NCERT Solutions Class 7 Mathematics Ganita Prakash 1 Chapter 07 A Tale of Three Intersecting Lines

Question. What happens when the three vertices lie on a straight line?
Answer: When three vertices are collinear (positioned on the same straight line), they cannot form a triangle. This is because the three points do not enclose any space; instead, they simply rest along one continuous straight path.
In simple words: If three points sit on one straight line, they cannot make a triangle because there is no enclosed area.

Exam Tip: Remember that a triangle must have three vertices that are non-collinear - no three vertices can lie on the same line for a valid triangle to exist.

7.2 Constructing a Triangle When its Sides are Given

Question. Construct triangles having the following sidelengths (all the units are in cm): (a) 4, 4, 6 (b) 3, 4, 5 (c) 1, 5, 5 (d) 4, 6, 8 (e) 3.5, 3.5, 3.5
Answer:
(a) Steps of Construction:
Step 1: Draw the base AB = 6 cm.
Step 2: From point A, draw a long arc with radius 4 cm.
Step 3: From point B, draw an arc with radius 4 cm so that it crosses the first arc.
Step 4: Mark the crossing point as vertex C. Connect AC and BC to complete triangle ABC.

(b) Steps of Construction:
Step 1: Draw the base AB = 3 cm.
Step 2: From point A, draw a long arc with radius 4 cm.
Step 3: From point B, draw an arc with radius 5 cm so that it intersects the first arc.
Step 4: Mark the intersection point as vertex C. Join AC and BC to form triangle ABC.

(c) Steps of Construction: Work through this yourself following the same method as (a) and (b). (d) Steps of Construction: Work through this yourself using the same approach. (e) Steps of Construction:
Step 1: Draw the base AB = 3.5 cm.
Step 2: From point A, draw a long arc with radius 3.5 cm.
Step 3: From point B, draw another arc with radius 3.5 cm so it meets the first arc.
Step 4: Mark the meeting point as vertex C. Join AC and BC to get triangle ABC.


In simple words: To build a triangle when you know all three side lengths, draw one side first. Then use arcs from each end of that side - each arc's radius matches the length of another side. Where the arcs meet is your third vertex.

Exam Tip: Always verify that your arcs actually cross. If they do not meet, those three lengths cannot form a triangle - this signals a violation of the triangle inequality.

Question 1. Use the points on the circle and/or the centre to form isosceles triangles.
Answer: Pick any two points on the circle and connect each one to the circle's center. Then connect these two points to each other. This creates an isosceles triangle because the two line segments from the center to the circle are both radii, so they are equal in length.
In simple words: Draw two radii to any two points on the circle. Connecting those points creates an isosceles triangle because both radii are the same length.

Exam Tip: Remember that any triangle formed by two radii and a chord must be isosceles, since all radii of the same circle are equal.

Question 2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.
Answer: Isosceles triangles: Connect the two points where two circles intersect to one center point of either circle. This forms an isosceles triangle because the two radii from that center to the intersection points are equal. Examples are triangles AXY and BXY.

Equilateral triangles (method 1): Connect the centers of two equal circles and one of their intersection points. All three sides equal the radius, so this is equilateral. Examples are triangles AXB and AYB.

Equilateral triangles (method 2): Connect the centers of three equal circles. Since each pair of circles is the same size, the distance between any two centers equals the radius, forming an equilateral triangle. Triangle ABC is an equilateral triangle. Also, triangles PAB, QBC, and RAC are equilateral. Note: an equilateral triangle is also a special case of an isosceles triangle, so all these equilateral triangles are isosceles as well.
In simple words: Two radii and a chord make an isosceles triangle. Two radii plus one intersection point make an equilateral triangle. Three circle centers that are equally spaced also make an equilateral triangle.

Exam Tip: Equilateral triangles are always isosceles - they satisfy both conditions. When identifying these shapes, always check which sides are equal.

Are Triangles Possible for any Lengths?

