Students often refer to Class 7 Maths Notes and Chapter 7 A Tale of Three Intersecting Lines Class 7 Notes during last-minute revisions.
Class 7 Maths Chapter 7 Notes A Tale of Three Intersecting Lines
Class 7 Maths Notes Chapter 7 – Class 7 A Tale of Three Intersecting Lines Notes
→ There are many interesting facts about triangles. Triangles have three sides and three angles, and the sum of their angles is always 180°. We know that equilateral triangles have all sides and angles equal, and it can be constructed with any side length. The measure of each angle of an equilateral triangle is 60°.
→ A triangle can be constructed when the following measurements are given:
- All the three sides
- Two of the sides and their included angle.
- Two angles and the included side.
→ Classification of Triangles based on sides:
- Equilateral Triangle: A triangle that has all its sides of equal length.
- Isosceles Triangle: A triangle that has two of its sides of equal length.
- ScalareTriangle: A triangle in which all sides arc unequal in length.
→ Triangle Inequality: The sum of the lengths of any two sides of a triangle is always greater than the third side.. ‘
→ The side lengths of a triangle must satisfy the triangle inequality. Ifa given set of lengths meets this criterion, a triangle can be constructed with those side lengths.
→ Angle Sum Property: The sum of all three angles of a triangle is 1800.
→ Classification of Triangles based on Angles:
- Acute-angled Triangle: A triangle that has all acute angles (less than 90e).
- Obtuse-angled Triangle: A triangle in which one angle is obtuse (greater than 90°).
- Right-angled Triangle: A triangle that has one right angle.
→ Exterior Angle: The angle formed between the extension of the side of a triangle and the other side is called an exterior angle of the triangle.
→ Exterior Angle Property: In a triangle, the measure of an exterior angle is equal to the sum of its two interior opposite angles.
→ Altitude of la Triangle: A perpendicular line from the vertex of a triangle to its opposite side.
→ Use of a compass simplifies the construction of triangles when the side lengths are given.
→ A set of three lengths where the length of each is smaller than the sum of the other two is said to satisfy the triangle inequality.
→ Sidelengths of a triangle satisfy the triangle inequality, and if a given set of lengths satisfies the triangle inequality, a triangle can be constructed with those sidelengths.
→ Triangles can be constructed when the following measurements are given:
- two of the sides and their included angle.
- two angles and the included side.
→ The sum of the angles of a triangle is always 180°.
→ The altitude of a triangle is a perpendicular line segment from a vertex to its opposite side.
→ Equilateral triangles have sides of equal length. Isosceles triangles have two sides of equal length. Scalene triangles have sides of three different lengths.
→ Triangles are classified based on their angle measures as acute-angled, right-angled, and obtuse-angled triangles.
→ A triangle is the most basic closed shape. As we know, it consists of:
- three corner points, that we call the vertices of the triangle, and
- three line segments or the sides of the triangle that join the pairs of vertices.
→ Triangles come in various shapes. Some of them are shown below.
→ Observe the symbol used to denote a triangle and how the triangles are named using their vertices. While naming a triangle, the vertices can come in any order.
→ The three sides meeting at the corners give rise to three angles that we call the angles of the triangle.
→ For example, in ∆ABC, these angles are ∠CAB, ∠ABC, ∠BCA, which we simply denote as ∠A, ∠B and ∠C, respectively.
Triangles and Its Classification
→ Let A, B and C be three points but not in a straight line. Then, the figure formed by the three line segments AB, BC and CA is called a triangle. Triangle ABC is denoted as △ ABC.
→ △ ABC has
- three vertices, namely A, B and C.
- three sides, namely, AB, BC and CA.
- three angles, namely ∠BAC, ∠ABC and ∠BCA which are also denoted by ∠A, ∠B and ∠C, respectively.
→ There are many familiar objects that have this triangular shape.
→ e.g. A set square, a road sign with a warning, etc.
