Measuring Space: Perimeter and Area | EOT

Question 27

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.

Question diagram 1
Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • Let the quarter circle AOBAOB have radius RR, with perpendicular radii OA=OB=ROA = OB = R and center OO.
  • The right-angled triangle AOBAOB has area 12R2\dfrac{1}{2}R^2.
  • By the Pythagoras theorem, the hypotenuse AB=R2AB = R\sqrt{2}. The semicircle drawn on diameter ABAB has an area equal to that of the quarter circle AOBAOB, which is 14πR2\dfrac{1}{4}\pi R^2.
  • By subtracting the circular segment from the semicircle, we can find the area of the crescent/lune region and compare it to the area of the shaded triangle region.

Step 1 · Calculate Areas of Triangle, Semicircle, and Quarter Circle

Let the radius of the quarter circle AOBAOB be RR, so OA=OB=ROA = OB = R.Diagram 1

Area of right triangle AOBAOB:

Area of ΔAOB=12×OA×OB=12×R×R=12R2\begin{aligned} \text{Area of } \Delta AOB &= \dfrac{1}{2} \times OA \times OB \\[0.6em] &= \dfrac{1}{2} \times R \times R \\[0.6em] &= \dfrac{1}{2} R^2 \end{aligned}

Using the Pythagoras theorem in ΔAOB\Delta AOB:

AB2=OA2+OB2=R2+R2=2R2AB=R2\begin{aligned} AB^2 &= OA^2 + OB^2 \\ &= R^2 + R^2 \\ &= 2R^2 \\ AB &= R\sqrt{2} \end{aligned}

Radius of the semicircle on diameter ABAB is R22=R2\dfrac{R\sqrt{2}}{2} = \dfrac{R}{\sqrt{2}}.

Area of the semicircle on ABAB:

Area of semicircle on AB=12π(R22)2=12π(2R24)=14πR2\begin{aligned} \text{Area of semicircle on } AB &= \dfrac{1}{2} \pi \left(\dfrac{R\sqrt{2}}{2}\right)^2 \\[0.6em] &= \dfrac{1}{2} \pi \left(\dfrac{2R^2}{4}\right) \\[0.6em] &= \dfrac{1}{4} \pi R^2 \end{aligned}

Area of the quarter circle AOBAOB: Area of quarter circle AOB=14πR2\text{Area of quarter circle } AOB = \dfrac{1}{4} \pi R^2

Step 2 · Calculate Area of the Outer Shaded Region (Lune)

The circular segment between chord ABAB and the quarter circle arc has area:

Area of segment AB=Area of quarter circle AOBArea of ΔAOB=14πR212R2\begin{aligned} \text{Area of segment } AB &= \text{Area of quarter circle } AOB - \text{Area of } \Delta AOB \\[0.6em] &= \dfrac{1}{4} \pi R^2 - \dfrac{1}{2} R^2 \end{aligned}

Now, calculate the area of the outer shaded region (lune on ABAB):

Area of outer shaded region=Area of semicircle on ABArea of segment AB=14πR2(14πR212R2)=14πR214πR2+12R2=12R2\begin{aligned} \text{Area of outer shaded region} &= \text{Area of semicircle on } AB - \text{Area of segment } AB \\[0.6em] &= \dfrac{1}{4} \pi R^2 - \left(\dfrac{1}{4} \pi R^2 - \dfrac{1}{2} R^2\right) \\[0.6em] &= \dfrac{1}{4} \pi R^2 - \dfrac{1}{4} \pi R^2 + \dfrac{1}{2} R^2 \\[0.6em] &= \dfrac{1}{2} R^2 \end{aligned}

Step 3 · Compare the Two Shaded Areas

The area of the shaded triangle AOBAOB is: Area of ΔAOB=12R2\text{Area of } \Delta AOB = \dfrac{1}{2} R^2

Since both shaded regions have an area equal to 12R2\dfrac{1}{2} R^2, their areas are equal.

Answer

Hence proved, both shaded regions have an equal area of 12R2\dfrac{1}{2} R^2.

Common Mistakes
  • Diameter vs. Radius of Semicircle: Forgetting to divide the hypotenuse AB=R2AB = R\sqrt{2} by 22 to obtain the radius of the semicircle on ABAB.
  • Sign Errors in Subtraction: Dropping the bracket when subtracting (14πR212R2)\left(\dfrac{1}{4}\pi R^2 - \dfrac{1}{2}R^2\right), which leads to an incorrect sign for the triangle's area.

More questions in EOT

Q1

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

The figure shown corresponds to the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Do you see how?

Draw figures corresponding to the identities (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2 and (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

Q2

An isosceles triangle has perimeter 40 cm40\text{ cm}; the equal sides are 15 cm15\text{ cm} each. Find the area of the triangle.

Q3

An isosceles triangle has base 10 cm10\text{ cm}, and its area is 60 cm260\text{ cm}^2. What are the lengths of the equal sides?

Q4

The area of a right-angled triangle is 54 sq. cm54\text{ sq. cm}. One of its legs has length 12 cm12\text{ cm}. Find its perimeter.

Q5

The sides of a triangle are in the ratio 2:3:42 : 3 : 4, and its perimeter is 45 cm45\text{ cm}. Find its area.

Q6

The sides of a triangle have lengths 7 cm7\text{ cm}, 24 cm24\text{ cm}, 25 cm25\text{ cm}. Find the area of the triangle in two different ways.

Q7

If the wheel of a bicycle has a diameter of 60 cm60\text{ cm}, find how far a cyclist will have travelled after the wheel has rotated 100100 times.

Q8

Find the area of a quadrant of a circle whose circumference is 66 cm66 \text{ cm}.

Q9

The wheel of a car has an outer radius of 28 cm28\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 km1\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., 12(a+b)h\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 aa, bb, and rectangle PQRS has sides 2a2a, 2b2b. 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) ΔABC\Delta ABC has sides aa, bb, cc, and ΔPQR\Delta PQR has sides 2a2a, 2b2b, 2c2c. Show that ΔPQR\Delta PQR has 4 times the area of ΔABC\Delta ABC. Does this mean that 4 copies of ΔABC\Delta ABC will fit into ΔPQR\Delta PQR? Check and see!

(iii) ΔABC\Delta ABC has sides aa, bb, cc, and ΔPQR\Delta PQR has sides 3a3a, 3b3b, 3c3c. Show that ΔPQR\Delta PQR has 9 times the area of ΔABC\Delta ABC. Does this mean that 9 copies of ΔABC\Delta ABC will fit into ΔPQR\Delta PQR? Check and see!

Q16
  1. 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 cm272\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 AA and BB.

Show that AA and BB have equal area.

Q22

In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units2 \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 OO. A chord BCBC of the larger circle is drawn, touching the smaller circle at AA. The length of BCBC is ll. Show that the area of the green region enclosed between the two circles is 14πl2\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 Area (A)+Area (B)=Area (C)\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 rr.

Q26

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

2(A+C)(B+C)C\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.

← Back to Measuring Space: Perimeter and Area