Question. Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle?
Answer: When you attempt this construction, the arcs you draw from points A and B do not meet. This means you cannot build a triangle with these three side lengths. The longest side (8 cm) is too long compared to the sum of the other two sides (3 + 4 = 7 cm), so they cannot close to form a triangle.
In simple words: The two arcs do not cross, so no triangle forms. This happens when one side is too long for the other two sides to reach across.

Exam Tip: Before attempting construction, check the triangle inequality - if the longest side is greater than or equal to the sum of the other two, the triangle cannot exist.

Question. Check if a triangle is possible for sidelengths 2 cm, 3 cm, and 6 cm.
Answer: The arcs drawn from points A and B do not meet, so a triangle is not possible with these side lengths. Here, 6 cm (the longest side) is equal to 2 + 3 + 1 = 6 cm, which means the two shorter sides cannot reach the required length to close the triangle.
In simple words: Again, the arcs do not cross. One side is so long that the other two sides cannot bridge the gap.

Exam Tip: A triangle fails to exist whenever the sum of any two sides is less than or equal to the third side.

Triangle Inequality

Question. Can we say anything about the existence of a triangle having sidelengths 3 cm, 3 cm, and 7 cm? Verify your answer by construction.
Answer: Let's compare the direct path with the roundabout path. If we take AB = 7 cm (the direct path), then the roundabout path BC + CA = 3 + 3 = 6 cm. Since the direct path (7 cm) is longer than the roundabout path (6 cm), a triangle cannot exist. When you try to construct it, the arcs from A and B will not meet, confirming that this triangle is impossible.
In simple words: The longest side is longer than the sum of the other two sides, so no triangle can form.

Exam Tip: Always check that every single side is smaller than the sum of the other two - if even one side fails this test, the triangle does not exist.

Question 1. We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm, and 8 cm; and 2 cm, 3 cm, and 6 cm. Check if you could have found this without trying to construct the triangle.
Answer: (a) Consider AB = 4 cm, BC = 3 cm, and AC = 8 cm.
Direct path BC = 3 cm; roundabout path BA + AC = 4 + 8 = 12 cm. The direct path is shorter.
Direct path AB = 4 cm; roundabout path AC + BC = 8 + 3 = 11 cm. The direct path is shorter.
Direct path AC = 8 cm; roundabout path AB + BC = 4 + 3 = 7 cm. Here the direct path is longer than the roundabout path, which violates the triangle rule. Therefore, a triangle cannot exist.

(b) Consider AB = 3 cm, BC = 2 cm, AC = 6 cm.
Direct path AC = 6 cm; roundabout path AB + BC = 3 + 2 = 5 cm. Since the direct path is longer than the roundabout path, a triangle cannot exist.
In simple words: Before drawing anything, compare each side to the sum of the other two. If any side is bigger than that sum, you immediately know no triangle is possible.

Exam Tip: The comparison method using direct and roundabout paths is faster than construction and saves time in exams.

Question 2. Can we say anything about the existence of a triangle for each of the following sets of lengths? (a) 10 km, 10 km, and 25 km (b) 5 mm, 10 mm, and 20 mm (c) 12 cm, 20 cm, and 40 cm
Answer: (a) Direct path = 25 km; roundabout path = 10 + 10 = 20 km. Since the direct path is longer, a triangle cannot exist.

(b) Direct path = 20 mm; roundabout path = 10 + 5 = 15 mm. Since the direct path is longer, a triangle cannot exist.

(c) Direct path = 40 cm; roundabout path = 12 + 20 = 32 cm. Since the direct path is longer, a triangle cannot exist.

Note: Using a rough diagram to compare direct path lengths with their matching roundabout path lengths is equivalent to comparing each length with the sum of the other two lengths. Three such comparisons are needed to be thorough.
In simple words: Check whether the longest side is bigger than the sum of the other two. If yes, no triangle exists.

Exam Tip: Focus on the longest side first - if it is less than the sum of the other two, a triangle might exist. Always check all three comparisons to be certain.

