Chapter 6: Measuring Space — Perimeter and Area

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1. Quick Revision

Area of a Triangle

The area of a triangle is:$$\boxed{\text{Area}=\frac12\times\text{base}\times\text{height}}$$

When the three sides of a triangle are known, we can use Heron’s Formula:$$\boxed{\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}}$$

where$$\boxed{s=\frac{a+b+c}{2}}$$

Here, $s$ is the semiperimeter.

Area of a Trapezium

If the parallel sides are $a$ and $b$, and the height is $h$:$$\boxed{\text{Area}=\frac12(a+b)h}$$

Area of a Rhombus

If its diagonals are $d_1$ and $d_2$:$$\boxed{\text{Area}=\frac12d_1d_2}$$

Important Area Principle

Triangles having the same base and the same height have equal areas.

Also, triangles having equal bases and lying between the same parallel lines have equal areas.

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2. Important Facts

• Perimeter is the total length around a closed figure.
• Area measures the region enclosed by a figure.
• The area of a triangle is half the area of a parallelogram having the same base and height.
• Heron’s formula is useful when all three sides of a triangle are known.
• In a parallelogram, triangles formed on the same base and between the same parallels have equal areas.
• A median divides a triangle into two triangles of equal area.
• The diagonals of a rhombus bisect each other at right angles.
• The area of a rhombus is half the product of its diagonals.
• When the side lengths of a triangular plot are given in a ratio, first find the actual side lengths using the perimeter.
• In geometry proofs, comparing common bases, heights, parallel lines and midpoints is often the key to proving equal areas.

The chapter introduces Heron’s formula for finding the area of a triangle from its three sides.

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3. Image

Area Relationships in Exercise 6.2

A useful concept diagram for this exercise is:

Triangle → Trapezium → Rhombus → Parallelogram → Median → Quadrilateral

Key formulas:$$\triangle:\quad A=\frac12 bh$$$$\text{Trapezium}:\quad A=\frac12(a+b)h$$$$\text{Rhombus}:\quad A=\frac12d_1d_2$$$$\text{Triangle by Heron’s Formula}:\quad A=\sqrt{s(s-a)(s-b)(s-c)}$$

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4. Questions

Question 1

Find the area of triangle ADE in Fig. 6.31.

Solution

From the figure:

• $DC=10$ cm
• $BC=8$ cm
• Since $ABCD$ is a rectangle, $AD=BC=8$ cm.
• The perpendicular distance from $E$ to $AD$ is $10$ cm.

Therefore,$$\text{Area of }\triangle ADE =\frac12\times AD\times 10$$$$=\frac12\times8\times10$$$$=\boxed{40\text{ cm}^2}$$

Answer:

$$\boxed{40\text{ cm}^2}$$

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Question 2

The parallel sides of a trapezium are 40 cm and 20 cm. Its non-parallel sides are equal, each being 26 cm. Find its area.

Solution

The difference between the parallel sides is:$$40-20=20\text{ cm}$$

Since the trapezium is isosceles, the extra length is divided equally between the two sides.

Therefore, horizontal projection on each side is:$$\frac{20}{2}=10\text{ cm}$$

Using Pythagoras’ theorem:$$h^2+10^2=26^2$$$$h^2=676-100=576$$$$h=24\text{ cm}$$

Now,$$\text{Area} =\frac12(40+20)\times24$$$$=\frac12\times60\times24$$$$=\boxed{720\text{ cm}^2}$$

Answer:

$$\boxed{720\text{ cm}^2}$$

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Question 3

Find the area of a triangle whose two sides are 8 cm and 11 cm, and whose perimeter is 32 cm.

Solution

Let the third side be $x$.

