NCERT Solutions Class 9 Maths Exercise 6.2 — Measuring Space: Perimeter and Area

NCERT Class 9 Mathematics — Ganita Manjari Part I, Measuring Space: Perimeter and Area, Exercise Set 6.2. All 11 questions solved.

Two halves again. Questions 1 to 5 are computations, using:

$$\text{triangle} = \tfrac{1}{2}bh \qquad \text{trapezium} = \tfrac{1}{2}(a+b)h \qquad \text{rhombus} = \tfrac{1}{2}d_1 d_2$$

and Heron’s formula, for when the three sides are known but no height is:

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \qquad s = \frac{a+b+c}{2}$$

Questions 6 to 11 are proofs, and all of them rest on one idea.

Key insight. Triangles on the same base, with their apexes on a line parallel to that base, have equal areas. Both the base and the height are then the same, so $\frac{1}{2}bh$ is unchanged — however different the triangles look. Every proof in the second half of this exercise is that observation applied once or twice.

Question 1

Find the area of triangle $ADE$ in Fig. 6.31.

Solution. The figure shows rectangle $ABCD$ with $A$ at the top left, $B$ top right, $C$ bottom right and $D$ bottom left. The rectangle is $10$ cm wide and $8$ cm tall, and $E$ is a point on the right-hand side $BC$.

Take $AD$ — the left side, of length $8$ cm — as the base of the triangle. The height is the perpendicular distance from $E$ to the line $AD$, and since $E$ is on the opposite side of the rectangle, that distance is the full width, $10$ cm.

$$\text{Area} = \frac{1}{2} \times 8 \times 10 = 40 \text{ cm}^2$$

Notice that the answer does not depend on where on $BC$ the point $E$ sits: the base and height are the same wherever it is. The triangle is always exactly half the rectangle.

Question 2

The parallel sides of a trapezium are $40$ cm and $20$ cm. If its non-parallel sides are both equal, each being $26$ cm, find the area of the trapezium.

Solution. The equal slant sides make this an isosceles trapezium, so it is symmetric and the height can be found from a right-angled triangle.

Drop perpendiculars from the ends of the shorter side to the longer one. The longer side is cut into three pieces: two equal overhangs and the middle piece of $20$ cm.

$$\text{each overhang} = \frac{40 – 20}{2} = 10 \text{ cm}$$

Each slant side is the hypotenuse of a right-angled triangle with base $10$ cm:

$$h = \sqrt{26^2 – 10^2} = \sqrt{676 – 100} = \sqrt{576} = 24 \text{ cm}$$

$$\text{Area} = \frac{1}{2}(40 + 20)(24) = \frac{1}{2}(60)(24) = 720 \text{ cm}^2$$

Question 3

Find the area of a triangle, given that its sides are $8$ cm and $11$ cm long, and its perimeter is $32$ cm.

Solution. The perimeter gives the third side:

$$c = 32 – 8 – 11 = 13 \text{ cm}$$

No height is given, so use Heron’s formula. The semi-perimeter is

$$s = \frac{32}{2} = 16$$

$$\text{Area} = \sqrt{16(16-8)(16-11)(16-13)} = \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{1920}$$

Simplify the surd by pulling out the largest square factor, $64$:

$$\sqrt{1920} = \sqrt{64 \times 30} = 8\sqrt{30} \approx 43.82 \text{ cm}^2$$

Question 4

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

Solution. Write the sides as $3k$, $5k$, $7k$. The perimeter gives $k$:

$$3k + 5k + 7k = 300 \quad\Longrightarrow\quad 15k = 300 \quad\Longrightarrow\quad k = 20$$

So the sides are $60$ m, $100$ m and $140$ m, and

$$s = \frac{300}{2} = 150$$

$$\text{Area} = \sqrt{150(150-60)(150-100)(150-140)} = \sqrt{150 \times 90 \times 50 \times 10}$$ $$= \sqrt{6{,}750{,}000} = \sqrt{2{,}250{,}000 \times 3} = 1500\sqrt{3} \approx 2598.08 \text{ m}^2$$

(A check worth doing: $60 + 100 = 160 > 140$, so these three lengths really do form a triangle.)

Question 5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area $128 \text{ cm}^2$, find the length of the shorter diagonal.

Solution. Let the shorter diagonal be $d$; then the longer is $2d$. The area of a rhombus is half the product of its diagonals:

$$\frac{1}{2} \times d \times 2d = 128$$ $$d^2 = 128$$ $$d = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt2 \approx 11.31 \text{ cm}$$

The longer diagonal is $16\sqrt2 \approx 22.63$ cm.

Check: $\frac{1}{2} \times 8\sqrt2 \times 16\sqrt2 = \frac{1}{2} \times 128 \times \ldots$ — more simply, $8\sqrt2 \times 16\sqrt2 = 128 \times 2 = 256$, and half of that is $128$ ✓

Question 6

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

Solution. Both triangles share the same base $CD$.

