Question

solve this question plz​

Answer 1

Given, ABCD is a square.

Clearly, AD=DC=CB=AB=14 cm.

In ∆ ABC,

AC² = BC² + AB²

or, AC² = (14)² + (14)²

or, AC² = 392

or, AC = √392

or, AC = 14√2 cm.

Now, AO = OC = 14√2/2 = 7√2 cm.

Now, in ∆ AOB,

AO is height and OB is base of the triangle.

Now, OB = OC = 7√2 cm

Ultimately, Area of ∆ AOB

= 1/2 × base × height

= 1/2 × 7√2 × 7√2

= 1/2 × 98

= 49 cm² [ANSWER]

Answer 2

According to the Question

Side of square = 14 cm

DA = CD = BC = BA = 14 cm

Now

In triangle ABC

AC^2 = BC^2 + BA^2

AC^2 = (14)^2 + (14)^2

AC^2 = 392

AC = √392

AC = 14 √2

OA = CO

$\bf\huge\frac {14\sqrt{2}}{2}$

= 7 √2

Therefore in Triangle AOB

$\bf\huge\frac {1}{2}\times b\times h$

$\bf\huge\frac {1}{2}\times 7\sqrt{2} \times 7\sqrt{2}$

$\bf\huge\frac {1}{2}\times 98$

= 49 cm