Classification of Triangles
→ On the basis of the length of their sides, triangles can be classified as
- Scalene triangle A triangle having all its sides of different lengths (unequal), is called a scalene triangle.
- Isosceles triangle A triangle having two of its sides equal, is called an isosceles triangle.
- Equilateral triangle A triangle having all its sides equal, is called an equilateral triangle.
→ On the basis of the measure of their angles, triangles can be classified as
- Acute angled triangle A triangle each of whose angles measures less than 90°, is called an acute angled triangle.
- Right angled triangle A triangle one of whose angles measures 90°, is called a right angled triangle.
- Obtuse angled triangle A triangle one of whose angles measures more than 90°, is called an obtuse angled triangle.
Triangle Inequality
→ The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
→ Let in △ ABC, BC = a units, AC = b units and AB = c units.
→ According to the rule, a + b > c, b + c > a and c + a > b.
→ This rule helps us to decide whether a triangle can be formed with given side lengths or not.
Note
- The difference between the lengths of any two sides of a triangle is always smaller than the length of the third side.
- When the sum of two given angles is less than 180°, a triangle exists with these angles.
- If the sum of two given angles is greater than or equal to 180°, there is no triangle with these angles.
Angle Sum Property of a Triangle
→ The sum of measures of all the three angles of a triangle is 180°.
→ In △ PQR, ∠P + ∠Q + ∠R = 180°
Exterior Angles
→ The angle formed between the extension of a side of a triangle and the other side is called an exterior angle of the triartgle.
If the side B C of △ ABC is extended to X, then ∠ACX is the exterior angle at vertex C of △ ABC.
By angle sum property of a triangle,
∠A + ∠B + ∠C = 180° …..(i)
and by linear pair,
∠C + ∠ACX = 180° …..(ii)
→ ∠C = 180° – ∠ACX
Now, from Eqs. (i) and (ii),
∠A + ∠B + 180° – ∠ACX = 180°
→ ∠A + ∠B = ∠ACX
Hence, an exterior angle of a triangle is equal to the sum of its interior opposite angles.
Altitude of a Triangle
→ The perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of a triangle.
e.g. In the given figure, the line segment AL is an altitude of the △ ABC.
Note The altitude is also called the height of the triangle.
Constructions
Constructions of the Altitudes of a Triangle
→ In order to construct an altitude of a triangle, the following steps should be followed
1. Keep the ruler aligned to the base of △ ABC on which the altitude is to be drawn. Place the set square on the ruler such that one of the edges of the right angle touches the ruler.
2. Slide the set square along the ruler till the vertical edge of the set square touches the vertex A.
3. Draw the altitude to BC through A using the vertical edge of the set square. Let it intersects BC at D. Then, AD is an altitude of a △ ABC.
Equilateral Triangles Class 7 Notes
Among all the triangles, the equilateral triangles are the most symmetric ones. These are triangles in which all the sides are of equal lengths. Let us try constructing them.
Construct a triangle in which all the sides are of length 4 cm How did you construct this triangle, and what tools did you use? Can this construction be done only using a marked ruler (and a pencil)?
Constructing this triangle using just a ruler is certainly possible. But this might require several trials. Say we draw the base — let us call it AB — of length 4 cm (see the figure below), and mark the third point C using a ruler such that AC = 4 cm. This may not lead to BC also having a length of 4 cm. If this happens, we will have to keep making attempts to mark C till we get BC to be 4 cm long.
How do we make this construction more efficient?
Recall solving a similar problem in the previous year using a compass (in the Chapter ‘Playing with Constructions’).
We had to mark the top point of a ‘house’ which is 5 cm from two other points. The method we used to get that point can also be used here.
After constructing AB = 4 cm, we can do the following.
Step 1: Using a compass, construct a sufficiently long arc of radius 4 cm from A, as shown in the figure. Point C is somewhere on this arc. How do we mark it?
