Page 238 - ICSE Math 7

Triangle A

Consider a triangle ABC.

Perimeter of triangle = Sum of the lengths of its sides
= AB + BC + CA
1 B C
Area of a triangle = × base × height
2
Any side of a triangle can be taken as its base (b) and the length of the corresponding altitude will
be its height (h).
A A A

D
h D
h b
b h
B b D C B C B C
Area of ∆ABC Area of ∆ABC Area of ∆ABC
1 1 1
= × BC × AD = × AC × BD = × AB × CD
2 2 2

Parallelogram D C

Consider a parallelogram ABCD.
Perimeter of parallelogram ABCD = 2(l + b) l
Area of a parallelogram = base × height A b B

Any side of a parallelogram can be taken as its base (b) and the length of the corresponding altitude
will be its height (h).
D C D C

h h E
b
A E b B A B
Area of parallelogram ABCD = AB × DE Area of parallelogram ABCD = BC × DE

Each diagonal of a parallelogram divides it into two congruent triangles. Thus, the area of a parallelogram
is equal to the sum of the areas of two triangles.
In the adjoining figure, diagonal AC divides parallelogram ABCD into two congruent triangles ABC
and ADC.
∴ Area of parallelogram ABCD = Area of ∆ABC + Area of ∆ADC D C
= 2(Area of ∆ABC)

= 2(Area of ∆ADC)
A B
Similarly, diagonal BD divides parallelogram ABCD into two congruent triangles D C
ABD and CBD.
∴ Area of parallelogram ABCD = 2(Area of ∆ABD)

= 2(Area of ∆CBD) A B

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