--- title: "NCERT Solutions Class 7 Maths A Tale of Three Intersecting Lines" url: https://www.swavid.com/maths/class/7/chapter/a-tale-of-three-intersecting-lines/ncert-solutions dateModified: 2026-10-07T15:07:27+00:00 --- # NCERT Solutions Class 7 Maths A Tale of Three Intersecting Lines This chapter's questions cover the fundamental properties of triangles, including construction methods, the triangle inequality, angle sum properties, exterior angles, and altitudes. Free PDF (40 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-7-a-tale-of-three-intersecting-lines-f70a5d9714.pdf ## In-text questions (Page 146) ### Question 1 *2 marks · Very short answer* What happens when the three vertices lie on a straight line? **Solution** 1. When the three vertices lie on a straight line, no closed shape is formed. 2. Instead of a triangle, we get a straight line segment, and no triangle is possible. **Answer:** No triangle is formed; it results in a straight line segment. > Common mistake: Thinking that a triangle can still be formed with zero area. ## In-text questions (Page 146) ### Question 1 *3 marks · Short answer* Construct a triangle in which all the sides are of length $4\text{ cm}$. **Solution** 1. Construct a line segment AB of length $4\text{ cm}$ using a ruler. 2. Using a compass, open it to a radius of $4\text{ cm}$, place the pointer at A, and draw a sufficiently long arc. 3. Keeping the same radius of $4\text{ cm}$, place the pointer at B and draw another arc to intersect the first arc at point C. 4. Join AC and BC using a ruler to get the required equilateral triangle ABC with each side of length $4\text{ cm}$. **Answer:** An equilateral triangle ABC with sides $4\text{ cm}$ each. > Common mistake: Taking incorrect radii or not using a compass, which leads to inaccurate side lengths. ## In-text questions (Page 147) ### Question 1 *3 marks · Short answer* 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)? **Solution** 1. Constructing the triangle using just a ruler and pencil is possible by drawing a base of length 4 cm and attempting to mark the third point C such that AC = 4 cm. 2. However, this method involves several trials because drawing BC of length 4 cm without a compass is not guaranteed on the first attempt. 3. Therefore, construction using only a marked ruler requires multiple trials to get all sides to be 4 cm long. **Answer:** Constructing the triangle with just a ruler is possible but requires several trials to get all sides of length 4 cm. > Common mistake: Assuming a ruler alone can directly position point C without trials. ### Question 2 *3 marks · Short answer* How do we make this construction more efficient? **Solution** 1. Construct the base AB of length 4 cm using a ruler. 2. Use a compass to draw an arc of radius 4 cm from point A, and another arc of radius 4 cm from point B. 3. Mark the point of intersection of the two arcs as C, and join AC and BC to get the required equilateral triangle efficiently. **Answer:** We make the construction more efficient by using a compass to draw arcs of radius 4 cm from both endpoints of the base. > Common mistake: Not setting the compass radius correctly to the required side length before drawing arcs. ## In-text questions (Page 148) ### Question 1 *3 marks · Short answer* The construction ensures that both $AC$ and $BC$ are of length $4\text{ cm}$. Can you see why? **Solution** 1. Point C is obtained by drawing arcs of radius $4\text{ cm}$ centered at points A and B respectively. 2. Since the arc from A is drawn with radius $4\text{ cm}$, any point on it, including C, is at a distance of $4\text{ cm}$ from A, so $AC = 4\text{ cm}$. 3. Similarly, the arc from B is drawn with radius $4\text{ cm}$, so the point of intersection C is at a distance of $4\text{ cm}$ from B, ensuring $BC = 4\text{ cm}$. **Answer:** Both $AC$ and $BC$ are $4\text{ cm}$ because they are radii of arcs of equal length $4\text{ cm}$ drawn from centres A and B. > Common mistake: Thinking the lengths depend on the angle of intersection rather than the fixed radius of the compass arcs. ### Question 2 *3 marks · Short answer* Construct a triangle of sidelength $4\text{ cm}$, $5\text{ cm}$ and $6\text{ cm}$. **Solution** 1. Step 1: Construct the base AB of length $4\text{ cm}$ using a ruler. 