Chapter 11 Constructions (Class 9 - Maths Old NCERT Textbook Solutions)
Welcome to Chapter 11: Constructions! This chapter bridges the gap between theoretical geometry and practical execution by emphasizing the creation of accurate figures using only a straightedge and compass. Mastering these classical construction techniques is a vital milestone in a student's mathematical journey, as it provides the essential framework for spatial reasoning and the sophisticated logical justification required in higher mathematical and scientific studies.
The chapter mainly focuses on fundamental geometric operations—including constructing angle bisectors, perpendicular bisectors, and specific angles such as $60^\circ, 90^\circ, 45^\circ$, and $120^\circ$ without the use of a protractor. Students will explore the more intricate task of constructing triangles based on specific data such as the base, a base angle, and the sum or difference of the other two sides, as well as constructions given the perimeter and two base angles. Crucially, the solutions provide systematic methods for each procedure, ensuring that every arc and line is drawn with mathematical precision.
Practical applications are highlighted through the requirement for geometric justification, where students must prove the validity of their constructions using theorems like SSS congruence and the properties of isosceles triangles. To support effective learning, this page provides clear step-by-step solutions, logical explanations, and practical examples for every exercise, reinforcing both theoretical structure and computational fluency. With well-structured content prepared by learningspot.co, students can confidently master constructions and build a solid foundation for future mathematical excellence.
| Content On This Page | ||
|---|---|---|
| Exercise 11.1 | Example 1 (Before Exercise 11.2) | Exercise 11.2 |
Exercise 11.1
Question 1. Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer:
Given:
A ray OA with initial point O.
To Construct:
An angle of $90^\circ$ at the initial point O of the ray OA.
Steps of Construction:
1. Draw a ray OA.
2. With O as centre and a convenient radius, draw an arc of a circle which intersects OA at a point P.
3. With P as centre and the same radius as before, draw an arc intersecting the previously drawn arc at point Q.
4. With Q as centre and the same radius as before, draw an arc intersecting the arc at point R.
5. With Q and R as centres and a radius greater than half of QR, draw two arcs intersecting each other at point S.
6. Draw the ray OS.
Then $\angle AOS$ is the required angle of $90^\circ$.
Justification:
Join OQ and OR.
By construction, OP = PQ = OQ. Therefore, $\triangle OPQ$ is an equilateral triangle, and $\angle POQ = 60^\circ$.
Similarly, OQ = QR = OR. Therefore, $\triangle OQR$ is an equilateral triangle, and $\angle QOR = 60^\circ$.
The ray OS is constructed as the bisector of $\angle QOR$.
So, $\angle QOS = \frac{1}{2} \angle QOR = \frac{1}{2} \times 60^\circ = 30^\circ$.
Now, the constructed angle is $\angle AOS$.
$\angle AOS = \angle POQ + \angle QOS$
$\angle AOS = 60^\circ + 30^\circ = 90^\circ$
Thus, the construction is justified.
Question 2. Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:
Given:
A ray AB with initial point A.
To Construct:
An angle of $45^\circ$ at the initial point A, i.e., $\angle CAB = 45^\circ$.
Steps of Construction:
1. Draw a ray AB.
2. With A as centre and any radius, draw an arc which intersects AB at point P.
3. With P as centre and the same radius, draw an arc intersecting the previous arc at point Q.
4. With Q as centre and the same radius, draw an arc intersecting the arc at point R.
5. With Q and R as centres and radius more than half of QR, draw two arcs intersecting each other at point D.
6. Join AD. $\angle DAB = 90^\circ$. Let the line AD intersect the original arc at point S.
7. With P and S as centres and radius more than half of PS, draw two arcs intersecting each other at point C.
8. Join AC. The angle $\angle CAB$ so formed is the required angle of $45^\circ$.
Justification:
By construction, AD is perpendicular to AB.
$\angle DAB = 90^\circ$
(By construction of perpendicular)
Also, AC is the angle bisector of $\angle DAB$.
