Question

please solve..
find the area of the enclosed figure​

Answer 1

$\bold{\large{\boxed{\red{ANSWER}}}}$

First of all , let us divide the figure into square and trapezium.

Now ,

$\boxed{\star \: {Given}}$

☆ Side of square = 24 cm

☆ Parallel sides of trapezium = 24 cm and 9 cm

☆ Height of trapezium = 10 cm

$\boxed{\star \: {Solution}}$

We know ,side of square = 24 cm

So ,

$\boxed{\purple{\bold{\tt{Area \: of \: square = {side}^{2} }}}}$

$\rightarrow {(24)}^{2}$

$\rightarrow576 \: {cm}^{2}$

$\bold{\boxed{\large{\pink{\tt{Area \: of \: trapezium = \frac{1}{2} \times sum \: of \: parallel \: sides \: \times height}}}}}$

$\rightarrow \: \frac{1}{2} \times (24 + 9) \times 10$

$\rightarrow \frac{1}{2} \times 33 \times 10$

$\rightarrow33 \times 5$

$\rightarrow165 \: {cm}^{2}$

$\bold{\large{\boxed{\blue{\tt{ \: Area \: of \: figure = Area \: of \: square \: + Area \: of \: trapezium}}}}}$

$\rightarrow576 + 165$

$\rightarrow741 \: {cm}^{2}$

$\bold{\green{\tt{I \: hope \: it \: helps \: u \: my \: Bestie \: !!!}}}$