Area - Chapter 7 Class 8 (Ganita Prakash II)

Identify the missing sidelengths.Here, from area of rectangle formula we need to find sides

The figure shows a path (the shaded portion) laid around a rectangular park EFGH. (i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. An example of a formula — Area of a rectangle = length × width. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.]Now,
Area of path
= Area of rectangle ABCD – Area of rectangle EFGH

The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.Let’s label the diagram and mark cross paths properly
Thus, we need to width of each path – p & q
Let’s assume width of each part is same, and it is 2 m

Find the area of the spiral tube shown in the figure. The tube has the same width throughout. [Hint: There are different ways of finding the area. Here is one method.] What should be the length of the straight tube if it is to have the same area as the bent tube on the left? Let’s look at our hint.
It tells us if we straighten our tube
Thus, if we straighten our tube
The area at the turn gets counted twice, so we remove it once
And, Area of turn = Area of Square of side 1
= 1 × 1
= 1 square unit

In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.Let side of square be a
Thus,
Original area of square = a × a = a2

Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside. You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials.Let’s do this activity
When you divide a square using two perpendicular lines, you can always rearrange those 4 pieces to form a larger, tilted square with a hole in the center.
The Math Behind It
Let's use a
square (Area
).
We drew our perpendicular lines offset by
units. The Pythagorean theorem tells us the cut length is
.
When rearranged, the long diagonal cuts form the outer boundary of a new
square (Area
).
Where does the extra area come from? The original rightangled corners are pushed to the center, perfectly forming a
hole (Area
)!

(i) Find the areas of the following triangles:Here, AE ⊥ BC

Find the length of the altitude BY.To find BY, we find Area of ∆ ABC using
Height AX, Base BC
Or Height BY, Base AC

Find the area of ΔSUB, given that it is isosceles, SE is perpendicular to UB, and the area of ΔSEB is 24 sq. units.We use the property:
In an Isosceles triangle, altitude bisects the side

In the Śulba-Sūtras, which are ancient Indian geometric texts that deal with the construction of altars, we can find many interesting problems on the topic of areas. When altars are built, they must have the exact prescribed shape and area. This gives rise to problems of the kind where one has to transform a given shape into another of the same area. The Śulba-Sūtras give solutions to many such problems. Such problems are also posed and solved in Euclid’s Elements. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.To turn it into a triangle with the exact same area, you can keep the base exactly the same length, but you must make the height of the triangle twice as tall as the rectangle.
(Because the triangle formula halves the area, doubling the height perfectly balances it out!).

In the Śulba-Sūtras, which are ancient Indian geometric texts that deal with the construction of altars, we can find many interesting problems on the topic of areas. When altars are built, they must have the exact prescribed shape and area. This gives rise to problems of the kind where one has to transform a given shape into another of the same area. The Śulba-Sūtras give solutions to many such problems. Such problems are also posed and solved in Euclid’s Elements. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.Keep the base of the triangle the same, but draw your rectangle so its height is exactly half the height of the triangle.

ABCD, BCEF, and BFGH are identical squares. (i) If the area of the red region is 49 sq. units, then what is the area of the blue region?Let’s assume side of each square is s
Area of Red Shape
Red Shape is a right angled triangle ∆ DCH with
Base = CD = s
Height = HC = s + s = 2s
Area = 49 sq units

If M and N are the midpoints of XY and XZ, what fraction of the area of ΔXYZ is the area of ΔXMN? [Hint: Join NY]This problem uses a fundamental rule of geometry: If two triangles have equal bases and share the same height, they have the exact same area.

Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.This is a real-world application of the "mirror proof" from the previous pages!

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.Quadrilateral ABCD is divided into two triangles ∆ ADC & ∆ ABC
So, we can write
Area of Quadrilateral ABCD = Area ∆ ADC + Area ∆ ABC

Find the area of the shaded region given that ABCD is a rectangle.Now, we can find Area like this
Area of shaded region = Area of ABCD − Area of ∆AEF − Area of ∆BEC
Area of ABCD

What measurements would you need to find the area of a regular hexagon?A "regular" hexagon is special because it can be divided perfectly into 6 identical equilateral triangles meeting at the center point.

What fraction of the total area of the rectangle is the area of the blue region?We label the rectangle and the Triangle
Let Length of Rectangle = l
Breadth of Rectangle = b

Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral. We follow these steps
Draw a diagonal line connecting two opposite corners. (Now you have two triangles).
Find the exact midpoint (center dot) of that diagonal line.
Draw a line from the other two corners to that center dot.

Find the area of a rhombus whose diagonals are 20 cm and 15 cm.Given Diagonals,
d1 = 20 cm
& d2 = 15 cm

Give a method to convert a rectangle into a rhombus of equal area using dissection.Let’s look at it Step-by-Step
Rectangle to Rhombus
A geometric dissection proving area equivalence by physically converting a rectangle into a rhombus.
STEP 1 OF 6
The Starting Rectangle
Start with a rectangle ABCD .
Let the total width be
, and the height be h.
The area of this rectangle is strictly
.
Previous
Next Step
Rectangle to Rhombus
A geometric dissection proving area equivalence by physically converting a rectangle into a rhombus.
STEP 2 OF 6
Mark and Cut
Mark the exact midpoint
on the top edge.
Draw diagonal lines from M to the bottom left (D) and bottom right (C) corners. This cleanly divides the rectangle into three separate triangles.
Previous
Next StepRectangle to Rhombus
A geometric dissection proving area equivalence by physically converting a rectangle into a rhombus.
STEP 3 OF 6
Isolate the Pieces
We now have three pieces:
A large central isosceles triangle (Indigo).
Two identical smaller right-angled triangles on the sides (Blue).
Notice the legs of the blue triangles are
and w/2.
Previous
Next StepRectangle to Rhombus
A geometric dissection proving area equivalence by physically converting a rectangle into a rhombus.
STEP 4 OF 6
The Cross-Slide
Now for the magic. We will take the left triangle and slide it down to the bottom right. We will take the right triangle and slide it down to the bottom left.
Their vertical edges (height
) will perfectly merge together in the center!
Previous
Next StepRectangle to Rhombus
A geometric dissection proving area equivalence by physically converting a rectangle into a rhombus.
STEP 5 OF 6
The Resulting Rhombus
The pieces interlock flawlessly to form an inverted isosceles triangle attached to the base of the first one.
Because all four outer edges are the hypotenuses of identical right-angled triangles, they are exactly the same length. This shape is a perfect Rhombus!
Previous
Next Step
Rectangle to Rhombus
A geometric dissection proving area equivalence by physically converting a rectangle into a rhombus.
STEP 6 OF 6
Area Equivalence Proof
No pieces were added or removed, so the Area remains exactly the same as the original rectangle.
Let's verify this using the Rhombus area formula.
Area of Rectangle
Rhombus Diagonals:
Area of Rhombus

(i) Find the areas of the following figures:Rotating it, we notice that this is a Trapezium with
Parallel sides 10 ft and 7 ft, i..e
a = 10 ft & b = 7ft

[Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.Method
In an isosceles trapezium, the two slanted sides are equal. Drop a perpendicular line from one top vertex to the base. Cut off the resulting right-angled triangle.

Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area — Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH? [Hint: If ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ, then the trapezium and rectangle have equal areas.]Let’s first construct it, and to prove Areas equal – we follow the hint
Trapezium to Rectangle
A geometric dissection converting a trapezium into a rectangle to prove the area formula.
STEP 1 OF 10
The Original Trapezium
Start with trapezium ABCD .
Let the top base be a and the bottom base be
. The height is
.
Our goal is to construct a rectangle of exactly the same area.
Previous
Next Step
Trapezium to Rectangle
A geometric dissection converting a trapezium into a rectangle to prove the area formula.
STEP 2 OF 10
Finding the Midpoints
Find the exact midpoint I on the left edge AD , and midpoint J on the right edge BC .
Draw straight, vertical lines down through these midpoints.
Previous
Next Step
Trapezium to Rectangle
A geometric dissection converting a trapezium into a rectangle to prove the area formula.
STEP 3 OF 10
The Target Rectangle
These vertical lines form the boundaries of our target rectangle EFGH.
Notice the pieces of the trapezium sticking out at the bottom (
and
), and the empty spaces missing from the rectangle at the top (
and
).
Previous
Next Step
Trapezium to Rectangle
A geometric dissection converting a trapezium into a rectangle to prove the area formula.
STEP 4 OF 10
Proving Congruence
Why will they fit perfectly? Focus on the left edge.
Because
is the midpoint, the segment AI = ID (indicated by the tick marks). The top and bottom bases are parallel, meaning their alternating inner angles are equal.
By geometric rules (ASA), this proves that the bottom piece
is exactly identical in size and shape to the empty top space ΔAHI!
Previous
Next Step
Trapezium to Rectangle
A geometric dissection converting a trapezium into a rectangle to prove the area formula.
STEP 5 OF 10
Conservation of Area
Since the triangles are identical,
.
Total Trapezium Area
Purple Center
Bottom Triangles.
Target Rectangle Area
(Purple Center)
Top Empty Spaces.
Because the pieces are equal, replacing the bottom pieces with the top spaces guarantees the total area will not change!
Previous
Next Step

Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm2.Let’s do it Step-by-Step
Rectangle to Trapezium
Reversing the dissection method to construct a trapezium with an exact area of
.
STEP 1 OF 6
The Target Area (
)
We want a trapezium with exactly 144
of area.
Let's work backward! We start by drawing a rectangle that already has an area of 144
.
Width

A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.We convert the Trapezium and the Rhombus into Equilateral triangles
Hexagon Division
Finding the ratio of the areas of a Trapezium, an Equilateral Triangle, and a Rhombus inside a Regular Hexagon.
STEP 1 OF 6
The Divided Hexagon
We are given a regular hexagon divided into three distinct shapes:
A Trapezium (Top)
A Rhombus (Bottom-Left)
An Equilateral Triangle (BottomRight)
Previous
Next StepHexagon Division
Finding the ratio of the areas of a Trapezium, an Equilateral Triangle, and a Rhombus inside a Regular Hexagon.
STEP 2 OF 6
The Geometric Secret
To solve this without complex math, we use a fundamental property:
Every regular hexagon can be perfectly divided into exactly 6 identical equilateral triangles by connecting its vertices to the center.
Let's say 1 triangle
Unit of Area.
Previous
Next StepHexagon Division
Finding the ratio of the areas of a Trapezium, an Equilateral Triangle, and a Rhombus inside a Regular Hexagon.
STEP 3 OF 6
The Equilateral Triangle
Look at the shape in the bottom right.
It perfectly perfectly outlines exactly one of the constituent equilateral triangles of the hexagon grid.
Therefore, its Area = 1 unit.
Previous
Next Step
Hexagon Division
Finding the ratio of the areas of a Trapezium, an Equilateral Triangle, and a Rhombus inside a Regular Hexagon.
STEP 4 OF 6
2. The Rhombus
Now look at the shape in the bottom left.
The grid shows us that this rhombus is formed by joining exactly two adjacent equilateral triangles together.
Therefore, its Area = 2 units.
Previous
Next Step
Hexagon Division
Finding the ratio of the areas of a Trapezium, an Equilateral Triangle, and a Rhombus inside a Regular Hexagon.
STEP 6 OF 6
3. The Trapezium
Finally, look at the large shape across the top.
This trapezium occupies exactly half of the entire hexagon. By counting the grid pieces, we see it is formed by three adjacent triangles.
Therefore, its Area = 3 units.
Previous
Next Step
Hexagon Division
Finding the ratio of the areas of a Trapezium, an Equilateral Triangle, and a Rhombus inside a Regular Hexagon.
STEP 6 OF 6
The Final Ratio
Because all the shapes are built from the exact same fundamental building blocks (the 6 identical triangles), their area ratio is simply the ratio of how many blocks they contain!
Final Ratio
Trapezium: 3 units
Triangle: 1 unit
Rhombus: 2 units
Ratio
Previous Restart Analysis

ZYXW is a trapezium with ZY‖WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ΔZWB.Let’s look at the answer in detail
Trapezium to Triangle
Show that Area of Trapezium ZYXW = Area of
.
STEP 1 OF 6
Step 1: Identifying the Midpoint
Look at
and
.
Point
is the midpoint of
, so
XA.
Previous
Next StepTrapezium to Triangle
Show that Area of Trapezium ZYXW = Area of
.
STEP 2 OF 6
Step 2: Angles & Congruence
Let's prove they are congruent using the ASA (Angle-Side-Angle) rule:
Angle (Red): Vertically opposite angles at A are equal.
Side (Ticks):
( A is the midpoint).
Angle (Blue): Since
, the alternate interior angles at Y and X are equal.
Because of ASA,
.
Previous
Next StepTrapezium to Triangle
Show that Area of Trapezium ZYXW = Area of
IWB.
STEP 3 OF 6
Step 3: Equal Area
Since they are congruent, they have the exact same area.
Previous
Next StepTrapezium to Triangle
Show that Area of Trapezium ZYXW = Area of
.
STEP 4 OF 6
Step 4: Trapezium Composition
The Trapezium ZYXW is made of the quadrilateral
.
Previous
Next Step
Trapezium to Triangle
Show that Area of Trapezium ZYXW = Area of
.
STEP 5 OF 6
Step 5: Triangle Composition
The large triangle
is made of the exact same quadrilateral
.
Previous
Next Step
Trapezium to Triangle
Show that Area of Trapezium ZYXW = Area of
.
STEP 6 OF 6
Conclusion
Since we just replaced one equal triangle with another, the total areas must be the same!
Trapezium