- For triangle OMN:
- Given: OM = 25 cm, ON = 30 cm, MP = 7 cm, and OP is perpendicular to MN.
- In right triangle OMP, by Pythagoras’ Theorem:
OP² = OM² - MP² = 25² - 7² = 625 - 49 = 576 → OP = √576 = 24 cm.
- In right triangle OPN, by Pythagoras’ Theorem:
PN² = ON² - OP² = 30² - 24² = 900 - 576 = 324 → PN = √324 = 18 cm.
- For trapezium ABCD and square:
- Trapezium sides: AB = 10 cm, BC = unknown, CD = 7 cm, DA = 4 cm.
- Perimeter of trapezium = AB + BC + CD + DA.
- To find BC, drop perpendicular from C to AB, meeting at E. Then AE = CD = 7 cm, so EB = AB - AE = 10 - 7 = 3 cm.
- CE = DA = 4 cm (height).
- In right triangle CEB: BC² = CE² + EB² = 4² + 3² = 16 + 9 = 25 → BC = 5 cm.
- Perimeter of trapezium = 10 + 5 + 7 + 4 = 26 cm.
- Square has same perimeter → side length = 26 ÷ 4 = 6.5 cm.
- Area of square = (6.5)² = 42.25 cm².
Final answers:
- Length of PN: 18 cm
- Area of the square: 42.25 cm²
Parent Tip: Review the logic above to help your child master the concept of pythagorean theorem perimeter worksheet.
