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प्रश्न
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A)\[7 + \sqrt{5}\]
(B) 5
(C) 10
(D) 12
उत्तर
Let us plot these coordinates, i.e. O(0, 0), A(0, 4) and B(3, 0), on the Cartesian plane.
We can observe that triangle AOB is a right-angled triangle with OB = 3 units and OA = 4 units.
Now, AB2 = OA2 + OB2 (By Pythagoras theorem)
⇒ AB2 = (42 + 32) sq. units
= (16 + 9) sq. units
= 25 sq. units
⇒ AB = 5 units
Perimeter of ∆AOB = OA + AB + BO
= (3 + 5 + 4) units
= 12 units
Hence, the correct option is D.
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