Question

Find the area of the polygon $MNOPQR$ if$MP = 9cm$,$MD = 7cm$, $MC = 6cm$, $MB = 4cm$,$MA = 2cm$ $NA,OC,QD$ and $RB$ are perpendiculars to diagonal $MP$.

Answer 1

Hint: In this problem, we have to find the area of the polygon. Polygon means that a flat shape with at least three and usually five or more angles and straight lines. Using the given data we can solve the problem. Understanding the figure and understanding the given data is important to solve the problem.
Formulae used in the problem:
Area of the triangle $ = \dfrac{1}{2} \times b \times h$
Here, $b = $ breadth
$h = $ height
Area of the trapezium$ = \left( {\dfrac{{a + b}}{2}} \right)h$
$a = base$

Complete step by step answer:

In the above step, the given picture is the figure of the polygon.
In this picture, we see that the,
$MP = 9cm $
$MD = 7cm $
$MC = 6cm $
$MB = 4cm $
$MA = 2cm $
And then, $NA,OC,QD,RB$ are perpendiculars to the diagonal $MP$ .
Now we are going to find the area of the given polygon by using the given data.
In this polygon, there are many triangles and trapezium.
Therefore we have to find the area of all triangles and trapezium and add the found values to find the area of a polygon.
Therefore from the figure,
Area of the polygon $ = $ area of the triangle $MAN$ $ + $ area of the trapezium $ACON$ $ + $ area of the triangle $CPO$ $ + $ area of the trapezium $BDQR + $ area of the triangle $DPQ$
Now we are going to find the area of required triangles,
Area of the triangle $MAN$=$\dfrac{1}{2} \times MA \times MN$
Now apply the values, we get
$ = \dfrac{1}{2} \times 2 \times 2.5$
Now simplify this we get,
$ = 2.5$
By following the above steps, find the area of other triangles.
Area of the triangle $CPO$ $ = \dfrac{1}{2} \times CP \times OC$
Now value of
$CP = MP - MC $
$= 9 - 6 $
$= 3 $
Now apply the values, we get
$ = \dfrac{1}{2} \times 3 \times 3$
Simplify this we get,
$= \dfrac{9}{2} $
$= 4.5 $
Now area of the triangle $MBR = \dfrac{1}{2} \times MB \times BR$
Now apply the values, we get
$ = \dfrac{1}{2} \times 4 \times 2.5$
$ = 2 \times 2.5$
Simplify this we get,
$ = 5$
Now area of the triangle $DPQ = \dfrac{1}{2} \times DP \times QD$
$DP = MP - MD $
$= 9 - 7 $
$= 2cm $
Now apply the values,
$\dfrac{1}{2} \times 2 \times 2$
$ = 2$
Now we are going to find the values of trapezium
Area of trapezium $ACON$ $ = \left( {\dfrac{{2.5 + 3}}{2}} \right)4$
Now simplify this we get,
$= 5.5 \times 2 $
$= 11 $
Area of the trapezium $BDQR = \left( {\dfrac{{2.5 + 2}}{2}} \right)3$
Simplify this we get,
$= \dfrac{{4.5 \times 3}}{2} $
$= \dfrac{{13.5}}{2} $
Now we get,
\[ = 6.75\]
Now the area of a polygon \[ = 2.5 + 11 + 4.5 + 5 + 6.75 + 2\]
Hence the area of a polygon \[ = 31.75c{m^2}\]

Note:
In geometry, a polygon is a plane Figure described by a finite number of a straight line. A trapezium which is a quadrilateral with one pair of opposite sides parallel. In this problem, understanding the data and figures is important to solve this problem.