Page 276 - Start Up Mathematics_8 (Non CCE)

6. Find the area of the given figure. 4 cm
20 cm 5 cm 20 cm

16 cm
7. The ratio of the lengths of parallel sides of a
trapezium is 3 : 5. The distance between them is 30 cm
2
12 cm. If the area of the trapezium is 720 cm , Y 10 cm X
find the lengths of the parallel sides.
7 cm
12 cm
T 5 cm U
8. Find the area of the given figure.
8 cm
S R
5 cm
P Q

Area of a Polygon
Area of a regular or irregular polygon can be found by using the formulae for finding the areas of a triangle,
parallelogram and trapezium. The idea is to divide the given polygon into non-overlapping rectilinear plane
figures whose areas can be found easily. The area of the polygon will be equal to the sum of the areas of the
non-overlapping parts.
Example 17: Find the area of a quadrilateral SING in which diagonal SN = 15 cm and the lengths of the
perpendiculars from G and I on SN are 3 cm and 5 cm respectively.
N
Solution: GH ^ SN and IJ ^ SN G
GH = 3 cm, IJ = 5 cm 3 cm J
SN = 15 cm H 15 cm 5 cm
1
Area of quadrilateral SINK = × SN × (GH + IJ) S
2 I
1 1
2
2
= × 15 × (3 + 5) cm = × 15 × 8 cm = 60 cm 2
2 2
R
Example 18: Find the area of hexagon STRAIN, if SA = 10 cm, T
SL = 8 cm, SP = 7 cm, SM = 5 cm, SU = 3 cm,
TU = 4 cm, RP = 6 cm, LI = 3 cm and MN = 5 cm. 4 cm 6 cm

Solution: Area of hexagon STRAIN = (Area of D SUT) S 3 cm U M P L 3 cm A
+ (Area of trapezium TUPR) + (Area of D RPA)
+ (Area of D SMN) + (Area of trapezium MNIL) 5 cm I
+ (Area of D ILA) N

=

Ê 1 ˜ { 1 } { 1 }
ˆ
¥
= Á 2 ¥SU TU ¯ + 2 ¥ TU RP( + ) ¥ SP SU( - ) + 2 ¥ RP ¥ SA SP( - )
Ë
Ê 1 ˜ { 1 } { 1 }
ˆ
¥
-
(
+ Á ¥SM MN ¯ + 2 ¥ MN LI( + ) ¥ SL SM( - ) ¥ 2 ¥ SA SL) ¥ LI
Ë 2
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