Class 9 - Chapter 6: Measuring Space: Perimeter and Area

Question 1.

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

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Solution:

From the figure,

$ DC=10\text{ cm} $ $ BC=8\text{ cm} $

Since $ABCD$ is a rectangle,

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

Also, point $E$ lies on side $BC$.

The perpendicular distance from $E$ to side $AD$ is equal to the width of the rectangle.

$ \text{Perpendicular distance}=DC=10\text{ cm} $

Area of a triangle is given by

$ \text{Area} = \frac12\times\text{Base}\times\text{Height} $

Here,

$ \text{Base}=AD=8\text{ cm} $ $ \text{Height}=10\text{ cm} $

Substitute the values.

$ \begin{aligned} \text{Area of }\triangle ADE &=\frac12\times8\times10\\[4pt] &=4\times10\\[4pt] &=40\text{ cm}^2 \end{aligned} $

Answer:

$ {\text{Area of }\triangle ADE=40\text{ cm}^2} $

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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.

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Solution:

Given,

$ \text{Parallel sides}=40\text{ cm and }20\text{ cm} $ $ \text{Non-parallel sides}=26\text{ cm each} $

Since both non-parallel sides are equal, the trapezium is an isosceles trapezium.

Step 1: Find the horizontal distance on each side.

The difference between the parallel sides is

$ 40-20=20\text{ cm} $

This difference is equally divided on both sides.

$ \frac{20}{2}=10\text{ cm} $

Step 2: Find the height of the trapezium.

Using Pythagoras' Theorem in the right triangle,

$ \begin{aligned} h^2+10^2&=26^2 \end{aligned} $ $ \begin{aligned} h^2+100&=676 \end{aligned} $ $ \begin{aligned} h^2&=676-100\\ &=576 \end{aligned} $ $ \begin{aligned} h&=\sqrt{576}\\ &=24\text{ cm} \end{aligned} $

Step 3: Find the area of the trapezium.

We know that

$ \text{Area} = \frac12 (\text{Sum of parallel sides}) \times \text{Height} $

Substitute the given values.

$ \begin{aligned} \text{Area} &=\frac12(40+20)\times24\\[4pt] &=\frac12\times60\times24\\[4pt] &=30\times24\\[4pt] &=720\text{ cm}^2 \end{aligned} $

Answer:

$ {\text{Area of the trapezium}=720\text{ cm}^2} $

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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.

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Solution:

Given,

$ a=8\text{ cm},\qquad b=11\text{ cm} $

Perimeter of the triangle

$ =32\text{ cm} $

Step 1: Find the third side.

$ \begin{aligned} c &=32-(8+11)\\[4pt] &=32-19\\[4pt] &=13\text{ cm} \end{aligned} $

Step 2: Find the semi-perimeter.

Using the formula,

$ s=\frac{a+b+c}{2} $ $ \begin{aligned} s &=\frac{8+11+13}{2}\\[4pt] &=\frac{32}{2}\\[4pt] &=16\text{ cm} \end{aligned} $

Step 3: Apply Heron's Formula.

$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $

Substitute the values.

$ \begin{aligned} \text{Area} &=\sqrt{16(16-8)(16-11)(16-13)}\\[4pt] &=\sqrt{16\times8\times5\times3}\\[4pt] &=\sqrt{1920}\\[4pt] &=\sqrt{64\times30}\\[4pt] &=8\sqrt{30}\text{ cm}^2 \end{aligned} $

Answer:

$ {\text{Area}=8\sqrt{30}\text{ cm}^2} $

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Question 4.

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

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Solution:

Let the three sides be

$ 3x,\qquad5x,\qquad7x. $

According to the question,

$ 3x+5x+7x=300 $ $ \begin{aligned} 15x &=300\\[4pt] x &=20 \end{aligned} $

Therefore, the sides are

$ 60\text{ m},\qquad100\text{ m},\qquad140\text{ m}. $

Step 1: Find the semi-perimeter.

$ \begin{aligned} s &=\frac{300}{2}\\[4pt] &=150\text{ m} \end{aligned} $

Step 2: Apply Heron's Formula.

$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $

Substitute the values.

$ \begin{aligned} \text{Area} &=\sqrt{150(150-60)(150-100)(150-140)}\\[4pt] &=\sqrt{150\times90\times50\times10} \end{aligned} $

Factorise the numbers.