2. Step 2: Using a compass, open it to a radius of $5\text{ cm}$ and draw a sufficiently long arc with centre A. 3. Step 3: Open the compass to a radius of $6\text{ cm}$ and draw an arc from centre B to intersect the first arc at point C. 4. Step 4: Join AC and BC to get the required triangle ABC with sides $4\text{ cm}$, $5\text{ cm}$, and $6\text{ cm}$. **Answer:** Triangle ABC with side lengths $AB = 4\text{ cm}$, $AC = 5\text{ cm}$, and $BC = 6\text{ cm}$ is constructed successfully. > Common mistake: Mixing up the compass radius values for sides AC ($5\text{ cm}$) and BC ($6\text{ cm}$). ## In-text questions (Page 149) ### Question 1 *3 marks · Short answer* How do we construct this triangle more efficiently? **Solution** 1. Choose one side length to be the base, say AB = 4 cm, and draw it using a ruler. 2. Using a compass, construct an arc of radius equal to the second side length (5 cm) from A. 3. Construct another arc of radius equal to the third side length (6 cm) from B to intersect the first arc at point C, and join AC and BC. **Answer:** We construct the base AB, draw arcs of radii 5 cm and 6 cm from A and B respectively to locate the third vertex C, and join the sides. > Common mistake: Taking incorrect radii lengths for the arcs from vertices A and B. ## Construct (Page 150) ### Question 1 *3 marks · Short answer* 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) $3.5, 3.5, 3.5$ **Part (a)** 1. Draw base AB = 6 cm. 2. Draw arcs of radius 4 cm from A and B to intersect at C. 3. Join AC and BC to form the triangle. Answer (a): Triangle with sides 4 cm, 4 cm, 6 cm **Part (b)** 1. Draw base AB = 5 cm. 2. Draw arcs of radius 3 cm from A and 4 cm from B to intersect at C. 3. Join AC and BC to form the triangle. Answer (b): Triangle with sides 3 cm, 4 cm, 5 cm **Part (c)** 1. Draw base AB = 5 cm. 2. Draw arcs of radius 1 cm from A and 5 cm from B to intersect at C. 3. Join AC and BC to form the triangle. Answer (c): Triangle with sides 1 cm, 5 cm, 5 cm **Part (d)** 1. Draw base AB = 8 cm. 2. Draw arcs of radius 4 cm from A and 6 cm from B to intersect at C. 3. Join AC and BC to form the triangle. Answer (d): Triangle with sides 4 cm, 6 cm, 8 cm **Part (e)** 1. Draw base AB = 3.5 cm. 2. Draw arcs of radius 3.5 cm from A and B to intersect at C. 3. Join AC and BC to form the equilateral triangle. Answer (e): Triangle with sides 3.5 cm, 3.5 cm, 3.5 cm **Answer:** Triangles of given sidelengths constructed using compass and ruler. > Common mistake: Not adjusting the compass to the exact required radius using the ruler before drawing arcs. ## Figure it Out (Page 150-151) ### Question 1 *3 marks · Short answer* Use the points on the circle and/or the centre to form isosceles triangles. **Solution** 1. Consider a circle with centre O and a few points marked on its boundary. 2. Join the centre O to any two points on the circle, say P and Q, to form triangle OPQ. 3. Since OP and OQ are radii of the same circle, $OP = OQ$, so triangle OPQ is an isosceles triangle. **Answer:** Triangle formed by joining the centre and two points on the circle is an isosceles triangle. > Common mistake: Confusing radii with chords when forming the triangles. ### Question 2 *3 marks · Short answer* Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size. **Solution** 1. Take points $C_1, C_2, C_3$ on the circle with centre B so that the triangles $ABC_1, ABC_2, ABC_3$ are isosceles triangles because $AB = BC_1 = BC_2 = BC_3$ (radii of the same circle). 2. Similarly, taking points $D_1, D_2, D_3$ on the circle with centre A gives isosceles triangles $BAD_1, BAD_2, BAD_3$. 3. The triangles $ABC$ and $ABD$ formed by the intersections and centres are equilateral triangles since all three sides are equal radii of the congruent circles. **Answer:** Triangles formed using the centres and intersection points are isosceles and equilateral triangles. > Common mistake: Not recognizing that radii of equal circles give equal side lengths. ## In-text questions (Page 151) ### Question 1 *3 marks · Short answer* Construct a triangle with sidelengths $3\text{ cm}$, $4\text{ cm}$, and $8\text{ cm}$. What is happening? Are you able to construct the triangle? **Solution** 1. Consider the given side lengths $3\text{ cm}$, $4\text{ cm}$, and $8\text{ cm}$. 2. Compare the sum of the two shorter lengths with the longest length: $3 + 4 = 7$. 3. Since $7 < 8$, the sum of two sides is less than the third side, so the triangle cannot be constructed. **Answer:** No, a triangle cannot be constructed because $3 + 4 < 8$. > Common mistake: Students may try to draw it with a ruler and get confused by intersecting arcs not meeting instead of checking the side lengths first. ### Question 2 *3 marks · Short answer* Here is another set of lengths: $2\text{ cm}$, $3\text{ cm}$, and $6\text{ cm}$. Check if a triangle is possible for these sidelengths. **Solution** 1. Consider the given side lengths $2\text{ cm}$, $3\text{ cm}$, and $6\text{ cm}$. 