Measuring Space Perimeter and Area

For a circle, the circumference is $C=2\pi r$.

Given $C=44$ cm and using $\pi=\frac{22}{7}$:

$$44=2\times \frac{22}{7}\times r$$

$$44=\frac{44}{7}r$$

$$r=44\times \frac{7}{44}=7$$

So the radius is 7 cm.

Use the circumference formula $C=2\pi r$ with $\pi=\frac{22}{7}$.

### (i) $r=7$ cm
$$C=2\times \frac{22}{7}\times 7=44\text{ cm}$$

### (ii) $r=10$ cm
$$C=2\times \frac{22}{7}\times 10=\frac{440}{7}=62.857\ldots$$
Correct to 3 significant figures: 62.8 cm.

### (iii) $r=12$ cm
$$C=2\times \frac{22}{7}\times 12=\frac{528}{7}=75.428\ldots$$
Correct to 3 significant figures: 75.4 cm.

Use the arc-length formula
$$l=2\pi r\times \frac{\theta}{360^\circ}$$
with $\pi=\frac{22}{7}$.

### (i) $r=3.5$ cm, $\theta=60^\circ$
$$l=2\times \frac{22}{7}\times 3.5\times \frac{60}{360}$$
$$=2\times \frac{22}{7}\times 3.5\times \frac{1}{6}$$
$$=11\times \frac{1}{6}=1.833\ldots$$
So the arc length is 1.83 cm.

### (ii) $r=6.3$ m, $\theta=120^\circ$
$$l=2\times \frac{22}{7}\times 6.3\times \frac{120}{360}$$
$$=2\times \frac{22}{7}\times 6.3\times \frac{1}{3}$$
$$=13.2\text{ m}$$
So the arc length is 13.2 m.

The computed answers are $1.83$ cm and $13.2$ m. If you want the first rounded differently, the exact calculation is $1.833\ldots$ cm.

A sector’s perimeter = two radii + arc length.

Given $r=14$ cm and $\theta=75^\circ$.

Arc length:
$$l=2\pi r\times \frac{\theta}{360^\circ} =2\times \frac{22}{7}\times 14\times \frac{75}{360}$$
$$=88\times \frac{75}{360}=18.333\ldots\text{ cm}$$

Perimeter of sector:
$$18.333\ldots + 2(14)=18.333\ldots + 28=46.333\ldots\text{ cm}$$
So the perimeter is 46.3 cm.

The tyre’s diameter is 56 cm, so one revolution covers the circumference:
$$C=\pi d=\frac{22}{7}\times 56=176\text{ cm}$$
So in one revolution the car travels 176 cm.

For 10 km:
$$10\text{ km}=10,000\text{ m}=1,000,000\text{ cm}$$
Number of revolutions:
$$\frac{1,000,000}{176}=5681.818\ldots$$
So the tyre makes about 5682 revolutions.

If the printed question in your book also asks for a second part, this is the computed result from the given data.

The circumference of a circle is $C=2\pi r$. Since $2\pi$ is the same for both circles, the ratio of circumferences equals the ratio of radii.

So if the ratio of perimeters is $5:4$, then the ratio of radii is also 5:4.

Use $C=2\pi r$ with $\pi=\frac{22}{7}$.

### (i) $r=7$ cm
$$C=2\times \frac{22}{7}\times 7=44\text{ cm}$$

### (ii) $r=10$ cm
$$C=2\times \frac{22}{7}\times 10=\frac{440}{7}=62.857\ldots$$
Correct to 3 significant figures: 62.8 cm.

### (iii) $r=12$ cm
$$C=2\times \frac{22}{7}\times 12=\frac{528}{7}=75.428\ldots$$
Correct to 3 significant figures: 75.4 cm.

Use the arc-length formula
$$l=2\pi r\times \frac{\theta}{360^\circ}$$
with $\pi=\frac{22}{7}$.

### (i) $r=3.5$ cm, $\theta=60^\circ$
$$l=2\times \frac{22}{7}\times 3.5\times \frac{60}{360} =\frac{44}{7}\times 3.5\times \frac{1}{6} =22\times \frac{1}{6}=3.666\ldots$$
So the arc length is 3.67 cm.

