A field is in the shape as shown in the figure. Find the area of the field if ZACD= LCDE = 90°. (use /399 = 19.97)
Answers
Answer:
Step-by-step explanation:
Solution
Join BD in ΔBCD, BC and DC are given.
So, we can calculate BD by applying Pythagoras theorem
⇒BD=
BC
2
+CD
2
=
12
2
+5
2
=
144+25
=13 m=BD
⇒Area of □ABCD= Area of ΔABD+ Area of ΔBCD
⇒Area of ΔBCD
=
2
1
×b×h=
2
1
×12×5
=30 m
2
⇒Area of ΔABD
=
s(s−a)(s−b)(s−c)
(Heron's formula)
⇒2S=9+8+13, S=
2
30
⇒S=15 m
⇒Area of ΔABD
=
15(15−9)(15−8)(15−13)
=
15×6×7×2
=
630×2
=6
1260
=35.49m
2
⇒Area of Park = Quad ABCD
=30+35.49
=65.49 m
2
≈65.5 m
2
Answer:
2
Step-by-step explanation:
Join BD in ΔBCD, BC and DC are given.
So, we can calculate BD by applying Pythagoras theorem
⇒BD=
BC
2
+CD
2
=
12
2
+5
2
=
144+25
=13 m=BD
⇒Area of □ABCD= Area of ΔABD+ Area of ΔBCD
⇒Area of ΔBCD
=
2
1
×b×h=
2
1
×12×5
=30 m
2
⇒Area of ΔABD
=
s(s−a)(s−b)(s−c)
(Heron's formula)
⇒2S=9+8+13, S=
2
30
⇒S=15 m
⇒Area of ΔABD
=
15(15−9)(15−8)(15−13)
=
15×6×7×2
=
630×2
=6
1260
=35.49m
2
⇒Area of Park = Quad ABCD
=30+35.49
=65.49 m
2
≈65.5 m
2