$\angle CAB = \frac{1}{2} \angle DAB$
Substituting the value of $\angle DAB$:
$\angle CAB = \frac{1}{2} \times 90^\circ$
$\angle CAB = 45^\circ$
Hence, the construction is justified.
Question 3. Construct the angles of the following measurements:
(i) 30°
(ii) $22\frac{1}{2}^\circ$
(iii) 15°
Answer:
(i) To construct an angle of $30^\circ$
Steps of Construction:
1. Draw a ray OA.
2. With O as centre and any radius, draw an arc intersecting OA at point P.
3. With P as centre and the same radius, draw an arc intersecting the previous arc at point Q. (This represents $60^\circ$)
4. With P and Q as centres and radius more than half of PQ, draw two arcs intersecting each other at point R.
5. Join OR. The angle $\angle ROA$ is the required angle of $30^\circ$.
Justification:
By construction, $\angle QOA = 60^\circ$.
OR is the bisector of $\angle QOA$.
$\angle ROA = \frac{1}{2} \angle QOA$
$\angle ROA = \frac{1}{2} \times 60^\circ = 30^\circ$
(ii) To construct an angle of $22\frac{1}{2}^\circ$
Steps of Construction:
1. Draw a ray OA and construct $\angle BOA = 90^\circ$ at the initial point O.
2. Draw the bisector OC of $\angle BOA$. Thus, $\angle COA = 45^\circ$.
3. Let the arc intersect ray OA at P and ray OC at S.
4. With P and S as centres and radius more than half of PS, draw two arcs intersecting each other at point D.
5. Join OD. The angle $\angle DOA$ is the required angle of $22\frac{1}{2}^\circ$.
Justification:
By construction, $\angle COA = 45^\circ$.
OD is the bisector of $\angle COA$.
$\angle DOA = \frac{1}{2} \angle COA$
$\angle DOA = \frac{1}{2} \times 45^\circ = 22\frac{1}{2}^\circ$
(iii) To construct an angle of $15^\circ$
Steps of Construction:
1. Draw a ray OA and construct $\angle QOA = 60^\circ$ at the initial point O.
2. Draw the bisector OR of $\angle QOA$. Thus, $\angle ROA = 30^\circ$.
3. Let the arc intersect ray OA at P and ray OR at T.
4. With P and T as centres and radius more than half of PT, draw two arcs intersecting each other at point S.
5. Join OS. The angle $\angle SOA$ is the required angle of $15^\circ$.
Justification:
By construction, $\angle ROA = 30^\circ$.
OS is the bisector of $\angle ROA$.
$\angle SOA = \frac{1}{2} \angle ROA$
$\angle SOA = \frac{1}{2} \times 30^\circ = 15^\circ$
Question 4. Construct the following angles and verify by measuring them by a protractor:
(i) $75^\circ$
(ii) $105^\circ$
(iii) $135^\circ$
Answer:
(i) Construction of $75^\circ$ angle
Steps of Construction:
1. Draw a ray OA.
2. With O as centre and any radius, draw a semi-circular arc intersecting OA at P.
3. With P as centre and same radius, draw an arc to intersect the previous arc at Q (representing $60^\circ$).
4. With Q as centre and same radius, draw another arc to intersect the semi-circle at R (representing $120^\circ$).
5. Bisect the arc QR to get a point S such that $\angle SOA = 90^\circ$. Let S be the point where the $90^\circ$ line intersects the semi-circle.
6. Bisect the arc QS (the angle between $60^\circ$ and $90^\circ$). With Q and S as centres, draw two arcs intersecting at T.
7. Join OT. The angle $\angle TOA$ is the required angle of $75^\circ$.
Verification:
On measuring $\angle TOA$ with a protractor, we find that the angle is $75^\circ$.
