Maharashtra Board Class 8 Maths part 2 Chapter 15 Area PDF Download

Area

Let's Recall

We know that, if sides of closed polygon are given in the units cm, m, km then their areas are in the units sq cm, sq m and sq km respectively; because the area is measured by squares.

Area of square = side²

Area of rectangle = length × breadth

Area of right angled triangle = \(\frac{1}{2}\) × product of sides making right angle

Area of triangle = \(\frac{1}{2}\) × base × height

Teacher's Note

Area is how much space a shape takes. Think of a square tile on your bathroom floor - that one tile has an area of 1 square unit.

Exam Trick

Remember: Area always has "square" in the unit - like sq cm, sq m. If you forget this in exam, your answer is wrong even if numbers are right.

Points to Remember

For a square, multiply side × side to get area.
For a rectangle, multiply length × breadth to get area.
For a triangle, multiply base × height and divide by 2.
Always write "sq" with your unit (sq cm, sq m, etc.).

Area of a Parallelogram

Activity:

Draw a big enough parallelogram ABCD on a paper as shown in the figure.

Draw perpendicular AE on side BC.

Cut the right angled triangle AEB. Join it with the remaining part of parallelogram ABCD as shown in the figure.

Note that the new figure formed is a rectangle.

The rectangle is formed from the parallelogram So areas of both the figures are equal.

Base of parallelogram is one side (length) of the rectangle and its height is the other side (breadth) of the rectangle.

Area of parallelogram = base × height

Teacher's Note

A parallelogram looks like a rectangle that is tilted. You can imagine pushing a rectangle to one side - that makes a parallelogram. It still has the same area.

Exam Trick

For a parallelogram, always use the height that is perpendicular (at 90 degrees) to the base. Do not use the slanted side as height.

Points to Remember

A parallelogram has opposite sides equal.
The height must be perpendicular to the base.
Area = base × height (same as rectangle).
Different bases can give different heights, but area stays the same.

Remember that, if we consider one of the parallel sides of a parallelogram as a base then the distance between these parallel sides is the height of the parallelogram corresponding to the base.

ABCD is a parallelogram.

seg DP ⊥ side BC, seg AR ⊥ side BC.

If side BC is a base then height = l(AR) = l(DP) = h.

If seg CQ ⊥ side AB and if we consider seg AB as a base then corresponding height is l(QC) = k.

A(parallelogram ABCD) = l(BC) × h = l(AB) × k.

Solved Examples

Ex. (1) If base of a parallelogram is 8 cm and height is 5 cm then find its area.

Solution: Area of a parallelogram = base × height = 8 × 5 = 40

Area of the parallelogram is 40 sq.cm

Ex. (2) If Area of a parallelogram is 112 sq cm and base of it is 10 cm then find its height

Solution: Area of a parallelogram = base × height

112 = 10 × height

\(\frac{112}{10}\) = height

Height of the parallelogram is 11.2 cm

Practice Set 15.1

1. If base of a parallelogram is 18 cm and its height is 11 cm, find its area.

2. If area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.

3. Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.

Teacher's Note

In real life, parallelogram shapes are seen in designs, tiles, and even in bridges. The Maharaja's palace doors sometimes have parallelogram patterns.

Exam Trick

If you know area and base, divide area by base to find height. If you know area and height, divide area by height to find base.

Points to Remember

Area = base × height for a parallelogram.
Height is always perpendicular to base.
You can choose any side as base.
When you change the base, the height also changes.

Area of a Rhombus

Activity: Draw a rhombus as shown in the adjacent figure. We know that diagonals of a rhombus are perpendicular bisectors of each other.

Let l(AC) = d₁ and l(BD) = d₂

ABCD is a rhombus. Its diagonals intersect in the point P. So we get four congruent right angled triangles. Sides of each right angled triangle are \(\frac{1}{2}\)l(AC) and \(\frac{1}{2}\)l(BD). Areas of all these four triangles are equal.

l(AP) = l(PC) = \(\frac{1}{2}\)l(AC) = \(\frac{d_1}{2}\),

and l(BP) = l(PD) = \(\frac{1}{2}\)l(BD) = \(\frac{d_2}{2}\)

Area of rhombus ABCD = 4 × A(triangle APB)

= \(4 × \frac{1}{2} × l(AP) × l(BP)\)

= \(2 × \frac{d_1}{2} × \frac{d_2}{2}\)

= \(\frac{1}{2} × d_1 × d_2\)

Area of a rhombus = \(\frac{1}{2}\) × product of lengths of diagonals.

Solved Examples

Ex. (1) Lengths of the diagonals of a rhombus are 11.2 cm and 7.5 cm respectively. Find the area of rhombus.

Solution: Area of a rhombus = \(\frac{1}{2}\) × product of lengths of the diagonals

= \(\frac{1}{2}\) × 11.2 × 7.5 = 5.6 × 7.5

= 42 sq cm

Teacher's Note

A rhombus is like a diamond shape. Think of a kite - it has two diagonals that cross each other at right angles, just like a rhombus.

Exam Trick

For a rhombus, you only need the two diagonals. Multiply them and divide by 2. You do not need the side length.

Points to Remember

All four sides of a rhombus are equal in length.
The two diagonals are perpendicular to each other.
Diagonals bisect each other (cut each other in half).
Area = \(\frac{1}{2}\) × diagonal₁ × diagonal₂.

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