Subjective Type

The sum of areas of two squares with sides 4a and 4b is:

Solution

Given,
Side of first square $$=4a$$
Side of second square $$=4b$$

We know that, area of a square is $$(side)^{2}$$
So, area of first square $$= (4a)^{2}$$
$$= 16a^{2}$$

Area of second square $$=(side)^{2}$$
$$= (4b)^{2}$$
$$= 16b^{2}$$

$$\therefore$$ Sum of the areas $$=16a^2+16b^{2} $$
$$= 16(a^{2}+b^{2})$$

Hence, the sum of areas of two squares with sides $$4a$$ and $$4b$$ is $$= 16(a^{2}+b^{2})$$.

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