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Class 9th Chapters
1. Number Systems	2. Polynomials	3. Coordinate Geometry
4. Linear Equations In Two Variables	5. Introduction To Euclid’s Geometry	6. Lines And Angles
7. Triangles	8. Quadrilaterals	9. Areas Of Parallelograms And Triangles
10. Circles	11. Constructions	12. Heron’s Formula
13. Surface Areas And Volumes	14. Statistics	15. Probability

Content On This Page
Exercise 11.1	Example 1 (Before Exercise 11.2)	Exercise 11.2

Chapter 11 Construction

This guide delves into the solutions for Chapter 11: Constructions, a practical and foundational area of geometry. Unlike previous chapters focused on theoretical proofs and properties, this chapter emphasizes the skill of creating accurate geometric figures using only the most basic, classical tools: an ungraduated ruler (a straightedge without markings) and a pair of compasses. This limitation forces a reliance on pure geometric principles rather than measurement. Building upon constructions learned in earlier classes, this chapter introduces more intricate procedures, demanding both precision in execution and, crucially, a clear understanding of the underlying justification for why each sequence of steps produces the desired figure.

The solutions provided offer meticulous, step-by-step instructions for each construction. Equally important, they often include the geometric reasoning or proof that validates the method. This justification typically relies on congruence criteria (like SSS) or established geometric properties previously studied. Mastering these constructions enhances spatial reasoning and provides a tangible connection to geometric theorems. Precision is key – neat arcs and clearly drawn lines are essential for successful constructions.

The key constructions covered in this chapter, along with their justifications as explained in the solutions, include:

Constructing the bisector of a given angle: The solutions detail how to draw arcs from the vertex and then intersecting arcs from points on the angle's arms. The justification usually involves proving two triangles congruent using the SSS criterion, thereby showing the equality of the angles created by the constructed line.
Constructing the perpendicular bisector of a given line segment: This involves drawing intersecting arcs of equal radius from both endpoints of the segment. The solutions explain the steps and justify the construction, again often using SSS congruence to show that the line joining the intersection points forms right angles and bisects the original segment (connecting to properties of rhombuses or kites).
Constructing specific angles without a protractor: Starting with the fundamental construction of a $60^\circ$ angle (by creating an equilateral triangle), the solutions demonstrate how to derive other key angles through bisection or addition/subtraction. This includes constructing angles such as $30^\circ$ (bisecting $60^\circ$), $90^\circ$ (bisecting the angle between $60^\circ$ and $120^\circ$, or constructing a perpendicular), $45^\circ$ (bisecting $90^\circ$), $120^\circ$ (adjacent $60^\circ$ angles), $75^\circ$ ($60^\circ + 15^\circ$ or $90^\circ - 15^\circ$), $105^\circ$ ($90^\circ + 15^\circ$), and $135^\circ$ ($90^\circ + 45^\circ$). Each sequence of steps is clearly outlined.
Constructing a triangle given its base, a base angle, and the sum of the other two sides: Solutions guide drawing the base, constructing the given angle, marking a point on the angle ray corresponding to the sum of the other two sides, connecting this point to the other end of the base, and finally constructing the perpendicular bisector of this connecting line segment. The intersection of the bisector with the initial ray gives the third vertex. Justification hinges on the properties of isosceles triangles created by the perpendicular bisector.
Constructing a triangle given its base, a base angle, and the difference of the other two sides: This construction presents two cases, depending on which of the other two sides is longer. The process is similar to the 'sum' case but involves marking the difference and requires careful attention to whether the difference is marked on the angle ray itself or its extension. The justification again relies on constructing a perpendicular bisector and the resulting isosceles triangle properties.
Constructing a triangle given its perimeter and its two base angles: The solutions detail drawing a line segment equal to the perimeter, constructing the given base angles at the ends of this segment, bisecting these angles, and finding the intersection point of the angle bisectors (which becomes the third vertex of the required triangle). The actual base vertices are then found by constructing perpendicular bisectors of the segments connecting this vertex to the ends of the perimeter line.