Step 2: Construct another arc of radius 4 cm from B.
Let C be the point of intersection of the arcs.
Step 3: Join AC and BC to get the required equilateral triangle.
Constructing a Triangle When its Sides are Given Class 7 Notes
How do we construct triangles that are not equilateral?
Construct a triangle of sidelength 4 cm, 5 cm, and 6 cm.
As in the previous case, this triangle can also be constructed using just a marked ruler. But it will involve several trials.
How do we construct this triangle more efficiently?
Choose one of the given lengths to be the base of the triangle: say 4 cm. Draw the base. Let A and B be the base vertices, and call the third vertex C. Let AC = 5 cm and BC = 6 cm.
Like we did in the case of equilateral triangles, let us first get all the points that are at a 5 cm distance from A. These points lie on the circle whose centre is A and has radius 5 cm. Point C must lie somewhere on this circle. How do we find it?
We will make use of the fact that point C is 6 cm away from B.
Construct an arc of radius 6 cm from B.
The required point C is one of the points of intersection of the two circles.
The reason why the point of intersection is the third vertex is the same as for equilateral triangles. This point lies on both circles.
Hence, its distance from A is the radius of the circle centred at A (5 cm), and its distance from B is the same as the radius of the circle centred at B (6 cm).
Let us summarise the steps of construction, noting that constructing full circles is not necessary to get the third vertex (See Figures).
- Step 1: Construct the base AB with one of the side lengths. Let us choose AB = 4 cm (see Fig.).
- Step 2: From A, construct a sufficiently long arc of radius 5 cm (see Fig.).
- Step 3: From B, construct an arc of radius 6 cm such that it intersects the first arc (see Fig.).
- Step 4: The point where both the arcs meet is the required third vertex C. Join AC and BC to get ∆ABC.
Construct
We have seen that triangles having all three equal sides are called equilateral triangles. Those having two equal sides are called isosceles triangles.
Are Triangles Possible for any Lengths?
Can one construct triangles having any given sidelengths? Are there lengths for which it is impossible to construct a triangle? Let us explore this.
Try to find more sets of lengths for which a triangle construction is impossible. See if you can find any pattern in them.
We see that a triangle is possible for some sets of lengths and not possible for others. How do we check if a triangle exists for a given set of lengths? One way is to actually try to construct the triangle and check if it is possible. Is there a more effient way to check this?
Triangle Inequality
Consider the lengths 10 cm, 15 cm, and 30 cm. Does there exist a triangle having these as side lengths?
To tackle this question, let us study a property of triangles. Imagine a small plot of plain land having a tent, a tree, and a pole. Imagine you are at the entrance of the tent and want to go to the tree. Which is the shorter path: (i) the straight-line path to the tree (the red path) or
(ii) the straight-line path from the tent to the pole, followed by the straight-line path from the pole to the tree (the yellow path)?
The direct straight-line path from the tent to the tree is shorter than the roundabout path via the pole. The direct straight-line path is the shortest possible path to the tree from the tent.
Will the direct path between any two points be shorter than the roundabout path via a third point? The answer is yes.
Can this understanding be used to tell something about the existence of a triangle having sidelengths 10 cm, 15 cm and 30 cm?
Let us suppose that there is a triangle for this set of lengths. Remember that at this point we are not sure about the existence of the triangle but we are only supposing that it exists. Let us draw a rough diagram.
Does everything look right with this triangle?
If this triangle were possible, then the direct path between any two vertices should be shorter than the roundabout path via the third vertex. Is this true for our rough diagram?
Let us consider the paths between B and C.
Direct path length = BC = 10 cm
What is the length of the roundabout path via the vertex A? It is the sum of the lengths of line segments BA and AC.
Roundabout path length = BA + AC = 15 cm + 30 cm = 45 cm
Is the direct path length shorter than the roundabout path length? Yes.
Let us now consider the paths between A and B.
