Page 239 - ICSE Math 7

Triangles on same base and between same parallel lines
A D l
Let l and m be two parallel lines.
Let ABC and DBC be any two triangles on the same base BC and

between the same parallel lines l and m. B C m
1
Area of each triangle = 2 × base × height
Since the triangles have a common base and are between the same parallels, therefore have equal heights.
Thus, we conclude that the triangles on the same base and between the same parallels are equal in area.

Example 8: Find the area of the following.
D C D X
G
6 cm 5 cm
(a) (b) (c) 3 cm
E 3.2 cm
4.8 cm
Y 4 cm Z L
A B E F
Solution: (a) Area of parallelogram = b × h = BC × AE = 6 cm × 4.8 cm = 28.8 cm 2
1
(b) We know, area of triangle = × base × height
2
1 1
ar (∆ DEF) = × DF × EG = × 5 cm × 3.2 cm = 8.0 cm 2
2 2
1 1 1
(c) ar (∆ XYZ) = × base × height = × YZ × XL = × 4 cm × 3 cm = 6 cm 2
2 2 2
Example 9: PQRS is a parallelogram. QM is the height from Q
Q to SR and QN is the height from Q to PS. If P
SR = 14 cm and QM = 6.4 cm, find:
(a) the area of parallelogram PQRS 6.4 cm
(b) QN, if PS = 8 cm N
Solution: (a) Area of parallelogram PQRS = base × height S M R
= SR × QM = 14 cm × 6.4 cm = 89.6 cm 2 14 cm
2
(b) Area of parallelogram PQRS = 89.6 cm
base × height = 89.6 cm 2 2 (  base = 8 cm)
2
⇒ 8 cm × QN = 89.6 cm ⇒ QN = 89.6 cm = 11.2 cm
8 cm D
Example 10: DL and BM are the heights on sides AB and AD M C
respectively of parallelogram ABCD. If the area of the
2
parallelogram is 300 cm , AB = 30 cm and AD = 15
cm, find the length of the two altitudes BM and DL. A L B
Solution: Area of parallelogram ABCD = base × height = AB × DL Try This
300 cm = 30 cm × DL
300 cm 2 ∆ DEF is right-angled at D. DG is
∴ DL = = 10 cm perpendi cular to EF. If DE = 8 cm,
30 cm EF = 10 cm and DF = 6 cm, find
Area of parallelogram ABCD = base × height = AD × BM the area of ∆ DEF. Also find the
2
300 cm = 15 cm × BM altitude DG.
300 cm 2
∴ BM = 15 cm = 20 cm

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