Ashifur August 8, 2023 - 17:47
Problem: In the figure ABCD is a square. 1f the length of the square is 10 ft, then what will be the area of the triangle OCD?

View all: RAJSHAHI KRISHI UNNAYAN BANK OFFICER (SO) | WRITTEN QUESTION (MATH) SOLVE | 2014

Correct Answer: 25 sq ft

Explanation:

Give, the length of a arm of a square = 10 ft

Diagonal of the square, BC = $\sqrt{10^{2}+10^{2}}$= $\sqrt{2\times 10^{2}}$= 10$\sqrt{2}$ ft

Half of the diagonal is, 0C =$\frac{10\sqrt{2}}{2}$= 5$\sqrt{2}$ ft

Here, OC = OD = 5$\sqrt{2}$ ft

As we know, area of triangle COD =$\frac{1}{2}$ x the length of two arm’s attached to the right angle.

=$\frac{1}{2}$$\times$ 5$\sqrt{2}$$\times$5$\sqrt{2}$ =25 ft

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