NCERT Solutions Class 7 Maths Chapter 10
NCERT Solutions for Class 7 Mathematics Chapter 10 Practical Geometry
By now, students should be familiar with the basic shapes and how to draw lines in geometry. The NCERT Mathematics Class 7 Chapter 10 builds on this learning by teaching them how to draw parallel lines and triangles with the various measurements. It explains in detail how to construct geometric figures.
Some questions in the Practical Geometry Class 7 chapter may seem difficult if they are not explained clearly. Extramarks strives to deliver its concepts in a straightforward and engaging manner. The NCERT Solution for Class 7 Mathematics Chapter 10 provided by our subject matter experts includes a step-by-step explanation of the answers that will make your learning experience interesting as well as comfortable.
NCERT Solutions for Class 7 Mathematics Chapter 10
Access NCERT Solutions for Class 7 Mathematics Chapter 10 – Practical Geometry
Chapter 10 – Practical Geometry Exercise | |
Exercise 10.1 | Questions & Solutions |
Exercise 10.2 | Questions & Solutions |
Exercise 10.3 | Questions & Solutions |
Exercise 10.4 | Questions & Solutions |
Exercise 10.5 | Questions & Solutions |
NCERT Solutions for Class 7 Mathematics –
NCERT Class 7 Mathematics Chapter 10 covers the topic of practical geometry. It includes the following topics:
UNIT | TOPIC |
10.1 | Introduction |
10.2 | Construction of a Line Parallel to a Given Line, and a Point Not on the Line |
10.3 | Construction of Triangles |
10.4 | Constructing a Triangle When the Lengths of its Three Sides are Known (SSS Criterion) |
10.5 | Constructing a Triangle When the Lengths of Two Sides and the Measure of the Angle Between Them are Known (SAS Criterion) |
10.6 | Constructing a Triangle When the Measures of Two of its Angles and the Length of the Side Included Between Them is Given (ASA Criterion) |
10.7 | Constructing a Right-Angled Triangle When the Length of One Leg and its Hypotenuse are Given (RHS Criterion) |
NCERT Solutions for Class 7 Mathematics Chapter 10
10.1 Introduction
This section of the chapter explains what Practical Geometry is and how to construct the different types of geometrical figures.
10.2 Construction of a Line Parallel to a Given Line, and a Point Not on The Line
In this section, students will learn how to draw a pair of lines that are parallel to each other. Parallel lines are those lines that always stay at some distance (never intersect) from each other. How to construct parallel lines is a basic concept for all and students should be well versed with it. It is critical in some fields like architecture.
The process for drawing two parallel lines is as follows:
- Two lines that are parallel to each other should not intersect even if those lines are extended to infinity in any direction.
- The distance between two parallel lines remains the same at all times.
For constructing a parallel line, you need some simple tools like a ruler and a compass. Apart from this, you also need a line segment and a point outside the line from where the parallel lines can be drawn. The steps for such construction are:
- Choose any point on the line segment, say AB and then join it to the outside point Z.
- Use X as the centre and with any suitable radius, draw an arc to cut the line segment ZX. Let this point be M on the line ZX, and N be the point that cuts AB.
- Now use Z as the centre and use the same radius used before drawing an arc. Let’s name this EF, that cuts the line segment ZX on the point say Y.
- Use Y as the centre and with the same radius used before drawing an arc. An arc so drawn will cut the arc EF at the point say R.
- Now join R and Z and it will give the line segment CD. Such a line segment will be the required parallel line.
10.3 Construction of Triangles
Before you proceed to this section, it is important for you to first recall the properties of triangles and the congruence of triangles taught in the previous chapters. Triangles can be classified based on their sides and angles. Here are some important features of triangles are:
- The exterior angle of any triangle is always equal to the sum of its opposite interior angles.
- The sum of the three angles of a triangle is always 180°.
- The sum of any two sides of the triangle is always more than the length of its third side.
- If a triangle is a right-angled triangle, then the square of its hypotenuse is equal to the sum of the square of its other two sides.
The Congruence of Triangles shows how a triangle can be drawn if any of the following measurements are given:
- The three sides of a triangle
- Two sides and the angle lying between them
- Values of two angles and the side that is between the angles
- The hypotenuse and the length of a right-angled triangle
10.4 Constructing a Triangle When the Lengths of its Three Sides are Known (SSS Criterion)
It is possible to construct a triangle when all its three sides are known by the following steps:
- When all three sides are given, the students first need to identify the side with the longest measure that is given.
