc with D and C as the centres, intersecting at P.
Connect the\ O and\ P points.
Q\ is the location where the arc intersects the ray OP.
Draw an arc with Q\ and C\ as the centre points, intersecting at R.
Connect the O and R\ points.
As a result, \angle AOE is the needed angle when OA is 75 degrees.

105^0
Construction Procedure:
An OA ray is created.
Draw an arc of any radius with O as the centre and intersect it at point B on the ray OA.
Draw an arc C with B as the centre and an arc D with C as the centre.
Draw an arc with D and C as the centres, intersecting at P.
Connect the O and P points.
Q is the location where the arc intersects the ray OP.
Draw an arc with Q and D as the centre points, intersecting at R.
Connect the O and R points.
As a result, ∠AOR is the needed angle, which is 105° with OA.

{135}^0
Construction Procedure:
Draw an AOA\prime line.
Draw an arc of any radius that intersects the line AOA\primeat B and B’.
Draw an arc of the same radius at point C with B as the center.
At point D, draw an arc with the same radius as C.
Draw an arc between D and C that intersects at point P.
Join OP
Q is the place where the arc intersects the ray OP, forming a 90° angle.
Draw an arc that connects at the point R with B’ and Q as the center.
As a result, \angle AOR is the needed angle, which is 135° with OA.

Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Construction Procedure:
Let’s draw a line segment with the length AB = 4 cm.
Draw two arcs on the line segment AB with A and B as the centres, and label the points D and E.
Draw the arcs that cut the previous arc and make a 60° angle each with D and E as the centres.
Connect the lines A and B at C.
As a result, ABC is the necessary triangle.

Justification
From construction, it is observed that
AB\ =\ 4\ cm,\ \angle A\ =\ 60° and ∠B = 60°
We know that, the sum of the interior angles of a triangle is equal to 180°
\angle A+\angle B+\angle C\ =\ 180°
Substitute the values
</p><p>\Rightarrow\ 60°+60°+∠C = 180° \Rightarrow\ 120°+∠C = 180° \Rightarrow\angle C\ =\ 60°
After measuring the sides, we get
BC\ =\ CA\ =\ 4\ cm (Sides opposite to equal angles are equal) AB\ =\ BC\ =\ CA\ =\ 4\ cm \angle A\ =\ \angle B\ =\ \angle C\ =\ 60°
Hence, justified.

Exercise 11.2

Question 1.
Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB+AC = 13 cm.
Construction Procedure:
To draw the triangle of a given measurement, follow these steps:
Draw a base BC\ =\ 7 cm line segment.
Draw the ray BX and measure and draw \angle B\ =\ 75°.
Using a compass, determine that AB+AC = 13 cm.
Draw an arc at the point D with B as the center.
Join DC
Draw the perpendicular bisector of the line DC, with A as the junction point.
Finally, join AC.
As a result, ABC is the necessary triangle.

Question 2.
Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB–AC = 3.5 cm.
Construction Procedure:
To draw the triangle of a given measurement, follow these steps:
Draw a base BC\ =\ 8\ cm line segment.
Draw the ray BX and measure and draw ∠B = 45°.
Using a compass, find\ AB-AC\ =\ 3.5\ cm.
Draw an arc at the point D on the ray BX with B as the centre.
Join DC
Draw the perpendicular bisector of the line\ CD, with A as the intersection point.
Finally, join AC.
As a result, ABC is the necessary triangle.

Question 3.
Construct a triangle PQR in which QR = 6cm, \angleQ = 60° and PR–PQ = 2cm.
Construction Procedure:
To draw the triangle of a given measurement, follow these steps:
Draw a 6 cm base QR line segment.
Measure and draw ∠Q = 60°, with QX as the ray.
Using a compass, find PR–PQ = 2cm.
QD will be below the line QR since PR–PQ is negative.
Draw an arc at the point D on the ray QX with Q as the centre.
Join DR
Now draw the line DR’s perpendicular bisector, with P as the junction point.
Finally, join PR.
As a result, PQR is the necessary triangle.

Question 4.
Construct a triangle XYZ in which \angle Y\ =\ 30°, \angle Z\ =\ 90° and XY+YZ+ZX\ =\ 11\ cm.
Construction Procedure:
To draw the triangle of a given measurement, follow these steps:
Draw an AB line segment with the length XY+YZ+ZX\ =\ 11\ cm.
From point A, make an angle of \angle LAB\ =\ 30°.
From point B, make an angle of \angle MBA\ =\ 90°.
At point X, divide \angle LAB\ and\ \angle MBA.
Let the perpendicular bisector of the lines XA and XB be\ Y and Z, respectively, and the intersection point be\ Z.
Bring\ XY and XZ together.
As a result, the needed triangle is XYZ.

Question 5.
Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.
Construction Procedure:
To draw the triangle of a given measurement, follow these steps:
Draw a base BC\ =\ 12\ cm line segment.
Draw the ray BX and measure and draw \angle B\ =\ 90°.
Using a compass, determine that AB+AC = 18 cm.
Draw an arc at the point D on the ray BX with\ B\ as the center.
Join DC
Draw the perpendicular bisector of the line CD, with A\ as the intersection point.
Finally, join AC.
As a result, ABC is the necessary triangle.