CBSE Class 9 Maths Chapter 11 Constructions NCERT Solutions

To construct an angle of 90° at the initial point of a given ray, follow these steps:

1. Let the given ray be AB, with A as the initial point.
2. Place the compass point at A and draw an arc with any convenient radius that intersects the ray AB at a point, say C.
3. Without changing the radius, place the compass point at C and draw another arc that intersects the first arc at a point, say E.
4. Again, without changing the radius, place the compass point at E and draw a third arc that intersects the first arc at a point, say F.
5. Now, place the compass point at E and draw an arc with a radius greater than the distance between E and F.
6. With the same radius as in the previous step, place the compass point at F and draw another arc that intersects the arc drawn in step 5 at a point, say G.
7. Draw a ray AG starting from A and passing through G.

The angle ∠BAG is the required 90° angle.

Justification:

Join AE, CE, and EF.

By construction, AC = CE = AE (radii of the same arc and distance between points).

Therefore, ΔACE is an equilateral triangle.

This implies that ∠CAE = 60°.

Similarly, by construction, CE = EF = CF (using same radius for arcs from C and E).

This implies that ∠CEF = 60°.

The angle ∠AEF is formed by two arcs of equal radius from E and F, intersecting at G. The angle ∠AEF is the sum of angles subtended by arcs CE and EF at the center A. However, the justification provided in the source seems to have a slight deviation. A more direct justification involves the properties of the constructed points.

Let's re-evaluate the justification based on standard construction:

1. AC = CE (radii) implies ∠CAE = 60°.

2. CE = EF (radii) implies ∠AEF = 60°.

3. The angle ∠CAF = ∠CAE + ∠EAF. However, the construction involves points E and F such that ∠CAE = 60° and ∠EAF = 60° (if F is constructed from E with same radius as C from A).

A more common justification for 90°:

Let the arc from A intersect AB at C. Let the arc from C intersect at E (∠CAE = 60°). Let the arc from E intersect at F (∠AEF = 60°). The angle ∠CAF = 120°.

Now, bisecting the angle ∠EAF (which is 60°) gives 30°. The ray bisecting ∠EAF will form an angle of 30° with AE. This ray, when drawn from A, will form an angle of 60° + 30° = 90° with AB.

Alternatively, using the points as described in the source:

AC = CE = AE implies ∠CAE = 60°.

CE = EF = CF implies ∠CEF = 60°.

The angle formed by the ray AG is typically found by bisecting the angle ∠EAF, where ∠EAF = 120° (formed by arcs from E and F). The bisector creates a 60° angle. This is not 90°.

Let's follow the source's justification more closely:

AC = CE = AE implies ∠CAE = 60°.

Similarly, consider the arc from E and F. Let G be the intersection. The source states FG = EG and implies G lies on the perpendicular bisector of EF. This is not directly evident from the steps.

A standard justification for 90°:

Points C and E are constructed such that ∠CAE = 60° and ∠CEA = 60°.

Points E and F are constructed such that ∠AEF = 60° and ∠AFE = 60°.

The angle ∠CAF = ∠CAE + ∠EAF = 60° + 60° = 120°.

The ray AG bisects the angle ∠EAF. The angle ∠EAG = 30°.

Therefore, ∠BAG = ∠CAE + ∠EAG = 60° + 30° = 90°.

The source's justification seems to be attempting a different approach or contains errors. The standard method confirms ∠BAG = 90°.

To construct an angle of 45° at the initial point of a given ray, follow these steps:

1. Let the given ray be AB, with A as the initial point.
2. First, construct an angle of 90° at A, such that ∠BAF = 90°, as described in Question 1.
3. Now, bisect the angle ∠BAF. To do this, place the compass point at B and draw an arc with a radius greater than half the distance between B and F.
4. Without changing the radius, place the compass point at F and draw another arc that intersects the arc drawn in step 3 at a point, say H.
5. Draw a ray AH starting from A and passing through H.

The angle ∠BAH is the required 45° angle.

Justification:

By construction in step 2, ∠BAF = 90°.

In step 4, the ray AH is constructed by bisecting ∠BAF.

Therefore, ∠BAH = ∠HAF = ½ ∠BAF.

Substituting the value of ∠BAF:

∠BAH = ½ × 90° = 45°.

Thus, ∠BAH is the required angle of 45°.

(i) Construction of 30° angle:

1. Draw a ray AB with initial point A.
2. With A as the centre and any convenient radius, draw an arc intersecting AB at C.
3. With C as the centre and the same radius, draw another arc intersecting the first arc at D. This makes ∠CAD = 60°.
4. With D as the centre and the same radius, draw an arc intersecting the first arc at E. This makes ∠CAE = 120°.
5. Now, bisect the angle ∠CAD (which is 60°). Place the compass point at C and draw an arc with a radius more than half the distance DC.
6. With D as the centre and the same radius, draw another arc intersecting the arc drawn in step 5 at a point, say F.
7. Draw the ray AF.

The angle ∠BAF is the required 30° angle, as it is half of ∠CAD (60°).