For every construction, the solutions emphasize the importance of providing the justification, explicitly linking the drawing steps back to established geometric theorems and ensuring the construction is mathematically sound, not just visually approximate. This reinforces the connection between theoretical geometry and practical application.

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 $90^\circ$.

Steps for constructing a 90-degree angle using a compass and straightedge. It shows a ray OA, an arc from O, arcs from P to get Q (60 deg), from Q to get R (120 deg), and then bisecting the angle between Q and R to get the 90-degree ray OS.

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 OA with initial point O.

To Construct:

An angle of $45^\circ$ at the initial point O of the ray OA.

Steps of Construction:

1. Draw a ray OA.

2. At the initial point O, construct an angle of $90^\circ$. Let the ray forming the $90^\circ$ angle be OB, so that $\angle AOB = 90^\circ$.

3. Now, bisect the angle $\angle AOB$. With O as centre and any radius, draw an arc that intersects OA at P and OB at Q.

4. With P as centre and a radius greater than half of PQ, draw an arc.

5. With Q as centre and the same radius, draw another arc intersecting the previous arc at point R.

6. Draw the ray OR.

Then $\angle AOR$ is the required angle of $45^\circ$.

Construction of a 45-degree angle. First, a 90-degree angle AOB is constructed. Then, the angle AOB is bisected by the ray OR, resulting in angle AOR = 45 degrees.

Justification:

First, we construct $\angle AOB = 90^\circ$.

Then, by construction, the ray OR is the bisector of $\angle AOB$.

To prove this, we can join PR and QR. In $\triangle OPR$ and $\triangle OQR$, we have OP = OQ (radii of the same arc), OR = OR (common), and PR = QR (arcs of equal radii). By SSS congruence, $\triangle OPR \cong \triangle OQR$.

By CPCTC, $\angle AOR = \angle BOR$.

Since OR bisects $\angle AOB$, we have:

$\angle AOR = \frac{1}{2}\angle AOB$

$\angle AOR = \frac{1}{2} \times 90^\circ = 45^\circ$

Thus, 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) Construction of $30^\circ$

Steps of Construction:

1. Draw a ray OA.

2. At O, construct a $60^\circ$ angle. With O as centre and any radius, draw an arc intersecting OA at P. With P as centre and the same radius, cut the arc at Q. Ray OQ makes a $60^\circ$ angle with OA.

3. Bisect the $60^\circ$ angle. With P and Q as centres and a radius greater than half of PQ, draw arcs to intersect at R.

4. Join OR. $\angle AOR$ is the required angle of $30^\circ$.

Construction of a 30-degree angle by first constructing a 60-degree angle and then bisecting it.

(ii) Construction of $22\frac{1}{2}^\circ$

Steps of Construction:

1. Draw a ray OA.

2. Construct a $90^\circ$ angle at O. Let this be $\angle AOB$.

3. Bisect the $90^\circ$ angle to get a $45^\circ$ angle. Let this be $\angle AOC = 45^\circ$.

4. Now, bisect the $45^\circ$ angle $\angle AOC$. Let the bisector be the ray OD.

5. $\angle AOD$ is the required angle of $22\frac{1}{2}^\circ$. ($45^\circ / 2 = 22.5^\circ$).

Construction of a 22.5-degree angle by successively bisecting a 90-degree angle to get 45 degrees, and then bisecting the 45-degree angle.

(iii) Construction of $15^\circ$

Steps of Construction:

1. Draw a ray OA.

2. Construct a $60^\circ$ angle at O. Let this be $\angle AOB$.

3. Bisect the $60^\circ$ angle to get a $30^\circ$ angle. Let this be $\angle AOC = 30^\circ$.

4. Now, bisect the $30^\circ$ angle $\angle AOC$. Let the bisector be the ray OD.

5. $\angle AOD$ is the required angle of $15^\circ$. ($30^\circ / 2 = 15^\circ$).

