CG Board Class 10 Maths Chapter 13 Solutions: Geometrical Constructions (ज्यामितीय रचनाएँ) (2026-27)

• सरल रेखा का अभिगृहीत: Given any two distinct points A and B, exactly one straight line passes through them.
• Thales Theorem / थेल्स प्रमेय / आधारभूत आनुपातिकता प्रमेय: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In △ ABC, if PQ ∥ BC ⇒ AP/PB = AQ/QC and AB/AP = AC/AQ

### 2. समरूप आकृतियाँ एवं रचना का अनुपात

• Scale Factor / मापनी गुणक / रचना का अनुपात: When constructing a similar triangle or polygon whose sides are m/n times the corresponding sides of a given figure:
• If m/n < 1 (e.g., 3/5 or 2/3), the constructed figure is smaller and lies inside the extended original figure.
• If m/n > 1 (e.g., 5/3 or 4/3), the constructed figure is larger and lies outside the original figure.

### 3. त्रिभुज से संबंधित वृत्त

• परिकेंद्र एवं परिवृत्त:
• The point of intersection of the perpendicular bisectors of the three sides of a triangle is called the circumcenter (O).
• The circle drawn with center O and radius equal to the distance from O to any vertex (OA = OB = OC) passes through all three vertices and is called the circumcircle.
• Location Rule: For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse.
• अंतःकेंद्र एवं अंतःवृत्त:
• The point of intersection of the internal angle bisectors of a triangle is called the incenter (O).
• The perpendicular distance from O to any side gives the inradius (r). The circle drawn touching all three sides internally is called the incircle.
• Equilateral Triangle Rule: In an equilateral triangle, the circumcenter and incenter coincide at the exact same point (centroid).

Two line segments of lengths a and b (where b > a).

• (a) Construction of line segment a + रेखाखंड a + b की रचना:

1. Draw a line ray LX with a ruler.
2. Using a compass opened to length a, place the pointer at L and cut an arc at point M on LX such that LM = a.
3. Now taking M as origin and compass radius equal to length b, cut an arc on MX at point N such that MN = b.
4. Line segment LN is the required line segment of length a + b.

• (b) Construction of line segment b – रेखाखंड b – a की रचना:

1. Draw a line ray LX with a ruler.
2. Place compass at L with radius b and cut an arc on LX at point P so that LP = b.
3. From point P, taking radius equal to a, cut an arc backwards towards L intersecting LP at point Q.
4. Line segment LQ is the required line segment of length b – a.

• (c) Construction of line segment 2b + रेखाखंड 2b + a की रचना:

1. Draw a line ray LX.
2. From point L, cut two consecutive arcs of length b each: LP = b and PQ = b along LX.
3. From point Q, cut an arc of length a to get point R such that QR = a.
4. Line segment LR is the required line segment of length 2b + a.

• (d) Construction of line segment 3a – रेखाखंड 3a – b की रचना:

1. Draw a ray LX.
2. From point L, cut three consecutive arcs of length a each to get points P, Q, R such that LR = 3a.
3. From point R, cut an arc of length b backwards towards L to intersect at point S.
4. Line segment LS is the required line segment of length 3a – b.

• दिए गए कोण: 15°, 45°, 105°, 75°.
• 1. Construction of 15°15°के कोण की रचना:
• पद: Draw a ray AB. Construct 60°using compass by taking an arc from A and cutting it with same radius. Bisect 60°to get 30°. Bisect 30°angle to get the required 15°angle.
• 2. Construction of 45°45°के कोण की रचना:
• पद: Draw a ray AB. Construct a perpendicular at A to get 90°. Bisect the 90°angle using compass. The resulting angle is 45°.
• 3. Construction of 75°75°के कोण की रचना:
• पद: Construct 60°and 90°angles on ray AB from point A. Bisect the 30°region between 60°and 90°(since 60°+ 15°= 75°). The resulting ray gives 75°.
• 4. Construction of 105°105°के कोण की रचना:
• पद: Construct 90°and 120°angles on ray AB from point A. Bisect the 30°region between 90°and 120°(since 90°+ 15°= 105°). The resulting ray gives 105°.

An obtuse angle X°and an acute angle Y°.

• (a) Construction of X°- Y°(कोण X°- Y°की रचना):
• पद: Construct angle ∠ AOB = X°on ray OA. From ray OB, construct an angle equal to Y°inwards towards OA as ∠ BOC = Y°. The remaining angle ∠ AOC is the required angle X°- Y°.
• (b) Construction of X°+ Y°(कोण X°+ Y°की रचना):
• पद: Construct angle ∠ AOB = X°on ray OA. Taking OB as base ray, construct angle ∠ BOC = Y°externally. The combined angle ∠ AOC = X°+ Y°.
• (c) Construction of (180 – X)°(कोण (180 – X)°की रचना):
• पद: Draw a straight line AOB representing 180°. Construct angle ∠ AOC = X°at center O. The supplementary angle ∠ BOC on the straight line represents (180 – X)°.
• (d) Construction of 2Y°(कोण 2Y°की रचना):
• पद: Construct angle ∠ AOB = Y°on ray OA. Taking ray OB as base, construct another congruent angle ∠ BOC = Y°adjacent to it. The total angle ∠ AOC = 2Y°.