$\angle TOA = 75^\circ$
(Measured by Protractor)
(ii) Construction of $105^\circ$ angle
Steps of Construction:
1. Draw a ray OA.
2. With O as centre and any radius, draw a semi-circular arc intersecting OA at P.
3. Mark points Q ($60^\circ$) and R ($120^\circ$) on the arc as done in the previous construction.
4. Construct $\angle SOA = 90^\circ$ by bisecting the arc between Q and R. Let S be the intersection point on the semi-circle.
5. Bisect the arc SR (the angle between $90^\circ$ and $120^\circ$). With S and R as centres, draw two arcs intersecting at U.
6. Join OU. The angle $\angle UOA$ is the required angle of $105^\circ$.
Verification:
On measuring $\angle UOA$ with a protractor, we find that the angle is $105^\circ$.
$\angle UOA = 105^\circ$
(Measured by Protractor)
(iii) Construction of $135^\circ$ angle
Steps of Construction:
1. Draw a line A'OA where O is the initial point of the ray OA.
2. With O as centre and any radius, draw a semi-circle intersecting the line at P (on ray OA) and B (on ray OA'). Note that $\angle BOA = 180^\circ$.
3. Construct $\angle SOA = 90^\circ$ at point O. Let S be the point on the semi-circle.
4. Now, $\angle BOS = 90^\circ$ (since $180^\circ - 90^\circ = 90^\circ$).
5. Bisect the angle $\angle BOS$. With B and S as centres, draw two arcs intersecting at V.
6. Join OV. The angle $\angle VOA$ is the required angle.
Verification:
Mathematically, $\angle VOA = \angle SOA + \angle VOS$.
$\angle VOA = 90^\circ + 45^\circ = 135^\circ$
On measuring $\angle VOA$ with a protractor, we find that the angle is $135^\circ$.
$\angle VOA = 135^\circ$
(Measured by Protractor)
Question 5. Construct an equilateral triangle, given its side and justify the construction.
Answer:
Given:
A line segment AB representing the side of the equilateral triangle.
To Construct:
An equilateral triangle with side length equal to AB.
Steps of Construction:
1. Draw a line segment AB of the given length.
2. With A as the centre and radius equal to AB, draw an arc.
3. With B as the centre and the same radius (equal to AB), draw another arc that intersects the first arc at a point C.
4. Join AC and BC.
Then $\triangle ABC$ is the required equilateral triangle.
Justification:
By construction, we have the base side AB.
The point C lies on the arc drawn from centre A with radius AB. Therefore, $AC = AB$.
The point C also lies on the arc drawn from centre B with radius AB. Therefore, $BC = AB$.
From these two statements, we have $AB = AC = BC$.
Since all three sides of the triangle $\triangle ABC$ are equal, it is an equilateral triangle.
Thus, the construction is justified.
Example 1 (Before Exercise 11.2)
Example 1. Construct a triangle ABC, in which ∠B = 60°, ∠ C = 45° and AB + BC + CA = 11 cm.
Answer:
Given:
In $\triangle ABC$, the base angles are $\angle B = 60^\circ$ and $\angle C = 45^\circ$.
The perimeter of the triangle is $AB + BC + CA = 11 \text{ cm}$.
To Construct:
A triangle ABC with the given measurements.
Steps of Construction:
