In the following figure it is given that $AB$ is parallel to $ED$, the distance between $AB$ and $ED$ is 10 cm, $\angle BCD=90^\circ$, $BC=9$ cm, $CD=8$ cm, $AB=18$ cm, and $ED=12$ cm. Find the area of the figure $ABCD$.
Show Hint
Whenever a figure can be split into standard geometric shapes such as triangles, rectangles, or trapeziums, calculate the area of each component separately and then add or subtract them appropriately.
For this question:
\[
\text{Area}
=
\text{Area of Trapezium}
+
\text{Area of Right Triangle}
\]
\[
=
150+36
=
186\text{ sq. cm.}
\]
This method is extremely useful in mensuration problems involving composite figures.
Concept:
The given figure is a composite geometric figure that can be divided into two simpler parts:
• A trapezium having parallel sides $AB$ and $ED$.
• A right-angled triangle $\triangle BCD$.
The area of the entire figure is obtained by adding the areas of these two simpler regions.
\[
\text{Area of Figure}
=
\text{Area of Trapezium}
+
\text{Area of Triangle}
\]
Step 1: Calculate the area of trapezium $ABDE$.
The parallel sides of the trapezium are:
\[
AB = 18 \text{ cm}
\]
and
\[
ED = 12 \text{ cm}
\]
The perpendicular distance between these two parallel sides is given as
\[
h = 10 \text{ cm}
\]
Using the formula:
\[
\text{Area of Trapezium}
=
\frac{1}{2}
(\text{Sum of Parallel Sides})
\times
\text{Height}
\]
Substituting the values:
\[
\text{Area of Trapezium}
=
\frac{1}{2}
(18+12)
\times
10
\]
\[
=
\frac{1}{2}
\times 30
\times 10
\]
\[
=
15\times10
\]
\[
=
150 \text{ sq. cm.}
\]
Thus,
\[
\boxed{\text{Area of Trapezium }ABDE=150\text{ sq. cm.}}
\]
Step 2: Calculate the area of right-angled triangle $BCD$.
It is given that
\[
\angle BCD=90^\circ
\]
Therefore, sides $BC$ and $CD$ are perpendicular.
Given:
\[
BC=9\text{ cm}
\]
\[
CD=8\text{ cm}
\]
The area of a right-angled triangle is
\[
\text{Area}
=
\frac{1}{2}
\times
\text{Base}
\times
\text{Height}
\]
Substituting the given values:
\[
\text{Area of }\triangle BCD
=
\frac{1}{2}
\times 9
\times 8
\]
\[
=
\frac{72}{2}
\]
\[
=
36 \text{ sq. cm.}
\]
Hence,
\[
\boxed{\text{Area of }\triangle BCD=36\text{ sq. cm.}}
\]
Step 3: Calculate the total area of the figure.
The required area is the sum of the trapezium and triangle areas.
\[
\text{Area of Figure}
=
150+36
\]
\[
=
186 \text{ sq. cm.}
\]
Therefore,
\[
\boxed{\text{Area of Figure }ABCD=186\text{ sq. cm.}}
\]
Hence the correct option is
\[
\boxed{\text{(C) }186\text{ sq. cm.}}
\]
