Expand the identity geometrically: $(2 t + 1) (2 t + 6)$

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Solution

Step 1: Draw a square and cut into $4$ parts.
Step 2: There are $4$ rectangle and $1$ square.
Step 3: Area of the full rectangle, $A B C D =$ $(2 t + 1) (2 t + 6)$
Step 4: Now we have to find the area of inside square and rectangle as shown in the figure.
Step 5: Consider the area of pink square $=$ $(2 t)^{2}$ and the area of yellow rectangle $=$ length $\times$ breadth $=$ $12 t$
Step 6: Area of blue rectangle $=$ $2 t$ and the area of green rectangle $=$ $6$
Step 7: Area of full rectangle $=$ area of pink square $+$ area of yellow rectangle $+$ area of blue rectangle $+$ area of green rectangle.
i.e., $(2 t + 1) (2 t + 6) = (2 t)^{2} + 12 t + 2 t + 6$
$(2 t + 1) (2 t + 6) = (2 t)^{2} + t (12 + 2) + 6$
Hence, geometrically we expanded the identity $(2 t + 1) (2 t + 6) = (2 t)^{2} + t (12 + 2) + 6$ .

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