2. Calculate the sum of the two smaller sides: $2 + 3 = 5$. 3. Compare this sum with the longest side: $5 < 6$, so the triangle is not possible. **Answer:** A triangle is not possible for these sidelengths because $2 + 3 < 6$. > Common mistake: Assuming any three random lengths can form a triangle without checking the sum. ### Question 3 *3 marks · Short answer* Try to find more sets of lengths for which a triangle construction is impossible. See if you can find any pattern in them. **Solution** 1. Look for sets of lengths where the sum of the two smaller lengths is less than or equal to the longest length. 2. Examples of impossible lengths include $1\text{ cm}$, $2\text{ cm}$, and $4\text{ cm}$ or $5\text{ cm}$, $5\text{ cm}$, and $12\text{ cm}$. 3. The common pattern is that the sum of any two side lengths must always be strictly greater than the third side length. **Answer:** Examples of impossible lengths are $1\text{ cm}$, $2\text{ cm}$, and $4\text{ cm}$. The pattern is that the sum of the two smaller lengths is less than the longest length. > Common mistake: Listing sets where the inequality holds true instead of failing. ## In-text questions (Page 152-153) ### Question 1 *3 marks · Short answer* Can this understanding be used to tell something about the existence of a triangle having sidelengths $10\text{ cm}$, $15\text{ cm}$ and $30\text{ cm}$? **Solution** 1. Let us consider the direct path lengths and the roundabout path lengths for the given side lengths $10\text{ cm}$, $15\text{ cm}$ and $30\text{ cm}$. 2. For the sides $10\text{ cm}$ and $15\text{ cm}$, the roundabout path via the third vertex is $30 + 15 = 45\text{ cm}$, which is greater than the direct path $10\text{ cm}$. 3. However, for the direct path of length $30\text{ cm}$, the roundabout path via the other two vertices is $10 + 15 = 25\text{ cm}$. 4. Since the direct path length ($30\text{ cm}$) is longer than the roundabout path ($25\text{ cm}$), which is impossible, a triangle with these sidelengths cannot exist. **Answer:** No, a triangle having sidelengths $10\text{ cm}$, $15\text{ cm}$ and $30\text{ cm}$ cannot exist because the sum of the two shorter sides is less than the longest side. > Common mistake: Students often assume any three numbers can form a triangle without checking if the sum of every pair of sides is greater than the third side. ### Question 2 *3 marks · Short answer* Can we say anything about the existence of a triangle having sidelengths $3\text{ cm}$, $3\text{ cm}$ and $7\text{ cm}$? Verify your answer by construction. **Solution** 1. Consider the given side lengths: $3\text{ cm}$, $3\text{ cm}$ and $7\text{ cm}$. 2. Check the triangle inequality by taking the sum of the two shorter sides: $3 + 3 = 6\text{ cm}$. 3. Since $6\text{ cm}$ is less than the third side ($7\text{ cm}$), the sum of the two sides is less than the third side ($3 + 3 < 7$). 4. Therefore, a triangle with these sidelengths cannot exist, and if we try to construct it using arcs of radii $3\text{ cm}$ from the ends of a $7\text{ cm}$ base, the arcs will not intersect. **Answer:** No, a triangle with sidelengths $3\text{ cm}$, $3\text{ cm}$ and $7\text{ cm}$ cannot exist. > Common mistake: Forgetting that isosceles triangles also must satisfy the fundamental triangle inequality. ### Question 3 *3 marks · Short answer* “In the rough diagram in Fig. 7.4, is it possible to assign lengths in a different order such that the direct paths are always coming out to be shorter than the roundabout paths? If this is possible, then a triangle might exist.” Is such rearrangement of lengths possible in the triangle? **Solution** 1. In a triangle with side lengths $10\text{ cm}$, $15\text{ cm}$ and $30\text{ cm}$, the lengths can be arranged in different orders for the three vertices. 2. However, whichever way we assign the lengths to the sides, the longest side ($30\text{ cm}$) will always be compared against the sum of the other two sides ($10\text{ cm}$ and $15\text{ cm}$). 3. Since $30$ is always greater than $10 + 15$ ($30 > 25$), the direct path will always be longer than the roundabout path in that combination. 