Question

Find the area of a rectangle whose vertices are A (–2, 6), B (5, 3), C (–1, –11) and D (–8, –8)

(a) \(4\sqrt{29}\) sq. units 

(b) 116 sq. units 

(c) \(29\sqrt{5}\) sq. units 

(d) \(58\sqrt{2}\) sq. units

Answer 1

(b) 116 Sq. units

AB = \(\sqrt{(5+2)^2+(3-6)^2}\)

= \(\sqrt{7^2+(-3)^2} = \sqrt{49+9}= \sqrt{58}\)

BC = \(\sqrt{(-1-5)^2+(-11-3)^2}\)

= \(\sqrt{(-6)^2+(-14)^2} = \sqrt{36+196}= \sqrt{232}\)

∴ Area of the rectangle ABCD = AB × BC

= \(\sqrt{58}\times\sqrt{232} = \sqrt{29\times2\times29\times8}\)

= 29 × 4 = 116 Sq. units