Measuring Space: Perimeter and Area | EOT

Question 20

Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Solution

Understand the Question

Area Equality: Two triangles have equal areas if they share the same base length and have the same perpendicular height, since $\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}$ .
Dissection Principle (Conservation of Area): If one figure can be cut into a set of pieces that can be rearranged without gaps or overlaps to form another figure, both figures must have the exact same area.

Step 1 · Compare Bases and Heights of Both Triangles

Let the base of the blue triangle be $b_1$ and its perpendicular height be $h_1$ . Let the base of the red triangle be $b_2$ and its perpendicular height be $h_2$ .

Since both triangles have equal bases ( $b_1 = b_2 = b$ ) and lie between the same parallel lines ( $h_1 = h_2 = h$ ): $\text{Area of blue triangle} = \dfrac{1}{2} \times b \times h$ $\text{Area of red triangle} = \dfrac{1}{2} \times b \times h$

Therefore: $\text{Area of blue triangle} = \text{Area of red triangle}$

Step 2 · Dissect and Rearrange the Blue Triangle

Drop a perpendicular line (altitude) from the top vertex of the blue triangle to its base, dividing it into smaller right-angled triangles.
Cut along the altitude line to obtain the individual triangular pieces.
Translate and rotate the cut pieces to align with the boundaries of the red triangle without any overlap or gap.

Since the blue pieces completely and exactly cover the red triangle, their areas are equal by dissection.

Answer

The areas are equal because both triangles have the same base and height ( $\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}$ ), and the blue triangle can be dissected along its altitude into congruent pieces that rearrange to cover the red triangle.

Common Mistakes

Assuming Different Shapes Mean Different Areas: Triangles with different slant heights or orientations can still have identical areas as long as their base and perpendicular height are equal.
Using Slant Height instead of Perpendicular Height: Always measure height as the perpendicular distance from the vertex to the line containing the base, not the length of a slanted side.

More questions in EOT

Identities in algebra can sometimes be shown as area relationships. For example:

The figure shown corresponds to the identity $(a + b)^2 = a^2 + 2ab + b^2$

Do you see how?

Draw figures corresponding to the identities $(a + b)(a - b) = a^2 - b^2$ and $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ .

An isosceles triangle has perimeter $40\text{ cm}$ ; the equal sides are $15\text{ cm}$ each. Find the area of the triangle.

An isosceles triangle has base $10\text{ cm}$ , and its area is $60\text{ cm}^2$ . What are the lengths of the equal sides?

The area of a right-angled triangle is $54\text{ sq. cm}$ . One of its legs has length $12\text{ cm}$ . Find its perimeter.

The sides of a triangle are in the ratio $2 : 3 : 4$ , and its perimeter is $45\text{ cm}$ . Find its area.

The sides of a triangle have lengths $7\text{ cm}$ , $24\text{ cm}$ , $25\text{ cm}$ . Find the area of the triangle in two different ways.

If the wheel of a bicycle has a diameter of $60\text{ cm}$ , find how far a cyclist will have travelled after the wheel has rotated $100$ times.

Find the area of a quadrant of a circle whose circumference is $66 \text{ cm}$ .

The wheel of a car has an outer radius of $28\text{ cm}$ . Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of $1\text{ km}$ .

Q10

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Q11

You know that the area of a parallelogram is base $\times$ height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides $\times$ height, i.e., $\dfrac{1}{2}(a + b)h$ .

Q12

By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).

Q13

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Q14

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

Q15

Three problems about fitting congruent shapes together:

(i) Rectangle ABCD has sides $a$ , $b$ , and rectangle PQRS has sides $2a$ , $2b$ . Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!

(ii) $\Delta ABC$ has sides $a$ , $b$ , $c$ , and $\Delta PQR$ has sides $2a$ , $2b$ , $2c$ . Show that $\Delta PQR$ has 4 times the area of $\Delta ABC$ . Does this mean that 4 copies of $\Delta ABC$ will fit into $\Delta PQR$ ? Check and see!

(iii) $\Delta ABC$ has sides $a$ , $b$ , $c$ , and $\Delta PQR$ has sides $3a$ , $3b$ , $3c$ . Show that $\Delta PQR$ has 9 times the area of $\Delta ABC$ . Does this mean that 9 copies of $\Delta ABC$ will fit into $\Delta PQR$ ? Check and see!

Q16

Find the fraction of the shaded region in each of the following figures:

(i) Fig. 6.43: What fraction of the triangle is shaded? (ii) Fig. 6.44: What fraction of the square is shaded?

Q17

Find the fraction of the rectangle covered by the circles in each of the following figures:

(i) Fig. 6.45: What fraction of the rectangle is covered by the circles?

(ii) Fig. 6.46: What fraction of the rectangle is covered by the circles?

Q18

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

Q19

*19. The figure shows nine identical rectangles fitted together to make a large rectangle whose area is $72\text{ cm}^2$ . Find the perimeter of each small rectangle.

Q20

Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Q21

The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions $A$ and $B$ .

Show that $A$ and $B$ have equal area.

Q22

In Fig. 6.50, four semicircles have been drawn within the given square whose side is $2 \text{ units}$ . The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

Q23

In Fig. 6.51 we see two concentric circles with a common centre $O$ . A chord $BC$ of the larger circle is drawn, touching the smaller circle at $A$ . The length of $BC$ is $l$ . Show that the area of the green region enclosed between the two circles is $\dfrac{1}{4} \pi l^2$ .

Q24

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that $\text{Area (A)} + \text{Area (B)} = \text{Area (C)}$ .

Q25

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius $r$ .

Q26

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are $A, B, C$ , as marked. Show that the area of the rectangle is

$\dfrac{2(A + C)(B + C)}{C}$

Q27

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.

Show that the areas of the two shaded regions are equal.