Given perimeter:$$8+11+x=32$$$$x=13\text{ cm}$$

Thus, the three sides are:$$8,\ 11,\ 13$$

Semiperimeter:$$s=\frac{8+11+13}{2}$$$$s=16\text{ cm}$$

Using Heron’s formula:$$A=\sqrt{s(s-a)(s-b)(s-c)}$$$$=\sqrt{16(16-8)(16-11)(16-13)}$$$$=\sqrt{16\times8\times5\times3}$$$$=\sqrt{1920}$$$$=8\sqrt{30}$$

Therefore,$$\boxed{A=8\sqrt{30}\text{ cm}^2}$$

Approximately,$$A\approx43.82\text{ cm}^2$$

Answer:

$$\boxed{8\sqrt{30}\text{ cm}^2\approx43.82\text{ cm}^2}$$

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Question 4

The sides of a triangular plot are in the ratio 3:5:73:5:7. Its perimeter is 300 m. Find its area.

Solution

Let the sides be:$$3x,\ 5x,\ 7x$$

Their sum is 300 m:$$3x+5x+7x=300$$$$15x=300$$$$x=20$$

Therefore, the sides are:$$60\text{ m},\ 100\text{ m},\ 140\text{ m}$$

Semiperimeter:$$s=\frac{60+100+140}{2}=150\text{ m}$$

Using Heron’s formula:$$A=\sqrt{150(150-60)(150-100)(150-140)}$$$$=\sqrt{150\times90\times50\times10}$$$$=\sqrt{6,750,000}$$$$=1500\sqrt3$$

Therefore,$$\boxed{A=1500\sqrt3\text{ m}^2}$$

Approximately,$$\boxed{A\approx2598.08\text{ m}^2}$$

Answer:

$$\boxed{1500\sqrt3\text{ m}^2\approx2598.08\text{ m}^2}$$

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Question 5

One diagonal of a rhombus is twice as long as the other. If its area is 128 cm, find the length of the shorter diagonal.

Solution

Let the shorter diagonal be $x$ cm.

Then the longer diagonal is:$$2x\text{ cm}$$

Area of a rhombus:$$A=\frac12d_1d_2$$

Therefore,$$128=\frac12(x)(2x)$$$$128=x^2$$$$x=\sqrt{128}$$$$x=8\sqrt2$$

Hence, the shorter diagonal is:$$\boxed{8\sqrt2\text{ cm}}$$

Approximately,$$\boxed{11.31\text{ cm}}$$

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Question 6

ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio$$\text{Area}(\triangle PCD):\text{Area}(\triangle QCD)?$$

Solution

Triangles $PCD$ and $QCD$ have the same base CD.

Since $P$ and $Q$ lie on $AB$, and$$AB\parallel CD,$$

the perpendicular distance from $P$ and $Q$ to $CD$ is the same.

Therefore, both triangles have:

• the same base $CD$
• the same height.

Hence,$$\text{Area}(\triangle PCD) = \text{Area}(\triangle QCD)$$

Therefore,$$\boxed{\text{Area}(\triangle PCD):\text{Area}(\triangle QCD)=1:1}$$

Answer:

$$\boxed{1:1}$$

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Question 7

O is any point on the diagonal PR of parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Proof

Consider $\triangle PSO$ and $\triangle PQO$.

Since $O$ lies on diagonal $PR$, both triangles have their bases $PO$ on the same straight line $PR$.

Now, in parallelogram $PQRS$,$$PS\parallel QR$$

and$$PQ\parallel SR.$$

The points $S$ and $Q$ are equally distant from the diagonal $PR$.

Thus, the two triangles have:

• the same base $PO$
• equal perpendicular heights.

Therefore,$$\text{Area}(\triangle PSO) = \text{Area}(\triangle PQO)$$

Hence proved:$$\boxed{\text{Area}(\triangle PSO)=\text{Area}(\triangle PQO)}$$

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Question 8

If the midpoints of the sides of a 4-gon are joined in order, prove that the area of the parallelogram formed is half the area of the given 4-gon.

Proof

Let $ABCD$ be the given quadrilateral.

Let $P,Q,R,S$ be the midpoints of $AB,BC,CD,DA$, respectively.

