Can anyone help with this problem

9514 1404 393

Step-by-step explanation:

For a quadrilateral in which the diagonals cross at right angles, such as a rhombus or kite, the area can be found as half the product of the diagonals.

For the purpose here, we can ignore the factor of 1/2, which is the same for both, and simply consider the products of the diagonals.

Kite A has an area proportional to (30 in)(30 in) = 900 in².

Kite B has an area proportional to (20 in)(40 in) = 800 in².

Kite A has a larger area, so will use more paper.

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Only Rudy knows why he chose answer C. Perhaps he guessed, rather than doing any math.

A. Kite A

Step-by-step explanation:

To determine which of the two kites will use need more paper to make, find the area of each kite.

Kite A:

Area = (d1*d2)/2

d1 = 30 in.

d2 = 30 in.

Area of kite A = (30*30)/2 = 450 in.²

Kite B:

Area = (d1*d2)/2

d1 = 20 in.

d2 = 40 in.

Area of kite B = (20*40)/2 = 400 in.²

✅Kite A will use more paper to make because it has a greater area of 450 in.² than Kite B (400 in.²).

So, Ruby was wrong. He most likely didn't calculate the area of both kites correctly.

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)

35 is an outlier

step-by-step explanation: