The sides of a quadrangular field ,taken in order are 26 metre,27 meter,7 meter are 24 metre respectively. the angle contained by the last two sides is a right angle. find its area.
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Hi there!
Here's your answer :-
Area of ΔABCD = ar ADC + ar ABC
ar ADC = 1 / 2 × 24 × 7 = 84 m²
Now,
AC = √24² + 7²
= √625
= 25 m
Now in ΔABC,
→ AB = 26 m
→ BC = 27 m
→ AC = 25 m
Thus,
Semi-perimeter = 26 + 27 + 25 / 2 = 78 / 2 = 39
by Heron's formula :-
Area of ΔABC = √s (s-a) (s-b) (s-c)
= √39 (39 - 26) (39 - 27) (39 - 25)
= √39 × 13 × 12 × 14
= √85176
= 291.85 m²
Therefore,
Area of ΔABCD = (84 + 291.85) m² = 375.85 m²
Hence, The required answer is :- 375.85 m²
Hope it helps!
Here's your answer :-
Area of ΔABCD = ar ADC + ar ABC
ar ADC = 1 / 2 × 24 × 7 = 84 m²
Now,
AC = √24² + 7²
= √625
= 25 m
Now in ΔABC,
→ AB = 26 m
→ BC = 27 m
→ AC = 25 m
Thus,
Semi-perimeter = 26 + 27 + 25 / 2 = 78 / 2 = 39
by Heron's formula :-
Area of ΔABC = √s (s-a) (s-b) (s-c)
= √39 (39 - 26) (39 - 27) (39 - 25)
= √39 × 13 × 12 × 14
= √85176
= 291.85 m²
Therefore,
Area of ΔABCD = (84 + 291.85) m² = 375.85 m²
Hence, The required answer is :- 375.85 m²
Hope it helps!
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