Question

find the area of quadrilateral ABCD in which AB=5cm BC =4cm CD =4cm DA =5cm AC =7cm​

Answer 1

Answer:

the area of quadrilateral ABCD is $8\sqrt{6}\ cm^{2}\end{array}$

Step-by-step explanation:

From the figure, the area of the quadrilateral is the sum of the area of the triangle ABC and the triangle ADC.

Firstly we have to find the areas of the triangles. We have given the length of sides, so we are using Heron's formula to find the area.

If a,b, and c are the length of sides of a triangle, then the area of the triangle by heron's formula is given as,

             $A=\sqrt{s\left( s-a\right) \left( s-b\right) \left( s-c\right) }$

Where, $s=\dfrac{a+b+c}{2}$

Area of the triangle ABC ⇒

Length of the sides are AB = 5 cm, BC = 4 cm, and AC = 7 cm.

Therefore, $s=\dfrac{5+4+7}{2}= 8$

Then, $A=\sqrt{8\left( 8-5\right) \left( 8-4\right) \left( 8-7\right) }$

         $\begin{array}{l}=\sqrt{8\times3\times4\times1}\\\\=\sqrt{96}\\\\=4\sqrt{6}\ cm^{2}\end{array}$

Area of the triangle ADC ⇒

Length of the sides are AD = 5 cm, DC = 4 cm, and AC = 7 cm.

Since the two triangles have equal sides, the area will also be the same. That is, $A=4\sqrt{6}\ cm^{2}\end{array}$

Hence the area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ADC

∴ Area of the quadrilateral $=4\sqrt{6}\ +\ 4\sqrt{6}$

                          $=8\sqrt{6}\ cm^{2}\end{array}$