### (ii) $r=6.3$ m, $\theta=120^\circ$
$$l=2\times \frac{22}{7}\times 6.3\times \frac{120}{360} =2\times \frac{22}{7}\times 6.3\times \frac{1}{3} =13.2\text{ m}$$
So the arc length is 13.2 m.

One revolution of the tyre covers its circumference.

Diameter $d=56$ cm, so
$$C=\pi d=\frac{22}{7}\times 56=176\text{ cm}$$
Thus the car travels 176 cm in one revolution.

For 10 km:
$$10\text{ km}=1,000,000\text{ cm}$$
Number of revolutions:
$$\frac{1,000,000}{176}=5681.818\ldots$$
So the tyre makes about 5682 revolutions.

The trapezium is isosceles with parallel sides 40 cm and 20 cm, and equal non-parallel sides 26 cm.

Difference of parallel sides:
$$40-20=20\text{ cm}$$
So each side overhang is
$$\frac{20}{2}=10\text{ cm}$$
Using one right triangle on the side, height $h$ is:
$$h^2=26^2-10^2=676-100=576$$
$$h=24\text{ cm}$$
Area of trapezium:
$$\frac{1}{2}(40+20)\times 24=30\times 24=720$$
So the area is 720 cm².

The triangle has sides 8 cm, 11 cm, and perimeter 32 cm, so the third side is
$$32-(8+11)=13\text{ cm}$$
Semi-perimeter:
$$s=\frac{32}{2}=16$$
By Heron’s formula:
$$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{16(16-8)(16-11)(16-13)}$$
$$=\sqrt{16\times 8\times 5\times 3} =\sqrt{1920} =8\sqrt{30} \approx 43.82$$
So the area is about 43.8 cm².

This is the computed value; it is not among any invented options if different. The exact answer is $8\sqrt{30}$ cm².

The sides are in the ratio $3:5:7$ and the perimeter is 300 m.

Let the common factor be $x$.
Then sides are $3x,5x,7x$.

$$3x+5x+7x=15x=300$$
$$x=20$$
So the sides are $60$ m, $100$ m, and $140$ m.

Semi-perimeter:
$$s=\frac{300}{2}=150$$
Heron’s formula:
$$\text{Area}=\sqrt{150(150-60)(150-100)(150-140)}$$
$$=\sqrt{150\times 90\times 50\times 10}$$
$$=\sqrt{6{,}750{,}000}=1500\sqrt{3}\approx 2598.1$$
So the area is about 2598 m².

Let the shorter diagonal be $x$ cm. Then the longer diagonal is $2x$ cm.

Area of a rhombus:
$$A=\frac{1}{2}d_1d_2$$
Given area $=128$ cm²:
$$128=\frac{1}{2}(x)(2x)=x^2$$
$$x^2=128$$
$$x=\sqrt{128}=8\sqrt{2}\approx 11.3$$
So the **shorter diagonal is $8\sqrt{2}$ cm, about 11.3 cm**.

If the question expects the exact value, that is the answer.

In a parallelogram, any triangle with vertex $C$ and base on side $AB$ has the same height from $C$ to line $AB$.

So
$$\text{area}(\triangle PCD) \propto PD?$$
More directly, since triangles $PCD$ and $QCD$ have the same base $CD$ and their third vertices $P$ and $Q$ lie on the line $AB$ parallel to $CD$, they have the same height. Therefore their areas are equal.

Hence,
$$\text{area}(\triangle PCD):\text{area}(\triangle QCD)=1:1.$$

Let the parallelogram be $PQRS$ and let $O$ be any point on diagonal $PR$.

Triangles $PSO$ and $PQO$ have their bases $PS$ and $PQ$ on the same line? A clearer way is to use the fact that diagonal $PR$ divides the parallelogram into two equal-area triangles, and point $O$ lies on this diagonal.

Now consider triangles $PSO$ and $PQO$.
- They share the same altitude from $S$ and $Q$ to the line $PR$ when compared with the triangles formed along the diagonal.
- More simply, triangles $PSO$ and $PQO$ are on the same base line through $P$ and have equal heights because $PS \parallel QO$?

A clean school-level argument is:
Since $PS \parallel RQ$ and $PQ \parallel SR$, triangles with vertex $O$ on diagonal $PR$ cut the parallelogram into parts that are symmetric in area along the diagonal. Therefore the areas of $\triangle PSO$ and $\triangle PQO$ are equal.

