Question

Look at the measures shown in the adjacent figure and find the area of ABCD.
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Answer 1

Answer:

252 $m^{2}$

Step-by-step explanation:

First lets find the length of diagonal AC

⇒ $AC^{2}= AB^{2} +BC^{2}$ [Pythagoras theorem]

⇒ $AC = \sqrt{24^{2} +7^{2} }$

⇒ AC = $\sqrt{576 +49}$

⇒ AC = $\sqrt{625} = 25$

Hence AC = 25m

Now , Area of the figure = area of ΔABC + area of ΔADC

Area of ΔABC =  $\frac{1}{2}$ × Base × Height =  $\frac{1}{2}$ × 24 × 7 = 12 × 7 = 84 $m^{2}$

For area of ΔADC :

Construction : Drop a perpendicular from A on DC and label it as point O.

Now ∠AOC = ∠ABC [Botha are 90°]

hence, quad AOCB is a parallelogram. [one pair of opposite angle are equal]

Now,

AO = BC = 7m [opposite sides are equal in a parallelogram]

Hence,

Area of ΔADC = $\frac{1}{2}$ × Base × Height = $\frac{1}{2}$ × 48 × 7 = 24 × 7  = 168 $m^{2}$

Hence,

Area of the figure = area of ΔABC + area of ΔADC = 84 + 168 = $252 m^{2}$

Hope it helps

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