1. Draw a line segment XY equal to the perimeter, i.e., $XY = 11 \text{ cm}$.
2. At point X, construct an angle $\angle LXY = 60^\circ$ (equal to $\angle B$).
3. At point Y, construct an angle $\angle MYX = 45^\circ$ (equal to $\angle C$).
4. Bisect $\angle LXY$ and $\angle MYX$. Let these bisectors intersect at point A.
5. Draw the perpendicular bisector of AX and let it intersect XY at point B.
6. Draw the perpendicular bisector of AY and let it intersect XY at point C.
7. Join AB and AC.
8. ABC is the required triangle.
Justification:
Since B lies on the perpendicular bisector of AX:
$AB = XB$
(Property of perpendicular bisector)
Similarly, since C lies on the perpendicular bisector of AY:
$AC = YC$
(Property of perpendicular bisector)
Now, consider the perimeter of $\triangle ABC$:
$Perimeter = AB + BC + CA$
$Perimeter = XB + BC + YC$
$Perimeter = XY = 11 \text{ cm}$
Now, in $\triangle ABX$:
$AB = XB$
$\angle BXA = \angle XAB$
(Angles opposite to equal sides)
Also, $\angle ABC$ is the exterior angle of $\triangle ABX$:
$\angle ABC = \angle BXA + \angle XAB$
$\angle ABC = 2 \angle BXA$
$\angle ABC = 2 \times (\frac{1}{2} \angle LXY) = \angle LXY = 60^\circ$
Similarly, for the other base angle:
$\angle ACB = \angle MYX = 45^\circ$
Thus, the construction is justified.
Exercise 11.2
Question 1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.
Answer:
Given:
In $\triangle ABC$, BC = 7 cm, $\angle B = 75^\circ$, and AB + AC = 13 cm.
To Construct:
A triangle ABC satisfying the given conditions.
Steps of Construction:
1. Draw the base line segment BC of length 7 cm.
2. At point B, construct an angle $\angle XBC = 75^\circ$.
3. From the ray BX, cut a line segment BD equal to the sum of the other two sides, i.e., BD = AB + AC = 13 cm.
4. Join the points D and C.
5. Construct the perpendicular bisector of the line segment DC.
6. Let the perpendicular bisector intersect the line segment BD at a point A.
7. Join AC.
Then $\triangle ABC$ is the required triangle.
Justification:
By construction, BC = 7 cm and $\angle B = 75^\circ$.
The point A lies on the perpendicular bisector of the segment DC. Therefore, A is equidistant from D and C.
So, AD = AC.
We constructed BD = 13 cm.
From the diagram, we can see that BD = BA + AD.
Substituting AD = AC, we get:
BD = BA + AC.
Since BD = 13 cm, we have:
AB + AC = 13 cm.
Thus, the construction is justified as all given conditions are met.
Question 2. Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Answer:
Given:
In $\triangle ABC$, BC = 8 cm, $\angle B = 45^\circ$, and AB – AC = 3.5 cm (implying AB > AC).
To Construct:
A triangle ABC satisfying the given conditions.
Steps of Construction:
1. Draw the base line segment BC of length 8 cm.
2. At point B, construct an angle $\angle XBC = 45^\circ$.
3. From the ray BX, cut a line segment BD equal to the difference of the other two sides, i.e., BD = AB – AC = 3.5 cm.
4. Join the points D and C.
5. Construct the perpendicular bisector of the line segment DC.
6. Let the perpendicular bisector intersect the ray BX at a point A.
7. Join AC.
Then $\triangle ABC$ is the required triangle.
Justification:
By construction, BC = 8 cm and $\angle B = 45^\circ$.
The point A lies on the perpendicular bisector of the segment DC. Therefore, A is equidistant from D and C.
So, AD = AC.
From the diagram, point D is on the segment AB. So, we can write AB = AD + DB.
Rearranging this gives: DB = AB - AD.
Substituting AD = AC, we get:
DB = AB - AC.
By construction, we made DB = 3.5 cm.
Therefore, AB - AC = 3.5 cm.
Thus, the construction is justified as all given conditions are met.
Question 3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.
Answer:
Given:
In $\triangle PQR$, QR = 6 cm, $\angle Q = 60^\circ$, and PR – PQ = 2 cm (implying PR > PQ).
To Construct:
A triangle PQR satisfying the given conditions.
Steps of Construction:
1. Draw the base line segment QR of length 6 cm.
2. At point Q, construct an angle $\angle XQR = 60^\circ$.
3. Extend the ray QX downwards to a point Y.
4. From the extended ray QY, cut a line segment QS equal to the difference of the other two sides, i.e., QS = PR – PQ = 2 cm.