Construction of a 15-degree angle by successively bisecting a 60-degree angle to get 30 degrees, and then bisecting the 30-degree angle.

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 $75^\circ$

Steps of Construction:

1. Draw a ray OA.

2. Construct angles of $60^\circ$ and $90^\circ$ at O. Let the rays be OQ (for $60^\circ$) and OR (for $90^\circ$).

3. The angle between OQ and OR is $90^\circ - 60^\circ = 30^\circ$.

4. Bisect the angle $\angle QOR$. Let the bisecting ray be OS.

5. $\angle AOS = \angle AOQ + \angle QOS = 60^\circ + 15^\circ = 75^\circ$.

Construction of a 75-degree angle by bisecting the angle between the 60-degree and 90-degree marks.

(ii) Construction of $105^\circ$

Steps of Construction:

1. Draw a ray OA.

2. Construct angles of $90^\circ$ and $120^\circ$ at O. Let the rays be OQ (for $90^\circ$) and OR (for $120^\circ$).

3. The angle between OQ and OR is $120^\circ - 90^\circ = 30^\circ$.

4. Bisect the angle $\angle QOR$. Let the bisecting ray be OS.

5. $\angle AOS = \angle AOQ + \angle QOS = 90^\circ + 15^\circ = 105^\circ$.

Construction of a 105-degree angle by bisecting the angle between the 90-degree and 120-degree marks.

(iii) Construction of $135^\circ$

Steps of Construction:

1. Draw a line A'OA.

2. Construct a $90^\circ$ angle at O. Let this be $\angle AOB = 90^\circ$.

3. The angle $\angle A'OB$ is also $90^\circ$.

4. Bisect the angle $\angle A'OB$. Let the bisecting ray be OC.

5. $\angle AOC = \angle AOB + \angle BOC = 90^\circ + 45^\circ = 135^\circ$.

Construction of a 135-degree angle by constructing a 90-degree angle and adding a 45-degree angle to it (by bisecting the adjacent 90-degree angle).

Verification:

After constructing each angle, use a protractor to measure it. Place the center of the protractor at the vertex O and align the baseline with the initial ray OA. The constructed ray should align with the corresponding degree mark on the protractor ($75^\circ$, $105^\circ$, or $135^\circ$).

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.

Construction of an equilateral triangle. Starting with a base segment AB, two arcs with radius AB are drawn from A and B to find the third vertex C.

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 $\triangle ABC$, $\angle B = 60^\circ$, $\angle C = 45^\circ$, and the perimeter AB + BC + CA = 11 cm.

To Construct:

A triangle ABC satisfying the given conditions.

Steps of Construction:

1. Draw a line segment XY equal to the perimeter, i.e., XY = 11 cm.

2. At point X, construct an angle $\angle YXL = \angle B = 60^\circ$.

3. At point Y, construct an angle $\angle XYM = \angle C = 45^\circ$.

4. Bisect the angles $\angle YXL$ and $\angle XYM$. Let the bisectors of these angles intersect at a point A.

5. Draw the perpendicular bisector of the line segment AX. Let it intersect XY at point B.

6. Draw the perpendicular bisector of the line segment AY. Let it intersect XY at point C.

7. Join AB and AC.

Then $\triangle ABC$ is the required triangle.

Construction of a triangle given its perimeter and base angles. A line XY of 11 cm is drawn. Angles of 60 and 45 degrees are made at X and Y. Their bisectors meet at A. Perpendicular bisectors of AX and AY meet XY at B and C. ABC is the required triangle.

Justification:

Since B lies on the perpendicular bisector of AX, AB = XB.

In $\triangle ABX$, $\angle BAX = \angle AXB = 30^\circ$.

