Question

find height of trapezium whose parallel sides are 7cm and 9cm and area is 24sq cm​

Answer 1

GIVEN:

• Parallel sides of trapezium = 7cm and 9cm

• Area of the trapezium = 24sq.cm

SOLUTION:

As we know that,

$\star \: {\underline{\boxed{\mathcal{\pmb{\quad Area \: of \: trapezium = \frac{1}{2} \: \times sum \: of \: it's \: parallel \: sides \times distance \: between \: parallel \: sides}}}}}$

Substituting values,

$\implies\sf{24 = \frac{1}{2} \times (7 + 9) \times h} \\ \\ \implies\sf{24 = \frac{1}{2} \times 16 \times h} \\ \\ \implies\sf{24 = 1 \times 8 \times h} \\ \\ \implies\sf{24 = 8h} \\ \\ \implies\sf{h = \frac{\cancel{24}}{\cancel{8}}} \\ \\ \implies{\boxed{\sf{\gray{h = 3 \: \bigstar}}}}$

Hence,

• The height of the trapezium is 3cm.

Answer 2

❍ Let the height of trapezium be x cm respectively.

__________________

$\underline{\bigstar\boldsymbol{According\;to\;the\;given\;Question:}}$

• Parallel sides of trapezium is 7 cm and 9cm.
• Area of trapezium is 24 sq.cm.

$\\ \underline{ \frak{ \dag \: As \: we \: know \: that : }} \\ \\ \star\underline{\boxed{\sf{Area\;of\;trapezium\;=\;\frac{1}{2}\times(a + b)h}}}$

Applying the values,

$\longrightarrow{\sf{24 = \frac{1}{2} \times (7 + 9) \times x}} \\ \\ \longrightarrow\sf{24 = \frac{1}{2} \times 16 \times x} \\ \\ \longrightarrow\sf{24 = \frac{16}{2} \times x} \\ \\ \longrightarrow\sf{24 = 8 \times x} \\ \\ \longrightarrow\sf{x = \frac{24}{8}} \\ \\ \longrightarrow\boxed{\sf{x = 3}}\star\\ \\ \\ \\ \sf{\therefore Hence,The \: height \: of \: trapezium \: is \: 3 cm.}$