- The side with the longest measure is made the base of the triangle.
- Use the other given measurements and mark the arcs of the triangle by using the endpoints of the base as vertices.
- By joining the arc intersection with the endpoints of the base, you will get the required triangle.
10.5 Constructing a Triangle When the Lengths of Two Sides and the Measure of the Angle Between Them are Known (SAS Criterion)
This section explains how to construct a triangle when any two sides of a triangle and one angle are given. The tools needed for such construction are a ruler and a compass.
To construct a triangle with the SAS provided, students need to follow the steps given below:
- Draw a straight line and mark the left endpoint as point D.
- Set the compass to the width of one of the sides and place the compass on point D, cut an arc on the line drawn.
- Name the point where the arc cuts the line as B.
- Construct an angle with the information provided on the line DB at point D.
- Now set the compass to the length of the second side provided.
- Then, placing the pointer head of the compass on point D, cut an arc on the angle line drawn.
- The point where the arc crosses the line is named as C.
- The points B and C should now be joined with a ruler and the required triangle will be obtained.
10.6 Constructing a Triangle When the Measures of Two of its Angles and the Length of the Side Included Between Them is Given (ASA Criterion)
This section deals with how to draw a triangle using the ASA criterion.
To satisfy such conditions, the given side has to be the one that is enclosed by the known angles. If any other side of the triangle has been given, a triangle with the ASA method cannot be drawn. To draw the required triangle, use a protractor and a ruler as follows:
- Use the ruler and draw a line segment with the length provided.
- Now using the protractor, draw a ray at point P, making one of the angles given.
- Similarly, draw the ray of the other angle provided at another point Q.
- The point where both the rays meet should be marked (in this case marking it as R).
Thus, the required triangle QPR will be obtained by joining all the points.
10.7 Constructing a Right-Angled Triangle When the Length of One Leg and its Hypotenuse are Given (RHS Criterion)
This section explains how to construct a right-angled triangle where a hypotenuse and one of the sides are given. To construct such a triangle, we require a ruler and a compass. Assuming the right angle is formed at point C, follow the given steps to obtain the required triangle:
- Draw a horizontal line of any length and mark the point of the right angle, say C, on it.
- Now set the compass according to the length of the side given. Place the head of the compass on point C and mark an arc on both sides. Name the points as Z and Q where the arc crosses the line.
- Now set the compass to the hypotenuse’s length and place the head of the compass on the point Z to mark an arc above C.
- Do the same step from point Q and name this point as S, which will be the point where both the arcs will cross each other.
- Join Q to C and S to C to obtain the required right-angled triangle.
Key Features of the NCERT Solution for Class 7 Mathematics Chapter 10
Students who are looking for correct answers to these practise questions have come to the right place. The NCERT Solutions provided by Extramarks will assist them in the following ways:
- Subject matter experts have meticulously written these solutions in a well-structured manner, making it easier for students to learn.
- These answers are written in accordance with the most recent CBSE curriculum and guidelines.
- Studying the solutions carefully will help students understand how to approach the various questions of Practical Geometry and boost their confidence.
NCERT Solutions for Class 7 Mathematics
Students can access the NCERT Solutions of Class 7 Mathematics from Extramarks and practise the NCERT textbook questions and even use them for last-minute preparation.. They need not look elsewhere for any help to supplement their studies. NCERT Solutions are reliable and accurate and they are complete in every way.
NCERT Solutions for Class 7
Extramarks provides textbook solutions, particularly for Class 7, which provide simplified answers to textbook questions for each chapter. These answers are reliable and in-depth as subject matter experts create our CBSE Class 7 study materials. Students can access the link given below to find solutions for all other subjects of Class 7.
Q.1 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line AB}\text{. Take a point P on it}\text{. Take a point C outside}\\ \text{this line}\text{. Join C to P}\text{.}\end{array}
$\text{(ii) Taking P as centre and with a convenient radius, draw arc intersecting line AB}\phantom{\rule{0ex}{0ex}}\text{at point D and PC at point E}\text{.}$
\text{(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H}\text{.}
\begin{array}{l}\text{(iv) Adjust the compasses up to the length of DE}\text{. Without}\\ \text{changing the opening of the compasses and taking H as the}\\ \text{cenre, draw an arc to intersect the previously draw arc}\\ \text{FG at po}\mathrm{int}\text{I}\text{.}\end{array}
\text{(v) Join the points C and I to draw a line}l.
