Question

find the area of the polygon ABCDE shown in the figure given aside​

Answer 1

Answer:

AE * ED = area of AEDC

∴ 4cm * 8cm = $32cm^{2}$

AC = ED/2

∴ AC = 8/2 = 4cm

take point of intersection of perpendicular B on AC as O.

Triangle ABO = (5 * 4)/2 = (20cm^2)/2 = 10cm^2

same for Triangle BOC.

Total area of the shape = 10cm^2 + 10cm^2 + 32cm^2

                                         = 52cm^2

PLS MARK AS BRAINLIEST

Answer 2

GivEn:

• Sides of ∆ ABC = 5cm, 5cm, 8cm (a)
• Sides of Rectangle ACDE = 8cm and 4cm (b)

To Find:

• Area of the polygon ABCDE ?

Solution:

(a) Area of the triangle ABC are;

A = √[s(s-a)(s-b)(s-c)]

Where as: s = a+b+c/2

s = (5+5+8)/2

s = 18/2

s = 9

After putting values in Heron's Formula;

a = √[9(9-5)(9-5)(9-8)]

a = √[9 × 4 × 4 × 1]

a = √[144]

a = 12 cm²

so,

The Area of ∆ ABC is 12cm².

b) Area of Rectangle ACDE;

Area = length × breadth

a = 8 × 4

a = 32 cm²

So,

The area of Rectangle ACDE is 32cm².

Again, After adding ∆ ABC and ACDE, we can get total area of the polygon ABCDE as;

=> Area of ABC + Area of ACDE

=> 12 + 32

=> 44 cm²

Hence,

The area of the polygon is 44cm².