Direct path length = AB = 15 cm
Finding the length of the roundabout path via the vertex C, we get
Roundabout path length = AC + CB = 30 cm + 10 cm = 40 cm
Is the direct path length shorter than the roundabout path length? Yes.
Finally, consider paths between C and A.
Direct path length = CA = 30 cm
Roundabout path length = CB + BA = 10 cm + 15 cm = 25 cm
Is the direct path length shorter than the roundabout path length? In this case, the direct path is longer, which is absurd. Can such a triangle exist? No.
Therefore, a triangle having sidelengths 10 cm, 15 cm, and 30 cm cannot exist.
We are thus able to see without construction why a triangle for the set of lengths 10 cm, 15 cm, and 30 cm cannot exist. We have been able to figure this out through spatial intuition and reasoning.
Recall how we used similar intuition and reasoning to discover properties of intersecting and parallel lines. We will continue to do this as we explore geometry.
Given three sidelengths, what do we need to compare to check for the existence of a triangle?
When each length is smaller than the sum of the other two, we say that the lengths satisfy the triangle inequality. For example, the set 3, 4, 5 satisfis the triangle inequality whereas, the set 10, 15, 30 does not satisfy the triangle inequality. We have seen that lengths such as 10, 15, 30 that do not satisfy the triangle inequality cannot be the sidelengths of a triangle.
Does a triangle exist with sidelengths 4 cm, 5 cm and 8 cm?
This satisfies the triangle inequality: 8 < 4 + 5 = 9
Why do we not need to check the other two sides?
This means that all the direct path lengths are less than the roundabout path lengths. Does this confim the existence of a triangle?
If one of the direct path lengths had been longer, we could have concluded that a triangle would surely not exist. But in this case, we can only say that a triangle may or may not exist.
For the triangle to exist, the arcs that we construct to get the third vertex must intersect. Is it possible to determine that this will happen without actually carrying out the construction?
Visualising the Construction of Circles
Let us imagine that we start the construction by constructing the longest side as the base. Let AB be the base of length 8 cm. The next step is the construction of sufficiently long arcs corresponding to the other two lengths: 4 cm and 5 cm.
Instead of just constructing the arcs, let us complete the full circles.
Suppose we construct a circle of radius 4 cm with A as the centre.
Now, suppose that a circle of radius 5 cm is constructed, centred at B.
Can you draw a rough diagram of the resulting figure?
Note that in the figure below, AX = 4 cm and AB = 8 cm. So, what is BX?
Does this length help in visualising the resulting figure?
Since BX = 4 cm, and the radius of the circle centred at B is 5 cm, the circles will intersect each other at two points.
What does this tell us about the existence of a triangle?
The points A and B along with either of the points of intersection of the circles will give us the required triangle. Thus, there exists a triangle having sidelengths 4 cm, 5 cm and 8 cm.
We observe from the previous problems that whenever there is a set of lengths satisfying the triangle inequality (each length < sum of the other two lengths), there is a triangle with those three lengths as sidelengths.
Will triangles always exist when a set of lengths satisfies the triangle inequality? How can we be sure?
We can be sure of the existence of a triangle only if we can show that the circles intersect internally (as in Fig.) whenever the triangle inequality is satisfied. But are there other possibilities when the two circles are constructed? Let us visualise and study them.
The following different cases can be conceived:
Case 1: Circles touch each other
Case 2: Circles do not intersect
Case 3: Circles intersect each other internally
Note that while constructing the circles, we take
(a) the length of the base AB = longest of the given length
(b) The radii of the circles are to be the smaller two lengths.
Which of the above-mentioned cases will lead to the formation of a triangle? Triangles are formed only when the circles intersect each other internally (Case 3).
Let us study each of these cases by finding the relation between the radii (the smaller two lengths) and AB (the longest length).
