Question 75
In quadrilateral PQRS, RM $$\perp$$ QS, PN $$\perp$$ QS and QS = 6 cm. If RM = 3 cm and PN = 2 cm, then the area of PQRS is
Solution
Area of PQRS = area of $$\triangle$$PQS + area of $$\triangle$$RQS
(Area of triangle = $$\frac{1}{2}\times base \times height$$)
= $$\frac{1}{2}\times 6 \times 2$$ + $$\frac{1}{2}\times 6 \times 3$$
= $$\frac{1}{2}\times 6(2 + 3)$$
= $$15 cm^2$$
Get AI Help
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
Free SSC Topicwise Questions
SSC Data InterpretationSSC GeometrySSC Linear EquationsSSC Number SystemsSSC Quadratic EquationsSSC TrigonometrySSC AgesSSC ArithmeticSSC Reading ComprehensionSSC VocabularySSC AnalogySSC Coding DecodingSSC DirectionsSSC Meaningful OrderSSC Number SeriesSSC PuzzlesSSC SyllogismsSSC Words FormationSSC Venn Diagram
