NCERT Solutions Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines

NCERT Solutions Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines introduces students to the basic ideas of angles formed when lines intersect. This chapter explains important concepts like complementary and supplementary angles, linear pairs, and vertically opposite angles. It also helps students understand how to identify different types of angles when three lines cross each other at a point.

The NCERT Solutions for this chapter follow the latest syllabus and are explained in a clear, step-by-step way. These solutions make it easy for students to practice questions, understand how the angles are formed, and solve geometry problems with confidence. With these NCERT Solutions, Class 7 students can build a strong base in geometry and improve their logical thinking.

1.0NCERT Solutions Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines – Download PDF

Our ALLEN-prepared PDF covers every problem in NCERT Solutions for Class 7 Maths Chapter 7 complete with clear diagrams, step-by-step explanations, and adherence to the updated NCERT syllabus.

2.0Key Concepts in Chapter 7:  A Tale of Three Intersecting Lines

1. Medians of a Triangle

• A median connects a vertex of the triangle to the midpoint of the opposite side.
• Every triangle has three medians, and they intersect at a point known as the centroid.
• The centroid divides each median in a 2:1 ratio, counting from the vertex.
• Construction of medians involves identifying midpoints and drawing lines accurately using a ruler and compass.

2. Altitudes of a Triangle

• An altitude is a perpendicular line drawn from a triangle’s vertex to its opposite side (or its extension).
• There are three altitudes, which intersect at the orthocenter.
• Depending on whether the triangle is acute, right, or obtuse, the orthocenter may lie inside, on, or outside the triangle.
• Constructing altitudes requires careful use of a set square or compass to draw precise right angles.

3. Angle Bisectors of a Triangle

• An angle bisector splits an angle of the triangle into two equal parts.
• The three angle bisectors intersect at a point called the incenter.
• The incenter is equidistant from all sides of the triangle and serves as the center of the inscribed circle (incircle).
• Constructing angle bisectors uses compass arcs and straightedge techniques to guarantee equal angles.

4. Circumcenter and Perpendicular Bisector

• A perpendicular bisector divides a side of a triangle into two equal halves at a right angle.
• The three perpendicular bisectors meet at the circumcenter, the center of the circumcircle that passes through the three vertices.
• The circumcenter’s position varies: inside an acute triangle, on a right triangle’s hypotenuse, or outside an obtuse triangle.
• Construction requires midpoint identification and perpendicular drawing using basic tools.

5. Relationships among the Four Centers

• Centroid (G), orthocenter (H), incenter (I), and circumcenter (O) are fundamental centers of the triangle.
• These points reveal deep geometric relationships—for instance, the line joining the orthocenter, centroid, and circumcenter (Euler line) in non-equilateral triangles.
• Chapter 7 introduces these relationships through descriptive explanations and problem-solving practice.

6. Constructions and Real-Life Applications

• Drawing accurate constructions helps build practical geometry skills, vital for design, architecture, and engineering.
• Applications include locating point of balance (centroid), designing structures (circumcenter), and inscribing circles using incenters.
• NCERT emphasizes hands-on practice to help students visualize these geometric ideas.

3.0NCERT Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines: Detailed Solutions

FIGURE IT OUT-01

• Use the point on the circle and/or the centre to form isosceles triangles.
  
  Sol. The steps of construction are given as (i) Draw a circle of any radius using compass and denote its centre by 0 .
  
  (ii) Take any two points A and B on the circle and join them to centre 0 . Then, $\mathrm{OA}=\mathrm{OB}$ [radii of a circle]
  
  (iii) Now, join AB . Then the $△\mathrm{ABC}$ is the required isosceles triangle.
  
• Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.
  
  Sol. (i) Given circles in figure (i) are
  
  Let these two circles intersect at point C and D . Now, join $\mathrm{AB},\mathrm{AC}$ and BC . Then, $\mathrm{AB}=\mathrm{AC}=\mathrm{AB}$ [radii of both the circles are same]
  
  Hence, the $△\mathrm{ABC}$ is the required equilateral triangle. We can also construct an isosceles triangle. Mark a point M on the circle whose centre is at A , out of the shaded region.
  
