Question

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Answer 1

Let us Consider that vertices are A(3,0), B(4,5), C(-1,4) & D(-2,-1).

We know that,

${\underline{\boxed{\sf{\red{Area\; of \; Rhombus\; =\; \dfrac{1}{2}\; \times\; (Product\; of\; its\; Diagonals)}}}}}$

Here, Diagonals are AC & BD

$\diamond\bold{\underline{\sf{Now,\; Finding\; AC}}}$

$\bold{\underline{\sf{Using\; Distance\; Formula}}}$

$\longrightarrow\large{\underline{\boxed{\sf{\pink{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}}}$

Now,

$:\implies\sf\; AC = \sqrt{(-1 - 3)^2 - (4 -0)^2}$

$:\implies\sf\;AC = \sqrt{(-4)^2 + (4)^2}$

$:\implies\sf\; AC = \sqrt{(4)^2 + (4)^2}$

$:\implies\sf\;AC = \sqrt{2(4)^2}$

$:\implies\sf\;AC = \sqrt{2} \times\; 4$

$:\implies\sf\; AC =4\sqrt{2}$

$\rule{150}3$

$\diamond\bold{\underline{\sf{Now,\; Finding\; BD}}}$

$:\implies\sf\;BD = \sqrt{(-2 -4)^2 + (-1 - 5)^2}$

$:\implies\sf\; BD = \sqrt{(-6)^2 - (-6)^2}$

$:\implies\sf\;BD = \sqrt{(6)^2 + (6)^2}$

$:\implies\sf\;BD = \sqrt{2(6)^2}$

$:\implies\sf\; BD = \sqrt{2} \times\; 6$

$:\implies\sf\; 6\sqrt{2}$

$\rule{150}3$

Area of Rhombus = $\sf\; \dfrac{1}{2} \times\; AC \times\; BD$

Here, AC = $\sf \; 4\sqrt{2}$

BD = $\sf\; 6\sqrt{2}$

Putting Values:-

$:\implies\sf\; \dfrac{1}{2}\; \times 4\sqrt{2} \times 6\sqrt{2}$

$:\implies\sf\; 2\sqrt{2} \;\times\; 6\sqrt{2}$

$:\implies\sf\; 2 \times \; 6 \times\; \sqrt{2} \times \; \sqrt{3}$

$:\implies\sf\; 12 \times\; 2$

$:\implies\large{\underline{\boxed{\sf{\pink{24\; Square\; units}}}}}$

$\bold{\underline{\sf{\blue{Hence\; Area\; of \; Rhombu\; is \;24\; Sq.\; units}}}}$

$\rule{150}3$

Answer 2

$\bf \boxed{ \bf \: Question:-}$

Find the area of a rhombus, if its vertices are (3,0), (4,5), (–1,4) and (–2, –1) taken in order.

$\bf \pink{Note : - } \bf \: Area \: of \: rhombus = \frac{1}{2} \: (Product \: of \: its \: diagonals)$

$\boxed{\bf{Solution:-}}$

Let A (3, 0), B (4, 5), C (–1, 4) and D (–2, –1) be the vertices of the rhombus ABCD.

$\therefore \: \bf \: Diagonal \: AC \: = \sqrt{( - 1 - 3) {}^{2} + (4 - 0) {}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [\because \: \bf \: distance \: = \sqrt{(x_2 \: - x_1 ) {}^{2} + (y_2 \: - y_1) {}^{2} }]$

$\: \: \: \: \: \: \: \: \: \bf = \sqrt{( - 4) {}^{2} + 4 {}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \bf = \sqrt{32} = 4 \sqrt{2} \: units$

$\bf \: Diagonal \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: BD = \sqrt{( - 2 - 4) {}^{2} + ( - 1 - 5) {}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \:\bf = \sqrt{( - 6) {}^{2} + ( - 6) { }^{2} } = \sqrt{36 + 36}$

$\: \: \: \: \: \: \: \: \: \: \: \: \ \bf = \: \sqrt{72} = 6 \sqrt{2 \: } \: units$

$\therefore \: \: \: \bf \: Area \: of \: the \: rhombus \: \red{ABCD}$

$\: \: \: \: \: \: \: \: \: \: \: \: \bf = \: \frac{1}{2} \times Product \: of \: its \: diagonals$

$\: \: \: \: \: \: \: \: \: \: \: \: \ \bf = \: \frac{1}{2} \times AC \times BD \:$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \bf = \: 2 \times 6 \times \sqrt{2} \times \sqrt{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \bf = \: 12 \times 2 = \red{24 \: sq \: units}.$