NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.2#

In Fig. 6.31, $ABCD$ is a rectangle with $AB = 10$ cm and $BC = 8$ cm, and $E$ is a point on $BC$.

Take $AD$ as the base of $\triangle ADE$. Since $AD = BC$:

$$AD = 8 \text{ cm}$$

The height is the perpendicular distance from $E$ to $AD$. $E$ lies on $BC$, which is parallel to $AD$ and 10 cm from it:

$$\text{Height} = AB = 10 \text{ cm}$$

So:

$$\text{Area of } \triangle ADE = \dfrac{1}{2} \times 8 \times 10$$

$$\text{Area of } \triangle ADE = 40$$

∴ The area of $\triangle ADE$ is 40 cm² — half the rectangle, wherever $E$ is on $BC$.

The non-parallel sides are equal, so the trapezium is symmetric. Dropping perpendiculars from the ends of the 20 cm side onto the 40 cm side cuts off two equal pieces:

$$\text{Each piece} = \dfrac{40 - 20}{2} = 10 \text{ cm}$$

Each perpendicular is the height $h$ of a right triangle with hypotenuse 26 cm and base 10 cm:

$$h^2 = 26^2 - 10^2$$

$$h^2 = 676 - 100$$

$$h^2 = 576$$

$$\Rightarrow h = 24 \text{ cm}$$

So:

$$\text{Area} = \dfrac{1}{2}(40 + 20) \times 24$$

$$\text{Area} = 30 \times 24$$

$$\text{Area} = 720$$

∴ The area of the trapezium is 720 cm².

The third side is:

$$c = 32 - 8 - 11 = 13 \text{ cm}$$

The semi-perimeter is:

$$s = \dfrac{32}{2} = 16 \text{ cm}$$

By Heron’s formula:

$$\text{Area} = \sqrt{16(16 - 8)(16 - 11)(16 - 13)}$$

$$\text{Area} = \sqrt{16 \times 8 \times 5 \times 3}$$

$$\text{Area} = \sqrt{1920}$$

$$\text{Area} = 8\sqrt{30}$$

∴ The area is $8\sqrt{30}$ cm² ≈ 43.82 cm².

Let the sides be $3k$, $5k$ and $7k$. The perimeter is 300 m, so:

$$3k + 5k + 7k = 300$$

$$\Rightarrow 15k = 300$$

$$\Rightarrow k = 20$$

So the sides are 60 m, 100 m and 140 m, and the semi-perimeter is:

$$s = \dfrac{300}{2} = 150 \text{ m}$$

By Heron’s formula:

$$\text{Area} = \sqrt{150(150 - 60)(150 - 100)(150 - 140)}$$

$$\text{Area} = \sqrt{150 \times 90 \times 50 \times 10}$$

$$\text{Area} = \sqrt{6750000}$$

$$\text{Area} = 1500\sqrt{3}$$

∴ The area of the plot is $1500\sqrt{3}$ m² ≈ 2598.08 m².

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

$$\dfrac{1}{2} \times d \times 2d = 128$$

$$\Rightarrow d^2 = 128$$

$$\Rightarrow d = \sqrt{128}$$

$$\Rightarrow d = 8\sqrt{2}$$

∴ The shorter diagonal is $8\sqrt{2}$ cm ≈ 11.31 cm (and the longer is $16\sqrt{2}$ cm).

$\triangle PCD$ and $\triangle QCD$ have the same base $CD$.

$P$ and $Q$ both lie on $AB$, and $AB \parallel CD$, so both are at the same distance $h$ from $CD$ — the height of the parallelogram. So both triangles have height $h$:

$$\text{Area}(\triangle PCD) = \dfrac{1}{2} \times CD \times h$$

$$\text{Area}(\triangle QCD) = \dfrac{1}{2} \times CD \times h$$$$\Rightarrow \text{Area}(\triangle PCD) = \text{Area}(\triangle QCD)$$

∴ The ratio is 1 : 1. (Each triangle is half the parallelogram.)

$\triangle PSO$ and $\triangle PQO$ have the same base $PO$, which lies along the diagonal $PR$. So it is enough to show that $S$ and $Q$ are at the same distance from the line $PR$.

Let the diagonals meet at $M$. The diagonals of a parallelogram bisect each other, so $MS = MQ$, and $S$, $M$, $Q$ lie on one line.

Draw perpendiculars $SX$ and $QY$ from $S$ and $Q$ to the line $PR$. In $\triangle SXM$ and $\triangle QYM$:

Part	Reason
$\angle SXM = \angle QYM = 90°$	perpendiculars
$\angle SMX = \angle QMY$	vertically opposite angles
$MS = MQ$	diagonals bisect each other

By AAS congruence, $\triangle SXM \cong \triangle QYM$, so:

$$SX = QY$$

So the two triangles have equal bases ($PO$) and equal heights:

$$\text{Area}(\triangle PSO) = \dfrac{1}{2} \times PO \times SX$$

$$\text{Area}(\triangle PQO) = \dfrac{1}{2} \times PO \times QY$$$$\Rightarrow \text{Area}(\triangle PSO) = \text{Area}(\triangle PQO)$$

Hence proved.