• (a) Condition for Triangle Construction with sides r, s, t:
• गणितीय नियम: A triangle can be constructed if and only if the sum of any two sides is strictly greater than the third side:

r + s > t, s + t > r, r + t > s

• Conclusion / उत्तर: Yes, if r, s, t satisfy the triangle inequality theorem, the triangle can be constructed by taking one side (say r) as base and drawing arcs of radii s and t from its endpoints.
• (b) Construction using sides s, t and r + t:
• गणितीय परीक्षण:

The sum of two given sides is s + t. The third side is r + t.

For a valid triangle, we must have s + t > r + t ⇒ s > r.

However, if s ≤ r, then s + t ≤ r + t, violating the triangle inequality theorem.

• Conclusion / उत्तर: It is not possible to construct a triangle in general because the sum of two sides s + t is not strictly greater than the third side r + t (unless s > r).

A triangle ABC.

1. Draw a triangle ABC with vertices A, B, C.
2. To draw a line parallel to BC through vertex A:

• Measure ∠ ABC at vertex B.
• At vertex A, draw an alternate interior angle ∠ DAB = ∠ ABC on the opposite side of AB.
• Extend line DA to E to form line DAE parallel to BC.
• Mathematical Verification / कोणों का गणितीय परीक्षण:
• Line DAE is a straight line passing through vertex A.
• Therefore, the sum of all angles formed on the straight line at vertex A is:

∠ DAB + ∠ BAC + ∠ CAE = 180°* Since DAE ∥ BC, by alternate interior angles property:

∠ DAB = ∠ B and ∠ CAE = ∠ C

• Substituting these in the line equation:

∠ B + ∠ BAC + ∠ C = 180° ⇒ ∠ A + ∠ B + ∠ C = 180°* Final Answer / उत्तर: Sum of angles at vertex A on the parallel line is 180° and the sum of interior angles of △ ABC is 180°.

A triangle ABC and scale factor 3/5.

1. Draw a given triangle ABC.
2. Draw a ray BX from vertex B making an acute angle with side BC on the side opposite to vertex A.
3. Mark 5 equal points on ray BX and name them B₁, B₂, B₃, B₄, B₅ such that BB₁ = B_1B₂ = B_2B₃ = B_3B₄ = B_4B₅.
4. Join B₅ to vertex C.
5. From point B₃, draw a line parallel to B_5C intersecting BC at point P.
6. From point P, draw a line parallel to side AC intersecting side AB at point Q.

• Result / परिणाम: △ QBP is the required triangle similar to △ ABC such that its sides are 3/5 of the corresponding sides of △ ABC.

An isosceles triangle ABC where scale relation is PQ = 3/4 AB ⇒ PQ/AB = 3/4.

1. Construct the base triangle ABC with AB = AC (isosceles triangle).
2. Draw a ray AX making an acute angle with side AB.
3. Mark 4 equal segments on AX: A₁, A₂, A₃, A₄.
4. Join A₄ to vertex B.
5. From A₃, draw a line parallel to A_4B intersecting AB at point P.
6. From P, draw a line parallel to BC intersecting AC at point R.

• Result / परिणाम: △ APR (renamed as △ PQR) is the required isosceles triangle such that PQ = 3/4 AB.

Triangle PQR and scale factor AB = 2/3 PQ.

1. Construct the given triangle PQR.
2. Draw a ray PX making an acute angle with side PQ.
3. Mark 3 equal points on PX: P₁, P₂, P₃.
4. Join P₃ to vertex Q.
5. From P₂, draw a line parallel to P_3Q intersecting PQ at point A.
6. From A, draw a line parallel to side QR intersecting side PR at point C.

• Result / परिणाम: △ PAC (or △ ABC) is the required similar triangle with AB = 2/3 PQ.

Scale factor 4/3 > 1 (constructed triangle is larger than given triangle).

1. Construct any base triangle ABC.
2. Extend rays AB to AX and AC to AY.
3. Draw a ray AZ making an acute angle with side AB.
4. Mark 4 equal divisions on AZ: A₁, A₂, A₃, A₄.
5. Join A₃ to vertex B.
6. From A₄, draw a line parallel to A_3B intersecting extended ray AX at point P.
7. From P, draw a line parallel to BC intersecting extended ray AY at point Q.

