Question

find the area of a rhombus if it's vertices are (3,0) (4,5)(-1,4) and (-2,-1) take in order.​

Answer 1

Step-by-step explanation:

1/2(15-0 + 16+5 + 1+8 )

1/2(45)

45/2

=24.5

Answer 2

SOLUTION:-

Given:

Vertices of rhombus ABCD are;

⚫A(3,0)

⚫B(4,5)

⚫C(-1,4)

⚫D(-2,-1)

Therefore,

Length of diagonal AC:

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

Length of diagonal BD:

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

Now,

Area of rhombus ABCD:

$\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2} \\ \\ = > 2 \sqrt{2} \times 6 \sqrt{2} \\ \\ = > 12 \times 2 \\ \\ = > 24 \: sq. \: units$

Hope it helps ☺️