Question

find the area of polygon ABCDE,whose measurements in metres are given in the figure. ans-244m²​

Answer 1

Step-by-step explanation:

ANSWER

→ There are three triangles comprising the pentagon

& area of triangle by heron's formula =

s(s−a)(s−b)(s−c)

△

1

a = b = 8 , c = 12

S=

2

8+8+12

=14

△

1

=

14(6)(6)(2)

=6×2

7

△

2

a = b = 12, b = 10

S = 17

△

2

17(5)(5)(7)

=5

119

△

3

a = 10, b = 10, c = 12

S = 16

△

3

=

16(6)(6)(4)

=6×4×2=48

⇒ Total area =48+12

7

+5

119

m

2

solution

Answer 2

Given:-

• $\sf{ABCDE\: is\: a\: polygon}$

To find:-

• $\sf{Area\: of \:the \:polygon}$

From the figure:-

$\sf{There \:are \:three \:triangle \:in \:the\: given \:figure\: of \:polygon}$

Names of triangle:-

1. $\sf{\triangle ABE}$
2. $\sf{\triangle DCE}$
3. $\sf{\triangle BEC}$

Solution:-

$\sf{In \triangle ABE,}$

$\sf{AB = 8m}$

$\sf{AE = 8m}$

$\sf{BE = 12m}$

$\sf{S = \dfrac{a + b + c}{2}}$

=>$\sf{S = \dfrac{8 + 8 + 12}{2}\:m}$

=>$\sf{S = \dfrac{28}{2}\:m}$

=>$\sf{S = 14\:m}$

Now According to heron's formula,

$\sf{Area = \sqrt{s(s-a)(s-b)(s-c)}\:sq.units}$

$\sf{Area = \sqrt{14(14-8)(14-8)(14-12)}\:{m}^{2}}$

= $\sf{Area = \sqrt{14 \times 6\times6\times2}\:{m}^{2}}$

= $\sf{ \:\:\:\:\:\:\: 6\sqrt{2\times7\times2}\:{m}^{2}}$

= $\sf{ \:\:\:\:\:\:\: 6\times2\sqrt{7}\:{m}^{2}}$

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

=$\sf{\:\:\:\:\:\:\: 12\times2.64 \: {m}^{2}}$

= $\sf{\:\:\:\:\:\:\: 31.68\:{m}^{2}}$

$\sf{In \triangle CED,}$

$\sf{CE = 12m}$

$\sf{CD = 10m}$

$\sf{DE = 10m}$

=$\sf{S = \dfrac{12+10+10}{2}\:m}$

=>$\sf{S = \dfrac{32}{2}\:m}$

=>$\sf{S = 16m}$

According to Heron's Formula,

=$\sf{Area = \sqrt{16(16-10)(16-10)(16-12)}\:{m}^{2}}$

=$\sf{\:\:\:\:\:\: \sqrt{16\times6\times6\times4}\:{m}^{2}}$

=$\sf{\:\:\:\:\:\: 6\sqrt{4\times4\times4}\:{m}^{2}}$

=$\sf{\:\:\:\:\:\: 6\times4\sqrt{2\times2}\:{m}^{2}}$

=$\sf{\:\:\:\:\:\: 24\times2\:{m}^{2}}$

= $\sf{\:\:\:\:\:\: 48\:{m}^{2}}$

$\sf{In \triangle BEC,}$

$\sf{BC = 8m}$

$\sf{CE = 12m}$

$\sf{BE = 12m}$

=$\sf{S = \dfrac{8+12+12}{2}\:m}$

=>$\sf{S = \dfrac{32}{2}\:m}$

=> $\sf{S = 16m}$

According to Heron's Formula,

$\sf{Area = \sqrt{16(16-8)(16-12)(16-12)}\:{m}^{2}}$

=$\sf{\:\:\:\:\:\: \sqrt{16\times8\times4\times4}\:{m}^{2}}$

= $\sf{\:\:\:\:\:\: 4\sqrt{4\times4\times2\times2\times2}\:{m}^{2}}$

=$\sf{\:\:\:\:\:\:\: 4\times4\times2\sqrt{2}\:{m}^{2}}$

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

= $\sf{\:\:\:\:\:\: 32\times1.41\:{m}^{2}}$

= $\sf{\:\:\:\:\:\: 45.12\:{m}^{2}}$

$\sf{Area\:of\:polygon = Sum\:of\:area\:of\:all\:three\:triangles}$

$\sf{Area\:of\:polygon = ( 31.68 + 48 + 45.12) \:{m}^{2}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: = 124.8\:{m}^{2} [Approx]}$