Question

find the area of polygons ​

Answer 1

AIM :-

To find the area of the figure

Given :-

• AB = 11cm
• AE = 60cm
• CD = 26cm
• Distance between CD and EB = 18cm

In ∆ABE :-

AB = 11cm

AE = 60cm

∠A = 90°

By using Pythagoras theorem,

(Base)² + (Height)² = (Hypotenuse)²

(AB)² + (AE)² = (EB)²

(11)² + (60)² = (EB)²

121 + 3600 = (EB)²

3721 = (EB)²

√3721 = EB

61cm = EB

Area of figure :- Area of BCDE + Area of ∆ABE

Area of ∆ABE :-

$\dfrac{1}{2} \times base \times height$

Substituting the values,

$\dfrac{1}{2} \times AB \times AE$

$\dfrac{1}{2} \times 11 \times 60$

By cancelling 2 in the denominator and 60 in the numerator,

$area = 11 \times 30$

$area = 330 {cm}^{2}$

Area of trapezium BCDE :-

$\dfrac{1}{2} \times height \times (sum \: of \: parallel \: sides)$

Substituting the values,

$\dfrac{1}{2} \times (CD + EB) \times 18$

$\dfrac{1}{2} \times (26 + 61) \times 18$

2 in the denominator and 18 in the numerator gets cancelled

$area = 87 \times 9$

$area = 783 {cm}^{2}$

Total area of the figure :- 330 + 783 =  1113cm²

Some more formulas :-

Area of square = (side)²

Area of parallelogram = base × height

Area of rectangle = length × breadth

Area of rhombus = ½ × Diagonal 1 × Diagonal 2

Answer 2

Answer:

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Step-by-step explanation:

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