A rhombus has sides 8 cm in length and its shortest diagonal is 10 cm in length. What is the area of the rhombus to the nearest cm^2?

2 Answers
Nov 19, 2017

Refer to explanation

Explanation:

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As you can see from the diagram, the area of a rhombus is the the two diagonals multiplied together and divided by 2.

From the information given in the problem, we know the each side of the rhombus is 8cm and the shortest diagonal( the vertical diagonal I presume) is 10cm. We have to work out what is the length of the horizontal diagonal before working out the area.

We are first going to cut in fours,as shown in the diagram, having the hypotenuse as 8cm and the opposite as 5cm. To find the adjacent, we will use Pythagora's theorem.

a^2+b^2=c^2

We want the adjacent so rearrange the equation to make a^2 the subject.

a^2= c^2-b^2

To remove the square, take the square root of the equation.

a = sqrt(c^2-b^2)

Now substitute the values for b and c to calculate a.

a= sqrt(8^2-5^2)
a= sqrt(39) (Keeping the surd for the sake of no rounding mistakes)

Since it is half the length, multiply sqrt39 by 2

sqrt39 * 2 = 2sqrt39

From the diagram, we now know both d_1 and d_2

d_1 = 2sqrt39cm
d_2 = 10cm

Now apply the formula for the area of a rhombus.

(10cm * 2sqrt39 cm)/2 = 10sqrt39cm^2 or 62.45cm^2 to 4 s.f

Nov 19, 2017

"Area" ~~ 62" cm"^2

Explanation:

The area is the cross product of the two sides.

One definition of the cross product is vecAxxvecB = |A||B|sin(theta) where theta is the angle between the two vectors.

In the case of a rhombus, the magnitudes are equal to the length of the equal sides and theta is the smallest angle between the two sides:

"Area" = (8" cm")^2sin(theta)" [1]"

The longest diagonal forms a triangle with the two sides and the largest angle between them, therefore, The Law of Cosines allows us to write:

(10" cm")^2= (8" cm")^2 + (8" cm")^2 - 2(8" cm")^2cos(pi-theta)

We know that cos(pi-theta) = -cos(theta):

(10" cm")^2= (8" cm")^2 + (8" cm")^2 + 2(8" cm")^2cos(theta)

(10" cm")^2= 2(8" cm")^2 + 2(8" cm")^2cos(theta)

(10" cm")^2= 2(8" cm")^2(1 + cos(theta))

100" cm"^2= 128" cm"^2(1+cos(theta))

100/128 = 1 + cos(theta)

cos(theta) = 100/128-1

cos(theta) = -7/32

theta = cos^-1(-7/32)" [2]"

Substitute equation [2] into equation [1]:

"Area" = (8" cm")^2sin(cos^-1(-7/32))

"Area" ~~ 62" cm"^2