The Ncert Solution For Class 9 Maths Chapter – 11 Constructions provides detailed, step-by-step instructions for all the exercises in the chapter. This chapter focuses on the practical application of geometric concepts through the construction of angles, triangles, and other geometric shapes using a compass and ruler. The solutions guide students in accurately constructing these shapes, helping them understand the underlying principles and techniques. Prepared by experts, these solutions are easy to follow and adhere to the latest NCERT guidelines. Perfect for students aiming to master the art of geometric constructions and perform well in their exams.
Here Are The Ncert Solution For Class 9 Maths Chapter – 11 Constructions
Exercise 11.1
1. Construct an angle of 90° at the initial point of a given ray and justify the construction.
Construction Procedure:
To construct an angle 90°, follow the given steps:
Draw a ray OA.
Take O as a center with any radius, and draw an arc DCB that cuts OA at B.
With B as a center with the same radius, mark a point C on the arc DCB.
With C as a center and the same radius, mark a point D on the arc DCB.
Take C and D as the centers, and draw two arcs which intersect each other with the same radius at P.
Finally, the ray OP is joined, which makes an angle of 90° with OA.
Justification
To prove ∠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, ∠BOC = 60°.
Similarly,
OD = DC = OC
Therefore, DOC is an equilateral triangle
So that, ∠DOC = 60°.
From SSS triangle congruence rule,
△OBC ≅ OCD
So, ∠BOC = ∠DOC [By C.P.C.T]
Therefore, ∠COP = ½ ∠DOC = ½ (60°).
∠COP = 30°
To find the ∠POA = 90°:
∠POA = ∠BOC+∠COP
∠POA = 60°+30°
∠POA = 90°
Hence, justified.
2. Construct an angle of 45° at the initial point of a given ray and justify the construction.
Construction Procedure:
Draw a ray OA.
Take O as a center with any radius, and draw an arc DCB that cuts OA at B.
With B as a center with the same radius, mark a point C on the arc DCB.
With C as a center and the same radius, mark a point D on the arc DCB.
Take C and D as the centers, and draw two arcs which intersect each other with the same radius at P.
Finally, the ray OP is joined, which makes an angle of 90° with OA.
Take B and Q as the centers, and draw the perpendicular bisector which intersects at the point R
Draw a line that joins the points O and R
So, the angle formed ∠ROA = 45°
Justification
From the construction,
∠POA = 90°
The perpendicular bisector from points B and Q divides the ∠POA into two halves. So it becomes
∠ROA = ½ ∠POA
∠ROA = (½)×90° = 45°
Hence, justified
Solution:
(i) 30°
Construction Procedure:
Draw a ray OA
Take O as a center with any radius, and draw an arc BC which cuts OA at B.
With B and C as centers, draw two arcs which intersect each other at point E, and the perpendicular bisector is drawn.
Thus, ∠EOA is the required angle making 30° with OA.
Construction Procedure:
Draw an angle ∠POA = 90°.
Take O as a center with any radius, and draw an arc BC which cuts OA at B and OP at Q
Now, draw the bisector from points B and Q, where it intersects at the point R such that it makes an angle ∠ROA = 45°.
Again, ∠ROA is bisected such that ∠TOA is formed, which makes an angle of 22.5° with OA
(iii) 15°
Construction Procedure:
- An angle ∠DOA = 60° is drawn.
2. Take O as the center with any radius, and draw an arc BC which cuts OA at B and OD at C
3. Now, draw the bisector from points B and C, where it intersects at point E such that it makes an angle ∠EOA = 30°.
4. Again, ∠EOA is bisected such that ∠FOA is formed, which makes an angle of 15° with OA.
5. Thus, ∠FOA is the required angle making 15° with OA.
4. Construct the following angles and verify by measuring them with a protractor:
(i) 75° (ii) 105° (iii) 135°
Solution:
(i) 75°
Construction Procedure:
A ray OA is drawn.
With O as the center, draw an arc of any radius and intersect at point B on the ray OA.
With B as the center, draw an arc C, and with C as the center, draw an arc D.
With D and C as the centers, draw an arc that intersects at point P.
Join points O and P.
The point that the arc intersects the ray OP is taken as Q.
With Q and C as the centers, draw an arc that intersects at point R.
Join points O and R.
Thus, ∠AOE is the required angle making 75° with OA.
(ii) 105°
Construction Procedure:
A ray OA is drawn.
With O as the center, draw an arc of any radius and intersect at point B on the ray OA.
With B as the center, draw an arc C, and with C as the center, draw an arc D.