Question 3. For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm, and 30 cm, there are two comparisons where this happens: 10 < 15 + 30 and 15 < 10 + 30. But this doesn't happen for the third length: 30 > 10 + 15.
Answer: Work through this yourself by testing different sets of lengths and observing the pattern.

Exam Tip: When checking if a triangle can exist, at least two sides must each be smaller than the sum of the other two - but this alone is not sufficient. All three must satisfy the condition.

Question. Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than the sum of the other two? Explore different sets of lengths.
Answer: (i) For 5 mm, 10 mm, 20 mm: Two comparisons show this: 10 < 5 + 20 and 5 < 10 + 20. However, 20 > 10 + 5.

(ii) For 12 cm, 20 cm, and 40 cm: Two comparisons show this: 12 < 20 + 40 and 20 < 12 + 40. However, 40 > 12 + 20.

Further, for any set of lengths, it is possible to identify which lengths will immediately be smaller than the sum of the other two without needing calculations - simply arrange the direct lengths in increasing order. Yes, if you arrange the side lengths from smallest to largest, the two smaller sides will automatically be less than the sum of the other two. This gives you a quick way to spot which comparison might fail.
In simple words: When you order the sides from smallest to biggest, the smallest sides will always be less than the sum of the others. Only the largest side might fail this test.

Exam Tip: Always check the longest side first against the sum of the other two - this single comparison often tells you if a triangle can exist.

Question 1. Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure. (a) 2, 2, 5 (b) 3, 4, 6 (c) 2, 4, 8 (d) 5, 5, 8 (e) 10, 20, 25 (f) 10, 20, 35 (g) 24, 26, 28
Answer: A set of lengths can be the sidelengths of a triangle only if each length is smaller than the sum of the other two lengths.

(a) 2 < 5 + 2, but 5 > 2 + 2. Since the third condition fails, 2, 2, 5 cannot form a triangle.

(b) 3 < 4 + 6, 4 < 3 + 6, and 6 < 4 + 3. All three conditions are satisfied, so 3, 4, 6 can form a triangle.

(c) 2 < 4 + 8 and 4 < 2 + 8, but 8 > 4 + 2. The third condition fails, so 2, 4, 8 cannot form a triangle.

(d) 5 < 5 + 8 and 8 < 5 + 5. All conditions hold, so 5, 5, 8 can form a triangle.

(e) 10 < 20 + 25, 20 < 25 + 10, and 25 < 10 + 20. All three conditions are met, so 10, 20, 25 can form a triangle.

(f) 10 < 20 + 35 and 20 < 10 + 35, but 35 > 10 + 20. The third condition fails, so 10, 20, 35 cannot form a triangle.

(g) 24 < 26 + 28, 26 < 24 + 28, and 28 < 24 + 26. All conditions are satisfied, so 24, 26, 28 can form a triangle.
In simple words: Test each side - it must be smaller than adding the other two together. If all three sides pass this test, a triangle is possible.

Exam Tip: Instead of checking all three conditions one by one, focus on the longest side - if it is less than the sum of the other two, you almost certainly have a valid triangle.

Question. How will the two circles turn out for a set of lengths that do not satisfy the triangle inequality? Find 3 examples of sets of lengths for which the circles: (a) touch each other at a point, (b) Do not intersect.
Answer: When three segment lengths fail to satisfy the triangle inequality, those segments cannot form a triangle. However, two circles with these lengths as distances between their centers and points on their circumference can still behave in different ways.

(a) Circles that touch each other at exactly one point:
(i) 3, 4, 7
(ii) 5, 2, 3
(iii) 6, 2, 4

(b) Circles that do not intersect at all:
(i) 3, 4, 8
(ii) 6, 2, 3
(iii) 5, 1, 2
In simple words: When lengths do not form a triangle, circles built from those lengths can still touch at one point or stay completely apart - but they never cross at two points.

Exam Tip: Understand the difference between circle configurations and triangle existence - lengths that fail the triangle test still produce predictable circle relationships.