4. Thus, no such rearrangement can make all direct paths shorter than the roundabout paths, and a triangle is not possible. **Answer:** No, such rearrangement of lengths is not possible because the longest side is always greater than the sum of the other two sides. > Common mistake: Thinking that rearranging the order of side labels can alter their numerical lengths or satisfy the inequality when it fails. ## Figure it Out (Page 154) ### Question 1 *3 marks · Short answer* We checked by construction that there are no triangles having sidelengths $3\text{ cm}$, $4\text{ cm}$ and $8\text{ cm}$; and $2\text{ cm}$, $3\text{ cm}$ and $6\text{ cm}$. Check if you could have found this without trying to construct the triangle. **Solution** 1. For the side lengths $3\text{ cm}$, $4\text{ cm}$ and $8\text{ cm}$, the sum of the two smaller sides is $3 + 4 = 7\text{ cm}$. 2. Since $7 < 8$, the sum of two sides is less than the third side, so a triangle cannot be formed. 3. Similarly, for $2\text{ cm}$, $3\text{ cm}$ and $6\text{ cm}$, the sum of the two smaller sides is $2 + 3 = 5\text{ cm}$, which is less than $6\text{ cm}$, so a triangle is not possible. **Answer:** We can check this without construction by using the triangle inequality: if the sum of any two sides is less than the third side, the triangle is not possible. > Common mistake: Comparing the largest side with the difference of the other two instead of their sum. ### Question 2 *3 marks · Short answer* Can we say anything about the existence of a triangle for each of the following sets of lengths? (a) $10\text{ km}$, $10\text{ km}$ and $25\text{ km}$ (b) $5\text{ mm}$, $10\text{ mm}$ and $20\text{ mm}$ (c) $12\text{ cm}$, $20\text{ cm}$ and $40\text{ cm}$ **Part (a)** 1. For the lengths $10\text{ km}$, $10\text{ km}$ and $25\text{ km}$, the sum of the two smaller lengths is $10 + 10 = 20\text{ km}$. 2. Since $20 < 25$, the sum of the two sides is less than the third side. Answer (a): A triangle does not exist. **Part (b)** 1. For the lengths $5\text{ mm}$, $10\text{ mm}$ and $20\text{ mm}$, the sum of the two smaller lengths is $5 + 10 = 15\text{ mm}$. 2. Since $15 < 20$, the sum of the two sides is less than the third side. Answer (b): A triangle does not exist. **Part (c)** 1. For the lengths $12\text{ cm}$, $20\text{ cm}$ and $40\text{ cm}$, the sum of the two smaller lengths is $12 + 20 = 32\text{ cm}$. 2. Since $32 < 40$, the sum of the two sides is less than the third side. Answer (c): A triangle does not exist. **Answer:** A triangle does not exist for any of the given sets of lengths. > Common mistake: Not checking all possible combinations of side sums. ### Question 3 *3 marks · Short answer* 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 for different sets of lengths. **Solution** 1. No, this will not always happen for any set of three arbitrary lengths. 2. For a valid set of triangle side lengths, each length is smaller than the sum of the other two, so all three comparisons hold true. 3. For sets of lengths that cannot form a triangle, like $10\text{ cm}$, $15\text{ cm}$ and $30\text{ cm}$, only two comparisons hold true while the third one fails ($30 > 10 + 15$). **Answer:** No, for sets of lengths that do not form a triangle, there may only be two comparisons where the direct length is less than the sum of the other two, and one comparison will fail. > Common mistake: Assuming all three comparisons always hold or always fail together. ## In-text questions (Page 154) ### Question 1 *3 marks · Short answer* Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations? [Hint: Consider the direct lengths in the increasing order.] **Solution** 1. Arrange the three given lengths in increasing order. 2. When the lengths are in increasing order, the sum of the two smaller lengths is always greater than or equal to the other two comparisons. 3. Therefore, we only need to check if the sum of the two smaller lengths (the first two numbers in increasing order) is strictly greater than the third and largest length. **Answer:** Yes, by arranging the lengths in increasing order, we only need to check if the sum of the two smaller lengths is greater than the largest length. > Common mistake: Checking all three combinations even when the lengths are already arranged, or checking the sum of the largest and a smaller side. ### Question 2 *3 marks · Short answer* Given three sidelengths, what do we need to compare to check for the existence of a triangle? **Solution** 1. Compare the sum of any two side lengths with the third side length. 2. Check that each length is strictly smaller than the sum of the other two lengths. 