## Chapter 4 – Constructions of Triangles

**Practice set 4.1**

**1. Construct ∆PQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm**

**Explanation:**

**Line l is perpendicular bisector of seg DR**

**∴ PD = PR …(i) … [Perpendicular bisector theorem]**

**QD = 8.5 cm …(ii)**

** **

**PQ + PD = QD … [Q – P – D]**

**∴ PQ + PR = 8.5 cm … [From (i) and (ii)]**

**This is the required construction.**

**2. Construct ∆XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. ∠XYZ = 50°.**

**Explanation:**

**Line l is perpendicular bisector of seg AZ**

**∴ XA = XZ …(i) [Perpendicular bisector theorem]**

**AY = 9 cm …(ii)**

** **

**XY + XA = AY … [A – X – Y]**

**∴ XY + XZ = 9 cm … [From (i) and (ii)]**

**This is the required construction.**

**3. Construct ∆ABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm.**

**Explanation:**

**Line l is perpendicular bisector of seg PB**

**∴ AP = AB …(i) [Perpendicular bisector theorem]**

**PC = 9.8 cm …(ii)**

** **

**AC + AP = PC … [P – A – C]**

**∴ AC + AB = 9.8 cm … [From (i) and (ii)]**

**This is the required construction.**

**4. Construct ∆ABC, in which BC = 3.2 cm, ∠ACB = 45° and perimeter of ∆ABC is 10 cm.**

**Solution:**

**Perimeter of ∆ABC = AB + BC + AC**

**∴ AB + AC + 5.2 = 15 cm**

**∴ AB + AC = 15 – 5.2**

**∴ AB + AC = 9.8 cm**

** **

**Explanation:**

**Line l is perpendicular bisector of seg PB**

**∴ AP = AB …(i) [Perpendicular bisector theorem]**

**PC = 9.8 cm …(ii) **

** **

**AP + AC = PC … [P – A – C]**

**∴ AB + AC = 9.8 cm … [From (i) and (ii)]**

**This is the required construction.**

**Practice set 4.2**

**Practice set 4.2**

**1. Construct ∆XYZ, such that YZ = 7.4 cm, ∠XYZ = 45° and XY – XZ = 2.7 cm.**

**Explanation:**

**Line l is perpendicular bisector of seg PZ**

**∴ XP = XZ …(i) [Perpendicular bisector theorem]**

**PY = 2.7 cm …(ii)**

** **

**XY = XP + PY .. .[X – P – Y]**

**∴ XY = XZ + 2.7 cm .. .[From (i) and (ii)]**

**∴ XY – XZ = 2.7 cm**

**This is the required construction.**

**2. Construct ∆PQR, such that QR = 6.5 cm, ∠PQR = 40° and PQ – PR = 2.5 cm.**

**Explanation:**

**Line l is perpendicular bisector of seg MR**

**∴ PM = PR …(i) [Perpendicular bisector theorem]**

**QM = 2.5 cm …(ii)**

** **

**PQ = PM + QM … [P – M – Q]**

**∴ PQ = PR + 2.5 … [From (i) and (ii)]**

**∴ PQ – PR = 2.5 cm**

**This is the required construction.**

**3. Construct ∆ABC, such that BC = 6 cm, ∠ABC = 100° and AC – AB = 2.5 cm.**

**Explanation:**

**Line l is perpendicular bisector of seg MC**

**∴ AM = AC …(i) [Perpendicular bisector theorem]**

**BM = 2.5 cm … (ii)**

** **

**AM = AB + BM … [A – B – M]**

**∴ AC = AB + 2.5 cm .. .[From (i) and (ii)]**

**∴ AC – AB = 2.5 cm**

**This is the required construction.**

**Practice set 4.3**

**Practice set 4.3****1. Construct ∆PQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.**

**Explanation:**

**Line l and m are perpendicular bisector of seg PA and PB respectively**

**∴ PQ = AQ and PR = RB …(i) [Perpendicular bisector theorem] **

** **

**PQ + QR + PR = 9.5 cm**

**∴ AQ + QR + RB = 9.5 cm**

**∴ AB = 9.5 cm**

** **

**In ∆PQA,**

**seg PQ ≅ seg AQ … [From (i)]**

**∴ ∠QPA ≅ ∠QAP … [Isosceles triangle theorem]**

** **

**Let ∠QPA = ∠QAP = x**

**∠PQR is an exterior angle of ∆PQA**

**∴ ∠QPA + ∠QAP = ∠PQR … [Remote interior angle theorem]**

**∴ x + x = 70**

**∴ 2x = 70**

**∴ x = 35**

**∴ ∠QPA = ∠QAP = 35°**

** **

**Similarly, we can prove **

**∠RPB = ∠RBP = 40°**

** **

**Now, draw ∆ PAB, with AB = 9.5 cm, ∠A = 35° and ∠B = 40°**

**This is the required construction.**

**2. Construct ∆XYZ, in which ∠Y = 58°, ∠X = 46° and the perimeter of the triangle is 10.5 cm.