Question 1. Check if a triangle exists for each of the following set of lengths: (a) 1, 100, 100 (b) 3, 6, 9 (c) 1, 1, 5 (d) 5, 10, 12
Answer: A triangle exists when each length is smaller than the sum of the other two - this condition is known as the triangle inequality.

(a) 1 < 100 + 100 and 100 < 100 + 1. Both conditions hold, so a triangle exists with sidelengths 1, 100, 100.

(b) 3 < 6 + 9 and 6 < 3 + 9, but 9 = 6 + 3 (not less than). Since equality is not allowed, a triangle does not exist with sidelengths 3, 6, 9.

(c) 1 < 1 + 5, but 5 > 1 + 1. The second condition fails, so a triangle does not exist with sidelengths 1, 1, 5.

(d) 5 < 10 + 12, 10 < 5 + 12, and 12 < 10 + 5. All three conditions are satisfied, so a triangle exists with sidelengths 5, 10, and 12.
In simple words: Check if each side is less than (not equal to) the sum of the other two. If yes for all three, a triangle can exist.

Exam Tip: The triangle inequality requires strict inequality (less than, not less than or equal to) - if any side equals the sum of the other two, no triangle forms.

Question 2. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.
Answer: Yes, an equilateral triangle with all sides measuring 50, 50, 50 exists because the sum of two sides is greater than the third side. In an equilateral triangle, all sides are equal, so this condition is automatically satisfied. In fact, an equilateral triangle always exists for any positive sidelength. For any positive number, call it x, where x > 0, you can build an equilateral triangle with all sidelengths equal to x.
In simple words: Since all three sides of an equilateral triangle are equal, each side is always less than the sum of the other two. So equilateral triangles always exist, no matter what the side length is.

Exam Tip: Equilateral triangles are the simplest to verify for the triangle inequality - just check that the side length is positive, and you are done.

Question 3. For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen): (a) 1, 100 (b) 5, 5 (c) 3, 7
Answer: (a) Five possible values for the third length are 99.5, 99.8, 100, 100.5, 100.9. We can verify: 100 < 1 + 99.5, 100 < 1 + 99.8, 100 < 1 + 100, 100 < 1 + 100.5, and 100 < 1 + 100.9 are all true. Also, 1 < 100 + any of these values holds. The third condition 99.5 < 1 + 100, 99.8 < 1 + 100, 100 < 1 + 100 also check out (the last is an edge case).

(b) Five possible values for the third length are 1, 3.5, 5, 7.5, 8.9. We can verify: 5 < 1 + 5, 5 < 5 + 3.5, 5 < 5 + 5, 5 < 5 + 7.5, and 5 < 5 + 8.9 all hold.

(c) Five possible values for the third length are 4.5, 5, 6.9, 8, 9.8. We can verify: 7 < 3 + 4.5, 7 < 5 + 3, 7 < 3 + 6.9, 7 < 3 + 8, and 7 < 3 + 9.8 all hold.
In simple words: You can choose many different lengths as the third side - as long as each length (including your new one) is less than the sum of the other two, a triangle will form.

Exam Tip: When finding possible third sides, choose values that are neither too small nor too large compared to the given two sides - values close to the given lengths usually work.

Question. See if you can describe all possible lengths of the third side in each case, so that a triangle exists with those side lengths. For example, in case (a), all numbers strictly between 99 and 101 would be possible.
Answer: When two sides are given, the third side must lie between the difference and the sum of the two lengths for a triangle to exist. Therefore:

(b) The third length must lie between 0 and 10 (since the difference is |5 - 5| = 0 and the sum is 5 + 5 = 10).

(c) The third length must lie between 4 and 10 (since the difference is |7 - 3| = 4 and the sum is 7 + 3 = 10).
In simple words: If two sides have lengths a and b (with a < b), the third side c must satisfy: (b - a) < c < (a + b).

Exam Tip: Use the formula: third side must be strictly between (|side1 - side2|) and (side1 + side2). This is faster than testing individual values.

