Question

3. Find the area of the polygon ABCDE, measurements in metre are given in the figure.

WITHOUT HERONS FORMULA FOR CLASS 8th​

Answer 1

Answer:

⇒ Draw a line segment AB=4.5cm

⇒ Take B as center and draw an angle of 60

⇒ Cut off BC=5.5cm

⇒ Take A as with radius 4.8cm and draw an arc.

⇒ Take C as with radius 5cm and draw an arc, which cuts off previous arc at point D

⇒ Join CD and AD.

Step-by-step explanation:

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Answer 2

Answer:

In ∆ABE :

a=8m

b=8m

c=12m

semi peri = (8+8+12)/2

=28/2

=14m

area by heron's formula :

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{14(14 - 8)(14 - 8)(14 - 12)}$

$= \sqrt{14 \times 6 \times 6 \times 2}$

$= \sqrt{2 \times 7 \times 3 \times 2 \times 3 \times 2 \times 2}$

$= 2 \times 2 \times 3 \sqrt{7}$

$= {12 \sqrt{7m} }^{2}$

In ∆ECD :

a=12m

b=10m

c=10m

semi peri = (12+10+10)/2

=32/2

=16m

area by heron's formula :

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{16(16 - 12)(16 - 10)(16 - 10)}$

$= \sqrt{16 \times 4 \times 6 \times 6}$

$= \sqrt{2304}$

$= {48m}^{2}$

In ∆EBC

a=8m

b=12m

c=12m

semi peri = (8+12+12)/2

=32/2

=16m

area by heron's formula =

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{16(16 - 12)(16 - 12)(16 - 8)}$

$= \sqrt{16 \times 4 \times 4 \times 8}$

$= \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$= 2 \times 2 \times 2 \times 2 \times 2 \ \times \sqrt{2}$

$= {32 \sqrt{2 \: m} }^{2}$

area of polygon ABCDE = sum of area of all three triangles

=12√7+48+32√2

=125.00385m²

Step-by-step explanation:

hope it helps...

do tell in the comment if it's right...

I also wanna confirm....

hope you understand sis....

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sorry I didn't read without heron's formula...

I will edit it tomorrow....