Case 1: Circles touch each other at a point
For this case to happen,
sum of the two radii = AB
or
sum of the two smaller lengths = longest length
Case 2: Circles do not Intersect Internally
For this case to happen, what should be the relation between the radii and AB?
It can be seen from the figure that,
sum of the two radii < AB
or
sum of the two smaller lengths < longest length
Case 3: Circles intersect each other
AB is composed of one radius and a part of the other. So,
sum of the two radii > AB,
or
sum of the two smaller lengths > longest length
Can we use this analysis to tell if a triangle exists when the lengths satisfy the triangle inequality?
If the given lengths satisfy the triangle inequality, then the sum of the two smaller lengths is greater than the longest length. This means that this will lead to Case 3 where the circles intersect internally, and so a triangle exists.
Conclusion
If a given set of three lengths satisfies the triangle inequality, then a triangle exists having those as side lengths. If the set does not satisfy the triangle inequality, then a triangle with those side lengths does not exist.
Construction of Triangles When Some Sides and Angles are Given Class 7 Notes
We have seen how to construct triangles when their sidelengths are given. Now, we will take up constructions where in place of some sidelengths, angle measures are given.
Two Sides and the Included Angle
How do we construct a triangle if two sides and the angle included between them are given? Here are some examples of measurements showing the included angle.
Construct a triangle ABC with AB = 5 cm, AC = 4 cm and ∠A= 45°.
Let us take one of the given sides, AB, as the base of the triangle.
Step 1: Construct a side AB of length 5 cm.
Step 2: Construct ∠A = 45° by drawing the other arm of the angle.
Step 3: Mark the point C on the other arm such that AC = 4 cm.
Step 4: Join BC to get the required triangle.
Two Angles and the Included Side
In this case, we are given two angles and the side that is a part of both angles, which we call the included side. Here are some examples of such measurements:
Construct a triangle ABC where AB = 5 cm, ∠A = 45°, and ∠B = 80°.
Let us take the given side as the base.
Step 1: Draw the base AB of length 5 cm.
Step 2: Draw ∠A and ∠B of measures 45°, and 80° respectively.
Step 3: The point of intersection of the two new line segments is the third vertex C.
Do triangles always exist?
Do triangles exist for every combination of two angles and their included side? Explore.
As in the case when we are given all three sides, it turns out that there is not always a triangle for every combination of two angles and the included side.
Find examples of measurements of two angles with the included side where a triangle is not possible.
Let us try to visualise such a situation. Once the base is drawn, try to imagine how the other sides should be so that they do not meet. Here are some obvious examples.
If the two angles are greater than or equal to a right angle (90°), then it is clear that a triangle is not possible.
Now we make one of the base angles an acute angle, say 40°. What are the possible values that the other angle should take so that the lines don’t meet?
It is clear that if the line from B is “inclined” sufficiently to the right, then it will not meet the line l.
(a) Try to find a possible ∠B (marked in the figure) for this to happen.
(b) What could be the smallest value of ∠B for the lines not to meet?
The blue line is the line with the least rightward bend that doesn’t meet the line ‘l’.
Visually, it is clear that the line that creates the smallest ∠B has to be the one parallel to 1. Let us call this parallel line m.
Can you tell the actual value of ∠B be in this case?
[Hint: Note that AB is the transversal.]
We have seen that when two lines are parallel, the internal angles on the same side of the transversal add up to 180°. So ∠B = 140°. So, for what values of ∠B, does a triangle not exist? Does the length AB play any part here?
From the discussion above, it can be seen that the length AB does not play any part in deciding the existence of a triangle. We can say that a triangle does not exist when ∠B is greater than or equal to 140°.
Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible? Can the sum of the two angles be used for framing this rule?
When the sum of two given angles is less than 180°, a triangle exists with these angles. If the sum is greater than or equal to 180°, there is no triangle with these angles.
Let us take two angles, say 60° and 70°, whose sum is less than 180°.
Let the included side be 5 cm.