\text{This is the required line which is parallel to AB}\text{.}
Q.2
$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{l}.\mathrm{Draw}\mathrm{a}\mathrm{perpendicular}\mathrm{to}\mathrm{l}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{l}.\\ \mathrm{On}\mathrm{this}\mathrm{perpendicular}\mathrm{choose}\mathrm{a}\mathrm{point}\mathrm{X},4\mathrm{cm}\mathrm{away}\mathrm{from}\mathrm{l}.\\ \mathrm{Through}\mathrm{X},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{m}\mathrm{parallel}\mathrm{to}\mathrm{l}.\end{array}$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line}l\text{and take a point P on the line}l\text{. Then draw a}\\ \text{perpendicular at point P}\text{.}\end{array}
\begin{array}{l}\text{(ii) Adjusting the compasses up to the length of 4 cm, draw an}\\ \text{arc to intersect this perpendicular at point X}\text{. Choose any point}\\ \text{Y on the line}l\text{. Join X to Y}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking Y as centre and with a convenient radius, draw}\\ \text{an arc intersecting}l\text{at A and XY at B}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iv) Taking X as centre and with same radius as before, draw}\\ \text{an arc CD cutting XY at E}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(v) Adjust the compasses upto the length of AB}\text{. Wothout}\\ \text{changing the opening of the compasses and taking E as the}\\ \text{centre, draw an arc intersecting the previous arc CD t F}\text{.}\end{array} \text{}
\text{(vi) Join the point X and F to draw a line}m\text{.} \text{Line}m\text{is the required line parallel to}l\text{.}
Q.3
$\begin{array}{l}\mathrm{Let}\mathrm{l}\mathrm{be}\mathrm{a}\mathrm{line}\mathrm{and}\mathrm{P}\mathrm{be}\mathrm{a}\mathrm{point}\mathrm{not}\mathrm{on}\mathrm{l}.\mathrm{Through}\mathrm{P},\mathrm{draw}\\ \mathrm{a}\mathrm{line}\mathrm{m}\mathrm{parallel}\mathrm{to}\mathrm{l}.\mathrm{Now}\mathrm{join}\mathrm{P}\mathrm{to}\mathrm{any}\mathrm{point}\mathrm{Q}\mathrm{on}\mathrm{l}.\mathrm{Choose}\\ \mathrm{any}\mathrm{other}\mathrm{point}\mathrm{R}\mathrm{on}\mathrm{m}.\mathrm{Through}\mathrm{R},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathrm{PQ}.\\ \mathrm{Let}\mathrm{this}\mathrm{meet}\mathrm{l}\mathrm{at}\mathrm{S}.\mathrm{What}\mathrm{shape}\mathrm{do}\mathrm{the}\mathrm{two}\mathrm{sets}\mathrm{of}\mathrm{parallel}\\ \mathrm{lines}\mathrm{enclose}?\end{array}$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line l and take a point A}\text{}\text{on it}\text{. Take a point P not}\\ \text{on}l\text{and join A to P}\text{.}\end{array}
\begin{array}{l}\text{(ii) Taking A as centre and with a convenient radius, draw an}\\ \text{arc cutting}l\text{at B and AP at C}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking P as centre and with the same radius as before,}\\ \text{draw an arc DE to intersect AP at F}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iv) Adust the compasses up to the length of BC}\text{. Without}\\ \text{changing the opening of compasses and taking F as the}\\ \text{centre, draw an arc to intersect the previously draw arc}\\ \text{DE at point G}\text{.}\end{array}
\text{(v) Join P to G to draw a line m}\text{. Line}m\text{will be parallel to line}l\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(vi) Join P to any point Q on line}\text{. Choose another point R on}\\ \text{line}m\text{. Similarly, a line can be drawn through point R and}\\ \text{parallel to PQ}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{Let it meet line}l\text{at point S}\text{. In quadrilateral PQRS, opposite}\\ \text{lines are parallel to each other}\text{. PQ}\parallel \text{RS and PR}\parallel \text{QS}\text{.}\\ \text{Thus PQRS is a parallelogram}\text{.}\end{array}
Q.4