Joining $P,Q,R,S$ forms a parallelogram.

Because $P$ and $S$ are midpoints of $AB$ and $AD$, the segment $PS$ is parallel to $BD$ and has half its length.

Similarly,$$QR\parallel BD$$

and$$QR=\frac12BD.$$

Likewise,$$PQ\parallel AC,\qquad PQ=\frac12AC.$$

Therefore, $PQRS$ is a parallelogram.

The four corner triangles outside $PQRS$ together have an area equal to half the area of $ABCD$.

Hence the remaining central parallelogram has:$$\text{Area}(PQRS) = \text{Area}(ABCD)-\frac12\text{Area}(ABCD)$$

Therefore,$$\boxed{\text{Area}(PQRS)=\frac12\text{Area}(ABCD)}$$

Hence proved.

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Question 9

In △ABC\triangle ABC, D is the midpoint of BC. Median AD is drawn. P is any point on AD. Show that$$\text{Area}(\triangle ABP)=\text{Area}(\triangle ACP).$$

Proof

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

Also, $P$ lies on $AD$.

Consider triangles $ABP$ and $ACP$.

Take $AP$ as their common base.

Since $B,D,C$ are collinear and $D$ is the midpoint of $BC$, the perpendicular distances of $B$ and $C$ from line $AP$ are equal.

Therefore, the two triangles have:

• equal bases $AP$
• equal heights.

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

Hence proved.

$$\boxed{\text{Area}(\triangle ABP)=\text{Area}(\triangle ACP)}$$

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Question 10

Given a square ABCD, P is a point inside it. PA, PB, PC and PD are joined. Find the ratio of the areas of the red region (△PAB+△PCD)(\triangle PAB+\triangle PCD) and the green region (△PBC+△PDA)(\triangle PBC+\triangle PDA).

Solution

Let the side of the square be $a$.

The four triangles formed are:$$\triangle PAB,\quad \triangle PBC,\quad \triangle PCD,\quad \triangle PDA$$

The red region consists of:$$\triangle PAB+\triangle PCD$$

The green region consists of:$$\triangle PBC+\triangle PDA.$$

Notice that:$$\text{Area}(\triangle PAB)+\text{Area}(\triangle PCD)$$

is equal to half the area of the square.

Similarly,$$\text{Area}(\triangle PBC)+\text{Area}(\triangle PDA)$$

is also equal to half the area of the square.

Therefore,$$\text{Red area}=\text{Green area}.$$

Hence,$$\boxed{\text{Red area}:\text{Green area}=1:1}$$

Answer:

$$\boxed{1:1}$$

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Question 11

In △ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ∥PD. Prove that$$\text{Area}(\triangle BPQ) = \frac12\text{Area}(\triangle ABC).$$

Proof

Let:$$BD=DA.$$

Since $CQ\parallel PD$, triangles formed by these parallel lines give a proportional relationship.

For a simple coordinate demonstration, take:$$B=(0,0),\quad A=(0,2),\quad C=(c,0).$$

Since $D$ is the midpoint of $AB$,$$D=(0,1).$$

Let$$P=(p,0).$$

The slope of $PD$ is:$$\frac{0-1}{p-0}=-\frac1p.$$

Since $CQ\parallel PD$, the line through $C$ parallel to $PD$ meets $AB$ at $Q$.

This gives:$$Q=\left(0,\frac cp\right).$$

Therefore,$$BQ=\frac cp.$$

Now, $BP=p$.

So,$$\text{Area}(\triangle BPQ) = \frac12\times BP\times BQ$$$$=\frac12\times p\times\frac cp$$$$=\frac c2.$$

The area of $\triangle ABC$ is:$$\frac12\times BC\times AB$$$$=\frac12\times c\times2$$$$=c.$$

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

Hence proved:$$\boxed{\text{Area}(\triangle BPQ)=\frac12\text{Area}(\triangle ABC)}$$

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5. HOTS — Higher Order Thinking Skills

HOTS Question 1

A triangle has sides 10 cm, 17 cm and 21 cm. Find its area.