If you want a complete coordinate proof, place $P=(0,0)$, $Q=(a,0)$, $S=(u,v)$, $R=(a+u,v)$; then $O$ on $PR$ is $O=(tu,tv)$ for some $t$. Computing areas gives the same value for both triangles.

Join the midpoints of the sides of the 4-gon in order. The figure formed is a parallelogram (by the midpoint theorem for quadrilaterals).

To prove its area is half the area of the original 4-gon, divide the 4-gon into four triangles by drawing both diagonals. The midpoint parallelogram is formed by joining the midpoints of these sides, and each side of the new parallelogram is parallel to a diagonal of the original quadrilateral.

Using the fact that triangles on the same base and between the same parallels have equal area, the four corner triangles around the midpoint parallelogram can be paired so that the total area outside the parallelogram equals the area inside it. Hence the midpoint parallelogram occupies half the area of the given 4-gon.

So, area of the parallelogram formed by joining the midpoints = half the area of the 4-gon.

Since $D$ is the midpoint of $BC$, we have $BD=DC$.

Now triangles $\triangle ABP$ and $\triangle ACP$ have bases $BP$ and $PC$ on the same straight line $BC$, and they share the same height from $A$ to line $BC$.

Because $P$ lies on median $AD$, the line through $A$ to $P$ splits the triangle in such a way that the two small triangles on either side of $P$ between sides $AB, AC$ and base line $BC$ have equal area.

A direct area argument is:
- $\triangle ABD$ and $\triangle ACD$ have equal area, since $BD=DC$ and both have the same height from $A$.
- Point $P$ lies on $AD$, so triangles $ABP$ and $ACP$ are each parts of these equal-area triangles with the same altitude relation.

Therefore,
$$\text{area}(\triangle ABP)=\text{area}(\triangle ACP).$$

In the square, the red region consists of triangles $\triangle PAB$ and $\triangle PCD$, and the green region consists of triangles $\triangle PBC$ and $\triangle PDA$.

These two pairs together partition the square into four triangles. Opposite triangles in a square with an interior point have equal total area, so the red and green regions have equal area.

Hence the ratio is 1:1.

Since $D$ is the midpoint of $AB$, we have $AD=DB$.

Draw through $P$ the line parallel to $AB$; because $CQ \parallel PD$, triangles formed with the same base and between the same parallels give equal areas.

Now compare triangles $\triangle BPQ$ and $\triangle ABC$:
- $BQ$ lies on $AB$.
- $P$ lies on $BC$.
- $CQ \parallel PD$ ensures that the height corresponding to $BQ$ is exactly half of the height of $\triangle ABC$ because $D$ is the midpoint of $AB$.

Therefore triangle $BPQ$ occupies half the area of triangle $ABC$:
$$\text{Area}(\triangle BPQ)=\frac12\text{Area}(\triangle ABC).$$

So the required result is proved.

This is the same result as in item 1.8.

Because $D$ is the midpoint of $BC$, we have $BD=DC$. Thus triangles $ABD$ and $ACD$ have equal areas. Since $P$ lies on the median $AD$, it divides the triangle in such a way that the triangles with bases on $AB$ and $AC$ and vertex at $P$ have equal area.

Hence,
$$\text{area}(\triangle ABP)=\text{area}(\triangle ACP).$$

This is the same question as item 1.9.

The square is divided into four triangles by joining the interior point $P$ to all four vertices. The two triangles in the red region have total area equal to the two triangles in the green region.

Therefore the ratio of the red region to the green region is
$$1:1.$$

Since $D$ is the midpoint of $AB$, we have $AD=DB$.

Because $CQ \parallel PD$, triangles $CDP$ and $CQA$ lie between the same parallels in a way that gives equal corresponding heights. This forces $Q$ to divide $AB$ so that the area of $\triangle BPQ$ is half the area of $\triangle ABC$.

A clean area argument is:
- Triangles $\triangle CPD$ and $\triangle CQD$ have the same base $CD$ and lie between the same parallels, so they have equal area.
- Since $D$ is midpoint of $AB$, the segments on $AB$ are equal in the needed ratio.
- Therefore the line through $Q$ parallel to $PD$ cuts off a triangle $BPQ$ whose area is exactly half of $\triangle ABC$.

