# Four Trapezia Formed by the Difference of Two Concentric Squares

**Problem***ABCD* is a square of side 10 cm. *PQRS* is a square inside *ABCD*. *PQBA*, *QRCB*, *RSDC*, and *SPAD* are identical trapezia, each of area 16 cm^{2}. What is the height of each trapezium if *PQ* is parallel to *AB* and *SR* is parallel to *DC*?

A. 3 cm | C. 2 cm |

B. 1.8 cm | D. 1.2 cm |

**Answer Key**

[ C ]

**Solution**

$10^2 - (10 - 2h)^2 = 4(16)$

$h = 2 ~ \text{cm}$ ← Answer: [ C ]

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