Question

Find the area of rhombus if its vertices are (3,0) (4,5) (-1,4) (-2,-1) taken in order .
[Hint : Area of rhombus =
1/2 (product of diagonals)


Solve the above question .
​

Answer 1

Let (3,0) (4,5) (-1,4) (-2,-1) are the vertices A,B,C,D of a rhombus ABCD.

ʟᴇɴɢᴛʜ ᴏf ᴅɪᴀɢᴏɴᴀʟ ᴀᴄ.

= √[3-(-1)]² + (0-4)²

= √16 + 16

= 4√2

ʟᴇɴɢᴛʜ ᴏf ᴅɪᴀɢᴏɴᴀʟ Bc

= √[4-(-2)]² + [5-(-1)]²

= √36 + 36

= 6√2

Therefore, Area of rhombus ABCD,

= ½ × 4√2 × 6√2

= 24 units ...✔

Answer 2

SOLUTION

Let (3,0), (4,5),(-1,4) & (-2,-1) are the vertices A,B,C,D of a rhombus ABCD.

Length of diagonal AC

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

Length of diagonal BD

$= > \sqrt{</u></strong><strong><u>[</u></strong><strong><u>4 - ( - 2)</u></strong><strong><u>]</u></strong><strong><u>{}^{2} + </u></strong><strong><u>[</u></strong><strong><u>5 - ( - 1)</u></strong><strong><u>]</u></strong><strong><u> {}^{2} } \\ \\ = > \sqrt{ {6}^{2} + {6}^{2} } \\ \\ = > \sqrt{36 + 36} = \sqrt{72} \\ \\ = > 6 \sqrt{2}$

Therefore,

Area of rhombus ABCD

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

Hope it helps ☺️