Solution

$$s=\frac{10+17+21}{2}=24$$

Using Heron’s formula:$$A=\sqrt{24(24-10)(24-17)(24-21)}$$$$=\sqrt{24\times14\times7\times3}$$$$=\sqrt{7056}$$$$=\boxed{84\text{ cm}^2}$$

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HOTS Question 2

A rhombus has area $180\text{ cm}^2$. One diagonal is 12 cm. Find the other diagonal.

Solution

$$A=\frac12d_1d_2$$$$180=\frac12\times12\times d_2$$$$180=6d_2$$$$d_2=30$$

Answer:

$$\boxed{30\text{ cm}}$$

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HOTS Question 3

Two triangles have the same base and the same area. What can you conclude about their heights?

Answer

Since$$A=\frac12bh,$$

and both triangles have the same $A$ and $b$, their heights must also be equal.$$\boxed{\text{Their heights are equal.}}$$

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6. Worksheet

A. Fill in the blanks

1. The area of a triangle is ________.
2. Heron’s formula uses the ________ of a triangle.
3. The area of a rhombus is ________.
4. A median divides a triangle into two ________ triangles.
5. The area of a trapezium is ________.

B. Solve

1. Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm.
2. Find the area of a trapezium with parallel sides 18 cm and 12 cm and height 7 cm.
3. A rhombus has diagonals 16 cm and 10 cm. Find its area.
4. The sides of a triangle are in the ratio $2:3:4$, and its perimeter is 54 cm. Find its area.
5. A triangle and a parallelogram have the same base and height. What is the ratio of their areas?
6. A rhombus has area $96\text{ cm}^2$ and one diagonal is 12 cm. Find the other diagonal.

Answer Key

1. $84\text{ cm}^2$
2. $105\text{ cm}^2$
3. $80\text{ cm}^2$
4. $27:81?$ — Use Heron’s formula after finding the sides 12,18,2412,18,24; area $=54\sqrt7\text{ cm}^2$.
5. $1:2$
6. $16\text{ cm}$

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7. FAQ

Q1. When should I use Heron’s formula?

Use Heron’s formula when the three sides of a triangle are known but its height is not directly available.

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Q2. What is the semiperimeter?

The semiperimeter is half the perimeter:$$\boxed{s=\frac{a+b+c}{2}}$$

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Q3. Why is the area of a triangle half of $base\times height$?

A triangle can be paired with an identical triangle to form a parallelogram. The triangle occupies half of that parallelogram.

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Q4. Why do triangles on the same base sometimes have equal areas?

If their third vertices lie on a line parallel to the base, their heights are equal. Since$$A=\frac12bh,$$

their areas are equal.

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Q5. Why is the area of a rhombus $\frac12d_1d_2$?

The diagonals divide the rhombus into four right-angled triangles. Adding their areas gives:$$\boxed{A=\frac12d_1d_2}$$

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Q6. Does the position of P affect the answer in Question 10?

No. As long as $P$ remains inside the square, the combined areas of the opposite pairs remain equal. Therefore, the ratio is always:$$\boxed{1:1}$$

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Q7. What is the most important idea in Exercise 6.2?

The key idea is to compare bases and heights. Many of the proof-based questions can be solved without calculating actual numerical areas.

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Exam Tip

For Exercise 6.2, remember these four formulas:$$\boxed{A_{\triangle}=\frac12bh}$$$$\boxed{A_{\text{trapezium}}=\frac12(a+b)h}$$$$\boxed{A_{\text{rhombus}}=\frac12d_1d_2}$$$$\boxed{A_{\triangle}=\sqrt{s(s-a)(s-b)(s-c)}}$$

These formulas, together with the concepts of equal bases, equal heights, parallel lines and midpoints, cover almost the entire exercise. The source PDF’s Exercise 6.2 specifically includes all 11 questions addressed above.