$\mathrm{Construct}\mathrm{\Delta XYZ}\mathrm{in}\mathrm{which}\mathrm{XY}=4.5\mathrm{cm},\mathrm{YZ}=5\mathrm{cm}\mathrm{and}\mathrm{ZX}=6\mathrm{cm}.$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment YZ of length 5 cm}\text{.}\end{array}
\begin{array}{l}\text{(ii) Point X is at a distance of 4}\text{.5 cm from point Y}\text{. Therefore}\\ \text{taking Y as centre, draw an arc of 4}\text{.5 cm radius}\text{.}\end{array} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Point X is at a distance of 6 cm from point Z}\text{.Therefore}\\ \text{taking Z as centre, draw an arc of 6 cm radius}\text{. Mark the}\\ \text{point of intersection of the arcs as X}\text{. Join XY and XZ}\text{.}\end{array} \text{}
\text{Therefore, XYZ is the required triangle}\text{.}
Q.5
$\mathrm{Construct}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{of}\mathrm{side}5.5\mathrm{cm}.$
Ans
\begin{array}{l}\text{Since, we need to construct an equilateral triangle so all sides}\\ \text{should be equal}\text{.}\\ \text{So, AB}=\text{BC}=\text{CA}=\text{5}\text{.5 cm}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 5}\text{.5 cm}\text{.}\end{array}
\text{(ii) Taking B as centre, draw an arc of 5}\text{.5 cm radius}\text{.}
\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 5}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array} \text{}
\text{(iv) Join A}\text{}\text{to B and C}\text{.}
\text{Therefore, ABC is the required triangle}\text{.}
Q.6
$\begin{array}{l}\mathrm{Draw}\mathrm{\Delta PQR}\mathrm{with}\mathrm{PQ}=4\mathrm{cm},\mathrm{QR}=3.5\mathrm{cm}\mathrm{and}\mathrm{PR}=4\mathrm{cm}.\\ \mathrm{What}\mathrm{type}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{this}?\end{array}$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 3}\text{.5 cm}\text{.}\end{array}
\text{(ii) Taking Q as centre, draw an arc of 4 cm radius}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 4 cm radius to}\\ \text{meet previous arc at point P}\text{.}\end{array} \text{}
\text{(iv) Join P to Q and R}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{Therefore, PQR is the required triangle}\text{.}\\ \text{Since two sides of required triangle are equal,}\\ \text{so it is an isosceles triangle}\text{.}\end{array}
Q.7
$\begin{array}{l}\mathrm{Construct}\mathrm{\Delta ABC}\mathrm{such}\mathrm{that}\mathrm{AB}=2.5\mathrm{cm},\mathrm{BC}=6\mathrm{cm}\mathrm{and}\\ \mathrm{AC}=6.5\mathrm{cm}.\mathrm{Measure}\angle \mathrm{B}.\end{array}$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 6 cm}\text{.}\end{array}
\text{(ii) Taking C as centre, draw an arc of 6}\text{.5 cm radius}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking B as centre, draw an arc of 2}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array}
\text{(iv) Join A}\text{}\text{to B and C}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{Therefore, ABC is the required triangle}\text{.}\\ \angle \text{B can be measured with the help of a protractor and}\\ \text{it comes out to be 90}\xb0\text{.}\end{array}
Q.8
$\mathrm{Construct}\mathrm{\Delta DEF}\mathrm{such}\mathrm{that}\mathrm{DE}=5\mathrm{cm},\mathrm{DF}=3\mathrm{cm}\mathrm{and}\mathrm{m}\angle \mathrm{EDF}=90\mathrm{\xb0}.$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment DE of length 5 cm}\text{.}\end{array}
\text{(ii) At point D, draw a ray DX making an angle of 90}\xb0\text{with DE}\text{.}
\begin{array}{l}\text{(iii) Taking D as centre, draw an arc of 3 cm radius}\text{. It will}\\ \text{intersect DX at point F}\end{array}
\text{(iv) Join F to E}
\text{Thus, DEF is the required triangle}\text{.}
Q.9
$\begin{array}{l}\mathrm{Construct}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}\mathrm{in}\mathrm{which}\mathrm{the}\mathrm{lengths}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{its}\mathrm{equal}\mathrm{sides}\mathrm{is}6.5\mathrm{cm}\mathrm{and}\mathrm{the}\mathrm{angle}\mathrm{between}\\ \mathrm{them}\mathrm{is}110\mathrm{\xb0}.\end{array}$
