Question

Find the area of the following figure.​

Answer 1

Answer:

18×15×30:8100m

Step-by-step explanation:

hope this is helpful to you

Answer 2

Construction:-

Draw a line segment parallel to PS and intersect line segment SR at O.

Solution:-

From construction PQOS will be a Rectangle and QRO will be a Triangle right-angled at O.

So, OR = SR - SO

            = 30 m - 18 m

            = 12 m

By Pythagoras theorem

OQ² = QR² - OR²

OQ² = 15² - 12²

OQ² = 225 - 144

OQ² = 81

OQ = √81

OQ = 9 m

So, OQ = PS = 9 m                [ PQOS is a Rectangle ]

now, Area of ΔQRO = $\frac{1}{2}$ × QO × OR

                                 = $\frac{1}{2}$ × 9 m × 12 m

                                 = 54 m²

and, Area of Rectangle PQOS = L × B

                                                  = 18 m × 12 m

                                                  = 216 m²

∴ Area of the given figure = Area of ΔQRO + Area of Rectangle PQOS

                                           = 54 m² + 216 m²

                                           = 270 m²

Area of the given figure = 270 m²

Some Important Formula:-

• Area of Triangle = $\frac{1}{2}$ × Base × Height

• Area of Rectangle = Length × Breadth

• Area of Square = Side²

• Area of Rhombus = $\frac{1}{2}$ × d₁ × d₂

• Area of Parallelogram = Base × Height

• Area of Trapezium = $\frac{1}{2}$ × (Sum of Parallel side) × height

• Perimeter of Square = 4 × side

• Perimeter of Rectangle = 2 ( Length + Breadth )

• Perimeter of Rhombus = 4 × Side

• Perimeter of Parallelogram = 3 ( Length + Breadth )

• Perimeter of Triangle = Sum of All sides

• Area of Equilateral Triangle = $\frac{\sqrt{3}}{4}$ × Side²