$ \begin{aligned} 150\times90\times50\times10 &=(15\times9\times5)\times10^4\\[4pt] &=6075\times10000 \end{aligned} $ $ \begin{aligned} \text{Area} &=100\sqrt{6075}\\[4pt] &=100\sqrt{81\times75}\\[4pt] &=900\sqrt{75}\\[4pt] &=900\sqrt{25\times3}\\[4pt] &=900\times5\sqrt3\\[4pt] &=4500\sqrt3 \end{aligned} $

Therefore,

$ {\text{Area}=4500\sqrt3\text{ m}^2} $

Approximate value:

$ \begin{aligned} 4500\sqrt3 &\approx4500\times1.732\\[4pt] &\approx7794\text{ m}^2 \end{aligned} $

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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.

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Solution:

Let the length of the shorter diagonal be

$ x\text{ cm}. $

Then the longer diagonal will be

$ 2x\text{ cm}. $

We know that the area of a rhombus is

$ \text{Area}=\frac12\times d_1\times d_2, $

where $d_1$ and $d_2$ are the lengths of the diagonals.

Given,

$ \text{Area}=128\text{ cm}^2. $

Substitute the values in the formula.

$ \begin{aligned} 128 &=\frac12\times x\times2x \end{aligned} $

Simplify the right-hand side.

$ \begin{aligned} 128 &=x^2 \end{aligned} $

Taking the positive square root of both sides (since length is always positive),

$ \begin{aligned} x &=\sqrt{128}\\[4pt] &=\sqrt{64\times2}\\[4pt] &=8\sqrt2\text{ cm} \end{aligned} $

Therefore, the shorter diagonal is

$ {8\sqrt2\text{ cm}} $

Verification:

The longer diagonal is

$ 2x=16\sqrt2\text{ cm}. $

Now,

$ \begin{aligned} \text{Area} &=\frac12\times8\sqrt2\times16\sqrt2\\[4pt] &=\frac12\times128\times2\\[4pt] &=128\text{ cm}^2 \end{aligned} $

Hence, the answer is verified.

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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)? $

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Solution:

Given,

• $ABCD$ is a parallelogram.
• $P$ and $Q$ are any two points on side $AB$.

Join $PC$, $PD$, $QC$ and $QD$.

Consider triangles $PCD$ and $QCD$.

Both triangles have the same base

$ CD. $

Also, the vertices $P$ and $Q$ lie on the line $AB$.

Since

$ AB\parallel CD, $

the perpendicular distance from every point on $AB$ to the line $CD$ is the same.

Hence, triangles $PCD$ and $QCD$ have

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

We know that

$ \text{Area of a triangle} = \frac12\times\text{Base}\times\text{Height}. $

Therefore,

$ \begin{aligned} \text{Area}(\triangle PCD) &=\frac12\times CD\times h, \end{aligned} $

and

$ \begin{aligned} \text{Area}(\triangle QCD) &=\frac12\times CD\times h. \end{aligned} $

Thus,

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

Hence, the required ratio is

$ {\text{Area}(\triangle PCD):\text{Area}(\triangle QCD)=1:1.} $

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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

Given: PQRS is a parallelogram and O is any point on its diagonal PR.

To prove:

Area of triangle PSO = Area of triangle PQO

Proof:

Since PQRS is a parallelogram,

PQ ∥ SR and PS ∥ QR.

O lies on the diagonal PR. Therefore, the points P, O and R are collinear.

Consider the triangles PSO and PQO. Both triangles have the same base PO.

The perpendicular distance from S to PO is equal to the perpendicular distance from Q to PO because S and Q lie on the parallel lines PS and QR respectively, and PO lies along the diagonal PR.

Therefore, the triangles PSO and PQO have equal bases and equal corresponding heights.

Using the formula,

Area of a triangle = ¹⁄₂ × base × height

we get,

Area(▵PSO) = ¹⁄₂ × PO × h

and

Area(▵PQO) = ¹⁄₂ × PO × h

Hence,

Area(▵PSO) = Area(▵PQO)

Therefore, the areas of triangles PSO and PQO are equal.

Hence proved.

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Question 8.

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it 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. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.).

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Solution:

Let $ABCD$ be the given quadrilateral.

Let $P$, $Q$, $R$ and $S$ be the mid-points of the sides $AB$, $BC$, $CD$ and $DA$, respectively.

Join $P$, $Q$, $R$ and $S$ in order.

The quadrilateral $PQRS$ is a parallelogram (Varignon's Theorem).