3. Specifically, when the three lengths are arranged in increasing order, check if the sum of the two smaller lengths is greater than the largest length. **Answer:** We need to compare each length with the sum of the other two lengths (or simply check if the sum of the two smaller lengths is greater than the largest length). > Common mistake: Comparing the difference of sides instead of the sum, or checking only one comparison. ## In-text questions (Page 155) ### Question 1 *3 marks · Short answer* Does a triangle exist with sidelengths $4\text{ cm}$, $5\text{ cm}$ and $8\text{ cm}$? **Solution** 1. Given the sidelengths are $4\text{ cm}$, $5\text{ cm}$, and $8\text{ cm}$. 2. Check the triangle inequality by taking the sum of the two smaller sides and comparing it with the longest side. 3. Sum of the two smaller sides = $4 + 5 = 9\text{ cm}$, which is greater than the longest side $8\text{ cm}$ ($8 < 4 + 5$). 4. Since each length is smaller than the sum of the other two, the given lengths satisfy the triangle inequality. 5. Therefore, a triangle with sidelengths $4\text{ cm}$, $5\text{ cm}$ and $8\text{ cm}$ exists. **Answer:** Yes, a triangle exists with sidelengths $4\text{ cm}$, $5\text{ cm}$ and $8\text{ cm}$. > Common mistake: Comparing only one pair of sides instead of checking the inequality for the longest side against the sum of the other two. ### Question 2 *3 marks · Short answer* Why do we not need to check the other two sides? **Solution** 1. We need to compare each side with the sum of the other two sides. 2. When we take the longest side as the direct length, checking if it is less than the sum of the other two smaller sides is sufficient. 3. If the longest side is smaller than the sum of the two smaller sides, then automatically every smaller side will also be less than the sum of the remaining two sides. 4. Thus, checking only the inequality for the longest side confirms that all direct path lengths are less than their corresponding roundabout path lengths. **Answer:** We do not need to check the other two sides because if the longest side is smaller than the sum of the two smaller sides, the other two automatic conditions are always satisfied. > Common mistake: Failing to realise that the inequality involving the longest side is the critical test for triangle existence. ### Question 3 *3 marks · Short answer* Now, suppose that a circle of radius $5\text{ cm}$ is constructed, centred at $B$. Can you draw a rough diagram of the resulting figure? **Solution** 1. Take the longest side $AB = 8\text{ cm}$ as the base. 2. Draw a circle of radius $4\text{ cm}$ centered at $A$ and a circle of radius $5\text{ cm}$ centered at $B$. 3. Mark the point $X$ on $AB$ such that $AX = 4\text{ cm}$, which means $BX = AB - AX = 8 - 4 = 4\text{ cm}$. 4. Since $BX = 4\text{ cm}$ is less than the radius $5\text{ cm}$ of the circle centered at $B$, the two circles intersect each other internally at two points. 5. Sketch the two intersecting circles with centres $A$ and $B$ overlapping such that they intersect at two points, forming the required rough diagram. **Answer:** The rough diagram shows two overlapping circles of radii $4\text{ cm}$ and $5\text{ cm}$ with centers $A$ and $B$ ($8\text{ cm}$ apart) intersecting each other at two points. > Common mistake: Drawing the two circles completely separate or touching externally without accounting for internal intersection. ## Figure it Out (Page 156) ### Question 1 *3 marks · Short answer* 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) $24, 26, 28$ **Part (a)** 1. Compare the lengths: $2 + 2 = 4$, which is less than $5$. 2. Since the sum of two sides is less than the third side, it violates the triangle inequality. Answer (a): Not a triangle **Part (b)** 1. Compare the lengths: $3 < 4 + 6$, $4 < 3 + 6$, and $6 < 3 + 4$. 2. Each length is smaller than the sum of the other two, so it satisfies the triangle inequality. Answer (b): Can be sidelengths of a triangle **Part (c)** 1. Compare the lengths: $2 + 4 = 6$, which is less than $8$. 2. Since the sum of two sides is less than the third side, it violates the triangle inequality. Answer (c): Not a triangle **Part (d)** 1. Compare the lengths: $5 < 5 + 8$ and $8 < 5 + 5$. 2. Each length is smaller than the sum of the other two, satisfying the triangle inequality. Answer (d): Can be sidelengths of a triangle **Part (e)** 1. Compare the lengths: $10 < 20 + 25$, $20 < 10 + 25$, and $25 < 10 + 20$. 2. Each length is smaller than the sum of the other two, satisfying the triangle inequality. Answer (e): Can be sidelengths of a triangle **Part (f)** 1. Compare the lengths: $10 + 20 = 30$, which is less than $35$. 