Question

Find the area of pentagon ABCDE in which $$BL\mathrm{\perp }AC,DM\mathrm{\perp }AC,EN\mathrm{\perp }AC$$ such that AC = 18cm, AM = 14cm, AN = 6cm, BL = 4cm, DM = 12cm, EN = 9cm.

Answer 1

Hint: In this problem, we have to find the area of the pentagon as we are given $$BL\mathrm{\perp }AC,DM\mathrm{\perp }AC,EN\mathrm{\perp }AC$$ such that AC = 18cm, AM = 14cm, AN = 6cm, BL = 4cm, DM = 12cm, EN = 9cm. We can first draw the diagram and mark the length given. We can then find the length required. We can find the area of triangles and trapezium in the diagram and we can add them to find the area of the pentagon.

Complete step by step solution:
Here, we have to find the area of pentagon, where we are given $$BL\mathrm{\perp }AC,DM\mathrm{\perp }AC,EN\mathrm{\perp }AC$$ such that AC = 18cm, AM = 14cm, AN = 6cm, BL = 4cm, DM = 12cm, EN = 9cm.
We can now draw the diagram and mark the length.

We can now find the area of each shape in the diagram and add them to find the area of the pentagon ABCDE.
We know that area of triangle formula is
$$\Rightarrow Area={\displaystyle \frac{1}{2}}\times b\times l$$
We can first find the area of triangle ANE as AN = 6cm and NE = 9cm.
$$\Rightarrow Area={\displaystyle \frac{1}{2}}\times 6\times 9=27c{m}^2$$
Area of triangle ANE = 27sq.cm ……. (1)
We can now find the area of ACB as AC= 18cm and AB = 4cm.
$$\Rightarrow Area={\displaystyle \frac{1}{2}}\times 18\times 4=36c{m}^2$$
Area of triangle ACB = 36sq.cm …….. (2)
We can now find the area of triangle DMC as MD = 12 and
$$\begin{array}{rl}& \Rightarrow MC=AC-AM=18-14\\ & \Rightarrow MC=4cm\end{array}$$
We can now substitute the length and height, we get
$$\Rightarrow Area={\displaystyle \frac{1}{2}}\times 4\times 12=24c{m}^2$$
Area of triangle ACB = 24sq.cm ……… (3)
We can now find the area of trapezium using the formula ,
$$\begin{array}{rl}& Area={\displaystyle \frac{1}{2}}\times h\times (b_1+b_2)\\ & \because b_1,b_2=bases\end{array}$$
We can see in the diagram that bases are 9cm and 12 cm and
$$\begin{array}{rl}& \Rightarrow h=NM=AM-AN\\ & \Rightarrow NM=14-6=8cm\end{array}$$
We can now substitute these values in area formula, we get
$$\Rightarrow Area={\displaystyle \frac{1}{2}}\times 8\times (9+12)=84c{m}^2$$
Area of trapezium = 84sq.cm …… (4)
We can now add (1), (2), (3), (4), we get
Area of the pentagon ABCDE,
$$\Rightarrow Area=36+27+84+24=171c{m}^2$$
Therefore, the area of the pentagon ABCDE is 171sq.cm.

Note: We should always remember that the area of the triangle formula is $${\displaystyle \frac{1}{2}}\times b\times l$$, where l is the length and b is the breadth. We should also remember that the formula to find the area of trapezium is $${\displaystyle \frac{1}{2}}\times h\times (b_1+b_2)$$, where $$b_1,b_2=bases$$.