Also, draw the diagonal

$ AC. $

This diagonal divides the quadrilateral into two triangles:

• $\triangle ABC$
• $\triangle ADC$

Step 1: Consider $\triangle ABC$.

In $\triangle ABC$,

• $P$ is the midpoint of $AB$.
• $Q$ is the midpoint of $BC$.

Therefore, by the Mid-point Theorem,

$ PQ\parallel AC $

and

$ PQ=\frac12AC. $

The triangle formed by joining the mid-points of two sides of a triangle has one-fourth of the area of the original triangle.

Hence,

$ \text{Area}(\triangle BPQ) = \frac14\, \text{Area}(\triangle ABC). $

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Step 2: Consider $\triangle ADC$.

Similarly, in $\triangle ADC$,

• $S$ is the midpoint of $AD$.
• $R$ is the midpoint of $CD$.

Therefore,

$ SR\parallel AC $

and

$ SR=\frac12AC. $

Also,

$ \text{Area}(\triangle DSR) = \frac14\, \text{Area}(\triangle ADC). $

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Step 3: Consider triangles $AQS$ and $CPR$.

Using the same midpoint property,

$ \text{Area}(\triangle AQS) = \frac14\, \text{Area}(\triangle ADC), $

and

$ \text{Area}(\triangle CPR) = \frac14\, \text{Area}(\triangle ABC). $

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Step 4: Find the area of parallelogram $PQRS$.

The quadrilateral $ABCD$ is divided into

• Parallelogram $PQRS$
• Four corner triangles: $\triangle BPQ$, $\triangle CPR$, $\triangle DSR$ and $\triangle AQS$.

Therefore,

$ \begin{aligned} \text{Area}(PQRS) &=\text{Area}(ABCD)\\ &\quad- \Big[ \text{Area}(\triangle BPQ) +\text{Area}(\triangle CPR)\\ &\qquad+ \text{Area}(\triangle DSR) +\text{Area}(\triangle AQS) \Big]. \end{aligned} $

Substituting the above results,

$ \begin{aligned} \text{Area}(PQRS) &=\text{Area}(ABCD)\\ &\quad- \left[ \frac14\text{Area}(\triangle ABC) +\frac14\text{Area}(\triangle ABC)\right.\\ &\qquad\left. +\frac14\text{Area}(\triangle ADC) +\frac14\text{Area}(\triangle ADC) \right]. \end{aligned} $

Taking common factors,

$ \begin{aligned} \text{Area}(PQRS) &=\text{Area}(ABCD)\\ &\quad- \frac12 \left[ \text{Area}(\triangle ABC) +\text{Area}(\triangle ADC) \right]. \end{aligned} $

Since,

$ \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = \text{Area}(ABCD), $

we get

$ \begin{aligned} \text{Area}(PQRS) &=\text{Area}(ABCD) -\frac12\text{Area}(ABCD)\\[4pt] &=\frac12\text{Area}(ABCD). \end{aligned} $

Hence,

Area of the parallelogram formed by joining
the mid-points of the sides of a quadrilateral

Hence Proved.

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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

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

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Solution:

Given,

• $D$ is the midpoint of $BC$.
• $AD$ is the median of $\triangle ABC$.
• $P$ is any point on $AD$.

We have to prove that

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

Step 1: Compare the areas of triangles $ABD$ and $ACD$.

Since $D$ is the midpoint of $BC$,

$ BD=DC. $

Also, triangles $ABD$ and $ACD$ have the same altitude from $A$ to the line $BC$. Therefore,

$ \begin{aligned} \text{Area}(\triangle ABD) &=\frac12\times BD\times h,\\[4pt] \text{Area}(\triangle ACD) &=\frac12\times DC\times h. \end{aligned} $

As

$ BD=DC, $

we get

$ {\text{Area}(\triangle ABD)=\text{Area}(\triangle ACD).} $

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Step 2: Compare triangles $ABP$ and $PBD$.

Triangles $ABP$ and $PBD$ have the same altitude from $B$ to the line $AD$. Therefore,

$ \frac{\text{Area}(\triangle ABP)} {\text{Area}(\triangle PBD)} = \frac{AP}{PD}. $

Hence,

$ { \text{Area}(\triangle ABP) = \frac{AP}{PD} \times \text{Area}(\triangle PBD). } $

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Step 3: Compare triangles $ACP$ and $PCD$.

