ICSE Class 9 Maths Chapter 16 Area

16 - Area

Points To Remember

1. Equal figures: Two plane figures having equal area are called equal figures.

2. Congruent figures: Two plane figures having the same shape and size are called congruent figures. But two plane figures having equal areas need not be congruent.

3. Results on Area of polygon regions

(i) Parallelograms on the same base and between the same parallels are equal in area.

(ii) The area of a parallelogram is equal to the area of the rectangle on the same base and of the same altitude i.e. between the same parallels.

(iii) Triangles are the same base and between the same parallels are equal in area.

4. Some more results:

(i) Area of a parallelogram = Base \(\times\) height

(ii) Area of a triangle = \(\frac{1}{2}\) \(\times\) Base \(\times\) height

(iii) Area of trapezium = \(\frac{1}{2}\) (sum of parallel sides) \(\times\) height

(iv) Area of rhombus = \(\frac{1}{2}\) \(\times\) Product of diagonals

5. (i) If a triangle and a parallelogram are on the same base and between the same parallels, then the area of triangle is half of the area of the parallelogram.

(ii) Parallelograms on equal bases and between the same parallels are equal in area.

Teacher's Note

Understanding area helps us calculate space for construction projects, garden planning, and painting walls in our homes.

Exercise 16

Question 1

In the adjoining figure, BD is a diagonal of quad. ABCD. Show that ABCD is a parallelogram and calculate the area of parallelogram ABCD.

Given: AB = 6 cm, CD = 6 cm and BD = 8 cm, \(\angle\)ABD = \(\angle\)BDC = 90-

To prove: (i) ABCD is a parallelogram. (ii) Find the area of parallelogram ABCD.

Proof: \(\because\) \(\angle\)ABD = \(\angle\)BDC (each = 90-)

But these are alternate angles.

\(\therefore\) AB \(\parallel\) DC

But AB = DC = 6 cm

\(\therefore\) ABCD is a parallelogram.

Now Area = Base \(\times\) Altitude = 6 \(\times\) 8 cm\(^2\) = 48 cm\(^2\) Ans.

Sol. Given: BD is the diagonal of quadrilateral ABCD

Teacher's Note

Parallelograms appear in bridge designs and floor tile patterns we see daily.

Question 2

In a parallelogram ABCD, it is given that AB = 16 cm and the altitudes corresponding to the sides AB and AD are 6 cm and 8 cm respectively. Find the length of AD.

Sol. In parallelogram ABCD, AB = 16 cm, altitudes on AB and AD are DE and BF are drawn and DE = 7 cm, BF = 8 cm

Area of parallelogram ABCD = Base \(\times\) altitude = AB \(\times\) DE = 16 \(\times\) 6 = 96 cm\(^2\) ...(i)

Again area of parallelogram = AD \(\times\) BF = AD \(\times\) 8 cm\(^2\) ...(ii)

From (i) and (ii)

8 AD = 96 \(\Rightarrow\) AD = \(\frac{96}{8}\) = 12 cm Ans.

In trapezium ABCD

AB \(\parallel\) DC and DL \(\perp\) AB

AB = 10 cm, DC = 8 cm and DL = 6 cm

Area of trapezium ABCD = \(\frac{\text{Sum of parallel sides}}{2}\) \(\times\) height = \(\frac{(10 + 8)}{2}\) \(\times\) 6cm\(^2\) = \(\frac{18}{2}\) \(\times\) 6 = 54 cm\(^2\) Ans.

Teacher's Note

Trapeziums are found in roof designs and road embankments in construction work.

Question 3

Find the area of a rhombus, the lengths of whose diagonals are 18 cm and 24 cm respectively.

Sol. Let the first diagonal of rhombus (d\(_1\)) = 18 cm and second diagonal (d\(_2\)) = 24 cm

\(\therefore\) Area = \(\frac{d_1 \times d_2}{2}\) = \(\frac{18 \times 24}{2}\) cm\(^2\) = 216 cm\(^2\) Ans.

Question 4

Find the area of a trapezium whose parallel sides measure 10 cm and 8 cm respectively and the distance between these sides is 6 cm.

Teacher's Note

Rhombus shapes appear in diamond lattice patterns and decorative tile work.

Question 5

Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram, divides it into two equal parallelograms.

Sol. Given: In parallelogram ABCD, P and Q are the mid points of sides AB and DC respectively. PQ is joined.

To prove: APQD and PBCQ are parallelograms of equal areas.

Proof: \(\because\) P and Q are mid points of AB and DC respectively.

\(\therefore\) AP = PB and DQ = QC

But AB \(\parallel\) DC (opposite sides of parallelogram)

\(\therefore\) AP \(\parallel\) DQ and AP = DQ

\(\therefore\) APQD is a parallelogram.

Similarly PBCQ is a parallelogram.

\(\therefore\) parallelograms APQD and PBCQ are on the equal bases and between the same parallel lines.

\(\therefore\) area of parallelogram APQD = area of parallelogram PBCQ

Hence APQD and PBCQ are parallelograms of equal areas.

Question 6

In the given figure, the area of parallelogram ABCD is 90 cm\(^2\). State giving reasons:

(i) ar (parallelogram ABEF) (ii) ar (\(\triangle\)ABD) (iii) ar (\(\triangle\)BEF).

Sol. Area of parallelogram ABCD = 90 cm\(^2\)

AF and BE are drawn and BD and BF are joined.

\(\therefore\) ABEF is a parallelogram.

(i) Now parallelogram ABCD and parallelogram ABEF are on the same base and between the same parallel lines.

\(\therefore\) area of parallelogram ABCD = area of parallelogram ABEF

But area of parallelogram ABCD = 90 cm\(^2\)

\(\therefore\) Area of parallelogram ABEF = 90 cm\(^2\)

(ii) \(\therefore\) BD and BF are the diagonals of parallelogram ABCD and parallelogram ABEF respectively and diagonals of a parallelogram bisect it into two triangles of equal area.

\(\therefore\) Area (\(\triangle\)ABD) = \(\frac{1}{2}\) area (parallelogram ABCD) = \(\frac{1}{2}\) \(\times\) 90 cm\(^2\) = 45 cm\(^2\)

(iii) and area (\(\triangle\)BEF) = \(\frac{1}{2}\) area (parallelogram ABEF) = \(\frac{1}{2}\) \(\times\) 90 cm\(^2\) = 45 cm\(^2\) Ans.

Teacher's Note

Understanding how lines divide parallelograms helps in designing floor tiles and patterned wallpapers.

This is a preview of the first 3 pages. To get the complete book, click below.