OABC is a rectangle whose vertices are A (0,3) , O (0,0), B (5,0). Find the length of its fourth vertice.
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A(0,3) B(5,0)
AB = √(5-0)^2 +(0-3)^2
=√(5)^2+(-3)^2
=√25+9
=√34
AB = √(5-0)^2 +(0-3)^2
=√(5)^2+(-3)^2
=√25+9
=√34
Answered by
0
Explanation:
= √(5 - (0)^2 + (3 - (0)^2)
= √25 + 9
= √34
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