Question

17. Find the area of the polygon ABCDE whose measurements
in metre are given in the following figure.​

Answer 1

ABCDE IS A PENTAGON

AREA OF THE GIVEN PENTAGON ABCDE = AREA OF Δ BCD + AREA OF  Δ ADB + AREA OF Δ EAD

GIVEN MEASURES

ED = 8 M

DC = 10 M

CB = 10 M

AB = 10 M

EA = 8 M

AD = 12 M

DB = 12 M

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

for Δ BCD

here A= 10 M (DC), B = 10 M(CB) , C = 12 M(DB)

$s = \frac{a+b+c}{2}$

$s = \frac{10+10+12}{2}$

$s = \frac{32}{2}$

s = 16 M

$AREA = \sqrt{16(16-10)(16-10)(16-12)} \\AREA = \sqrt{16 * (6 * 6 * 4)}$

$AREA = \sqrt{4*4*6*6*2*2}$

AREA = 4 × 6 × 2 = 48 m²

FOR Δ ABD

HERE A = 12 M (AD) , B = 12 M (DB) C = 10 M (AB)

$S = \frac{A+B+C}{2}$

$S = \frac{12+12+10}{2}$

$S = \frac{34}{2}$

S = 17 M

$AREA = \sqrt{S(S-A)(S-B)(S-C)} \\AREA = \sqrt{17(17-12)(17-12)(17-10)} \\AREA = \sqrt{17*5*5*7}$

AREA = 5 √17 × 7

AREA = 5√119 m²

FOR Δ EAD

HERE A = 12 M ( AD) , B = 8 M (ED) , C = 8 M (EA)

$S = \frac{A+B+C}{2} \\\\S = \frac{12+8+8}{2} \\\\S = \frac{28}{2}$

S = 14 M

$AREA = \sqrt{S(S-A)(S-B)(S-C)]}$

$AREA = \sqrt{14(14-12)(14-8)(14-8)} \\\\AREA = \sqrt{14* 2* 6*6} \\\\AREA = 6 \sqrt{7*2*2} \\\\AREA = 6 * 2 \sqrt{7}$

AREA = 12√7 m²

AREA OF THE GIVEN PENTAGON ABCDE = AREA OF Δ BCD + AREA OF  Δ ADB + AREA OF Δ EAD

AREA OF THE GIVEN PENTAGON ABCDE = 48 + 5√119 + 12 √7

AREA OF THE GIVEN PENTAGON ABCDE= 134.292576306 = 134.29 m²

HOPE IT HELPS !!

Answer 2

Step-by-step explanation:

hey mate the answer is 134.29

REGARDS...❤