Ans
\begin{array}{l}\text{An isosceles triangle PQR has to be constructed with}\\ \text{PQ}=\text{QR}=\text{6}\text{.5 cm}\text{.}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 6}\text{.5 cm}\end{array}
\text{(ii) At point Q, draw a ray QX making an angle 110}\xb0\text{with QR}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking Q as centre, draw an arc of 6}\text{.5 cm radius}\text{. It}\\ \text{intersects QX at point P}\text{.}\end{array} \text{}
\text{(iv) Join P to R to obtain the required triangle PQR}\text{.}
Q.10
$\mathrm{Construct}\mathrm{\Delta ABC}\mathrm{with}\mathrm{BC}=7.5\mathrm{cm},\mathrm{AC}=5\mathrm{cm}\mathrm{and}\mathrm{m}\angle \mathrm{C}=60\mathrm{\xb0}.$
Ans
\begin{array}{l}\text{The steps of constructions are as follows:}\\ \text{(i) Draw a line segment BC of length 7}\text{.5 cm}\text{.}\end{array}
\text{(ii) At point C, draw a ray CX making 60 with BC}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 5 cm radius}\text{.}\\ \text{It intersect CX at point A}\text{.}\end{array}
\text{(iv) Join A to B to obtain the required triangle ABC}\text{.}
Q.11
$\mathrm{Construct}\mathrm{\Delta ABC},\mathrm{given}\mathrm{m}\angle \mathrm{A}=60\mathrm{\xb0},\mathrm{m}\angle \mathrm{B}=30\mathrm{\xb0}\mathrm{and}\mathrm{AB}=5.8\mathrm{cm}.$
Ans
\text{(i) Draw a line segment AB of length 5}\text{.8 cm}\text{.}
\text{(ii) At point A, draw a ray AX making 60}\xb0\text{angle with AB}\text{.}
\text{(iii) At point B, draw a ray AX making 30}\xb0\text{angle with AB}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iv) Point C has to lie on both the rays, AX and BY}\text{. Therefore,}\\ \text{C is the point of intersection of these two rays}\text{.}\end{array} \text{}
\text{Thus, ABC is the required triangle}\text{.}
Q.12
$\begin{array}{l}\mathrm{Construct}\mathrm{\Delta PQR}\mathrm{if}\mathrm{PQ}=5\mathrm{cm},\mathrm{m}\angle \mathrm{PQR}=105\mathrm{\xb0}\mathrm{and}\mathrm{m}\angle \mathrm{QRP}=40\mathrm{\xb0}.\\ (\mathrm{Hint}:\mathrm{Recallangle}\u2013\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{a}\mathrm{triangle}).\end{array}$
Ans
\begin{array}{l}\text{To construct triangle PQR, we need to find}\angle \text{RPQ}\text{.}\\ \text{Using angle sum property of triangles to get}\\ \angle \text{PQR}+\angle \text{PRQ}+\angle \text{RPQ}=180\xb0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}105\xb0+40\xb0+\angle \text{RPQ}=180\xb0\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{RPQ}=180\xb0-145\xb0\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=35\xb0\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment PQ of length 5 cm}\text{.}\end{array}
\text{(ii) At P, draw a ray PX making an angle of 35}\xb0\text{with PQ}\text{.} \text{}
\text{(iii) At Q, draw a ray QY making an angle of 105}\xb0\text{with PQ}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iv) Point R has to lie on both the ray, PX and QY}\text{. Therefore,}\\ \text{R is the point of intersectin of these two rays}\text{.}\end{array} \text{}
\text{Thus, PQR is the required triangle}\text{.}
Q.13
$\begin{array}{l}\mathrm{Examine}\mathrm{whether}\mathrm{you}\mathrm{can}\mathrm{construct}\mathrm{\Delta DEF}\mathrm{such}\mathrm{that}\\ \mathrm{EF}=7.2\mathrm{cm},\mathrm{m}\angle \mathrm{E}=110\mathrm{\xb0}\mathrm{and}\mathrm{m}\angle \mathrm{F}=80\mathrm{\xb0}.\mathrm{Justify}\mathrm{your}\mathrm{answer}.\end{array}$
Ans
\begin{array}{l}\text{We are given,}\\ \angle \text{E}=\text{110}\xb0\text{and}\angle \text{F}=\text{80}\xb0\\ \text{Using angle sum property to get}\\ \angle \text{D+}\angle \text{E+}\angle \text{F=180}\xb0\\ \angle D+110\xb0+80\xb0=180\xb0\\ \angle D+190\xb0=180\xb0\\ \text{Since}\angle \text{D}\text{\hspace{0.17em}}\text{can not be zero and it does not satisfy angle sum}\\ \text{property}\text{.}\\ \text{Thus, we can not construct}\Delta \text{DEF}\text{.}\end{array}
