Question

in the Adjoining quadrilateral ABCD,diagonal AC=48cm DP and BQ are perpendiculars To AC. DP=17.5Cm and BQ=12cm Find the area of quadrilateral ABCD.

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Answer 1

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

Here ,

☆ Diagonal AC = 48 cm

☆ Perpendicular 1 i.e DP = 17.5 cm

☆ Perpendicular 2 i.e BQ = 12 cm

As we know ,

$\bold{ \large{\boxed{\purple{\tt{Area \: of \: quadrilateral \: = \: \frac{1}{2} \times Diagonal × sum \: of \: perpendiculars}}}}}</p><p>$

$\rightarrow{\large{\tt{ \frac{1}{2} \times AC × ( DP + BQ )}}}$

$\rightarrow{\large{\tt{ \frac{1}{2} \times 48 \times (17.5 + 12)}}}$

$\rightarrow{\large{\tt{ \frac{1}{2} \times 48 \times 29.5}}}$

$\rightarrow{\large{\tt{24 \times 29.5}}}$

$\rightarrow{\large{\tt{708 \: {cm}^{2} }}}$

$\large{\bold{\blue{\tt{I \: hope \: it \: helps \: u \: my \: bestie \: !!!}}}}$

Answer 2

Answer:

Here ,

☆ Diagonal AC = 48 cm

☆ Perpendicular 1 i.e DP = 17.5 cm

☆ Perpendicular 2 i.e BQ = 12 cm

As we know ,

\bold{ \large{\boxed{\purple{\tt{Area \: of \: quadrilateral \: = \: \frac{1}{2} \times Diagonal × sum \: of \: perpendiculars}}}}}

Areaofquadrilateral=

2

1

×Diagonal×sumofperpendiculars

\rightarrow{\large{\tt{ \frac{1}{2} \times AC × ( DP + BQ )}}}→

2

1

×AC×(DP+BQ)

\rightarrow{\large{\tt{ \frac{1}{2} \times 48 \times (17.5 + 12)}}}→

2

1

×48×(17.5+12)

\rightarrow{\large{\tt{ \frac{1}{2} \times 48 \times 29.5}}}→

2

1

×48×29.5

\rightarrow{\large{\tt{24 \times 29.5}}}→24×29.5

\rightarrow{\large{\tt{708 \: {cm}^{2} }}}→708cm

2