In Fig. 15.77, ∠AOB=90°, AC=BC, OA=12 cm and OC=6,5 cm. Find the area of Δ AOB.
Answers
Given : ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm.
Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.
Therefore , CB = CA = OC = 6.5 cm
AB = 2 CB
AB = 2 x 6.5 cm
AB = 13 cm
In right ΔOAB ,
By using Pythagoras theorem :
AB² = OB² + OA²
13² = OB² + 12²
169 = OB² + 144
OB² = 169 – 144
OB² = 25
OB = √25
OB = 5 cm
Now, Area of ΔAOB ,A = ½(Base x height)
Area of ΔAOB ,A = ½(OB x OA)
A = ½ × (5 x 12)
A = 5 × 6
A = 30 cm²
Area of ΔAOB = 30 cm²
Hence, the Area of ΔAOB is 30 cm².
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Answer:-
Given :
∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm.
Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.
Therefore , CB = CA = OC = 6.5 cm
AB = 2 CB
AB = 2 x 6.5 cm
AB = 13 cm
In right ΔOAB ,
By using Pythagoras theorem :
AB² = OB² + OA²
13² = OB² + 12²
169 = OB² + 144
OB² = 169 – 144
OB² = 25
OB = √25
OB = 5 cm
Now, Area of ΔAOB ,A = ½(Base x height)
Area of ΔAOB ,A = ½(OB x OA)
A = ½ × (5 x 12)
A = 5 × 6
A = 30 cm²
Area of ΔAOB = 30 cm²
Hence, the Area of ΔAOB is 30 cm².
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