# Area Maze Math Puzzle | Find the Area of Rectangle

## Geometry math problem: Find the area of rectangle from the area maze math puzzle

*This puzzle is made with inspiration from Japanese puzzle creator Naoki Inaba*

Find the value of the question mark in the following diagrams (Area of the shaded rectangle).

## Solution

To solve this puzzle, we can name the figure

Let MN = *x* and FG = *y*, then

MY = XF

⇒ MN + NY = XE + FE

⇒ *x* + 2 = 2 + FE

so, FE* = x*

MX = FY

⇒ XL + LM = FG + GY

⇒ 2 + LM = *y *+ 3

so, LM = *y *+ 1

Area of the shaded rectangle = XM × MY

Now extent CN and AL then they meet at D. and form rectangle ABCD and rectangle MNDL

Area of the rectangle ABCD = AB × BC = 8 × 7 = 56 cm^{2}

Area of the rectangle MNDL = Area of the rectangle ABCD − Area of ABCNML

⇒ Area of the rectangle MNDL = 56 − 36 = 20 cm^{2}

Also we have, Area of the rectangle MNDL = MN × ML = *x*(* y *+ 1)

⇒ *x*( *y *+ 1) = 20

⇒ *xy *+* x* = 20……………………………………….*eq*(1)

Now extent CN and AL then they meet at D. and form rectangle PQRS and rectangle EFGS

Area of the rectangle PQRS = PQ × QR = 7 × 6 = 42 cm^{2}

Area of the rectangle EFGS = Area of the rectangle PQRS − Area of PQRGFE

⇒ Area of the rectangle EFGS = 42 − 26 = 16 cm^{2}

Also, we have, Area of the rectangle EFGS = EF × GF = *xy*

⇒ *xy* = 16……………………………………….*eq*(2)

### Now we need to solve the equation system 1 and 2

That is from *equation* (1)* & equation* (2)

*xy + x* = 20

⇒ 16 + *x* = 20

⇒ *x* = 4 cm ⇒ *y *= 4 cm

Area of the grey rectangle = XM × MY = (*y* + 3) × (*x +* 2)

⇒ Area of the shaded rectangle = 7 × 6

so we got the solution to this area maze math puzzle as

**Area of the shaded rectangle = 42 cm ^{2}**