GSEB Class 9 Maths Solutions 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.
Answer:
Required: To make an angle of 90° at O and explain the steps involved.
Steps of construction:
(1) Using O as the center and keeping the same radius, draw a circular arc that cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw another arc that crosses the arc drawn earlier, let's say at point C.
(3) Taking C as the center and with the same radius, draw an arc that cuts the first arc drawn in step 1, let's call that point D.
(4) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(5) Draw the ray OF that goes through D. Then, \( \angle \)FOE = 60°.
(6) Next, using C and D as centers and with a radius larger than CD, draw arcs that cross each other, let's say at G.
(7) Draw the ray OG. This ray OG divides the angle \( \angle \)FOE exactly in half, so \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
Therefore, \( \angle \)GOA = \( \angle \)GOE + \( \angle \)EOA = 30° + 60° = 90°.
Justification:
(i) Join BC.
Then, OC = OB = BC (Based on how it's drawn).
\( \implies \) \( \triangle \)COB is an equilateral triangle.
\( \implies \)\( \angle \)COB = 60°.
\( \implies \)\( \angle \)EOA = 60°.
(ii) Join CD.
Then, OD = OC = CD (Based on how it's drawn).
\( \implies \) \( \triangle \)DOC is an equilateral triangle.
\( \implies \)\( \angle \)DOC = 60°.
\( \implies \)\( \angle \)FOE = 60°.
(iii) Join CG and DG. In \( \triangle \)ODG and \( \triangle \)OCG,
OD = OC [Radii from the same arc]
DG = CG [Arcs made with equal radii]
OG = OG [Shared side]
\( \implies \) \( \triangle \)ODG \( \cong \) \( \triangle \)OCG [SSS rule]
\( \implies \) \( \angle \)DOG = \( \angle \)COG [CPCT]
From the earlier calculation, \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
Therefore, \( \angle \)GOA = \( \angle \)GOE + \( \angle \)EOA = 30° + 60° = 90°.
In simple words: To make a 90-degree angle, you first draw a ray and mark points to get 60-degree angles. Then, you bisect the angle between two 60-degree markings to add 30 degrees, making a total of 90 degrees. The justification shows how these steps create equilateral triangles and bisected angles, proving the angle is indeed 90 degrees.

Exam Tip: Remember to clearly label all points and rays. The justification must link each step of the construction to a geometric property (like forming equilateral triangles or bisecting angles).

Question 2. Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:
Given: A ray OA.
Required: To make an angle of 45° at O and explain the steps involved.
Steps of construction:
(1) Using O as a center and the same radius, draw an arc of a circle, which cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw an arc that crosses the arc drawn earlier, let's say at point C.
(3) Taking C as the center and with the same radius as before, draw an arc that crosses the arc drawn in step 1, let's say at point D.
(4) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(5) Draw the ray OF that goes through D. Then, \( \angle \)FOE = 60°.
(6) Next, using C and D as centers and with a radius larger than CD, draw arcs that cross each other, let's say at G.
(7) Draw the ray OG. This ray OG is the line that divides \( \angle \)FOE exactly in half, meaning \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
Therefore, \( \angle \)GOA = \( \angle \)GOE + \( \angle \)EOA = 30° + 60° = 90°.
(8) Now, using O as the center and any radius, draw an arc that cuts the rays OA and OG, let's say at H and I respectively.
(9) Next, using H and I as centers and with a radius greater than \( \frac {1}{2} \)HI, draw arcs that cross each other, let's say at J.
(10) Draw the ray OJ. This ray OJ is the needed line that divides \( \angle \)GOA exactly in half. Thus, \( \angle \)GOJ = \( \angle \)AOJ = \( \frac {1}{2} \)\( \angle \)GOA = \( \frac {1}{2} \)(90°) = 45°.
Justification:
(i) Join BC.
Then, OC = OB = BC (Based on how it's drawn).
\( \implies \) \( \triangle \)COB is an equilateral triangle.
\( \implies \)\( \angle \)COB = 60°.
\( \implies \)\( \angle \)EOA = 60°.
(ii) Join CD.
Then, OD = OC = CD (Based on how it's drawn).
\( \implies \) \( \triangle \)DOC is an equilateral triangle.
\( \implies \)\( \angle \)DOC = 60°.
\( \implies \)\( \angle \)FOE = 60°.
(iii) Join CG and DG. In \( \triangle \)ODG and \( \triangle \)OCG,
OD = OC [Radii from the same arc]
DG = CG [Arcs made with equal radii]
OG = OG [Shared side]
\( \implies \) \( \triangle \)ODG \( \cong \) \( \triangle \)OCG [SSS rule]
\( \implies \) \( \angle \)DOG = \( \angle \)COG [CPCT]
\( \implies \) \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
Therefore, \( \angle \)GOA = \( \angle \)GOE + \( \angle \)EOA = 30° + 60° = 90°.
(iv) Join HJ and IJ.
In \( \triangle \)OIJ and \( \triangle \)OHJ,
OI = OH [Radii from the same arc]
IJ = HJ [Arcs made with equal radii]
OJ = OJ [Shared side]
\( \implies \) \( \triangle \)OIJ \( \cong \) \( \triangle \)OHJ [SSS rule]
\( \implies \) \( \angle \)AOJ = \( \angle \)GOJ [CPCT]
Thus, \( \angle \)GOJ = \( \angle \)AOJ = \( \frac {1}{2} \)\( \angle \)GOA = \( \frac {1}{2} \)(90°) = 45°.
In simple words: To make a 45-degree angle, you first build a 90-degree angle. Then, you simply cut that 90-degree angle exactly in half using another set of arcs. The line that divides the 90-degree angle will give you the 45-degree angle you need.

