Question

The area of parallelogram whose diagonal is \[6.8{\rm{cm}}\] and the perpendicular distance of this diagonal from an opposite vertex is \[7.5{\rm{cm}}\] is
\[25.5{\rm{c}}{{\rm{m}}^2}\]
\[11.9{\rm{c}}{{\rm{m}}^2}\]
\[12.5{\rm{c}}{{\rm{m}}^2}\]
\[51{\rm{c}}{{\rm{m}}^2}\]

Answer 1

Hint: Here, we will use the fact that a diagonal of a parallelogram divides it into two congruent triangles. We will find the area of one of the triangles and multiply it by 2 to get the required area of the area of parallelogram.

Formula Used:
We will use the formula Area of triangle \[ = \dfrac{1}{2} \times {\rm{base}} \times {\rm{height}}\]

Complete step-by-step answer:
First of all we will draw a figure showing the above information,

Now, from the figure,
We have a parallelogram \[ABCD\] with a diagonal \[AC = 6.8{\rm{cm}}\]
The perpendicular distance of this diagonal from the opposite vertex, \[DE = 7.5{\rm{cm}}\]
We can observe that the diagonal has divided the parallelogram into two triangles which are congruent to each other as the opposite sides of a parallelogram are equal and the base of the triangle, i.e. the diagonal is same.
Now we will find the area of the triangle\[ADC\].
Substituting base \[ = 6.8{\rm{cm}}\] and height \[ = 7.5{\rm{cm}}\] in the formula area of triangle \[ = \dfrac{1}{2} \times {\rm{base}} \times {\rm{height}}\], we get
Area of triangle \[ADC\] \[ = \dfrac{1}{2} \times 6.8 \times 7.5\]
Converting the decimal into fraction, we get
\[ \Rightarrow \] Area of triangle \[ADC\] \[ = \dfrac{1}{2} \times \dfrac{{68}}{{10}} \times \dfrac{{75}}{{10}}\]
Multiplying the terms, we get
\[ \Rightarrow \] Area of triangle \[ADC\] \[ = \dfrac{{2550}}{{100}}\]
Dividing the terms, we get
\[ \Rightarrow \] Area of triangle \[ADC\] \[ = 25.5{\rm{c}}{{\rm{m}}^2}\]
Since, both the triangles have the same base and the same height as the perpendiculars from both the vertices to the diagonal will be of the same length.
Hence, $ar\left( {\vartriangle ABC} \right) = ar\left( {\vartriangle ADC} \right)$
\[ \Rightarrow \] Area of the parallelogram $ABCD = 2 \times ar\left( {\vartriangle ADC} \right)$
\[ \Rightarrow \] Area of the parallelogram \[ABCD = 2 \times 25.5{\rm{c}}{{\rm{m}}^2} = 51{\rm{c}}{{\rm{m}}^2}\]
Therefore, the area of parallelogram is \[51{\rm{c}}{{\rm{m}}^2}\].
Hence, option D is the correct answer.

Note: A parallelogram is a type of quadrilateral having opposite sides parallel and equal. Also, in a parallelogram, the opposite angles are congruent or equal and the consecutive angles are supplementary. Supplementary angles are the angles whose sum is equal to \[180^\circ \]. Thus, the sum of the consecutive angles in a parallelogram is equal to \[180^\circ \]. If we state the properties of diagonals of a parallelogram then, they bisect each other and each diagonal divides the parallelogram into two congruent triangles.