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  • The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (A) 5 (B) 12 (C) 11 (D) 7+ root 5

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(A) 5

(B) 12

(C) 11

(D) 7+ \sqrt5

Answers (1)

Given: Vertices of the triangle are A (0, 4), B(0, 0) and C (3, 0)

\\ \mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}} \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,4) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,0) \\ \mathrm{AB}=\sqrt{(0-0)^{2}+(0-4)^{2}}=\sqrt{(4)^{2}}=4 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(3,0) \\ \mathrm{BC}=\sqrt{(3-0)^{2}+(0-0)^{2}}=\sqrt{(3)^{2}}=3 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(3,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,4) \\ \mathrm{CA}=\sqrt{(0-3)^{2}+(4-0)^{2}}=\sqrt{9+16}

\\ \mathrm{CA}=\sqrt{25}=5

Perimeter of  \DeltaABC = AB + BC + AC = 4 + 3 + 5 = 12

Hence perimeter of \DeltaABC is 12

Posted by

infoexpert21

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