Similarly, triangles $ACP$ and $PCD$ have the same altitude from $C$ to the line $AD$. Therefore,

$ \frac{\text{Area}(\triangle ACP)} {\text{Area}(\triangle PCD)} = \frac{AP}{PD}. $

Hence,

$ { \text{Area}(\triangle ACP) = \frac{AP}{PD} \times \text{Area}(\triangle PCD). } $

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Step 4: Compare triangles $PBD$ and $PCD$.

Triangles $PBD$ and $PCD$ have

• the same altitude from $P$ to the line $BC$, and
• equal bases since $BD=DC$.

Therefore,

$ { \text{Area}(\triangle PBD) = \text{Area}(\triangle PCD). } $

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Step 5: Complete the proof.

Using the results obtained above,

$ \begin{aligned} \text{Area}(\triangle ABP) &=\frac{AP}{PD}\times\text{Area}(\triangle PBD)\\[4pt] &=\frac{AP}{PD}\times\text{Area}(\triangle PCD)\\[4pt] &=\text{Area}(\triangle ACP). \end{aligned} $

Hence,

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

Hence Proved.

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Question 10.

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

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Solution:

Let the side of the square be

$ a. $

Therefore, the area of the square is

$ \text{Area}(ABCD)=a^2. $

The point $P$ divides the square into four triangles:

• $\triangle PAB$
• $\triangle PBC$
• $\triangle PCD$
• $\triangle PDA$

The red region consists of

$ \triangle PAB+\triangle PCD. $

The green region consists of

$ \triangle PBC+\triangle PDA. $

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Step 1: Find the area of the red region.

Triangles $PAB$ and $PCD$ have the same base

$ AB=CD=a. $

Let the perpendicular distances from $P$ to $AB$ and $CD$ be

$ h_1\quad\text{and}\quad h_2. $

Since $AB$ and $CD$ are opposite sides of a square,

$ h_1+h_2=a. $

Therefore,

$ \begin{aligned} \text{Area}(\triangle PAB) +\text{Area}(\triangle PCD) &=\frac12ah_1+\frac12ah_2\\[4pt] &=\frac12a(h_1+h_2)\\[4pt] &=\frac12a(a)\\[4pt] &=\frac{a^2}{2}. \end{aligned} $

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Step 2: Find the area of the green region.

Similarly, triangles $PBC$ and $PDA$ have bases

$ BC=AD=a. $

Let the perpendicular distances from $P$ to $BC$ and $AD$ be

$ h_3\quad\text{and}\quad h_4. $

Since $BC$ and $AD$ are opposite sides of the square,

$ h_3+h_4=a. $

Hence,

$ \begin{aligned} \text{Area}(\triangle PBC) +\text{Area}(\triangle PDA) &=\frac12ah_3+\frac12ah_4\\[4pt] &=\frac12a(h_3+h_4)\\[4pt] &=\frac12a(a)\\[4pt] &=\frac{a^2}{2}. \end{aligned} $

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Step 3: Compare the two regions.

From Steps 1 and 2,

$ \text{Red Region} = \frac{a^2}{2}, $

and

$ \text{Green Region} = \frac{a^2}{2}. $

Therefore,

$ \text{Red Region} = \text{Green Region}. $

Hence, the required ratio is

$ { (\triangle PAB+\triangle PCD): (\triangle PBC+\triangle PDA) = 1:1. } $

Answer:

$ {1:1} $

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

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Given:

D is the midpoint of AB, P lies on BC, and CQ ∥ PD.

To Prove:

Area (ΔBPQ) = 1/2 Area (ΔABC)

Construction:

Join DC.

Proof:

Since D is the midpoint of AB, CD is a median of ΔABC.

A median divides a triangle into two triangles of equal areas.

Therefore,

Area (ΔBDC) = 1/2 Area (ΔABC) ........ (1)

Now, ΔDPC and ΔDPQ have the same base DP.

Also, CQ ∥ PD. Therefore, these triangles lie between the same parallel lines and have equal heights.

Hence,

Area (ΔDPC) = Area (ΔDPQ) ........ (2)

Now,

Area (ΔBDC) = Area (ΔBPD) + Area (ΔDPC)

Using equation (2),

Area (ΔBDC) = Area (ΔBPD) + Area (ΔDPQ)

Since ΔBPQ consists of ΔBPD and ΔDPQ,

Area (ΔBDC) = Area (ΔBPQ) ........ (3)

From equations (1) and (3),

Area (ΔBPQ) = 1/2 Area (ΔABC)

Hence proved.

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