Question

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

Answer 1

840

Explanation:

The four sides of a rhombus all have equal length, so if the perimeter is 148, then each side has length 148/4 = 37. Also, the diagonals of a rhombus bisect each other at right angles, so the diagonal of length 24 is cut into two pieces of length 12. We can show this information in a diagram (shown below.)

Applying the Pythagorean Theorem to any of the four right triangles in our diagram, we have

12² + x² = 37².

Solving this equation for positive x, we get x = √37² - 12² = √1369 - 144 = √1225 = 35. The length of the long diagonal is x + x = 70.

The area of a rhombus is half the product of its diagonals. In this case, that is 24 ∙ 70/2 = 12 ∙ 70 = 840.

Answer 2

Area of the rhombus=840

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