D
C
Fig. 12.14
herefore, area of A BCD - x160x60 in' = 4800 m
160
2
EXERCISE 12.2
1. A park, in the shape of a quadrilateral ABCD, has 2C-90°, AB - 9m, BC = 12 m,
CD-5 mand AD-8m. How much area does it occupy?
2. Find the area of a quadrilateral ABCD in which AB - 3 cm, BC = 4 cm, CD=4 cm,
DA -5 cm and AC-5 cm.
3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find
the total area of the paper used.
Som
6cm
IV
1.5cm
Answers
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Answer:
Join BD in ΔBCD, BC and DC are given.
So, we can calculate BD by applying Pythagoras theorem
⇒BD=BC2+CD2
=122+52 =144+25=13 m=BD
⇒Area of □ABCD= Area of ΔABD+ Area of ΔBCD
⇒Area of ΔBCD
=21×b×h=21×12×5
=30 m2
⇒Area of ΔABD
=s(s−a)(s−b)(s−c) (Heron's formula)
⇒2S=9+8+13, S=230
⇒S=15 m
⇒Area of ΔABD
15(15−9)(15−8)(15−13)
=s(s−a)(s−b)(s−c) (Heron's formula)
⇒2S=9+8+13, S=230
⇒S=15 m
⇒Area of ΔABD
=15(15−9)(15−8)(15−13)
=15×6×7×2=630×2
=61260=35.49m2
⇒Area of Park = Quad ABCD
=30+35.49
=65.49 m2≈65.5 m2
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