Can somebody help me with this??

Question

The sides of a square and a rhombus are congruent to each other, but the area of the square is twice the area of the rhombus. Find the ratio of the smaller diagonal to the larger diagonal in the rhombus.
An explanation would be great. Thank you so much

Answer 1

Here is one way to solve this.

Let s = the length of the side of the square and therefore of the side of the rhombus too.

Let p = length of short diagonal of rhombus

Let q = length of long diagonal of rhombus.

That means we want to find p/q.

A_s = s²

A_r = pq/2

s² = 2(pq/2) = pq (area square twice area of rhombus)

The diagonals of a rhombus intersect at right angles meaning there are 4 right triangles created by the intersecting diagonals. Each of those right triangles has a hypotenuse of s and legs of p/2 and q/2.

That means by the Pythagorean Theorem: s² = (p/2)² + (q/2)². We know s² = pq, so substitute to get:

pq = p²/4 + q²/4

4pq = p² + q²

p² - 4pq + q² = 0

At this point since we are looking for p/q, divide every term by q² to get a quadratic in terms of p/q.

(p/q)² - 4(p/q) + 1 = 0

Use the quadratic formula with a = 1, b = -4, c = 1 or complete the square to solve for p/q.

For now let u = p/q.

u² - 4u = -1

u² - 4u + 4 = 3

(u - 2)² = 3

u - 2 = ±√3

u = 2 ± √3

Since p is the shorter diagonal the ratio must be less than one:

p/q = 2 - √3

Details are not provided here, but the rhombus that satisfies these conditions has an acute angle of 30 degrees.

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