Exam Tip: Constructing 45° relies on the 90° construction. Make sure your 90° angle is accurate before attempting to bisect it. Every arc must be precise.

Question 3. Construct the angles of the following measurements:
(i) 30°
(ii) \( 22\frac {1}{2}° \)
(iii) 15°
Answer:
(i) 30°
Given: A ray OA.
Required: To make an angle of 30° at O.
Steps of construction:
(1) Using O as a center and the same radius, draw an arc of a circle, which cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw an arc that crosses the arc drawn earlier, let's say at point C.
(3) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(4) Taking B and C as centers and with a radius larger than \( \frac {1}{2} \)BC, draw arcs that cross each other, let's say at point D.
(5) Draw the ray OD. This ray OD divides the angle EOA exactly in half, meaning \( \angle \)EOD = \( \angle \)AOD = \( \frac {1}{2} \)\( \angle \)EOA = \( \frac {1}{2} \)(60°) = 30°.
(ii) \( 22\frac {1}{2}° \)
Given: A ray OA.
Required: To make an angle of \( 22\frac {1}{2}° \) at O.
Steps of construction:
(1) Using O as a center and the same radius, draw a circular arc that cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw an arc that crosses the arc drawn earlier, let's say at point C.
(3) Taking C as the center and with the same radius as before, draw an arc that crosses the arc drawn in step 1, let's say at point D.
(4) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(5) Draw the ray OF that goes through D. Then, \( \angle \)FOE = 60°.
(6) Next, taking C and D as centers and with a radius larger than CD, draw arcs that cross each other, let's say at G.
(7) Draw the ray OG. This ray OG divides the angle \( \angle \)FOE exactly in half, meaning \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
Therefore, \( \angle \)GOA = \( \angle \)GOE + \( \angle \)EOA = 30° + 60° = 90°.
(8) Now, using O as the center and any radius, draw an arc that cuts rays OA and OG, let's say at H and I respectively.
(9) Next, taking H and I as centers and with a radius larger than \( \frac {1}{2} \)HI, draw arcs that cross each other, let's say at J.
(10) Draw the ray OJ. This ray OJ is the needed line that divides \( \angle \)GOA exactly in half, meaning \( \angle \)GOJ = \( \angle \)AOJ = \( \frac {1}{2} \)\( \angle \)GOA = \( \frac {1}{2} \)(90°) = 45°.
(11) Now, using O as the center and any radius, draw an arc that cuts rays OA and OJ, let's say at K and L respectively.
(12) Next, taking K and L as centers and with a radius larger than \( \frac {1}{2} \)KL, draw arcs that cross each other, let's say at M.
(13) Draw the ray OM. This ray OM divides the angle \( \angle \)AOJ exactly in half, meaning \( \angle \)JOM = \( \angle \)AOM = \( \frac {1}{2} \)\( \angle \)AOJ = \( \frac {1}{2} \)(45°) = \( 22\frac {1}{2}° \).
(iii) 15°
Given: A ray OA.
Required: To make an angle of 15° at O.
Steps of construction:
(1) Using O as a center and the same radius, draw a circular arc that cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw an arc that crosses the arc drawn earlier, let's say at point C.
(3) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(4) Now, taking B and C as centers and with a radius larger than \( \frac {1}{2} \)BC, draw arcs that cross each other, let's say at point D.
(5) Draw the ray OD, which cuts the arc drawn in step 1 at F. This ray OD divides the angle EOA exactly in half, meaning \( \angle \)EOD = \( \angle \)AOD = \( \frac {1}{2} \)\( \angle \)EOA = \( \frac {1}{2} \)(60°) = 30°.
(6) Now, taking B and F as centers and with a radius larger than \( \frac {1}{2} \)BF, draw arcs that cross each other, let's say at point G.
(7) Draw the ray OG. This ray OG divides the angle AOD exactly in half, meaning \( \angle \)DOG = \( \angle \)AOG = \( \frac {1}{2} \)\( \angle \)AOD = \( \frac {1}{2} \)(30°) = 15°.
In simple words: To create these angles, you start with basic 60-degree angles and then repeatedly bisect them. Bisecting a 60-degree angle gives 30 degrees. Bisecting a 90-degree angle (which is made from a 60-degree angle and a bisected 60-degree angle) gives 45 degrees. Bisecting a 30-degree angle gives 15 degrees. Each new angle comes from dividing a known angle in half.