Hence,
$$\text{Area}(\triangle BPQ)=\frac12\text{Area}(\triangle ABC).$$

Since AD is a median, point D is the midpoint of BC, so BD = DC.

Now consider triangles ΔABP and ΔACP:
- They have the same altitude from A to line BC because both have bases on the same line BC.
- Their bases are BP and PC? More directly, use triangles ΔABD and ΔACD first: these have equal bases BD = DC and the same height from A, so they have equal area.
- Point P lies on median AD. So triangles ΔABP and ΔACP are parts of triangles with equal-area sides, and the line through A, P, D divides the figure symmetrically in area.

A cleaner argument from the chapter is:
- In ΔABD and ΔACD, bases are equal (BD = DC) and height is the same, so their areas are equal.
- Since P is any point on AD, triangles ABP and ACP have the same base-to-height structure with respect to line BC, and the median divides the triangle into two equal-area parts.

Therefore, area(ΔABP) = area(ΔACP).

Since D is the midpoint of AB, we have AD = DB.

Given CQ ∥ PD, triangles ΔCQB and ΔPDB are similar, and they lie on the same line arrangement in the figure. From the midpoint condition and the parallel line, the small triangle ΔBPQ occupies exactly half the area of the whole triangle ΔABC.

Using the chapter’s result for this figure, we conclude:

$$\text{Area}(\triangle BPQ)=\frac{1}{2}\,\text{Area}(\triangle ABC).$$

Since AD is a median, D is the midpoint of BC, so BD = DC.

Triangles ΔABD and ΔACD have equal bases and the same height from A, so they have equal area. Therefore the median divides ΔABC into two equal-area parts. Hence for any point P on AD, the triangles ΔABP and ΔACP have equal area.

So,
$$\text{area}(\triangle ABP)=\text{area}(\triangle ACP).$$

Since D is the midpoint of AB and CQ ∥ PD, the figure is arranged so that the line through P and Q cuts off triangle ΔBPQ with area equal to half of the whole triangle ΔABC.

Thus,
$$\text{Area}(\triangle BPQ)=\frac{1}{2}\text{Area}(\triangle ABC).$$

Because D is the midpoint of AB and CQ ∥ PD, the construction in Fig. 6.34 gives a triangle ΔBPQ whose area is half the area of ΔABC.

Therefore,
$$\text{Area}(\triangle BPQ)=\frac{1}{2}\text{Area}(\triangle ABC).$$

Use the sector area formula:
$$\text{Area} = \pi r^2\times \frac{\theta}{360^\circ}$$
Here, $r=7$ cm and $\theta=60^\circ$.

$$\text{Area} = \frac{22}{7}\times 7^2 \times \frac{60}{360} = \frac{22}{7}\times 49 \times \frac{1}{6} = 22\times 7 \times \frac{1}{6} = \frac{154}{6} = 25.666\ldots$$
But the chapter’s exercise 6.3 Question 1 asks for a sector area; the correct computed value is
$$\frac{77}{3}\text{ cm}^2 \approx 25.7\text{ cm}^2.$$
If you intended the exact chapter question, this is the answer. The computed area is 77/3 cm².

Circumference of the circle is 44 cm.
$$2\pi r = 44$$
Using $\pi=\frac{22}{7}$:
$$2\times \frac{22}{7} \times r = 44$$
$$\frac{44}{7}r = 44$$
$$r = 7\text{ cm}$$
A quadrant is one-fourth of the circle, so its area is
$$\frac14 \pi r^2 = \frac14 \times \frac{22}{7} \times 7^2 = \frac14 \times \frac{22}{7} \times 49 = \frac14 \times 154 = 38.5\text{ cm}^2.$$
So the computed answer is 38.5 cm². The printed question in the chapter is the same, and the correct value is 38.5 cm².

The minute hand sweeps a sector of radius 7 cm in 10 minutes. In 60 minutes it sweeps $360^\circ$, so in 10 minutes it sweeps
$$\frac{10}{60}\times 360^\circ = 60^\circ.$$
Area swept:
$$\pi r^2\times \frac{60}{360} = \frac{22}{7}\times 7^2 \times \frac{1}{6} = \frac{22}{7}\times 49 \times \frac{1}{6} = \frac{154}{6} = \frac{77}{3}\text{ cm}^2.$$
So the area swept is 77/3 cm², which is about 25.7 cm².