Question

find the area of a quadrilateral ABCD in which AB=50 cm, BC=60 CM, CD=30 cm DA=90 cm and BD = 70 cm

Answer 1

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ BDC

Area of Δ ABD

By Heron's formula :

Area = $\sqrt{s(s-a)(s-b)(s-c)}$

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

a = 50

b = 70

c = 90

$s=\dfrac{50+70+90}{2}\\\\\implies s=\dfrac{210}{2}\\\\\implies s=105$

$A=\sqrt{105(105-90)(105-70)(105-50)}\\\\\implies A=\sqrt{105\times 15\times 35 \times 55}\\\\\implies A=\sqrt{3031875}\\\\\implies A=1741.228$

Area of Δ BCD

a = 70

b = 60

c = 30

$s=\dfrac{60+30+70}{2}\\\\\implies s=\dfrac{160}{2}\\\\\implies s=80$

$A=\sqrt{80(80-70)(80-60)(80-30)}\\\\\implies A=\sqrt{80\times 10\times 20 \times 50}\\\\\implies A=\sqrt{800000}\\\\\implies A=894.427$

Area of ABCD = 1741.228 + 894.427

= 2635.655

= 2635.66 approx.

ANSWER : 2635.66 aprox.

Answer 2

Answer

s= 2a+b+c

a = 50

b = 70

c = 90

$\begin{lgathered}s=\dfrac{50+70+90}{2}\\\\\implies s=\dfrac{210}{2}\\\\\implies s=105\end{lgathered}$

s= 250+70+90

s= 2210

s=105

$\begin{lgathered}A=\sqrt{105(105-90)(105-70)(105-50)}\\\\\implies A=\sqrt{105\times 15\times 35 \times 55}\\\\\implies A=\sqrt{3031875}\\\\\implies A=1741.228\end{lgathered}$

A= 105(105−90)(105−70)(105−50)

⟹A= 105×15×35×55

⟹A= 3031875

⟹A=1741.228

Area of triangle BCD

a = 70

b = 60

c = 30

$\begin{lgathered}s=\dfrac{60+30+70}{2}\\\\\implies s=\dfrac{160}{2}\\\\\implies s=80\end{lgathered}$

s= 2.60+30+70

⟹s= 2160

⟹s=80

$\begin{lgathered}A=\sqrt{80(80-70)(80-60)(80-30)}\\\\\implies A=\sqrt{80\times 10\times 20 \times 50}\\\\\implies A=\sqrt{800000}\\\\\implies A=894.427\end{lgathered}$

A= 80(80−70)(80−60)(80−30)

⟹A= 80×10×20×50

⟹A= 800000

⟹A=894.427

Area of ABCD = 1741.228 + 894.427

= 2635.655

= 2635.655= 2635.66