  Joint $\mathrm{AB},\mathrm{AM}$ and BM . Then, $\mathrm{AM}=\mathrm{AB}$ [radii of a circle]
  
  Hence, the $△\mathrm{ABM}$ is the required isosceles triangle. (ii) Given three circles in figure (ii) are
  
  Join points A to $\mathrm{B},\mathrm{B}$ to C and A to C . Observe that $\mathrm{AB}=\mathrm{BC}=\mathrm{CA}$.
  
  Hence, $△\mathrm{ABC}$ is the required equilateral triangle. We can also construct an isosceles triangle. Mark a point $M$ on the intersection of the circles whose centre is at $A$ and $C$, out of the shaded region.
  
  Joint $\mathrm{AB},\mathrm{AM}$ and BM . Then, $\mathrm{AM}=\mathrm{AB}$
  
  Hence, the $△\mathrm{ABM}$ is the required isosceles triangle.

FIGURE IT OUT-02

• We checked by construction that there are no triangles having side lengths $3\mathrm{cm},4\mathrm{cm}$ and $8\mathrm{cm};2\mathrm{cm},3\mathrm{cm}$ and 6 cm . Check if you could have found this without trying to construct the triangle. Sol. Yes, without constructing the triangles, we could have determined that they cannot exist using the triangle inequality. i.e. $3+4<8$ and $2+3<6$ Hence, both the triangles cannot be formed.
• Can we say anything about the existence of a triangle for each of the following sets of lengths? (i) $10\mathrm{km},10\mathrm{km}$ and 25 km (ii) $5\mathrm{mm},10\mathrm{mm}$ and 20 mm (iii) $12\mathrm{cm},20\mathrm{cm}$ and 40 cm Sol. We know that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. (i) Here, $10+10<25$ So, a triangle cannot be formed. (ii) Here, $5+10<20$ So, a triangle cannot be formed. (iii) Here, $12+20<40$ So, a triangle cannot be formed.

FIGURE IT OUT-03

• Which of the following lengths can be the side lengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure. (i) $2,2,5$ (ii) $3,4,6$ (iii) $2,4,8$ (iv) $5,5,8$ (v) $10,20,25$ (vi) $10,20,35$ (vii) $24,26,28$ Sol. We know that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. (i) Here, $2+2<5$ So, a triangle cannot be formed. (ii) Here, $3+4>6$, $4+6>3$ and $3+6>4$ So, a triangle is possible. (iii) Here, $2+4<8$ So, a triangle cannot be formed. (iv) Here, $5+5>8$ and $5+8>5$ So, a triangle is possible. (v) Here, $10+20>25$ $20+25>10$ and $10+25>20$ So, a triangle is possible. (vi) Here, $10+20<35$ So, a triangle cannot be formed. (vii) Here, $24+26>28$ and $24+28>26$ So, a triangle is possible. $26+28>24$

FIGURE IT OUT-04

• Check if a triangle exists for each of the following set of lengths. (i) $1,100,100$ (ii) $3,6,9$ (iii) $1,1,5$ (iv) $5,10,12$ Sol. Triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. (i) Here, $1+100>100$ and $100+100>1$ So, triangle is possible. (ii) Here, $3+6=9$ So, triangle cannot be formed. (iii) Here, $1+1<5$ So, triangle cannot be formed. (iv) Here, $5+10>12$, $10+12>5$ and $12+5>10$ So, triangle is possible.
• Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any side length? Justify your answer. Sol. Yes, an equilateral triangle is possible with side lengths 50,50 and 50 . Since, an equilateral triangle is a triangle in which all three sides are equal. Hence, there exist an equilateral triangle of any side length.
• For each of the following, give atleast 5 possible values for the third length, so there exists a triangle having these as side lengths (decimal values could also be chosen). (i) 1,100 (ii) 5,5 (iii) 3,7 Sol. By the triangle inequality, the sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Also, the difference between the lengths of any two sides of a triangle is always smaller than the length of the third side. Let x be the length of third side of the triangle. (i) Given, two sides of a triangle are 1 and 100. Then, $x$ satisfies $\begin{array}{rl}& 100-1<x<100+1\\ \Rightarrow & 99<x<101\end{array}$ $\therefore$ The five possible values for x are 99.5, 100, 100.2, 100.5, 100.9. (ii) Given, two sides of a triangle are 5 and 5. Then, $x$ satisfies $\begin{array}{rl}& 5-5<x<5+5\\ \Rightarrow & 0<x<10\end{array}$ $\therefore$ The five possible values of $x$ are $1,3,5,7$ and 9 . (iii) Given, two sides of a triangle are 3 and 7. Then, $x$ satisfies $\begin{array}{rl}& 7-3<x<7+3\\ \Rightarrow & 4<x<10.\end{array}$ $\therefore$ The five possible values for x are $4.5,5,6,6.8$ and 6.9.

