Question

Show that the points A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus. Find its area.

Answer 1

The given points are A (3,0), B(4,5) and C(-1,4), D(-2,-1). Then

$AB=\sqrt{(3-4{)}^2+(0-5{)}^2}=\sqrt{(-1{)}^2+(-5{)}^2}=\sqrt{1+25}=\sqrt{26}units$

$BC=\sqrt{(4+1{)}^2+(5-4{)}^2}=\sqrt{(5{)}^2+(1{)}^2}=\sqrt{25+1}=\sqrt{26}units$

$CD=\sqrt{(-1+2{)}^2+(4+1{)}^2}=\sqrt{(1{)}^2+(5{)}^2}=\sqrt{1+25}=\sqrt{26}units$

$AD=\sqrt{(3+2{)}^2+(0+1{)}^2}=\sqrt{(5{)}^2+(1{)}^2}=\sqrt{25+1}=\sqrt{26}units$

$AC=\sqrt{(3+1{)}^2+(0-4{)}^2}=\sqrt{(4{)}^2+(-4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}units$

$BD=\sqrt{(4+2{)}^2+(5+1{)}^2}=\sqrt{(6{)}^2+(6{)}^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}units$

Therefore AB = BC = CD = DA = $\sqrt{26}$ units and diagonal AC is not equal to diagonal BD

Hence, ABCD is a rhombus

Area of a rhombus $=\frac{1}{2}\times$(Product of its diagonals)

$=\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}$

= 24

= 24 square units