5. Join the points S and R.
6. Construct the perpendicular bisector of the line segment SR.
7. Let the perpendicular bisector intersect the ray QX at a point P.
8. Join PR.
Then $\triangle PQR$ is the required triangle.
Justification:
By construction, QR = 6 cm and $\angle Q = 60^\circ$.
The point P lies on the perpendicular bisector of the segment SR. Therefore, P is equidistant from S and R.
So, PS = PR.
From the diagram, we can see that PS = PQ + QS.
Substituting PS = PR, we get:
PR = PQ + QS.
Rearranging this gives: PR - PQ = QS.
By construction, we made QS = 2 cm.
Therefore, PR - PQ = 2 cm.
Thus, the construction is justified as all given conditions are met.
Question 4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer:
Given:
In $\triangle XYZ$, the base angles are $\angle Y = 30^\circ$ and $\angle Z = 90^\circ$.
The perimeter of the triangle is $XY + YZ + ZX = 11 \text{ cm}$.
To Construct:
A triangle XYZ with the given perimeter and base angles.
Steps of Construction:
1. Draw a line segment AB equal to the perimeter of the triangle, i.e., $AB = 11 \text{ cm}$.
2. At point A, construct an angle $\angle PAB = 30^\circ$.
3. At point B, construct an angle $\angle QBA = 90^\circ$.
4. Bisect $\angle PAB$ and $\angle QBA$. Let these bisectors intersect at point X.
5. Draw the perpendicular bisector of AX and let it intersect AB at point Y.
6. Draw the perpendicular bisector of BX and let it intersect AB at point Z.
7. Join XY and XZ.
8. XYZ is the required triangle.
Justification:
Point Y lies on the perpendicular bisector of AX.
$AY = XY$
(Any point on the perpendicular bisector is equidistant from the endpoints)
Similarly, point Z lies on the perpendicular bisector of BX.
$BZ = XZ$
(Property of perpendicular bisector)
Now, consider the perimeter of $\triangle XYZ$:
$XY + YZ + ZX = AY + YZ + ZB$
$XY + YZ + ZX = AB = 11 \text{ cm}$
For the angles, in $\triangle AYX$:
$AY = XY$
$\angle YAX = \angle YXA$
(Angles opposite to equal sides are equal)
$\angle XYZ$ is the exterior angle of $\triangle AYX$:
$\angle XYZ = \angle YAX + \angle YXA$
$\angle XYZ = 2 \angle YAX$
$\angle XYZ = 2 \times (\frac{1}{2} \times 30^\circ) = 30^\circ$
Similarly, for $\angle XZY$:
$\angle XZY = 2 \angle ZBX$
$\angle XZY = 2 \times (\frac{1}{2} \times 90^\circ) = 90^\circ$
Thus, the construction is justified.
Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:
Given:
A right triangle with base = 12 cm, and the sum of its hypotenuse and the other side (perpendicular) = 18 cm.
To Construct:
A right triangle, say $\triangle ABC$, where the base BC = 12 cm, $\angle B = 90^\circ$, and AB + AC = 18 cm.
Steps of Construction:
1. Draw the base line segment BC of length 12 cm.
2. At point B, construct a right angle $\angle XBC = 90^\circ$.
3. From the ray BX, cut a line segment BD equal to the sum of the other two sides, i.e., BD = AB + AC = 18 cm.
4. Join the points D and C.
5. Construct the perpendicular bisector of the line segment DC.
6. Let the perpendicular bisector intersect the line segment BD at a point A.
7. Join AC.
Then $\triangle ABC$ is the required right triangle.
Justification:
By construction, BC = 12 cm and $\angle B = 90^\circ$.
The point A lies on the perpendicular bisector of the segment DC. Therefore, A is equidistant from D and C.
So, AD = AC.
We constructed BD = 18 cm.
From the diagram, we can see that BD = BA + AD.
Substituting AD = AC, we get:
BD = BA + AC.
Since BD = 18 cm, we have:
AB + AC = 18 cm.
Thus, the construction is justified as all given conditions are met.