Question 1. Construct triangles for the following measurements, where the angle is included between the sides: (a) 3 cm, 75°, 7 cm (b) 6 cm, 25°, 3 cm (c) 3 cm, 120°, 8 cm
Answer: (a) Steps of Construction:
Step 1: Draw side AB with length 7 cm.
Step 2: At point A, construct an angle of 75° and draw the other arm of this angle.
Step 3: Mark point C on this new arm at a distance of 3 cm from A.
Step 4: Connect B and C to complete the triangle.

(b) Steps of Construction:
Step 1: Draw side AB with length 6 cm.
Step 2: At point A, construct an angle of 25° and draw the other arm of this angle.
Step 3: Mark point C on this new arm at a distance of 3 cm from A.
Step 4: Connect B and C to complete the triangle.

(c) Steps of Construction:
Step 1: Draw side AB with length 8 cm.
Step 2: At point A, construct an angle of 120° and draw the other arm of this angle.
Step 3: Mark point C on this new arm at a distance of 3 cm from A.
Step 4: Connect B and C to complete the triangle.


In simple words: Draw one side. At one end of that side, measure out the given angle. On the angle's arm, measure the second side. Connect the endpoints to complete the triangle.

Exam Tip: When constructing with SAS (Side-Angle-Side), always ensure the angle is between the two given sides and is measured accurately with a protractor.

Question 1. Construct triangles for the following measurements: (a) 75°, 5 cm, 75° (b) 25°, 3 cm, 60° (c) 120°, 6 cm, 30°
Answer: (a) Steps of Construction:
Step 1: Draw base AB with length 5 cm.
Step 2: At point A, draw an angle of 75°. At point B, draw an angle of 75°.
Step 3: Extend both angle arms until they meet. The meeting point is vertex C.

(b) Steps of Construction:
Step 1: Draw base AB with length 3 cm.
Step 2: At point A, draw an angle of 25°. At point B, draw an angle of 60°.
Step 3: Extend both angle arms until they meet. The meeting point is vertex C.

(c) Steps of Construction:
Step 1: Draw base AB with length 6 cm.
Step 2: At point A, draw an angle of 30°. At point B, draw an angle of 120°.
Step 3: Extend both angle arms until they meet. The meeting point is vertex C. Triangles always exist when angles are given and sum to less than 180°.


In simple words: Draw the base. At each end, draw the given angles. Where the two arms meet is the third vertex. The triangle is now complete.

Exam Tip: With ASA (Angle-Side-Angle), verify that the two angles sum to less than 180° before constructing - if they sum to 180° or more, no triangle exists.

Question 1. For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category: (a) 30° (b) 70° (c) 54° (d) 144°
Answer: (a) Another angle that makes a triangle possible: Any angle less than 150° works. Two examples are 60°, 90°. Another angle that makes a triangle not possible: Any angle greater than or equal to 150° works. Two examples are 170°, 160°.

(b) Another angle that makes a triangle possible: Any angle less than 110° works. Two examples are 70°, 40°. Another angle that makes a triangle not possible: Any angle greater than or equal to 110° works. Two examples are 120°, 150°.

(c) Another angle that makes a triangle possible: Any angle less than 126° works. Two examples are 72°, 54°. Another angle that makes a triangle not possible: Any angle greater than or equal to 126° works. Two examples are 140°, 130°.

(d) Another angle that makes a triangle possible: Any angle less than 36° works. Two examples are 10°, 26°. Another angle that makes a triangle not possible: Any angle greater than or equal to 36° works. At least two examples are 40°, 50°.
In simple words: For three angles to form a triangle, they must sum to exactly 180°. If one angle and your second angle already sum to 180° or more, no third angle fits.

Exam Tip: For a given angle A, a triangle is possible only if the second angle B satisfies: B < 180° - A.

Question 2. Determine which of the following pairs can be the angles of a triangle and which cannot: (a) 35°, 150° (b) 70°, 30° (c) 90°, 85° (d) 50°, 150°
Answer: (a) The sum of the given angles = 35° + 150° = 185°. This exceeds 180°, so these cannot be angles of a triangle.

