In a Δ ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC.
Answers
Given : In a Δ ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm.
Let the given sides be a = 15 cm, b = 13 cm , c = 14 cm
Semi Perimeter of the ∆,s = (a + b + c) /2
s = (15 + 13 + 14) / 2
s = 42/2
s = 21 cm
Semi Perimeter of the ∆ = 21 cm
Using Heron’s formula :
Area of the wall , A = √s (s - a) (s - b) (s - c)
A = √21(21 - 15) (21 - 13) (21 - 14)
A = √21 × 6 × 8 × 7
A = √(3 × 7) × (2 × 3) × (2 × 2 × 2) × (7)
A = √(3 × 3) × (7 × 7) × (2 × 2 × 2 × 2)
A = 3 × 7 × 2 × 2
A = 84 cm²
Area of triangle of ΔABC = 84 cm²
Now, area of triangle , A = ½ x Base x altitude
84 = ½ × 14 × altitude
Altitude = (84 × 2)/14
Altitude = 6 × 2
Altitude = 12 cm
Hence, the area of triangle of ΔABC is 84 cm² and its altitude on AC is 12 cm.
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Answer:
Step-by-step explanation:
Given :-
Sides of Triangle = 15 cm, 13 cm and 14 cm.
To Find :-
Area of triangle.
Formula to be sued :-
Area of the triangle = √s(s - a)(s - b)(s - c)
Solution :-
Let the sides be a, b and c.
s = (a + b + c) /2
s = (15 + 13 + 14) / 2
s = 42/2
s = 21 cm
Area of the triangle = √s(s - a)(s - b)(s - c)
⇒ Area of the triangle = √21(21 - 15) (21 - 13) (21 - 14)
⇒ Area of the triangle = √21 × 6 × 8 × 7
⇒ Area of the triangle = √(3 × 7) × (2 × 3) × (2 × 2 × 2) × (7)
⇒ Area of the triangle = √(3 × 3) × (7 × 7) × (2 × 2 × 2 × 2)
⇒ Area of the triangle = 3 × 7 × 2 × 2
⇒ Area of the triangle = 84 cm²
Area of triangle = ½ x Base x altitude
⇒ 84 = ½ × 14 × altitude
⇒ Altitude = (84 × 2)/14
⇒ Altitude = 6 × 2
⇒ Altitude = 12 cm
Hence, the area of ΔABC is 84 cm² and hence its altitude on AC is 12 cm.