Question

In the given figure, \[ABCD\] is a rectangle where \[AB = 8cm\] and \[BC = 6cm\] and the diagonals bisect each other at \[O\].Find the area of the shaded region by Heron’s formula.

Answer 1

Hint:Here, we look at the right angled triangle formed inside the rectangle and use the Pythagoras theorem to find the length of the diagonal, using the fact that diagonals bisect each other we find the lengths of sides of \[\vartriangle AOB\]. Now we use Heron’s formula on \[\vartriangle AOB\] to find its area.
* Pythagoras theorem gives us the length of the third side in any right angled triangle if we are given the lengths of two sides of the triangle. If we have a right triangle \[\vartriangle ABC\] with base \[BC\] , perpendicular \[AB\] and hypotenuse \[AC\]then Pythagoras theorem tells us that sum of square of base and square of perpendicular will be equal to square of the hypotenuse i.e. \[A{C^2} = A{B^2} + B{C^2}\]
* If a line is bisected at a point then it means it is breaking into two equal parts at that point.
* Heron’s formula gives us the formula for the area of a triangle whose three side lengths are given. Given \[a,b,c\] as three sides of triangle, we find the semi-perimeter first \[S = \dfrac{1}{2}(a + b + c)\] and then
Area of triangle \[ = \sqrt {s(s - a)(s - b)(s - c)} \]

Complete step-by-step answer:
Since we know all the angles of the rectangle are right angles.
Therefore, we consider the right angle triangle \[\vartriangle ABC\] with \[\angle B = {90^ \circ }\]
Given the length of sides \[AB = 8cm\] and \[BC = 6cm\] , we can find the length of side \[AC\] using Pythagora's theorem.
\[A{C^2} = A{B^2} + B{C^2}\]
Substitute the values of \[AB = 8cm\] and \[BC = 6cm\]
\[A{C^2} = {(8)^2} + {(6)^2}\]
Opening the squares by multiplication of the number to itself.
\[
  A{C^2} = 8 \times 8 + 6 \times 6 \\
  A{C^2} = 64 + 36 \\
  A{C^2} = 100 \\
 \]
Taking square root on both sides of the equation
\[
  \sqrt {A{C^2}} = \sqrt {100} \\
  \sqrt {A{C^2}} = \sqrt {{{(10)}^2}} \\
 \]
Cancel square root with the square power on both sides
\[AC = 10\]
Therefore, length of the diagonal \[AC = 10cm\]
Since, both the diagonals bisects each other at point \[O\], therefore we can write
\[
  AC = AO + OC \\
  AC = AO + AO \\
  AC = 2AO \\
 \]
Dividing both sides by 2
\[
  \dfrac{{AC}}{2} = \dfrac{{2AO}}{2} \\
  \dfrac{1}{2}AC = AO \\
 \]
Substituting value of \[AC = 10cm\]
\[
  \dfrac{1}{2}(10) = AO \\
  5 = AO \\
 \]
Therefore, length \[AO = 5cm\]
Similarly, if we consider the another right angle triangle \[\vartriangle DAB\] with \[\angle A = {90^ \circ }\]
Given the length of sides \[AB = 8cm\] and \[BC = AD = 6cm\]( since opposite sides of a rectangle are equal) , we can find the length of side \[BD\] using Pythagora's theorem.
\[B{D^2} = A{B^2} + A{D^2}\]
Substitute the values of \[AB = 8cm\] and \[AD = 6cm\]
\[B{D^2} = {(8)^2} + {(6)^2}\]
Opening the squares by multiplication of the number to itself.
\[
  B{D^2} = 8 \times 8 + 6 \times 6 \\
  B{D^2} = 64 + 36 \\
  B{D^2} = 100 \\
 \]
Taking square root on both sides of the equation
\[
  \sqrt {B{D^2}} = \sqrt {100} \\
  \sqrt {B{D^2}} = \sqrt {{{(10)}^2}} \\
 \]
Cancel square root by square power on both sides.
\[BD = 10\]
Therefore, length of the diagonal \[BD = 10cm\]
Since, both the diagonals bisects each other at point \[O\], therefore we can write
\[
  BD = BO + OD \\
  BD = BO + BO \\
  BD = 2BO \\
 \]
Divide both sides of the equation by 2
\[
  \dfrac{{BD}}{2} = \dfrac{{2BO}}{2} \\
  \dfrac{1}{2}BD = BO \\
 \]
Substituting value of \[BD = 10cm\]
\[
  \dfrac{1}{2}(10) = BO \\
  5 = BO \\
 \]
Therefore, length \[BO = 5cm\]
Now taking the \[\vartriangle AOB\]
We have \[AB = 8cm,AO = 5cm,OB = 5cm\]
Using the semi-perimeter formula \[S = \dfrac{1}{2}(a + b + c)\], where \[a,b,c\]are sides of a triangle.
Substituting the values \[a = AB,b = AO,c = OB\]
\[
  S = \dfrac{1}{2}(AB + AO + OB) \\
  S = \dfrac{1}{2}(8 + 5 + 5) \\
  S = \dfrac{1}{2} \times 18 \\
  S = 9 \\
 \]
Now applying Heron’s formula, area of \[\vartriangle ABC = \sqrt {s(s - a)(s - b)(s - c)} \]
Substitute the values \[a = AB,b = AO,c = OB,s = 9\] in the formula
Area of \[\vartriangle AOB = \sqrt {s(s - AB)(s - AO)(s - OB)} \]
                           \[
   = \sqrt {9(9 - 8)(9 - 5)(9 - O5)} \\
   = \sqrt {9(1)(4)(4)} \\
   = \sqrt {{{(3)}^2} \times {{(4)}^2}} \\
   = \sqrt {{{(3 \times 4)}^2}} \\
   = \sqrt {{{(12)}^2}} \\
   = 12 \\
 \] {\[{a^m} \times {b^m} = {(ab)^m}\] and square root is cancelled by square}
Therefore, the area of the shaded region is \[12c{m^2}\].

Note:Students usually get confused with squares and take all sides equal giving them diagonals equal to the length of the sides. A rectangle has only opposite sides equal and has a length and a breadth whereas square just has length of side. We can check the value of the semi-perimeter as it will always be greater than the lengths of three sides of the triangle.