\begin{array}{l}\text{Also, it can be observed that point D should lie on both rays,}\\ \text{EX and FY, for the constructing the required triangle}\text{.}\\ \text{However, both rays are not intersecting each other}\text{.}\\ \text{Therefore, the required triangle can not be formed}\text{.}\end{array}
Q.14
$\mathrm{Construct}\mathrm{the}\mathrm{right}\mathrm{angled}\mathrm{\Delta PQR},\mathrm{where}\mathrm{m}\angle \mathrm{Q}=90\mathrm{\xb0},\mathrm{QR}=8\mathrm{cm}\mathrm{and}\mathrm{PR}=10\mathrm{cm}.$
Ans
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 8 cm}\text{.}\end{array}
\text{(ii) At point Q, draw a ray QX making 90}\xb0\text{with QR}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 10 cm radius to}\\ \text{intersect ray QX at point P}\text{.}\end{array}
\text{(iv) Join P to R}\text{.}
\text{Thus,}\Delta \text{PQR is the required right-angled triangle}\text{.}
Q.15 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Ans
\begin{array}{l}\text{A right-angled triangle ABC with hypotenuse 6 cm and one}\\ \text{of the legs as 4 cm has to be constructed}\text{.}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 4 cm}\text{.}\end{array}
\text{(ii) At point B, draw a ray BX making 90}\xb0\text{with BC}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 6 cm radius to}\\ \text{intersect ray BX at point A}\text{.}\end{array} \text{}
\text{(iv) Join A to C}\text{.}
\text{} \text{Thus,}\Delta \text{ABC is the required right-angled triangle}\text{.}
Q.16
$\mathrm{Construct}\mathrm{an}\mathrm{isosceles}\mathrm{right}\u2013\mathrm{angled}\mathrm{triangle}\mathrm{ABC},\phantom{\rule{0ex}{0ex}}\mathrm{where}\mathrm{m}\angle \mathrm{ACB}=90\mathrm{\xb0}\mathrm{and}\mathrm{AC}=6\mathrm{cm}.$
Ans
\begin{array}{l}\text{Since, in a isosceles triangle, the length of any two side}\\ \text{are equal}\text{. So, AB}=\text{BC}=\text{6cm}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment AC of length 6 cm}\text{.}\end{array}
\text{(ii) At point C, draw a ray CX making 90}\xb0\text{with AC}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 6 cm radius to}\\ \text{intersect ray CX at point B}\text{.}\end{array}
\text{(iv) Join A to B}\text{.}
\text{} \text{Thus,}\Delta \text{ABC is the required triangle}\text{.}
FAQs (Frequently Asked Questions)
1. Are there any tips for performing better in practical geometry?
Students can follow the tips given below to perform better in practical geometry:
- The pencil used for construction must be sharp. It’s good to use HB or 2B pencil which has a dark lead.
- Avoid drawing thick lines as much as possible. Drawing dark lines does not mean the same thing as drawing thick lines.
- Always make sure you only use a small pencil in a compass (to avoid miscalculation).
- Remember that all the measurements marked in your protractor and scale are clear to get exact calculations.
- You should mark clearly for better results.
- While writing proofs, you must ensure that all the related theorems are specified as part of your proof.
- While giving proof you must include all these points- given data, theorems and a concluding statement.
2. Is practical geometry easy?
The chapter on Practical Geometry will be easy for you if your basics are clear. Therefore, you must work hard towards building a strong foundation in geometry. Once you are familiar with all the principles of geometry taught in your previous chapters, you will be able to deal with this chapter easily.
In case your basics are not clear yet, it is not too late to start working on them. Having a strong command of the geometry section will surely make a difference in your performance.