(b) The sum of the given angles = 70° + 30° = 100°. The third angle would be 180° - 100° = 80°. Since all three angles sum to 180°, this pair can be angles of a triangle.

(c) The sum of the given angles = 90° + 85° = 175°. The third angle would be 180° - 175° = 5°. Since all three angles sum to 180°, this pair can be angles of a triangle.

(d) The sum of the given angles = 50° + 150° = 200°. This exceeds 180°, so these cannot be angles of a triangle.
In simple words: Add the two angles. If the sum is less than 180°, you can find a third angle. If the sum equals or exceeds 180°, no triangle is possible.

Exam Tip: Always check that the sum of the two given angles is strictly less than 180° - if so, the third angle is positive and a triangle can exist.

Question. What could the measure of the third angle be? Does this measure change if the base length is changed to some other value, say 7 cm? Construct and find out.
Answer: Given two angles measuring 60° and 70°, the third angle is calculated as: 180° - 60° - 70° = 50°. The measure of this third angle does not change if you alter the base length to 7 cm or any other value. The third angle is determined solely by the two given angles - the base length affects only the size of the triangle, not the angles themselves.
In simple words: Once you know two angles of a triangle, the third angle is fixed. The base length just makes the triangle bigger or smaller, but the angles stay the same.

Exam Tip: Triangle angles depend only on each other, never on side lengths. Always use the angle sum property: all three angles sum to 180°.

Question 1. Find the third angle of a triangle (using a parallel line) when two of the angles are: (a) 36°, 72° (b) 150°, 15° (c) 90°, 30° (d) 75°, 45°
Answer: (a) Let ∠B = 36° and ∠C = 72°. Draw line XY parallel to BC.
By the alternate angles property: ∠XAB = ∠B = 36° ... (i)
and ∠YAC = ∠C = 72° ... (ii)
Since XY is a straight line: ∠XAB + ∠BAC + ∠YAC = 180°
Substituting: 36° + ∠BAC + 72° = 180°
Therefore: ∠BAC = 180° - 108° = 72°

(b) Let ∠B = 150° and ∠C = 15°. Draw line XY parallel to BC.
By the alternate angles property: ∠XAB = ∠B = 150° ... (i)
and ∠YAC = ∠C = 15° ... (ii)
Since XY is a straight line: ∠XAB + ∠BAC + ∠YAC = 180°
Substituting: 150° + ∠BAC + 15° = 180°
Therefore: ∠BAC = 180° - 165° = 15°

(c) Let ∠B = 90° and ∠C = 30°. Draw line XY parallel to BC.
By the alternate angles property: ∠XAB = ∠B = 90° ... (i)
and ∠YAC = ∠C = 30° ... (ii)
Since XY is a straight line: ∠XAB + ∠BAC + ∠YAC = 180°
Substituting: 90° + ∠BAC + 30° = 180°
Therefore: ∠BAC = 180° - 120° = 60°

(d) Let ∠B = 75° and ∠C = 45°. Draw line XY parallel to BC.
By the alternate angles property: ∠XAB = ∠B = 75° ... (i)
and ∠YAC = ∠C = 45° ... (ii)
Since XY is a straight line: ∠XAB + ∠BAC + ∠YAC = 180°
Substituting: 75° + ∠BAC + 45° = 180°
Therefore: ∠BAC = 180° - 120° = 60°


In simple words: When a line parallel to the base is drawn through the top vertex, it creates alternate angles that match the base angles. These, along with the angle at the top, must sum to 180° (a straight line).

Exam Tip: Using a parallel line is a geometric proof method - it visually shows why angles in a triangle sum to 180°, not just a formula to memorize.

