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. |
From Application of Congruency of Triangles to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Mar 02, 24 05:31 PM
Mar 02, 24 04:36 PM
Mar 02, 24 03:32 PM
Mar 01, 24 01:42 PM
Feb 29, 24 05:12 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.