Mathematics solution NCERT
Class 9 - Chapter 6: Measuring Space: Perimeter and Area
Question 1.
Find the area of triangle $ADE$ in Fig. 6.31.
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} $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:
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} $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:
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} $Question 4.
The sides of a triangular plot are in the ratio $3:5:7$; its perimeter is $300$ m. Find its area.
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} $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 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.
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:
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.} $Question 7.
The area of quadrilateral $ABCD$ is $74\text{ cm}^2$. The area of triangle $ACD$ is $32\text{ cm}^2$. Find the area of triangle $ABC$.
Solution:
The diagonal $AC$ divides the quadrilateral $ABCD$ into two triangles:
- $\triangle ABC$
- $\triangle ACD$
Therefore,
$ \text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ACD). $Given,
$ \text{Area}(ABCD)=74\text{ cm}^2, $ $ \text{Area}(\triangle ACD)=32\text{ cm}^2. $Substitute these values.
$ \begin{aligned} 74 &=\text{Area}(\triangle ABC)+32. \end{aligned} $Subtract $32$ from both sides.
$ \begin{aligned} \text{Area}(\triangle ABC) &=74-32\\[4pt] &=42\text{ cm}^2. \end{aligned} $Answer:
$ {\text{Area}(\triangle ABC)=42\text{ cm}^2} $Question 8.
If the mid-points of the sides of a $4$-gon (quadrilateral) are joined in order, prove that the area of the parallelogram thus formed is half of the area of the given $4$-gon.
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). $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). $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). $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.
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
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).} $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). } $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). } $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). } $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.
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 $\triangle PDA)$?
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. $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} $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} $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} $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
Solution:
Given,
- $D$ is the midpoint of $AB$.
- $CQ \parallel PD$.
- $P$ lies on $BC$.
We have to prove that
$ {\text{Area}(\triangle BPQ)=\frac12\text{Area}(\triangle ABC).} $Step 1: Consider triangles $QAC$ and $QDC$.
Since $CQ \parallel PD$, triangles $QAC$ and $QDC$ lie on the same base $QC$.
Also, points $A$ and $D$ lie on the same line $AB$.
Therefore,
$ \frac{\text{Area}(\triangle QDC)} {\text{Area}(\triangle QAC)} = \frac{QD}{QA}. $Since $D$ is the midpoint of $AB$,
$ AD=DB. $Hence,
$ QD=\frac12QA. $Therefore,
$ \begin{aligned} \text{Area}(\triangle QDC) &=\frac12\text{Area}(\triangle QAC). \end{aligned} $Step 2: Compare triangles $QDC$ and $PDC$.
Since $CQ \parallel PD$, triangles $QDC$ and $PDC$ are on the same base $DC$
and between the same parallel lines.
Hence,
$ { \text{Area}(\triangle QDC) = \text{Area}(\triangle PDC). } $Using Step 1,
$ { \text{Area}(\triangle PDC) = \frac12 \text{Area}(\triangle QAC). } $Step 3: Compare triangles $BPQ$ and $PDC$.
Triangles $BPQ$ and $PDC$ have the same altitude from $P$ to the line $AB$.
Also,
so their corresponding bases are equal.
Therefore,
$ { \text{Area}(\triangle BPQ) = \text{Area}(\triangle PDC). } $Step 4: Complete the proof.
From Steps 2 and 3,
$ \begin{aligned} \text{Area}(\triangle BPQ) &=\text{Area}(\triangle PDC)\\[4pt] &=\frac12 \text{Area}(\triangle QAC). \end{aligned} $Since triangles $QAC$ and $ABC$ have the same altitude from $C$ to the line $AB$,
and $QA=AB$, we get
$ \text{Area}(\triangle QAC) = \text{Area}(\triangle ABC). $Hence,
$ \begin{aligned} \text{Area}(\triangle BPQ) &=\frac12 \text{Area}(\triangle ABC). \end{aligned} $Therefore,
$ { \text{Area}(\triangle BPQ) = \frac12 \text{Area}(\triangle ABC). } $Hence Proved.