**

**Explanation:**

**Line l and m are perpendicular bisector of seg PZ and QZ respectively**

**∴ PX = ZX and ZY = QY …(i) [Perpendicular bisector theorem]**

**XY + YZ + XZ = 10.5 cm … [Given]**

**∴ XY + QY + PX = 10.5 cm .. .[From (i) and (ii)]**

**∴ PQ = 10.5 cm … [P – X – Y – Q]**

** **

**In ∆PXZ,**

**seg PX ≅ seg ZX … [From (i)]**

**∴ ∠XPZ ≅ ∠XZP … [Isosceles triangle theorem]**

** **

**Let ∠XPZ = ∠XZP = x**

**∠ZXY is an exterior angle of ∆PXZ**

**∴ ∠ZXY = ∠XPZ + ∠XZP … [Remote interior angle theorem]**

**∴ 46 = x + x**

**∴ 46 = 2x**

**∴ x = 23**

**∴ ∠XPZ = ∠XZP = 23°**

** **

**Similarly, **

**we can prove ∠YQZ = ∠YZQ = 29°.**

** **

**Now, draw ∆ZPQ, with PQ = 10.5 cm, ∠P = 23° and ∠Q = 29°**

**This is the required construction.**

**3. Construct ∆LMN, in which ∠M = 60°, ∠N = 80° and LM + MN + NL = 11 cm.**

**Explanation:**

**Line l and m are perpendicular bisector of seg PL and seg LQ respectively**

**∴ MP = ML and NL = NQ …(i) [Perpendicular bisector theorem]**

** **

**LM + MN + NL = 11 cm … [Given]**

**∴ MP + MN + NQ = 11 cm .. .[From (i) and (ii)]**

**∴ PQ = 11 cm … [P – M – N – Q]**

** **

**In ∆PML,**

**∴ seg MP ≅ seg ML … [From (i)]**

**∠MPL ≅ ∠MLP … [Isosceles triangle theorem]**

** **

**Let ∠MPL = ∠MLP = x**

**∠LMN is an exterior angle of ∆PML**

**∴ ∠LMN = ∠MPL + ∠MLP … [Remote interior angle theorem]**

**∴ 60 = x + x**

**∴ 60 = 2x**

**∴ x = 30**

**∴ ∠MPL = ∠MLP = 30°**

** **

**Similarly, **

**we can prove ∠NQL = ∠NLQ = 40°.**

** **

**Now, draw ∆LPQ, with PQ = 11 cm, ∠P = 30° and ∠Q = 40°**

**This is the required construction.**

**This is the required construction.**

**Problem Set 4**

**Problem Set 4****1. Construct ∆XYZ, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, ∠XYZ.**

**Explanation:**

**Line l is perpendicular bisector of seg PZ**

**∴ XP = XZ …(i) [Perpendicular bisector theorem]**

**PY = 10.3 cm …(ii)**

** **

**XY + XP = PY … [P – X – Y]**

**∴ XY + XZ = 10.3 cm … [From (i) and (ii)]**

**This is the required construction.**

**2. Construct ∆ABC, in which ∠B = 70°, ∠C = 60°, AB + BC + AC = 11.2 cm.**

**Explanation:**

**Line l and m are perpendicular bisector of seg AP and AQ respectively**

**∴ BP = BA …(i) and CQ = CA …(ii) [Perpendicular bisector theorem] **

** **

**AB + BC + AC = 11.2 cm … [Given]**

**∴ BP + BC + CQ = 11.2 cm … [From (i), (ii)]**

**∴ PQ = 11.2 cm … [P – B – C – Q]**

** **

**In ∆ABP,**

**seg BP ≅ seg BA .. .[From (i)]**

**∴ ∠BPA ≅ ∠BAP … [Isosceles triangle theorem]**

** **

**Let ∠BPA = ∠BAP = x**

**∠ABC is an exterior angle of ∆ABP**

**∴ ∠ABC = ∠BPA + ∠BAP … [Remote interior angle theorem]**

**∴ 70 = x + x**

**∴ 70 = 2x**

**∴ x = 35**

**∴ ∠BPA = ∠BAP = 35°**

** **

**Similarly, **

**we can prove ∠CAQ = ∠CQA = 30°.**

** **

**Now, draw ∆APQ, with PQ = 11.2 cm, ∠P = 35° and ∠Q = 30°**

**This is the required construction.**

**3. The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4. Construct the triangle.**

**Explanation:**

**Let the required triangle be ∆ABC.**

**AB + BC + AC = 14.4 … [Given]**

**AB : BC : AC = 2 : 3 : 4 … [Given]**

** **

**Let the common multiple be x**

**∴ 2x + 3x + 4x = 14.4 **

**∴ 9x = 14.4**

**∴ x = \(\large \frac {14.4}{9}\)**

**∴ x = 1.6**

** **

**∴ AB = 2x = 2 × 1.6 = 3.2 cm**

**∴ BC = 3x = 3 × 1.6 = 4.8 cm**

**∴ AC = 4x = 4 × 1.6 = 6.4 cm**

**This is the required construction.**

**4. Construct ∆PQR, in which PQ – PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55°.**

**Explanation:**

**Line l is perpendicular bisector of seg TR**

**∴ PT = PR …(i) and QT = 2.4 cm …(ii) [Perpendicular bisector theorem]**

** **

**PQ = PT + QT .. .[P – T – Q]**

**∴ PQ = PR + 2.4 … [From (i) and (ii)]**

**∴ PQ – PR = 2.4 cm**

**This is the required construction.**