Here we will prove some Application
of congruency of triangles.
1. PQRS is a rectangle and POQ an equilateral triangle. Prove
that SRO is an isosceles triangle.
Solution:
Given:
PQRS is a rectangle. POQ is an equilateral triangle to prove ∆SOR is an isosceles triangle.
Proof:
Statement |
Reason |
1. ∠SPQ = 90° |
1. Each angle of a rectangle is 90° |
2. ∠OPQ = 60° |
2. Each angle of an equilateral triangle is 60° |
3. ∠SPO = ∠SPQ – ∠OPQ = 90° – 60° = 30° |
3. Using statements 1 and 2. |
4. Similarly, ∠RQO = 30° |
4. Proceeding as above. |
5. In ∆POS and ∆QOR, (i) PO = QO (ii) PS = QR (iii) ∠SPO = ∠RQO = 30° |
5. (i) Sides of an equilateral triangle are equal. (ii) Opposite sides of a rectangle are equal. (iii) From statements 3 and 4. |
6. ∆POS ≅ ∆QOR |
6. By SAS criterion of congruency. |
7. SO = RO |
7. CPCTC. |
8. ∆SOR is an isosceles triangle. (Proved) |
8. From statement 7. |
2. In the given figure,
triangle XYZ is a right angled at Y. XMNZ and YOPZ are squares. Prove that XP =
YN.
Solution:
Given:
In ∆XYZ, ∠Y = 90°, XMNZ and YOPZ are squares.
To prove: XP = YN
Proof:
Statement |
Reason |
1. ∠XZN = 90° |
1. Angle of square XMNZ. |
2. ∠YZN = ∠YZX + ∠XZN = x° + 90° |
2. Using statement 1. |
3. ∠YZP = 90° |
3. Angle of square YOPZ. |
4. ∠XZP = ∠XZY + ∠YZP = x° + 90° |
4. Using statement 3. |
5. In ∆XZP and ∆YZN, (i) ∠XZP = ∠YZN (ii) ZP = YZ (iii) XZ = ZN |
5. (i) Using statements 2 and 4. (ii) Sides of square YOPZ. (iii) Sides of square XMNZ. |
6. ∆XZP ≅ ∆YZN |
6. By SAS criterion of congruency. |
7. XP = YN. (Proved) |
7. CPCTC. |
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