FIGURE IT OUT-05

• Construct triangles for the following measurements where the angle is included between the sides. (i) $3\mathrm{cm},75^{\circ },7\mathrm{cm}$ (ii) $6\mathrm{cm},25^{\circ },3\mathrm{cm}$ (iii) $3\mathrm{cm},120^{\circ },8\mathrm{cm}$ Sol. (i) $3\mathrm{cm},75^{\circ },7\mathrm{cm}$ The steps of construction are given as (a) Draw a line segment AB of length 7 cm using a ruler and a pencil.
  
  (b) At point A , a draw an angle of $75^{\circ }$ using protector and mark X .
  
  (c) At point A , mark a point C along A X such that $\mathrm{AC}=3\mathrm{cm}$.
  
  (d) Join the point C to B . Then, $△\mathrm{ABC}$ is the required triangle.
  
  (ii) $6\mathrm{cm},25^{\circ },3\mathrm{cm}$ The steps of construction are given as (a) Draw a line segment AB of length 3 cm using a ruler and a pencil.
  
  (b) At point A , draw an angle of $25^{\circ }$ using protector and mark X . (c) At point A , mark a point C along A X such that $\mathrm{AC}=6\mathrm{cm}$.
  
  (d) Join the point C to B . Then, $△\mathrm{ABC}$ is the required triangle. (iii) $3\mathrm{cm},120^{\circ },8\mathrm{cm}$ The steps of construction are given as (a) Draw a line segment AB of length 8 cm using a ruler and a pencil. $\mathrm{A}\hookrightarrow 8\mathrm{cm}⟶\mathrm{B}$ (b) At point A , draw an angle of $120^{\circ }$ using protector and mark X . $\underset{\mathrm{A}120^{\circ }}{\underbrace{\mathrm{X}\mathrm{cm}}}\mathrm{B}$ (c) At point A , mark a point C along A X such that $\mathrm{AC}=3\mathrm{cm}$.
  
  (d) Join the point C to B . Then, $△\mathrm{ABC}$ is the required triangle.
  

FIGURE IT OUT-06

• Construct triangles for the following measurements. (i) $75^{\circ },5\mathrm{cm},75^{\circ }$ (ii) $25^{\circ },3\mathrm{cm},60^{\circ }$ (iii) $120^{\circ },6\mathrm{cm},30^{\circ }$ Sol. (i) $75^{\circ },5\mathrm{cm},75^{\circ }$ The steps of construction are given as (a) Draw a line segment AB of length 5 cm using a ruler and a pencil.
  
  (b) Draw angles of $75^{\circ }$ and $75^{\circ }$ at points A and B respectively using protector.
  
  (c) Let these two rays intersect at point C . Then, $△\mathrm{ABC}$ is the required triangle.
  
  (ii) $25^{\circ },3\mathrm{cm},60^{\circ }$ The steps of construction are given as (a) Draw a line segment AB of length 3 cm using a ruler and a pencil.
  
  (b) Draw angles of $25^{\circ }$ and $60^{\circ }$ at points A and B respectively using protector. (c) Let these two rays intersect at point C . Then, $△\mathrm{ABC}$ is the required triangle. (iii) $120^{\circ },6\mathrm{cm},30^{\circ }$ The steps of construction are given as (a) Draw a line segment AB of length 6 cm using a ruler and a pencil.
  
  (b) Draw angles of $120^{\circ }$ and $30^{\circ }$ at points A and B respectively using protector.
  
  (c) Let these two rays intersect at point C . Then, $△\mathrm{ABC}$ is the required triangle.
  

