Answer:
c. 10 cm
Explanation:
The parallel sides of trapizium are the bases so
base 1(B1) =9cm , base(B2)= 12cm
![area(a) = 105 {cm}^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/96q1quw2yrjtueh4heoltled2obsyt1vzw.png)
the perpendicular distance means the height so lets find the height
area of trapizium= B1 + B2 × h
2
![{105cm}^(2)(the \: given \: area) = (9cm(the \: given \: base) + 12cm(the \: given \: base))/(2) * h](https://img.qammunity.org/2021/formulas/mathematics/high-school/591cza2mzmd2s9x4lbpw6wn0e9aaqz7vge.png)
![{105cm}^(2) = (21cm)/(2) * h](https://img.qammunity.org/2021/formulas/mathematics/high-school/q6r4z8dgiyp6qp7rrc2eybmoxu9m3u37q4.png)
now we crisscross
![{105cm}^(2) * 2 = 21cm * h](https://img.qammunity.org/2021/formulas/mathematics/high-school/glutspw8oxqh8jbmmcjg99s4ywo3kkd4cz.png)
![{210cm}^(2) = 21cm * h](https://img.qammunity.org/2021/formulas/mathematics/high-school/5r5oagoaq9okcke8a80tydwj19uxqcpmpb.png)
![\frac{ {210cm}^(2) }{21cm} = (21cm \: )/(21cm) * h](https://img.qammunity.org/2021/formulas/mathematics/high-school/2hl77ge3v3i6ykk1h9pphuxj4xj0ttvoz3.png)
h = 10 cm
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