Area of Polygons | IT

Question 11

Line $l \parallel \text{BC}$ . Consider the different triangles that have BC as their base, and with their third vertex lying anywhere on $l$ .

(i) Which of these triangles has the maximum area, and which has the minimum area?

(ii) Which of these triangles has the maximum perimeter, and which has the minimum perimeter?

Solution

The area of a triangle is half of its base multiplied by its height. The perimeter is the sum of its three sides.

Step 1 — Understanding Area

Let us consider any triangle with base BC and its third vertex on line $l$ . The base of all these triangles is the segment BC. This length is constant. Line $l$ is parallel to line BC. This means the perpendicular distance between $l$ and BC is always the same. This constant perpendicular distance is the height of all such triangles. The formula for the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$ . Since both the base (BC) and the height (distance between $l$ and BC) are constant, the area of all these triangles will be the same. Therefore, there is no maximum or minimum area. All triangles have the same area.

$\boxed{\text{All triangles have the same area}}$

Diagram 1

Step 2 — Understanding Perimeter

Let us consider the perimeter of these triangles. The perimeter of a triangle is the sum of the lengths of its three sides. For any triangle with base BC and a third vertex P on line $l$ , the perimeter is $\text{BC} + \text{PB} + \text{PC}$ . The length of the base BC is constant. So, we need to find when the sum of the other two sides, PB + PC, is maximum or minimum.

Step 3 — Minimum Perimeter

Let M be the midpoint of the base BC. Let $P_M$ be the point on line $l$ that is directly above M. This means the line segment $MP_M$ is perpendicular to both $l$ and BC. In triangle $P_M BC$ , the sides $P_M B$ and $P_M C$ are equal in length. This makes $\triangle P_M BC$ an isosceles triangle. The sum of the lengths of the two sides $PB + PC$ is smallest when P is this point $P_M$ . Any other point P' on line $l$ will result in $P'B + P'C$ being greater than $P_M B + P_M C$ . So, the minimum perimeter occurs when the third vertex is directly above the midpoint of BC.

$\boxed{\text{Minimum perimeter when vertex is above midpoint of BC}}$

Step 4 — Maximum Perimeter

Line $l$ extends infinitely in both directions. The third vertex P can be anywhere on line $l$ . If we move the vertex P further and further away from the segment BC along line $l$ (either to the left or to the right), the lengths of the sides PB and PC will become larger and larger. There is no limit to how far P can be from B and C. This means the sum PB + PC can be arbitrarily large. Therefore, there is no maximum perimeter for these triangles.

$\boxed{\text{No maximum perimeter}}$

Answer

(i) All these triangles have the same area. Therefore, there is no maximum area and no minimum area. (ii) The triangle with the minimum perimeter is the one whose third vertex is directly above the midpoint of BC. There is no maximum perimeter because the third vertex can be arbitrarily far along line $l$ .

More questions in IT

Try to think of different creative ways to divide a square into 4 parts of equal area.

Why Can't Perimeter be a Measure of Area?

Why do we count the number of unit squares to assign measures for area? Couldn't we have just used the perimeter of a region, i.e., the length of its boundary as a measure of its area?

Context: Consider two regions, Region 1 and Region 2, such that Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 < Area of Region 2.

Q. Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this.

Also give an example of two regions of other shapes, where the region with the larger perimeter has the smaller area! This property should be visually clear in your example.

In the given figure, which triangle has a greater area: $\Delta$ XDC or $\Delta$ YDC, if both the rectangles are identical?

In the given figure, which triangle has a greater area: $\Delta$ XDC or $\Delta$ YBC, if both the rectangles are identical?

Find the area of $\Delta$ XDC.

To find the area of a triangle, what measurements do we need?

How do we get the outer rectangle from the given triangle?

Q10

Will this formula hold for the kind of triangle, around which we cannot draw a rectangle with BC as the base?

Q11

Line $l \parallel \text{BC}$ . Consider the different triangles that have BC as their base, and with their third vertex lying anywhere on $l$ .

(i) Which of these triangles has the maximum area, and which has the minimum area?

(ii) Which of these triangles has the maximum perimeter, and which has the minimum perimeter?

Q12

Analyse whether A lies on the perpendicular bisector of BC.

Q13

Area of any Polygon

How do we find the area of this quadrilateral? What measurements do we need for this?

Q14

How do we find the area of this pentagon?

Q15

Can any polygon be divided into triangles?

Q16

Give a method to convert a parallelogram into a rectangle of equal area.

You can try this using a cut-out of a parallelogram.

Q17

Can $\Delta\text{AXD}$ and $\text{ABCX}$ fit together, as shown in the figure, to get a rectangle?

Q18

Try working this out!

Q19

What are the sidelengths of the rectangle WXYZ?

Q20

Area of rhombus ABCD can also be determined by finding the areas of $\Delta\text{ADB}$ and $\Delta\text{CDB}$ . What formula does this give us?

Q21

Context: The area of rhombus ABCD can be written as: $\text{Area of rhombus ABCD} = \text{Area}(\Delta\text{ADB}) + \text{Area}(\Delta\text{CDB}) = \frac{1}{2} \times \text{AO} \times \text{BD} + \frac{1}{2} \times \text{CO} \times \text{BD}$

Q. Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals.

Q22

Find the areas of the following trapeziums by breaking them into figures whose areas can be computed.

Q23

Will this formula hold for a trapezium that looks like this?

Q24

Will Approach 2 work for any type of trapezium?

Q25

What figure will we get when the two trapeziums are joined along BC?

Q26

What type of a quadrilateral is this?

Q27

What do you think is the area of an A4 sheet? Its sidelengths are 21 cm and 29.7 cm. Now find its area.

Q28

What do you think is the area of the tabletop that you use at school or at home? You could perhaps try to visualise how many A4 sheets can fit on your table.

Q29

Express the following lengths in centimeters:

(i) 5 in

(ii) 7.4 in

Q30

Express the following lengths in inches:

(i) 5.08 cm

(ii) 11.43 cm

Q31

How many $\text{cm}^2$ is $1\text{ in}^2$ ?

Q32

Context: Convert $161.29\text{ cm}^2$ to $\text{in}^2$ . Every $6.4516\text{ cm}^2$ gives an $\text{in}^2$ . Hence, $161.29\text{ cm}^2 = \frac{161.29}{6.4516}\text{ in}^2$ .