Mathematics: Hellp me solve this thanks

Hellp me solve this thanks - id7375984, rex1578

thil54451

01.02.2022

c. 10 cm

Step-by-step explanation:

The parallel sides of trapizium are the bases so

base 1(B1) =9cm , base(B2)= 12cm

$area(a) = 105 {cm}^{2}$

the perpendicular distance means the height so lets find the height

area of trapizium= B1 + B2 × h

${105cm}^{2}(the \: given \: area) = \frac{9cm(the \: given \: base) + 12cm(the \: given \: base)}{2} \times h$

${105cm}^{2} = \frac{21cm}{2} \times h$

now we crisscross

${105cm}^{2} \times 2 = 21cm \times h$

${210cm}^{2} = 21cm \times h$

$\frac{ {210cm}^{2} }{21cm} = \frac{21cm \: }{21cm} \times h$

h = 10 cm

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