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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

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#### Solution

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, AB^{2} = OA^{2}^{ }+ OB^{2} (By Pythagoras theorem)

⇒ AB^{2} = (4^{2} + 3^{2}) 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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