Page 242 - Start Up Mathematics_7

1
(c) area = × 80 × 31.4 = 1,256 mm 2
2

1 341
(d) × b × 15.5 = 170.5 ⇒ b = = 22 cm
2 15.5
Example 12: PQRS is a parallelogram. QM is the height from
Q to SR and QN is the height from Q to PS. If P Q
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
= 14 × 6.4 = 89.6 cm 2
2
(b) Area of parallelogram PQRS = 89.6 cm
base × height = 89.6 cm 2 (  base = 8 cm)
89.6
⇒ 8 × QN = 89.6 ⇒ QN = = 11.2 cm D C
8
Example 13: DL and BM are the heights on sides AB and AD M

respectively of parallelogram ABCD. If the area
2
of the parallelogram is 300 cm , AB = 30 cm
and AD = 15 cm, find the length of the two A L B
altitudes BM and DL.

Solution: Area of parallelogram ABCD = base × height = AB × DL
300 = 30 × DL

∴ DL = 300 = 10 cm
30
Area of parallelogram ABCD = base × height = AD × BM
300 = 15 × BM
300
∴ BM = = 20 cm
15
Example 14: ∆ DEF is right-angled at D. DG is perpendi cular to EF. If DE = 8 cm, EF = 10 cm
and DF = 6 cm, find the area of ∆ DEF. Also find the altitude DG.
D
Solution: In right-angled ∆ DEF, let DE be the base and DF its
corresponding altitude.
1 8 cm 6 cm
∴ Area of ∆ DEF = × DE × DF
2
1
= × 8 × 6 = 24 cm 2 E G F
2 10 cm
1
Area of ∆ DEF = × EF × DG (Base = EF, altitude = DG)
2
1
⇒ 24 = × 10 × DG
2
24 × 2 48
∴ DG = 10 = 10 = 4.8 cm

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