Question:
The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) $7+\sqrt{5}$
(b) 5
(c) 10
(d) 12
Solution:
Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$A B=\sqrt{(0-0)^{2}+(4-0)^{2}}=\sqrt{16}=4$
$B C=\sqrt{(0-3)^{2}+(0-0)^{2}}=\sqrt{9}=3$
$A C=\sqrt{(0-3)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.