Question

In a given figure ,abcd is a rectangle in ab 8cm bc 6cn and the diagonal intersect each other at o. find the area of shaded region by using heron ,s formula?

Answer 1

Where is the figure or just tell what is the shaded region of the rectangle

Answer 2

$\huge\underline\mathfrak{Answer:}$

AC and BD are the diagonals of a rectangle ABCD.

$\bf{In\:a\:right\:triangle,\:we\:have}$

= $\bf{AC^2=AB^2+BC^2}$

=$\bf{[(8)^2+(6)^2]cm^2}$

=$\bf{(64+36)cm^2}$

=$\bf\underline{100cm^2}$

$\implies$$\bf{AC=\sqrt100\:cm}$

$\implies$$\bf\underline\pink{AC=10cm}$

$\therefore$$\bf{OA=}$$\bf\frac{1}{2}diagonal\:AC=$

$\bf\frac{1}{2}\times10cm\:=5cm$

$\bf\underline\purple{Similarly,\:Diagonal\:BD=10cm}$

$\bf{OA=}$$\bf\frac{1}{2}diagonal\:BD=$

$\bf\frac{1}{2}\times10cm\:=5cm$

In ∆OAB, we have

a = OA = 5cm, b = OB = 5 cm and c = AB = 8 cm

Semi - perimeter of ∆OAB (S)

$\huge\bf\frac{5+5+8}{2}=\frac{18}{2}cm=9cm$

$\therefore$ s - a = 9 - 5 = 4cm, s - b = 9 - 5 = 4 and s - c = 9 - 8 = 1cm

$\bf\underline\red{Area\:of\:a\:triangle\:OAB}$

$\leadsto$$\bf\sqrt{s(s-a)(s-b)(s-c)}$

$\leadsto$$\bf\sqrt{9\times4\times4\times1}cm^2$

$\leadsto$$\bf\sqrt{(3\times4)}cm^2$

12 cm².