Question

2.In a pentagon ABCDE AB= 28 cm, BC=15cm, CD=9 cm, DE=24 cm, At=32 cm, AC=41 cm, AD=40 cm. Find the area of the pentagon.
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Answer 1

Answer:

Area of the pentagon is 690$cm^2$

Step-by-step explanation:

The pentagon is shown here. we can find the area of pentagon by dividing it into different triangle.

Area of the triangle can be found by using Heron's formula which is given by

$A=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$ and a,b,c are the sides of a triangle.

From the figure it can be seen that the pentagon is divided into 3 triangles and named as $A_1,A_2,A_3$

For triangle 1, the sides are 28,15 and 41 so $s=\frac{28+41+15}{2}=42$, hence $A_1=\sqrt{42(42-28)(42-15)(42-41)} =126$

Similarly for triangle 2 Sides are 40,41,9 and $s=\frac{40+41+9}{2}=45$, hence

$A_2=\sqrt{45*4*5*36}=180$

for triangle 3, sides are 32,40,24 and $s=48$

$A_3=\sqrt{48*16*24*8}=384$

Hence the area of the pentagon is =$A_1+A_2+A_3=690cm^2$