Exam Tip: These constructions are based on repeated angle bisection. Ensure you understand how to construct 60°, then 30°, then how 90° is formed, and how bisection is applied systematically for smaller angles.

Question 4. Construct the following angles and verify by measuring them by a protractor:
(i) 75°
(ii) 105°
(iii) 135°
Answer:
(i) 75°
Required: To make an angle of 75° at O.
Steps of construction:
(1) Using O as a center and the same radius, draw a circular arc that cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw an arc that crosses the arc drawn earlier, let's say at point C.
(3) Taking C as the center and with the same radius as before, draw an arc that crosses the arc drawn in step 1, let's say at point D.
(4) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(5) Draw the ray OF that goes through D. Then, \( \angle \)FOE = 60°.
(6) Next, taking C and D as centers and with a radius larger than \( \frac {1}{2} \)CD, draw arcs that cross each other, let's say at G.
(7) Draw the ray OG, which cuts the arc drawn in step 1 at H. This ray OG divides the angle FOE exactly in half, meaning \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
(8) Next, taking C and H as centers and with a radius larger than \( \frac {1}{2} \)CH, draw arcs that cross each other, let's say at I.
(9) Draw ray OI. This ray OI is the needed line that divides the angle GOE exactly in half, meaning \( \angle \)GOI = \( \angle \)EOI = \( \frac {1}{2} \)\( \angle \)GOE = \( \frac {1}{2} \)(30°) = 15°.
Therefore, \( \angle \)IOA = \( \angle \)IOE + \( \angle \)EOA = 15° + 60° = 75°.
By measuring \( \angle \)IOA with a protractor, we find that \( \angle \)IOA = 75°. Thus, the construction is confirmed.
(ii) 105°
Given: A ray OA.
Required: To make an angle of 105° at O.
Steps of construction:
(1) Using O as a center and the same radius, draw a circular arc that cuts ray OA, let's say at point B.
(2) With B as the center and using the same radius as before, draw an arc that crosses the arc drawn earlier, let's say at point C.
(3) Taking C as the center and with the same radius as before, draw an arc that crosses the arc drawn in step 1, let's say at point D.
(4) Draw the ray OE that goes through C. Then, \( \angle \)EOA = 60°.
(5) Draw the ray OF that goes through D. Then, \( \angle \)FOE = 60°.
(6) Next, taking C and D as centers and with a radius larger than \( \frac {1}{2} \)CD, draw arcs that cross each other, let's say at G.
(7) Draw the ray OG, which cuts the arc drawn in step 1 at H. This ray OG divides the angle FOE exactly in half, meaning \( \angle \)FOG = \( \angle \)EOG = \( \frac {1}{2} \)\( \angle \)FOE = \( \frac {1}{2} \)(60°) = 30°.
Therefore, \( \angle \)GOA = \( \angle \)GOE + \( \angle \)EOA = 30° + 60° = 90°.
(8) Next, taking H and D as centers and with a radius larger than \( \frac {1}{2} \)HD, draw arcs that cross each other, let's say at I.
(9) Draw the ray OI. This ray OI divides the angle FOG exactly in half, meaning \( \angle \)FOI = \( \angle \)IOG = \( \frac {1}{2} \)\( \angle \)FOG = \( \frac {1}{2} \)(30°) = 15°.
Therefore, \( \angle \)IOA = \( \angle \)IOG + \( \angle \)GOA = 15° + 90° = 105°.
By measuring \( \angle \)IOA with a protractor, we find that \( \angle \)IOA = 105°. Thus, the construction is confirmed.
(iii) 135°
Given: A ray OA.
Required: To make an angle of 135° at O.
Steps of construction:
(1) Draw a ray OA. Extend OA backwards to a point A' to form a straight line A'OA. This creates an angle of 180° along the line.
(2) Using O as a center and any convenient radius, draw a large arc that cuts the line A'OA at points B' (on the left of O) and B (on the right of O).
(3) With B as the center and the same radius, draw an arc to intersect the large arc at C. Then with C as center and the same radius, intersect the large arc at D. This creates points for 60° and 120° from OA.
(4) Taking C and D as centers, and with a radius greater than \( \frac {1}{2} \)CD, draw arcs that intersect each other at G. Draw ray OG. This ray OG forms an angle of 90° with OA (i.e., \( \angle \)GOA = 90°).
(5) Now, \( \angle \)A'OG represents the angle between the 180° ray and the 90° ray. This angle is 90° (180° - 90°). To get 135°, we need to add 45° to 90°, or subtract 45° from 180°.
(6) Taking B' (the point on the extended ray A'O) and G (the point for 90°) as centers, and with a radius greater than \( \frac {1}{2} \)B'G, draw arcs that intersect each other at I.
(7) Draw the ray OI. This ray OI is the bisector of \( \angle \)A'OG, so \( \angle \)A'OI = \( \angle \)IOG = \( \frac {1}{2} \)\( \angle \)A'OG = \( \frac {1}{2} \)(90°) = 45°.
Therefore, the angle formed by ray OI with ray OA is \( \angle \)IOA = \( \angle \)IOG + \( \angle \)GOA = 45° + 90° = 135°.
By measuring \( \angle \)IOA with a protractor, we find that \( \angle \)IOA = 135°. Thus, the construction is confirmed.
In simple words: To make 75 degrees, you combine a 60-degree angle with a 15-degree angle (which is half of 30). For 105 degrees, you combine a 90-degree angle with a 15-degree angle. For 135 degrees, you combine a 90-degree angle with a 45-degree angle, typically by extending the initial line and bisecting the angle between the 90-degree line and the 180-degree line. Each involves building on previous angle constructions.

