ABCD rectangle. K-middle BC. the perimeter of ABCD is 36 cm. AD is 2 times more than AB.

| August 22, 2021 | education

ABCD rectangle. K-middle BC. the perimeter of ABCD is 36 cm. AD is 2 times more than AB. find: a) find the area of the rectangle ABCD. b) find the area of the triangle ADM.

Let the length of the side AB be equal to X cm, then, by condition, the length of the side AD is equal to 2 * X cm.

Since the opposite sides of the rectangle are equal, the perimeter of the rectangle will be equal to: Ravsd = 2 * (AB + BC) = 2 * (X + 2 * X) = 36.

2 * X + 4 * X = 36.

6 * X = 36.

X = AB = 36/6 = 6 cm.

Then AD = 2 * 6 = 12 cm.

Determine the area of the rectangle ABCD.

Savsd = AB * AD = 6 * 12 = 72 cm2.

Let’s draw the height of the KN of the ACD triangle.

KN = AB = 6 cm.

Determine the area of the triangle AKD. Sacd = AD * KN / 2 = 12 * 6/2 = 36 cm2.

Answer: The area of the rectangle ABCD is 72 cm2, the area of the triangle ACD is 36 cm2.



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