With D and C as the centers, draw an arc that intersects at point P.
Join the points O and P
The point that the arc intersects the ray OP is taken as Q.
With Q and D as the centers, draw an arc that intersects at point R.
Join points O and R.
Thus, ∠AOR is the required angle making 105° with OA.
(iii) 135°
Construction Procedure:
Draw a line AOA‘
Draw an arc of any radius that cuts the line AOA‘ at points B and B‘
With B as the center, draw an arc of the same radius at point C.
With C as the center, draw an arc of the same radius at point D.
With D and C as the centers, draw an arc that intersects at point P.
Join OP.
The point that the arc intersects the ray OP is taken as Q, and it forms an angle of 90°.
With B‘ and Q as the center, draw an arc that intersects at point R.
Thus, ∠AOR is the required angle making 135° with OA.
5. Construct an equilateral triangle, given its side and justify the construction.
Construction Procedure:
- Let us draw a line segment AB = 4 cm.
2. With A and B as centers, draw two arcs on the line segment AB and note the point as D and E.
3. With D and E as centers, draw the arcs that cut the previous arc respectively that forms an angle of 60° each.
4. Now, draw the lines from A and B that are extended to meet each other at point C.
5. Therefore, ABC is the required triangle.
Justification:
From the construction, it is observed that,
AB = 4 cm, ∠A = 60° and ∠B = 60°
We know that the sum of the interior angles of a triangle is equal to 180°
∠A+∠B+∠C = 180°
Substitute the values
⇒ 60°+60°+∠C = 180°
⇒ 120°+∠C = 180°
⇒∠C = 60°
While measuring the sides, we get,
BC = CA = 4 cm (Sides opposite to equal angles are equal)
AB = BC = CA = 4 cm
∠A = ∠B = ∠C = 60°
Hence, justified.
Ncert Solution For Class 9 Maths Chapter – 11 Constructions
Exercise 11.2
1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB+AC = 13 cm.
Construction Procedure:
The steps to draw the triangle of the given measurement are as follows:
Draw a line segment of base BC = 7 cm.
Measure and draw ∠B = 75° and draw the ray BX.
Take a compass and measure AB+AC = 13 cm.
With B as the center, draw an arc at the point D.
Join DC.
Now draw the perpendicular bisector of the line DC, and the intersection point is taken as A.
Now join AC.
Therefore, ABC is the required triangle.
2. Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB–AC = 3.5 cm.
Construction Procedure:
The steps to draw the triangle of the given measurement are as follows:
Draw a line segment of base BC = 8 cm
Measure and draw ∠B = 45° and draw the ray BX
Take a compass and measure AB-AC = 3.5 cm.
With B as the center, draw an arc at point D on the ray BX.
Join DC.
Now draw the perpendicular bisector of the line CD, and the intersection point is taken as A.
Now join AC.
Therefore, ABC is the required triangle.
3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR–PQ = 2cm.
Construction Procedure:
The steps to draw the triangle of the given measurement are as follows:
Draw a line segment of base QR = 6 cm
Measure and draw ∠Q = 60° and let the ray be QX.
Take a compass and measure PR–PQ = 2cm.
Since PR–PQ is negative, QD will be below the line QR.
With Q as the center, draw an arc at point D on the ray QX.
Join DR.
Now draw the perpendicular bisector of the line DR and the intersection point is taken as P.
Now join PR.
Therefore, PQR is the required triangle.
4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY+YZ+ZX = 11 cm.
Construction Procedure:
The steps to draw the triangle of the given measurement are as follows:
Draw a line segment AB which is equal to XY+YZ+ZX = 11 cm.
Make an angle ∠LAB = 30° from the point A.
Make an angle ∠MBA = 90° from the point B.
Bisect ∠LAB and ∠MBA at point X.
Now, take the perpendicular bisectors of the lines XA and XB, and the intersection points are Y and Z, respectively.
Join XY and XZ.
Therefore, XYZ is the required triangle
5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Construction Procedure:
The steps to draw the triangle of the given measurement are as follows:
Draw a line segment of base BC = 12 cm
Measure and draw ∠B = 90° and draw the ray BX.
Take a compass and measure AB+AC = 18 cm.
With B as the center, draw an arc at point D on the ray BX.
Join DC.
Now draw the perpendicular bisector of the line CD, and the intersection point is taken as A.
Now join AC.
Therefore, ABC is the required triangle.
Ncert Solution For Class 9 Maths Chapter – 11 Constructions
For the Next Chapter Solution Click Below
Chapter 3 – Coordinate Geometry
Chapter 4 – Linear Equations in Two Variables
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