• Result / परिणाम: Triangle APQ is the required similar triangle which is 4/3 times △ ABC.

(i) Perpendicular at point P on line l (रेखा l पर बिंदु P पर लंब):**

• (ii) Perpendicular from external point S on line l (बाह्य बिंदु S से रेखा l पर लंब):
• (iii) Perpendicular bisector of line segment JK (रेखाखंड JK का लंब समद्विभाजक):
• (iv) Parallel line through point T to line l (बिंदु T से होकर रेखा l के समांतर रेखा):
• (v) Line parallel to ED passing through point F (बिंदु F से ED के समांतर रेखा):
• (vi) Perpendicular from point G on side HJ (बिंदु G से भुजा HJ पर लंब):

(i) Rectangle of sides a and b (भुजाओं a तथा b का आयत):**

• (ii) Square of Perimeter 4b (परिमाप 4b का वर्ग):
• (iii) Square with diagonal equal to a (विकर्ण a का वर्ग):
• (iv) Cyclic quadrilateral with sides a, b and included angle θ (चक्रीय चतुर्भुज की रचना):

A right-angled triangle ABC where ∠ B = 90°.

1. Construct a right-angled triangle ABC with ∠ B = 90°.
2. Draw perpendicular bisectors of side AB and side BC.
3. Let the two perpendicular bisectors intersect at point O. Point O lies exactly on the midpoint of hypotenuse AC.
4. Taking O as center and radius R = OA = OB = OC, draw a circle passing through vertices A, B, C.

• त्रिज्या का गणितीय मान:

R = (Hypotenuse AC)/(2)

Circumcenter O lies at the midpoint of hypotenuse AC, and radius R = 1/2 × Hypotenuse AC.

Right-angled triangle ABC with sides base b, perpendicular p, and hypotenuse h.

1. Construct right-angled triangle ABC with ∠ B = 90°.
2. Draw internal angle bisectors of ∠ B and ∠ C.
3. Let the angle bisectors intersect at incenter O.
4. From O, draw perpendicular OQ ⊥ BC intersecting BC at point Q.
5. Taking O as center and radius r = OQ, draw a circle touching all three sides.

• अंतःवृत्त की त्रिज्या का सूत्र:

r = (p + b – h)/(2)

Where p = लंब, b = आधार, and h = कर्ण.

An equilateral triangle ABC with side length a.

1. Construct an equilateral triangle ABC of side length a.
2. Draw perpendicular bisectors of sides AB, BC, CA. Their intersection point is circumcenter O.
3. Draw angle bisectors of ∠ A, ∠ B, ∠ C. Their intersection point is incenter I.
4. Observe that circumcenter O and incenter I coincide at the exact same point.

• त्रिज्याओं का परिकलन:
• परिवृत्त की त्रिज्या, R: R = a/√3.
• अंतःवृत्त की त्रिज्या, r: r = a/2√3.
• संबंध: R = 2r.

Yes, both the incentre and circumcentre are located at the exact same point (the centroid of the equilateral triangle).

Three non-collinear points A, B, C.

1. Plot three non-collinear points A, B, C on paper.
2. Join point A to B and point B to C to form line segments AB and BC.
3. Draw the perpendicular bisector of line segment AB.
4. Draw the perpendicular bisector of line segment BC.
5. Let both perpendicular bisectors intersect at point O.
6. Taking O as center and radius R = OA, draw a circle.

• Verification / प्रमाण: Since O lies on the perpendicular bisector of AB, OA = OB. Since O lies on the perpendicular bisector of BC, OB = OC. Thus, OA = OB = OC = R. The circle passes through all three points A, B, C.

• ज्यामितीय अवधारणा: The center of a circle can be uniquely located by finding the point of intersection of the perpendicular bisectors of any two non-parallel chords on the circular boundary.
• Step-by-चरणबद्ध कार्ययोजना:

1. Mohan, Rahul, and Zoya choose three distinct non-collinear points A, B, C along the boundary fence of the circular school ground.
2. They stretch a measuring rope between points A and B (Chord 1) and another measuring rope between points B and C (Chord 2).
3. Rahul finds the exact midpoint of rope AB and uses a set-square/rope method to walk along a line perpendicular to rope AB (Perpendicular bisector 1).
4. Zoya finds the exact midpoint of rope BC and walks along a line perpendicular to rope BC (Perpendicular bisector 2).
5. The point O where Rahul’s line and Zoya’s line cross each other is the exact center of the circular ground.

• Conclusion / उत्तर: By constructing the perpendicular bisectors of two chords on the circular boundary, they successfully determined the exact center point O where the flag pole should be fixed.

💡 Teacher Nudge / अध्ययन की अगली सलाह: