# What is the length of a diagonal of a square if its area is 98 square feet?

**Solution:**

A diagonal of a square divides it into two congruent right-triangles.

Let's find the length of a diagonal using the Pythagoras theorem.

Let's draw a square ABCD of area 98 square feet as shown below

Join points B and C to construct the diagonal BC.

Area of the square = Side^{2}

⇒ AC^{2} = 98 (given)

⇒ AC = √98

⇒ AC ≈ 9.89949493

Thus, AC ≈ 9.9 feet

Also, AC = AB = CD = BD = 9.9 feet (since, they are the sides of square ABCD)

From the figure, we observe that △BAC is a right-angled triangle with ∠A = 90°.

Applying Pythagoras theorem to △BAC we get,

(BC)^{2} = (AC)^{2 }+ (AB)^{2}

(BC)^{2 } = 9.9^{2} + 9.9^{2} (Since, AC = AB)

(BC)^{2} = 98.01 + 98.01

(BC)^{2} =196.02

BC = √196.02

BC ≈ 14.00071427

BC ≈ 14.0 feet

Thus, the length of a diagonal of the square with an area of 98 square feet is equal to 14 feet.

## What is the length of a diagonal of a square if its area is 98 square feet?

**Summary:**

The length of a diagonal of the square with an area of 98 square feet is equal to 14 feet.

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