Both apexes, $P$ and $Q$, lie on the side $AB$ — and in a parallelogram $AB \parallel CD$. So both apexes lie on a line parallel to the common base, which means both triangles have the same height: the perpendicular distance between the parallel lines $AB$ and $CD$.

$$\text{area}(\triangle PCD) = \frac{1}{2} \times CD \times h = \text{area}(\triangle QCD)$$

So the ratio is $1 : 1$ — the two areas are equal, wherever $P$ and $Q$ are on $AB$.

Each is also exactly half the parallelogram, since the parallelogram’s area is $CD \times h$.

Question 7

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

Solution. The key is that the diagonal $PR$ splits the parallelogram into two congruent halves, and $O$ divides them proportionally.

Step 1 — the diagonal halves the parallelogram. $\triangle PQR \cong \triangle RSP$ by SSS ($PQ = SR$, $QR = PS$ as opposite sides, and $PR$ common). So

$$\text{area}(\triangle PQR) = \text{area}(\triangle PSR)$$

Step 2 — compare triangles on the same base. Take $PO$ as a common base for $\triangle PSO$ and $\triangle PQO$. Their apexes are $S$ and $Q$.

Let $M$ and $N$ be the feet of the perpendiculars from $S$ and $Q$ to the line $PR$. The diagonals of a parallelogram bisect each other, so $S$ and $Q$ are the same distance from the line $PR$, on opposite sides of it:

$$SM = QN$$

(This follows because $\triangle SMT \cong \triangle QNT$, where $T$ is the centre of the parallelogram — vertically opposite angles, right angles, and $ST = QT$.)

Step 3 — conclude. Both triangles have base $PO$ and equal heights $SM$ and $QN$:

$$\text{area}(\triangle PSO) = \frac{1}{2}(PO)(SM) = \frac{1}{2}(PO)(QN) = \text{area}(\triangle PQO) \qquad \blacksquare$$

Question 8

If the mid-points of the sides of a $4$-gon are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given $4$-gon.

Solution. Let the quadrilateral be $ABCD$ with midpoints $P$, $Q$, $R$, $S$ of $AB$, $BC$, $CD$, $DA$. Draw the diagonal $AC$.

Step 1 — the midpoint theorem. In $\triangle ABC$, $P$ and $Q$ are midpoints of two sides, so $PQ \parallel AC$ and $PQ = \frac{1}{2}AC$. In $\triangle ACD$, likewise $SR \parallel AC$ and $SR = \frac{1}{2}AC$. (This is why $PQRS$ is a parallelogram — a question the book defers to the chapter on quadrilaterals.)

Step 2 — measure the corner triangles. The four corners cut off are $\triangle APS$, $\triangle BPQ$, $\triangle CQR$, $\triangle DRS$. Take $\triangle APS$: its sides $AP$ and $AS$ are half of $AB$ and $AD$, with the same angle at $A$. Halving two sides about a fixed angle quarters the area:

$$\text{area}(\triangle APS) = \frac{1}{4}\,\text{area}(\triangle ABD)$$

and similarly for the other three, each a quarter of the triangle cut off by a diagonal at that vertex.

Step 3 — add them up. Adding the four corner triangles:

$$\frac{1}{4}\Big[ \text{area}(\triangle ABD) + \text{area}(\triangle BAC) + \text{area}(\triangle CBD) + \text{area}(\triangle DAC) \Big]$$

Now, $\triangle ABD$ and $\triangle CBD$ together make the whole quadrilateral (split along $BD$), and $\triangle BAC$ and $\triangle DAC$ together make it too (split along $AC$). So the bracket is $2 \times$ the area of $ABCD$, and

$$\text{four corners} = \frac{1}{4} \times 2\,\text{area}(ABCD) = \frac{1}{2}\,\text{area}(ABCD)$$

Step 4 — subtract. The parallelogram is what remains:

$$\text{area}(PQRS) = \text{area}(ABCD) – \frac{1}{2}\text{area}(ABCD) = \frac{1}{2}\,\text{area}(ABCD) \qquad \blacksquare$$

Question 9

In $\triangle ABC$, the midpoint of $BC$ is $D$ (Fig. 6.32). Median $AD$ is drawn. $P$ is any point on $AD$. Show that area $(\triangle ABP) =$ area $(\triangle ACP)$.

Solution. Use the fact that a median bisects the area, twice over.

Step 1 — the median of $\triangle ABC$. $\triangle ABD$ and $\triangle ACD$ have equal bases $BD = DC$ (as $D$ is the midpoint) and the same height, the perpendicular from $A$ to $BC$. So

$$\text{area}(\triangle ABD) = \text{area}(\triangle ACD)$$

Step 2 — the median of $\triangle PBC$. The point $P$ lies on $AD$, and $PD$ is a median of $\triangle PBC$ for exactly the same reason: bases $BD = DC$, common height from $P$ to $BC$. So

$$\text{area}(\triangle PBD) = \text{area}(\triangle PCD)$$

Step 3 — subtract. The triangle $ABP$ is what remains of $\triangle ABD$ after removing $\triangle PBD$, and similarly on the other side:

$$\text{area}(\triangle ABP) = \text{area}(\triangle ABD) – \text{area}(\triangle PBD)$$ $$\text{area}(\triangle ACP) = \text{area}(\triangle ACD) – \text{area}(\triangle PCD)$$

Both right-hand sides are differences of equal quantities, so

$$\text{area}(\triangle ABP) = \text{area}(\triangle ACP) \qquad \blacksquare$$

Question 10

Given a square $ABCD$, let $P$ be a point within it. Join $PA$, $PB$, $PC$, $PD$ (Fig. 6.33). What is the ratio of the areas of the red region ($\triangle PAB$ and $\triangle PCD$) and the green region ($\triangle PBC$ and $\triangle PDA$)?