FIGURE IT OUT-07

• For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find atleast two different angles for each category. (i) $30^{\circ }$ (ii) $70^{\circ }$ (iii) $54^{\circ }$ (iv) $144^{\circ }$ Sol. We know that when the sum of two angles is less than $180^{\circ }$ then the triangle exists and when the sum of two angles is greater than or equal to $180^{\circ }$ then there exists no triangle. (i) (a) Two possible angles are $70^{\circ }$ and $60^{\circ }$ (b) Two not possible angles are $150^{\circ }$ and $155^{\circ }$ (ii) (a) Two possible angles are $105^{\circ }$ and $50^{\circ }$ (b) Two not possible angles are $110^{\circ }$ and $120^{\circ }$ (iii) (a) Two possible angles are $120^{\circ }$ and $110^{\circ }$ (b) Two not possible angles are $126^{\circ }$ and $130^{\circ }$ (iv) (a) Two possible angles are $20^{\circ }$ and $30^{\circ }$ (b) Two not possible angles are $36^{\circ }$ and $40^{\circ }$
• Determine which of the following pairs can be the angles of a triangle and which cannot. (i) $35^{\circ },150^{\circ }$ (ii) $70^{\circ },30^{\circ }$ (iii) $90^{\circ },85^{\circ }$ (iv) $50^{\circ },150^{\circ }$ Sol. We know that if the sum of two angles of a triangle is greater than or equal to $180^{\circ }$ then triangle is not possible with these angles. (i) Given, the two angles of a triangle are $35^{\circ }$ and $150^{\circ }$. Here, $35^{\circ }+150^{\circ }=185^{\circ }>180^{\circ }$. Therefore, triangle is not possible. (ii) Given, the two angles of a triangle are $70^{\circ }$ and $30^{\circ }$. Here, $70^{\circ }+30^{\circ }=100^{\circ }<180^{\circ }$. Therefore, triangle is possible. (iii) Given, the two angles of a triangle are $90^{\circ }$ and $85^{\circ }$. Here, $90^{\circ }+85^{\circ }=175^{\circ }<180^{\circ }$. Therefore, triangle is possible. (iv) Given, the two angles of a triangle are $50^{\circ }$ and $150^{\circ }$. Here, $50^{\circ }+150^{\circ }=200^{\circ }>180^{\circ }$. Therefore, triangle is not possible.

FIGURE IT OUT-08

• Find the third angle of a triangle (using a parallel line) when two of the angles are (i) $36^{\circ },72^{\circ }$ (ii) $150^{\circ },15^{\circ }$ (iii) $90^{\circ },30^{\circ }$ (iv) $75^{\circ },45^{\circ }$ Sol. (i) Given, the two angles of a triangle are $36^{\circ }$ and $72^{\circ }$. Let the third angle be a. Let $△\mathrm{ABC}$ be any triangle with angles $\angle \mathrm{B}=36^{\circ }\angle \mathrm{C}=72^{\circ }$ and $\angle \mathrm{A}=\mathrm{a}$. Now, draw a line XY parallel to BC , passing through A .
  
  Then, $\angle \mathrm{XAB}=36^{\circ }$ and angle $\mathrm{YAC}=72^{\circ }[\because$ alternate interior angles $]$ We know that $\angle \mathrm{XAB},\angle \mathrm{YAC}$ and $\angle \mathrm{BAC}$ together form a straight line. So, $\angle \mathrm{XAB}+\angle \mathrm{YAC}+\angle \mathrm{BAC}=180^{\circ }$ $\Rightarrow 36^{\circ }+72^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow 108^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow \mathrm{a}=72^{\circ }$ Hence, the third angle of a triangle is $72^{\circ }$. (ii) Given, the two angles of a triangle are $150^{\circ }$ and $15^{\circ }$. Let the third angle be a. Let $△\mathrm{ABC}$ be any triangle with angles $\angle \mathrm{B}=150^{\circ }\angle \mathrm{C}=15^{\circ }$ and $\angle \mathrm{A}=\mathrm{a}$. Now, draw a line XY parallel to BC , passing through A .
  