In general, once the two angles are fixed, does the third angle depend on the included sidelength? Try with different pairs of angles and lengths.
The measurements might show that the sidelength has no effct (or a very small effect) on the third angle. With this observation, let us see if we can find the third angle without carrying out the construction and measurement.
Consider a triangle ABC with ∠B = 50° and ∠C = 70°. Let us see how we can find ∠A without construction.
We saw that the notion of parallel lines was useful to determine that the sum of any two angles of a triangle is less than 180°. Parallel lines can be used to find the third angle, ∠BAC as well.
Let us suppose we construct a line XY parallel to BC through vertex A.
We can see new angles being formed here: ∠XAB and ∠YAC. What are their values?
Since the line XY is parallel to BC, ∠XAB = ∠B and ∠YAC = ∠C, because they are alternate angles of the transversals AB and AC.
Therefore, ∠XAB = 50°, and ∠YAC = 70°. Can we find ∠BAC from this?
We know that ∠XAB, ∠YAC, and ∠BAC together form 180°.
So, ∠XAB + ∠YAC + ∠BAC = 180°
50° + ∠BAC + 70° = 180°
120° + ∠BAC = 180°
Thus, ∠BAC = 60°
Now construct a triangle (taking BC to be of any suitable length) and verify if this is indeed the case.
Angle Sum Property
What can we say about the sum of the angles of any triangle?
Consider a triangle ABC. To find the sum of its angles, we can use the same method of drawing a line parallel to the base: construct a line through A that is parallel to BC.
We need to find ∠A + ∠B + ∠C.
We know that ∠B = ∠XAB, ∠C = ∠YAC.
So, ∠A + ∠B + ∠C = ∠A + ∠XAB + ∠YAC = 180°, as together they form a straight angle.
Thus, we have proved that the sum of the three angles in any triangle is 180°! This rather surprising result is called the angle sum property of triangles.
Take some time to reflect upon how we figured out the angle sum property. In the beginning, the relationship between the third angle and the other two angles was not at all clear. However, a simple idea of drawing a line parallel to the base through the top vertex (as in Fig.) suddenly made the relationship obvious. This ingenious idea can be found in a very inflential book in the history of mathematics called ‘The Elements’. This book is attributed to the Greek mathematician Euclid, who lived around 300 BCE. This solution is yet another example of how creative thinking plays a key role in mathematics!
There is a convenient way of verifying the angle sum property by folding a triangular cut-out of a paper. Do you see how this shows that the sum of the angles in this triangle is 180°?
Exterior Angles
The angle formed between the extension of a side of a triangle and the other side is called an exterior angle of the triangle. In this figure, ∠ACD is an exterior angle.
Find ∠ACD, if ∠A = 50°, and ∠B = 60°.
From the angle sum property, we know that
50° + 60° + ∠ACB = 180°
110° + ∠ACB = 180°
So, ∠ACB = 70°
So, ∠ACD = 180 ° – 70° = 110°,
Since ∠ACB and ∠ACD together form a straight angle.
Find the exterior angle for different measures of ∠A and ∠B. Do you see any relation between the exterior angle and these two angles?
[Hint: From the angle sum property, we have ∠A + ∠B + ∠ACB = 180°.]
We also have ∠ACD + ∠ACB = 180°, since they form a straight angle. What does this show?
Constructions Related to Altitudes of Triangles Class 7 Notes
There is another set of useful measurements concerning a triangle — the height of each of its vertices concerning the opposite sides.
In the world around us, we talk of the heights of various objects: the height of a person, the height of a tree, the height of a building, etc. What do we mean by the word ‘height’?
Consider a triangle ABC. What is the height of the vertex A from its opposite side BC, and how can it be measured?
Let AD be the line segment from A drawn perpendicular to BC. The length of AD is the height of the vertex A from BC. The line segment AD is said to be one of the ‘altitudes’ of the triangle. The other altitudes are BE and CF in the figure below: the perpendiculars drawn from the other vertices to their respective opposite sides.
