Question

find area of this quadrilateral::​

Answer 1

Answer:400√5

Step-by-step explanation:in triangle DOC,

                                                                          DO=√30²-20²

                                                                      ⇒√900-400

                                                                      ⇒√500

                                                                      ⇒10√5

area =1/2*20*2*10√5*2

       =20*20√5

      =400√5

Answer 2

$\large\sf\pink{ANSWER}$

$\bigstar\sf\red{GIVEN}\bigstar$

• AC=20cm

• CD=30cm

$\bigstar\sf\blue{HOW\:TO\:SOLVE}$

• There are two triangle in this figure $\triangle{ADC}\:and\triangle{ACB}$

• AC is Common base in both triangle.

Now, In triangle ADC

• AC=20cm.(Base)

• CD=30cm(Height)

$\huge\tt{Area\:of\:triangle=\frac{1}{2}×Base×Height}$

$\implies\tt{Area\:of\:triangle=\frac{1}{2}×20×30}$

$\implies\tt{Area\:of\:triangle=\frac{1}{\cancel{2}}×{\cancel{20}×30}}$

$\implies\tt{Area\:of\:triangle=300cm}$

As we know it is a square because one of it's angle is 90°.So, All sides are equal and bases are common.So, From this we came to knlw that-

AREA OF TRIANGLE ABD=ABC

Hence,

Area of quadrilateral ABCD = 300+300=600cm^2.