  Then, $\angle \mathrm{XAB}=150^{\circ }$ and angle $\mathrm{YAC}=15^{\circ }[\because$ alternate angles $]$ We know that $\angle \mathrm{XAB},\angle \mathrm{YAC}$ and $\angle \mathrm{BAC}$ together form a straight line. So, $\angle \mathrm{XAB}+\angle \mathrm{YAC}+\angle \mathrm{BAC}=180^{\circ }$ $\Rightarrow 150^{\circ }+15^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow 165^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow \mathrm{a}=15^{\circ }$ Hence, the third angle of a triangle is $15^{\circ }$. (iii) Given, the two angles of a triangle are $90^{\circ }$ and $30^{\circ }$. Let the third angle be a. Let $△\mathrm{ABC}$ be any triangle with angles $\angle \mathrm{B}=90^{\circ }\angle \mathrm{C}=30^{\circ }$ and $\angle \mathrm{A}=\mathrm{a}$. Now, draw a line XY parallel to BC , passing through A .
  
  Then, $\angle \mathrm{XAB}=90^{\circ }$ and angle $\mathrm{YAC}=30^{\circ }[\because$ alternate angles $]$ We know that $\angle \mathrm{XAB},\angle \mathrm{YAC}$ and $\angle \mathrm{BAC}$ together form a straight line. So, $\angle \mathrm{XAB}+\angle \mathrm{YAC}+\angle \mathrm{BAC}=180^{\circ }$ $\Rightarrow 90^{\circ }+30^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow 120^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow \mathrm{x}=60^{\circ }$ Hence, the third angle of a triangle is $60^{\circ }$. (iv) Given, the two angles of a triangle are $75^{\circ }$ and $45^{\circ }$. Let the third angle be a. Let $△\mathrm{ABC}$ be any triangle with angles $\angle \mathrm{B}=75^{\circ }\angle \mathrm{C}=45^{\circ }$ and $\angle \mathrm{A}=\mathrm{a}$. Now, draw a line XY parallel to BC , passing through A .
  
  Then, $\angle \mathrm{XAB}=75^{\circ }$ and angle $\mathrm{YAC}=45^{\circ }[\because$ alternate angles $]$ We know that $\angle \mathrm{XAB},\angle \mathrm{YAC}$ and $\angle \mathrm{BAC}$ together form a straight line. So, $\angle \mathrm{XAB}+\angle \mathrm{YAC}+\angle \mathrm{BAC}=180^{\circ }$ $\Rightarrow 75^{\circ }+45^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow 120^{\circ }+\mathrm{a}=180^{\circ }$ $\Rightarrow \mathrm{a}=60^{\circ }$ Hence, the third angle of a triangle is $60^{\circ }$.
• Can you construct a triangle all of whose angles are equal to $70^{\circ }$ ? If two of the angles are $70^{\circ }$, what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out. Sol. We know that the sum of the angles of a triangle is $180^{\circ }$. Here, $70^{\circ }+70^{\circ }+70^{\circ }=210^{\circ }>180^{\circ }$. So, we cannot construct a triangle whose all of angles are equal to $70^{\circ }$. If two angles are $70^{\circ }$ each, then the sum of these two angles is $70+70^{\circ }=140^{\circ }$ Therefore, the third angle $=180^{\circ }-140^{\circ }=40^{\circ }$. Let all the angles of a triangle to be x . Then, by angle sum property, $\mathrm{x}+\mathrm{x}+\mathrm{x}=180^{\circ }$ $\Rightarrow 3\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}=60^{\circ }$ Hence, if all the angles of a triangle are equal, then each of them measures to $60^{\circ }$.
• Here is a triangle in which we know $\angle \mathrm{B}=\angle \mathrm{C}$ and $\angle \mathrm{A}=50^{\circ }$. Can you find $\angle \mathrm{B}=\angle \mathrm{C}$ ? Sol. We know that the sum of all angles of a triangle is $180^{\circ }$. $\begin{array}{ll}\therefore & \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }\\ \Rightarrow & 50^{\circ }+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }\\ \Rightarrow & \angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }-50^{\circ }=130^{\circ }\\ \Rightarrow & \angle \mathrm{B}+\angle \mathrm{B}=130^{\circ }\\ \Rightarrow & 2\angle \mathrm{B}=130^{\circ }\\ \Rightarrow & \angle \mathrm{B}=\frac{130}{2}=65^{\circ }\end{array}$ Therefore, the values of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ are $65^{\circ }$.

FIGURE IT OUT-9

• Construct a $△\mathrm{ABC}$ with $\mathrm{BC}=5\mathrm{cm},\mathrm{AB}=6\mathrm{cm},\mathrm{CA}=5\mathrm{cm}$. Construct an altitude from A to BC . Sol. The steps of construction are given as (i) Draw a line segment AB of length 6 cm using a ruler and a pencil.
  