(ii) Construction of 22½° angle:

1. First, construct an angle of 45° at the initial point A of a ray AB. Let this angle be ∠BAH = 45°, as constructed in Question 2.
2. Now, bisect the angle ∠BAH. Place the compass point at B and draw an arc with a radius greater than half the distance BH.
3. With H as the centre and the same radius, draw another arc intersecting the arc drawn in step 2 at a point, say J.
4. Draw the ray AJ.

The angle ∠BAJ is the required 22½° angle, as it is half of ∠BAH (45°).

(iii) Construction of 15° angle:

1. First, construct an angle of 30° at the initial point A of a ray AB. Let this angle be ∠BAE = 30°, as constructed in part (i) of this question.
2. Now, bisect the angle ∠BAE. Place the compass point at B and draw an arc with a radius greater than half the distance BE.
3. With E as the centre and the same radius, draw another arc intersecting the arc drawn in step 2 at a point, say G.
4. Draw the ray AG.

The angle ∠BAG is the required 15° angle, as it is half of ∠BAE (30°).

(i) Construction of 75° angle:

1. Draw a ray AB with initial point A.
2. Construct a 90° angle at A. Let the ray forming the 90° angle be AC, so ∠BAC = 90°.
3. Now, bisect the angle ∠BAC. Place the compass point at B and draw an arc with a radius greater than half the distance BC.
4. With C as the centre and the same radius, draw another arc intersecting the arc drawn in step 3 at a point, say D.
5. Draw the ray AD.

The angle ∠BAD is the required 75° angle, as it is half of ∠BAC (90°), making it 45°, and adding it to the 30° part of the 90° angle (which is ∠BAE where E is the 60° point) gives 75°. A more direct method is to construct 60° and 30° angles adjacent to each other.

Alternative method for 75°:

1. Construct a 60° angle ∠BAE.
2. Construct a 30° angle ∠EAC by bisecting the remaining 120° part or by constructing 90° and taking the difference. A simpler way is to construct 60° and then bisect the angle between 60° and 90°.
3. Construct ∠BAE = 60° and ∠EAC = 90° - 60° = 30°. The ray AC forms 90°. Bisect the angle between the 60° ray and the 90° ray.
4. Let's construct 60° and 90° angles. Draw ray AB. Construct ∠ABC = 60° and ∠ABD = 90°. The angle between the 60° ray and the 90° ray is 30°. Bisect this 30° angle.
5. A more standard approach: Construct ∠BAE = 60°. Construct ∠CAF = 90°. The angle ∠EAF = 30°. Bisect ∠EAF. Let the bisector be AG. Then ∠BAG = ∠BAE + ∠EAG = 60° + 15° = 75°.

Verification: Measure ∠BAD with a protractor. It should be 75°.

(ii) Construction of 105° angle:

1. Draw a ray AB with initial point A.
2. Construct a 90° angle at A. Let the ray forming the 90° angle be AC, so ∠BAC = 90°.
3. Construct a 60° angle adjacent to ∠BAC. Extend the ray AB to D. Construct ∠DAC' = 60° where C' is on the other side of AB. This is not correct.
4. Let's construct 90° and 15° angles. Construct ∠BAC = 90°. Bisect the angle adjacent to 90° (i.e., 180°-90° = 90°). This is not helpful.
5. Correct method: Construct ∠BAE = 60° and ∠EAF = 90°. The angle ∠BAF = 150°. This is not 105°.
6. Standard method: Construct ∠BAE = 90°. Construct ∠EAF = 60° adjacent to it. This gives 150°.
7. Let's construct 90° and 15°. Construct ∠BAC = 90°. Bisect the angle adjacent to 90° (e.g., the angle formed by extending AB backwards). This is not practical.
8. Correct method: Construct ∠BAE = 60° and ∠EAF = 45° adjacent to it. This gives 105°.
9. Let's use 90° and 15°. Construct ∠BAC = 90°. Bisect the angle formed by extending AB backwards and AC. This is complex.
10. Simpler method: Construct ∠BAE = 60°. Construct ∠EAF = 45° (by bisecting 90°). Then ∠BAF = ∠BAE + ∠EAF = 60° + 45° = 105°.

Verification: Measure ∠BAF with a protractor. It should be 105°.

(iii) Construction of 135° angle:

1. Draw a ray AB with initial point A.
2. Construct a 90° angle at A. Let the ray forming the 90° angle be AC, so ∠BAC = 90°.
3. Extend the ray AB to a point D, forming a straight line. The angle ∠CAD = 180°.
4. Bisect the angle ∠CAD. Place the compass point at C and draw an arc.
5. With D as the centre and the same radius, draw another arc intersecting the arc drawn in step 4 at a point, say E.
6. Draw the ray AE.

The angle ∠CAE is 90°. The angle ∠DAE is 90°. This is not 135°.

Correct method for 135°:

1. Construct a 90° angle ∠BAC.
2. Extend the ray AB to D. The angle ∠CAD = 180° - 90° = 90°.
3. Bisect the angle ∠CAD. Place the compass point at C and draw an arc.
4. With D as the centre and the same radius, draw another arc intersecting the arc drawn in step 3 at a point, say E.
5. Draw the ray AE.

The angle ∠CAE = 45° (half of ∠CAD). Therefore, ∠BAE = ∠BAC + ∠CAE = 90° + 45° = 135°.

Verification: Measure ∠BAE with a protractor. It should be 135°.