The exterior angle $\angle ABC$ of $\triangle ABX$ is $\angle BAX + \angle AXB \ $$ = 30^\circ + 30^\circ = 60^\circ$. This matches the required $\angle B$.

Since C lies on the perpendicular bisector of AY, AC = YC.

In $\triangle ACY$, $\angle CAY = \angle AYC = 22.5^\circ$.

The exterior angle $\angle ACB$ of $\triangle ACY$ is $\angle CAY + \angle AYC \ $$ = 22.5^\circ + 22.5^\circ = 45^\circ$. This matches the required $\angle C$.

For the perimeter:

AB + BC + AC = XB + BC + YC = XY = 11 cm.

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.

Construction of a triangle with a given base, base angle, and sum of the other two sides. Shows base BC=7cm, angle 75 deg at B, a point D on the ray such that BD=13cm. The perpendicular bisector of CD intersects BD at A.

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.

Construction of a triangle with a given base, base angle, and difference of the other two sides (AB > AC). Shows base BC=8cm, angle 45 deg at B, a point D on the ray such that BD=3.5cm. The perpendicular bisector of CD intersects the ray at A.

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.

Construction of a triangle with a given base, base angle, and difference of the other two sides (PR > PQ). Shows base QR=6cm, angle 60 deg at Q. The ray from Q is extended backwards, and a point S is marked such that QS=2cm. The perpendicular bisector of SR intersects the original ray at P.

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 $\triangle XYZ$, $\angle Y = 30^\circ$, $\angle Z = 90^\circ$, and perimeter XY + YZ + ZX = 11 cm.

To Construct:

A triangle XYZ satisfying the given conditions.

Steps of Construction:

1. Draw a line segment AB of length 11 cm (equal to the perimeter).

2. At point A, construct an angle $\angle PAB = 30^\circ$ (equal to $\angle Y$).

3. At point B, construct an angle $\angle QBA = 90^\circ$ (equal to $\angle Z$).

4. Bisect the angles $\angle PAB$ and $\angle QBA$. Let the bisectors of these angles intersect at a point X.

5. Draw the perpendicular bisector of the line segment AX. Let it intersect AB at point Y.

6. Draw the perpendicular bisector of the line segment BX. Let it intersect AB at point Z.

7. Join XY and XZ.

Then $\triangle XYZ$ is the required triangle.

Construction of a right triangle given its perimeter and one base angle. Shows base line AB=11cm, angle 30 at A, angle 90 at B. Their bisectors meet at X. Perpendicular bisectors of AX and BX meet AB at Y and Z.

Justification:

Point Y lies on the perpendicular bisector of AX, so AY = XY.

In $\triangle AXY$, $\angle YXA = \angle YAX$. Since $\angle YAX = \frac{1}{2} \times 30^\circ = 15^\circ$, we have $\angle YXA = 15^\circ$.

The exterior angle $\angle XYZ$ of $\triangle AXY$ is equal to $\angle YAX + \angle YXA = 15^\circ + 15^\circ = 30^\circ$. This matches the given $\angle Y$.

Point Z lies on the perpendicular bisector of BX, so BZ = XZ.

In $\triangle BXZ$, $\angle ZXB = \angle ZBX$. Since $\angle ZBX = \frac{1}{2} \times 90^\circ = 45^\circ$, we have $\angle ZXB = 45^\circ$.

The exterior angle $\angle XZY$ of $\triangle BXZ$ is equal to $\angle ZBX + \angle ZXB = 45^\circ + 45^\circ = 90^\circ$. This matches the given $\angle Z$.

For the perimeter:

XY + YZ + ZX = AY + YZ + BZ = AB = 11 cm.

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.

Construction of a right triangle with a given base and sum of hypotenuse and the other side. Shows base BC=12cm, 90-degree angle at B. A point D on the perpendicular ray such that BD=18cm. The perpendicular bisector of CD intersects BD at A.

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.