  (ii) Place the compass pointer at point A and open it to 5 cm . Draw a circle.
  
  (iii) Place the compass pointer at point B and open it to 5 cm . Draw another circle.
  
  (iv) Let point C be one of the points of intersection of these two circles. Now, join AC and BC . Then, $△\mathrm{ABC}$ is the required triangle.
  
  (v) Construction of an altitude Keep the ruler aligned to the base of $△\mathrm{ABC}$. Place the set square on the ruler such that one of the edges of the right angle touches the ruler and slide it along ruler till the vertical edge touches $C$. Then, draw an altitude to $AB$ through $C$.
  
• Construct a triangle TRY with $\mathrm{RY}=4\mathrm{cm},\mathrm{TR}=7\mathrm{cm}\angle \mathrm{R}=140^{\circ }$. Construct an altitude from T to RY. Sol. The steps of construction are given as (i) Draw a line segment TR of length 7 cm using a ruler and a pencil.
  
  (ii) At point R , draw an angle $140^{\circ }$ using protector and mark X .
  
  (iii) At point R , mark a point Y along RX such that $\mathrm{RY}=4\mathrm{cm}$
  
  (iv) Join the point Y to T . Then, $\Delta \mathrm{TRY}$ is the required triangle.
  
  (v) Construction of an altitude Keep the ruler aligned to the base RY. Place the set square on the ruler such that one of the edges of the right angle touches the ruler and slide it along ruler till the vertical edge touches T. Then, draw the altitude TM on extended RY.
  
  Hence, an altitude from $T$ to $RY$ is outside of a $△TRY$.
• Through construction, explore if it is possible to construct an equilateral triangle that is (i) right angled (ii) obtuse angled. Also, construct an isosceles triangle that is (i) right angled (ii) obtuse angled Sol. (i) Consider the length of sides of an equilateral triangle to be 3 cm . The steps of the construction are given as 1.Draw a line segment AB of length 3 cm using pencil and ruler.
  
  2.At point A , draw an angle of $90^{\circ }$ using protector and mark X .
  
  3.At point A , mark a point C along AX such that $\mathrm{AC}=3\mathrm{cm}$.
  
  4.Join BC. Here, we observe that the length of BC is no longer 3 cm . Hence, it is not possible to construct an equilateral triangle that is right angled but can construct an isosceles triangle. (ii) Consider the length of sides of an equilateral triangle to be 3 cm and an obtuse angle to be $110^{\circ }$. The steps of the construction are given as 1.Draw a line segment AB of length 3 cm using ruler and pencil. $A-3\mathrm{cm}B$

2.At point A , draw an angle of $110^{\circ }$ using protector and mark X .

3.At point A mark a point C along AX such that $\mathrm{AC}=3\mathrm{cm}$.

4.Join BC.

4.0Key Features of NCERT Solutions Class 7 Maths Chapter 7 :  A Tale of Three Intersecting Lines

• Step-by-Step Problem-Solving: All exercises from the NCERT Ganita Prakash textbook are solved with detailed, easy-to-follow steps. This methodical approach ensures students understand the reasoning and logical progression required to arrive at the correct answer.
• Visual Aids and Diagrams: Geometry is best understood visually. Our solutions frequently incorporate accurate diagrams and illustrations to help students visualize the lines, angles, and points of intersection, making abstract concepts concrete.
• Property and Theorem Reinforcement: The solutions emphasize and reinforce the key properties and theorems associated with each type of special line and point of concurrency (e.g., centroid dividing medians in a 2:1 ratio, incenter being equidistant from sides).
• Alignment with Updated NCERT Syllabus: The content strictly adheres to the latest NCERT Ganita Prakash curriculum for Class 7, ensuring that students are well-prepared for their examinations and aligned with current academic standards.
• Practice-Oriented Approach: With numerous solved examples and exercises, students get ample opportunity to practice and apply the learned concepts, building confidence and proficiency in geometric problem-solving.
• Downloadable PDF Format: Conveniently available as a PDF, these NCERT Solutions can be accessed anytime, anywhere, facilitating flexible and self-paced learning for all students.