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Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \huge\mathcal{\fcolorbox{aqua}{azure}{\red{✿ \: Answer ❖}}}$

GIVEN :

• Quadrilateral sides

TO FIND :

• Area of the Quadrilateral = ?

SOLUTION :

CONSTRUCTION :

• Draw AD ll CE So that CE parallel to AB.

IN TRIANGLE BEC , ANGLE E = 90°

${ \boxed{ \blue {by \: pythagoreas \: theoram = (h) ^{2} = (b) ^{2} + (h) ^{2} }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: = (BC)^{2} = (EB)^{2} + (EC) ^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: = (13)^{2} = (5)^{2} + (EC) ^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 169 = 25 + (EC)^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 169 - 25 = (EC)^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 144 = (EC)^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sqrt{144} = EC$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 12 \: m \: = EC$

Area of Triangle = 1/2 × b ×h

= 1/2 × 5 × 12

= 30 m2

Area of rectangle AECD = l × b

= 13 × 12

= 156 m2

Area of Quadrilateral = Area of Reactangle AECD + Area of Triangle

= 156 + 30

= 186 m2

HENCE, THE AREA OF QUADRILATERAL = 186 M

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \huge\mathcal{\fcolorbox{aqua}{azure}{\red{✿ \: Answer ❖}}}$

CONSTRUCTION :

• Draw AD ll CE So that CE parallel to AB.

IN TRIANGLE BEC , ANGLE E = 90°

${ \boxed{ \blue {by \: pythagoreas \: theoram = (h) ^{2} = (b) ^{2} + (h) ^{2} }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: = (BC)^{2} = (EB)^{2} + (EC) ^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: = (13)^{2} = (5)^{2} + (EC) ^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 169 = 25 + (EC)^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 169 - 25 = (EC)^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 144 = (EC)^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sqrt{144} = EC$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 12 \: m \: = EC$

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