Let $ABCD$ be the 4-gon (convex), and let $P$, $Q$, $R$, $S$ be the midpoints of $AB$, $BC$, $CD$, $DA$. The region $PQRS$ is what is left when the four corner triangles $APS$, $BQP$, $CRQ$, $DSR$ are removed from $ABCD$.

Corner triangle at A. $\triangle APS$ and $\triangle ABD$ share the angle $A$, and $AP = \dfrac{1}{2}AB$, $AS = \dfrac{1}{2}AD$. Taking $AB$ as the base: $AP$ is half of $AB$, and the height of $S$ above $AB$ is half the height of $D$ (because $S$ is halfway along $AD$). So:

$$\text{Area}(\triangle APS) = \dfrac{1}{2} \times \dfrac{1}{2} \times \text{Area}(\triangle ABD)$$

$$\text{Area}(\triangle APS) = \dfrac{1}{4}\text{Area}(\triangle ABD)$$

In the same way:

$$\text{Area}(\triangle CRQ) = \dfrac{1}{4}\text{Area}(\triangle CDB)$$

$$\text{Area}(\triangle BQP) = \dfrac{1}{4}\text{Area}(\triangle BCA)$$

$$\text{Area}(\triangle DSR) = \dfrac{1}{4}\text{Area}(\triangle DAC)$$

The diagonal $BD$ splits $ABCD$ into $\triangle ABD$ and $\triangle CDB$, and the diagonal $AC$ splits it into $\triangle BCA$ and $\triangle DAC$. Adding the four corner triangles:

$$\text{Corners} = \dfrac{1}{4}\text{Area}(ABCD) + \dfrac{1}{4}\text{Area}(ABCD)$$

$$\text{Corners} = \dfrac{1}{2}\text{Area}(ABCD)$$

So:

$$\text{Area}(PQRS) = \text{Area}(ABCD) - \dfrac{1}{2}\text{Area}(ABCD)$$

$$\text{Area}(PQRS) = \dfrac{1}{2}\text{Area}(ABCD)$$

Hence proved.

Step 1. In $\triangle ABC$, $D$ is the midpoint of $BC$, so $BD = DC$. $\triangle ABD$ and $\triangle ACD$ have equal bases and the same height (from $A$), so:

$$\text{Area}(\triangle ABD) = \text{Area}(\triangle ACD) \quad \cdots (1)$$

Step 2. In the same way, in $\triangle PBC$, $PD$ is a median, so:

$$\text{Area}(\triangle PBD) = \text{Area}(\triangle PCD) \quad \cdots (2)$$

Step 3. $\triangle ABP$ is what is left of $\triangle ABD$ after removing $\triangle PBD$, and $\triangle ACP$ is what is left of $\triangle ACD$ after removing $\triangle PCD$.

Subtracting equation (2) from equation (1):

$$\text{Area}(\triangle ABD) - \text{Area}(\triangle PBD) = \text{Area}(\triangle ACD) - \text{Area}(\triangle PCD)$$

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

Hence proved.

Let the side of the square be $a$. Let $P$ be at distance $h_1$ from $AB$ and $h_2$ from $CD$. $AB$ and $CD$ are opposite sides, $a$ apart, so:

$$h_1 + h_2 = a$$

The red region is:

$$\text{Red} = \dfrac{1}{2} \times a \times h_1 + \dfrac{1}{2} \times a \times h_2$$

$$\text{Red} = \dfrac{1}{2}a(h_1 + h_2)$$

$$\text{Red} = \dfrac{1}{2}a^2$$

So the red region is half the square, and the green region is the other half:

$$\text{Green} = a^2 - \dfrac{1}{2}a^2 = \dfrac{1}{2}a^2$$

∴ The ratio of the red region to the green region is 1 : 1, wherever $P$ is inside the square.

Join $DP$ and $DC$.

Step 1. $\triangle DPQ$ and $\triangle DPC$ have the same base $DP$. Their third vertices $Q$ and $C$ lie on the line $CQ$, which is parallel to $DP$, so they have equal heights:

$$\text{Area}(\triangle DPQ) = \text{Area}(\triangle DPC) \quad \cdots (1)$$

Step 2. $\triangle BPQ$ is made of $\triangle BPD$ and $\triangle DPQ$, and $\triangle BDC$ is made of $\triangle BPD$ and $\triangle DPC$. Using equation (1):

$$\text{Area}(\triangle BPQ) = \text{Area}(\triangle BPD) + \text{Area}(\triangle DPQ)$$

$$\text{Area}(\triangle BPQ) = \text{Area}(\triangle BPD) + \text{Area}(\triangle DPC)$$

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

Step 3. $D$ is the midpoint of $AB$, so $CD$ is a median of $\triangle ABC$, and a median divides a triangle into two equal areas:

$$\text{Area}(\triangle BDC) = \dfrac{1}{2}\text{Area}(\triangle ABC)$$

∴ $\text{Area}(\triangle BPQ) = \dfrac{1}{2}\text{Area}(\triangle ABC)$. Hence proved.