Solution. Let the square have side $a$, and let $P$ be at perpendicular distances $h_1$ from $AB$ and $h_2$ from the opposite side $CD$. Since $AB$ and $CD$ are opposite sides of the square, those two distances together span the square:

$$h_1 + h_2 = a$$

The red region.

$$\text{area}(\triangle PAB) + \text{area}(\triangle PCD) = \frac{1}{2}a h_1 + \frac{1}{2}a h_2 = \frac{1}{2}a(h_1 + h_2) = \frac{1}{2}a^2$$

— exactly half the square, whatever the position of $P$.

The green region. The same argument with the other pair of opposite sides, $BC$ and $DA$, whose distances also add to $a$:

$$\text{area}(\triangle PBC) + \text{area}(\triangle PDA) = \frac{1}{2}a^2$$

Both regions are half the square, so

$$\text{red} : \text{green} = 1 : 1$$

The striking part is that this holds for every position of $P$ inside the square, even though the four individual triangles change size as $P$ moves.

Question 11

In $\triangle ABC$, $D$ is the midpoint of $AB$. $P$ is any point on $BC$, and $Q$ is a point on $AB$ such that $CQ \parallel PD$. $PQ$ is joined (Fig. 6.34). Prove that $\text{Area}(\triangle BPQ) = \frac{1}{2}\,\text{Area}(\triangle ABC)$.

Solution. The parallel line is there to swap one triangle for another of equal area.

Step 1 — the median. $D$ is the midpoint of $AB$, so $CD$ is a median of $\triangle ABC$ and

$$\text{area}(\triangle BDC) = \frac{1}{2}\,\text{area}(\triangle ABC)$$

Step 2 — use the parallel. $\triangle DPQ$ and $\triangle DPC$ share the base $DP$, and their apexes $Q$ and $C$ both lie on the line $QC$, which is parallel to $DP$. Triangles on the same base with apexes on a line parallel to it have equal areas:

$$\text{area}(\triangle DPQ) = \text{area}(\triangle DPC)$$

Step 3 — add the common piece. Add $\triangle BDP$ to each side. On the left, $\triangle BDP$ and $\triangle DPQ$ together make $\triangle BPQ$; on the right, $\triangle BDP$ and $\triangle DPC$ together make $\triangle BDC$:

$$\text{area}(\triangle BPQ) = \text{area}(\triangle BDC)$$

Step 4 — combine with Step 1.

$$\text{area}(\triangle BPQ) = \frac{1}{2}\,\text{area}(\triangle ABC) \qquad \blacksquare$$

Common mistakes

• Question 1, using $10$ as the base and $8$ as the height without checking. It happens to give the same answer here, but the reasoning matters: the base must be a side of the triangle and the height measured perpendicular to it. $AD$ is a side; $DC$ is not.
• Question 2, taking the height as $26$ cm. The slant side is not the height. Dropping perpendiculars gives a $10$–$24$–$26$ right triangle, and $24$ is the height.
• Question 3, applying Heron’s formula with only two sides. The third side must be found from the perimeter first. Also, $s$ is the semi-perimeter, $16$, not $32$.
• Questions 3 and 4, giving a rounded answer only. $8\sqrt{30}$ and $1500\sqrt3$ are exact; the decimals are approximations of them.
• Question 5, taking the diagonals as $d$ and $d + 2$. “Twice as long” is multiplication, so the diagonals are $d$ and $2d$.
• Question 5, using $d_1 d_2$ without halving. The rhombus area is half the product of the diagonals, which would give $d = 8$ instead of $8\sqrt2$.
• Question 6, thinking the ratio depends on where $P$ and $Q$ are. It does not. Both apexes are on $AB$, parallel to the common base $CD$, so both heights are the same.
• Questions 7, 9 and 11, comparing triangles that do not share a base. Each proof works by finding a common base first and then arguing that the heights are equal. Without the common base there is nothing to compare.
• Question 10, computing the four triangles separately. They cannot be found individually — $P$ is an arbitrary point. It is the pairs that have a fixed total, because the distances to opposite sides always add to the side length.

Practise next

• Exercise Set 6.3 — areas of sectors and segments, applying the same ideas to curved regions.
• The End-of-Chapter Exercises, Questions 2 to 6 — five more triangle-area problems, including one asking for the area to be found in two different ways.
• The End-of-Chapter Exercises, Questions 11 to 14 — the trapezium and kite formulas derived geometrically, which is the proof half of this exercise continued.