Exam Tip: For angles like 75°, 105°, and 135°, think of them as combinations of basic angles you can construct (like 60°, 90°, 180°) and their bisectors (30°, 45°, 15°). Always verify your constructions with a protractor if allowed.

Question 5. Construct an equilateral triangle, given its side, and justify the construction.
Answer:
Required: To construct an equilateral triangle with a side length of 6 cm and justify the construction.
Steps of construction:
(1) Draw a ray AX. From point A on the ray, cut off a line segment AB of length 6 cm.
(2) Using A as the center and a radius equal to AB (6 cm), draw an arc above the line segment AB.
(3) Using B as the center and the same radius (6 cm), draw another arc that crosses the first arc at point C.
(4) Join A to C and B to C.
(5) \( \triangle \)ABC is the required equilateral triangle.
Justification:
In \( \triangle \)ABC:
- AB = 6 cm (By construction, based on the given side length).
- AC = 6 cm (By construction, because the arc was drawn with radius AB from A).
- BC = 6 cm (By construction, because the arc was drawn with radius AB from B).
Thus, AB = BC = CA = 6 cm.
Since all three sides of \( \triangle \)ABC are equal, it is an equilateral triangle. Hence, the construction is explained.
In simple words: To draw a triangle with all equal sides (equilateral), you first draw one side. Then, using that side's length, you draw two arcs from each end of the first side. Where these arcs meet is the third corner of your triangle. Connect the corners, and all three sides will be the same length.

Exam Tip: The key to constructing an equilateral triangle is using the exact same radius (equal to the side length) for all arcs. This ensures all three sides are equal by definition.