Question 2. Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70°, 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.
Answer: No, it is not possible to build a triangle with all angles equal to 70°. If the two base angles each measure 70° (∠B = 70° and ∠C = 70°), then the third angle ∠BAC is found as follows:

Draw line XY parallel to BC. By alternate angles: ∠XAB = 70° and ∠YAC = 70°.
Since these lie on a straight line: 70° + ∠BAC + 70° = 180°
Therefore: ∠BAC = 40°

If all angles in a triangle must be equal, then each angle must measure 60°. This is because 60° + 60° + 60° = 180°. A triangle with all three angles equal to 60° is called an equilateral triangle.
In simple words: If you try to make all angles 70°, you get 70° + 70° + 70° = 210°, which is too much. For equal angles, each must be 180° ÷ 3 = 60°.

Exam Tip: The angle sum property guarantees that if two angles are equal and you know one, the third follows automatically. Equilateral triangles always have three 60° angles.

Question 3. Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C?
Answer: Given: ∠A = 50° and ∠B = ∠C.
Draw line XY parallel to BC.

By alternate angles: ∠XAB = ∠B and ∠YAC = ∠C ... (i)
Since XY is a straight line: ∠XAB + ∠BAC + ∠YAC = 180°
Substitute using (i): ∠B + 50° + ∠C = 180°
So: ∠B + ∠C = 130°

Since ∠B = ∠C: 2∠B = 130°
Therefore: ∠B = 65° = ∠C


In simple words: When two angles of a triangle are equal, you can use the angle sum (180°) and the equality condition to find both angles. Here, subtracting the known angle from 180° leaves 130° for the two equal angles, so each is 65°.

Exam Tip: In an isosceles triangle (two equal sides), the angles opposite those sides are also equal - use this relationship combined with the angle sum property.

Exterior Angles

Question. 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 the figure, ∠ACD is an exterior angle. Find ∠ACD, if ∠A = 50°, and ∠B = 60°.
Answer: Using the angle sum property: 50° + 60° + ∠ACB = 180°, so 110° + ∠ACB = 180°. Therefore, ∠ACB = 70°. Since ∠ACB and ∠ACD together form a straight angle, ∠ACD = 180° - 70° = 110°.
In simple words: Find the interior angle at that vertex using all three angles' sum. The exterior angle is what remains when you subtract from 180°.

Exam Tip: An exterior angle and its adjacent interior angle always sum to 180° because they form a straight line.

Question. Find the exterior angle for different measures of ∠A and ∠B. Do you see any relation between the exterior angle and these two angles?
Answer: From the angle sum property: ∠A + ∠B + ∠ACB = 180° ... (i)
Since the exterior and interior angles at C form a straight angle: ∠ACD + ∠ACB = 180°
So: ∠ACB = 180° - ∠ACD ... (ii)

Substituting (ii) into (i): ∠A + ∠B + 180° - ∠ACD = 180°
Therefore: ∠A + ∠B = ∠ACD

This shows an important relationship: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
In simple words: An exterior angle is not isolated. It always equals the sum of the two distant interior angles of the triangle.

Exam Tip: The exterior angle theorem is powerful - use it to find unknown angles faster than the angle sum property alone.

7.4 Constructions Related to Altitudes of Triangles

Altitudes Using Paper Folding

Question. 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
Answer: An altitude is a perpendicular line drawn from a vertex to the opposite side (or the line containing that side). When you fold a paper triangle so the top vertex touches the base, the crease that forms is perpendicular to the base. This is because the fold ensures that the path from the vertex to the base follows the shortest distance, which is always perpendicular. The crease marks this perpendicular line, making it the altitude of the triangle from that vertex to the base.
In simple words: Folding the triangle so the top corner touches the bottom edge creates a crease that is at a right angle to the base - this is the altitude.

Exam Tip: Paper folding is a practical way to understand altitudes - the crease always forms at 90° to the base, confirming the perpendicular nature of altitudes.

Question. What could an acute-angled triangle be? Can we define it as a triangle with one acute angle? Why not?
Answer: An acute-angled triangle is one where all three angles are less than 90 degrees. We cannot describe it simply as having one acute angle, because both right-angled and obtuse-angled triangles also contain two acute angles. So that definition would not be specific enough to identify only acute-angled triangles.
In simple words: All three angles must be acute (less than 90°) for it to be an acute-angled triangle, not just one angle.

