Question

10. Find the area of a rhombus if its vertices are (3,0),(4,5).(-1,4) and (-2.-1) taken in
ordez (Hint: Area of a rhombus = (product of its diagonals)]​

Answer 1

Answer:

construction a rhombus ABCD in which diagonals are AC and BD

Step-by-step explanation:

At AC

(3,0)(-1,4)

apply distance formula =

√(X2-x1) + (Y2-y1)

(4)2 + (4)2

16+16

4√2

.

AT BD

.(-2,-11)(4,5)

apply formula

we get :- 6√2

multiply AC AND BD

48

MARK AS BRAILIST PLZZ

Answer 2

Answer:

$ac = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

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

$= \sqrt{ { - 4}^{2} + {4}^{2} }$

$= \sqrt{16 + 16}$

$= \sqrt{32}$

$= 4 \sqrt{2}$

$bd = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

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

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

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

$= \sqrt{ {( - 2 - 4)}^{2} + {( - 1 - 5)}^{2} }$

$= \sqrt{ { - 6}^{2} + { - 6}^{2} }$

$= \sqrt{36 + 36}$

$= \sqrt{72}$

$= 6 \sqrt{2}$

$area \: of \: rombus = d1d2$

$= ac \times bd$

$4 \sqrt{2} \times 6 \sqrt{2 } = 48$