Whenever we use the word height of the triangle, we generally refer to the length of the altitude to whatever side we take as base (this altitude is AD in the case of Fig.).
What would the altitude from A to BC be in this triangle?
We extend BC and then drop the perpendicular from A to this line.
Altitudes Using Paper Folding
Cut out a paper triangle. Fix one of the sides as the base. Fold it in such a way that the resulting crease is an altitude from the top vertex to the base. Justify why the crease formed should be perpendicular to the base.
Construction of the Altitudes of a Triangle
Construct an arbitrary triangle. Label the vertices A, B, C, taking BC to be the base.
Construct the altitude from A to BC Constructing the altitude using just a ruler is not accurate. To get a more precise angle of 90°, we use a set square along with a ruler. Can you see how to do this?
Step 1: Keep the ruler aligned with the base. Place the set square on the ruler as shown, such that one of the edges of the right angle touches the ruler.
Step 2: Slide the set square along the ruler till the vertical edge of the set square touches the vertex A.
Step 3: Draw the altitude to BC through A using the vertical edge of the set square.
Does there exist a triangle in which a side is also an altitude?
Visualise such a triangle and draw a rough diagram. We see that this happens in triangles where one of the angles is a right angle.
Triangles having one right angle are called right-angled triangles or simply right triangles.
Altitude from A to BC
Types of Triangles Class 7 Notes
In our study of triangles, we have encountered the following types of triangles: equilateral, isosceles, scalene, and right-angled triangles.
Did you spot any other type of triangle?
The classification of triangles as equilateral and isosceles was based on the equality of sides.
- Equilateral triangles have sides of equal length.
- Isosceles triangles have two sides of equal length.
- Scalene triangles have sides of three different lengths.
Can a similar classification be done based on equality of angles? Is there any relation between these two classifiations? We will answer these questions in a later chapter.
We used angle measures when classifying a triangle as a rightangled triangle.
What are the other types of triangles based on angle measures?
A classification of triangles based on their angle measures is acuteangled, right-angled and obtuse-angled triangles. We have already seen what a right-angled triangle is. It is a triangle with one right angle. Similarly, an obtuse-angled triangle has one obtuse angle.
What could an acute-angled triangle be? Can we define it as a triangle with one acute angle? Why not?
In an acute-angled triangle, all three angles are acute angles.
The post A Tale of Three Intersecting Lines Class 7 Notes Maths Chapter 7 appeared first on Learn CBSE.
📚 NCsolve - Your Global Education Partner 🌍
Empowering Students with AI-Driven Learning Solutions
Welcome to NCsolve — your trusted educational platform designed to support students worldwide. Whether you're preparing for Class 10, Class 11, or Class 12, NCsolve offers a wide range of learning resources powered by AI Education.
Our platform is committed to providing detailed solutions, effective study techniques, and reliable content to help you achieve academic success. With our AI-driven tools, you can now access personalized study guides, practice tests, and interactive learning experiences from anywhere in the world.
🔎 Why Choose NCsolve?
At NCsolve, we believe in smart learning. Our platform offers:
- ✅ AI-powered solutions for faster and accurate learning.
- ✅ Step-by-step NCERT Solutions for all subjects.
- ✅ Access to Sample Papers and Previous Year Questions.
- ✅ Detailed explanations to strengthen your concepts.
- ✅ Regular updates on exams, syllabus changes, and study tips.
- ✅ Support for students worldwide with multi-language content.
🌐 Explore Our Websites:
🔹 ncsolve.blogspot.com
🔹 ncsolve-global.blogspot.com
🔹 edu-ai.blogspot.com
📲 Connect With Us:
👍 Facebook: NCsolve
📧 Email: ncsolve@yopmail.com
😇 WHAT'S YOUR DOUBT DEAR ☕️
🌎 YOU'RE BEST 🏆