Rhombus $ABCD$ has perimeter $148$, and one of its diagonals has length $24$. What is the area of $ABCD$?

Area of the rhombus=840

Step-by-step explanation:

Perimeter of a rhombus=4a

148=4a

a=148/4

=37

a=37

The diagonal divides the rhombus into two congruent triangle,

Each congruent triangle= 37 x 37 x 24.

To get the area of the rhombus, we will find the area of one of the congruent triangle, then multiply by 2.

Using Hero's formula to find the area of a triangle, we will use the three sides

Area = √[ s(s-a)(s-b)(s-c) ]

where a, b, and c are the lengths of the three sides: a = 37, b = 37, and c = 24.

s=semiperimeter

s = (a + b + c) / 2

= (37 + 37 + 24)/2

= 98/2

= 49.

s=49

Substitute all the values into the formula

Area = √[ s(s-a)(s-b)(s-c) ]

= √[ 49(49-37)(49-37)(49-24) ]

=√[ 49(12)(12)(25) ]

=√[49(3600)]

=√(176,400)

= 420

Area of one triangle=420

Area of a rhombus=Area of one triangle×2

=420×2

=840

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.