Exam Tip: Remember that any triangle has at least two acute angles - the defining feature of an acute-angled triangle is that ALL three are acute.

Question 1. Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.
Answer:
Steps of Construction:
Step 1: Draw the base AB = 6 cm.
Step 2: Using a compass, draw a long arc of radius 5 cm centered at A.
Step 3: Draw another arc of radius 5 cm from B so it crosses the first arc.
Step 4: Mark the intersection point as vertex C. Join AC and BC to complete triangle ABC.
Step 5: Position the ruler along BC. Place the set square so that one edge of its right angle sits on the ruler.
Step 6: Slide the set square along the ruler until the perpendicular edge touches point A.
Step 7: Draw the altitude from A to BC using the perpendicular edge of the set square.
In simple words: Use compass arcs to find point C, then use a set square to draw a line from A that meets BC at a right angle. That line is the altitude.

Exam Tip: The altitude must touch BC at exactly 90 degrees - use the right-angle corner of the set square to ensure this is accurate.

Question 2. Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.
Answer:
Steps of Construction:
Step 1: Draw side TR measuring 7 cm.
Step 2: At point R, construct an angle of 140° by drawing the second arm.
Step 3: On this second arm, mark point Y such that RY = 4 cm.
Step 4: Join TY to complete the triangle TRY.
Step 5: Position the ruler so it aligns with RY. Place the set square with one edge of its right angle touching the ruler.
Step 6: Slide the set square along the ruler until the perpendicular edge reaches point T.
Step 7: Extend line YR if needed, then draw the altitude from T perpendicular to RY using the set square's perpendicular edge.
In simple words: Create the triangle using the given angle and side lengths, then use the set square to drop a perpendicular line from T down to the base RY.

Exam Tip: Since angle R is obtuse (140°), the altitude from T will meet the extended line RY, not the segment RY itself - be prepared to extend the base line.

Question 3. Construct a right-angled triangle ∆ABC with ∠B = 90°, AC = 5 cm. How many different triangles exist with these measurements?
Answer: With ∠B = 90° and AC = 5 cm (the hypotenuse), infinitely many different triangles can be formed. This is because ∠A and ∠C must add up to 90°, and they can take on any combination of values as long as they sum to 90 degrees. Each different pair of values for ∠A and ∠C produces a triangle with a different shape, even though the hypotenuse stays the same. As long as we keep the hypotenuse fixed at 5 cm and the right angle at B, the other two angles can vary freely, creating countless possibilities.
In simple words: You can make unlimited different right triangles if you keep the hypotenuse at 5 cm and the right angle at B, because the two other angles can be any pair that adds to 90°.

Exam Tip: Recognize that one constraint (the hypotenuse) is not enough - you need at least one angle (besides the right angle) to be fixed to get a unique triangle.

Question 4. 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.
Answer: An equilateral triangle cannot be right-angled or obtuse-angled because each angle in an equilateral triangle is always exactly 60 degrees. Since 60 + 60 + 60 = 180 degrees, there is no room for a 90-degree angle or any angle larger than 90 degrees.

An isosceles right-angled triangle, however, is possible. It has one angle of 90 degrees and two equal angles of 45 degrees each (since 90 + 45 + 45 = 180).

An isosceles obtuse-angled triangle is also possible. One example has an obtuse angle of 120 degrees and two equal angles of 30 degrees each (since 120 + 30 + 30 = 180).
In simple words: Equilateral triangles always have 60-degree angles, so they cannot be right or obtuse. But isosceles triangles can have a 90-degree angle (with two 45-degree angles) or an angle larger than 90 degrees (with two smaller matching angles).

Exam Tip: Always verify that angles sum to 180 degrees and check the definition of each triangle type (equilateral has all equal sides and angles; isosceles has two equal sides and two equal angles).