2. Since the sum of two sides is less than the third side, it violates the triangle inequality. Answer (f): Not a triangle **Part (g)** 1. Compare the lengths: $24 < 26 + 28$, $26 < 24 + 28$, and $28 < 24 + 26$. 2. Each length is smaller than the sum of the other two, satisfying the triangle inequality. Answer (g): Can be sidelengths of a triangle **Answer:** Sets (b), (d), (e), and (g) can form a triangle; sets (a), (c), and (f) cannot. > Common mistake: Forgetting to check all three combinations of side sums. ## In-text questions (Page 156) ### Question 1 *3 marks · Short answer* Will triangles always exist when a set of lengths satisfies the triangle inequality? How can we be sure? **Solution** 1. When a set of lengths satisfies the triangle inequality, the sum of the two smaller lengths is greater than the longest length. 2. If we take the longest length as the base AB and draw circles of radii equal to the two smaller lengths from A and B, this condition ensures that the two circles intersect each other internally. 3. The points of intersection of these circles give the third vertex, confirming that a triangle always exists. **Answer:** Yes, triangles will always exist because the triangle inequality guarantees that the two construction circles intersect internally. > Common mistake: Thinking that satisfying the triangle inequality only means the lengths form an open path without checking circle intersections. ## In-text questions (Page 157-158) ### Question 1 *3 marks · Short answer* Let us study each of these cases by finding the relation between the radii (the smaller two lengths) and $AB$ (longest length). **Solution** 1. Take the longest side as the base AB of length equal to the longest length. 2. Construct two circles with centers A and B, taking the other two sides as their radii. 3. A triangle is formed only when these two circles intersect each other internally (Case 3), meaning the sum of the two radii is greater than AB. **Answer:** Sum of the two smaller lengths > Longest length (or sum of two radii > AB) > Common mistake: Confusing internal intersection of circles with external touch or non-intersection. ### Question 2 *3 marks · Short answer* For this case to happen, what should be the relation between the radii and $AB$? **Solution** 1. Observe the figure for Case 2 where the two circles do not intersect internally. 2. The distance between the centers AB is greater than the sum of the radii of the two circles. 3. Therefore, the relation between the radii and AB is that the sum of the two radii is less than AB, which corresponds to the sum of the two smaller lengths being less than the longest length. **Answer:** Sum of the two radii < AB (or sum of the two smaller lengths < longest length) > Common mistake: Writing greater than instead of less than for non-intersecting circles. ## In-text questions (Page 159) ### Question 1 *3 marks · Short answer* Can we use this analysis to tell if a triangle exists when the lengths satisfy the triangle inequality? **Solution** 1. If the given lengths satisfy the triangle inequality, the sum of the two smaller lengths is greater than the longest length. 2. When the base is taken as the longest length and the smaller two lengths are taken as the radii of circles from the base vertices, this condition leads to Case 3 where the circles intersect internally at two points. 3. Thus, the intersection point serves as the third vertex, confirming that a triangle always exists. **Answer:** Yes, if lengths satisfy the triangle inequality, the circles intersect internally, ensuring a triangle exists. > Common mistake: Confusing internal intersection with external touching or non-intersection. ### Question 2 *3 marks · Short answer* 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. **Part (a)** 1. For circles to touch each other at a point, the sum of the two smaller lengths must equal the longest length. 2. Example sets where sum of two smaller lengths equals the longest length: (i) $2, 3, 5$, (ii) $4, 4, 8$, (iii) $5, 10, 15$. Answer (a): The circles touch each other at a point when the sum of the two smaller lengths equals the longest length (e.g., 2, 3, 5). **Part (b)** 1. For circles not to intersect, the sum of the two smaller lengths must be strictly less than the longest length. 2. Example sets where sum of two smaller lengths is less than the longest length: (i) $2, 3, 6$, (ii) $3, 4, 8$, (iii) $10, 15, 30$. Answer (b): The circles do not intersect when the sum of the two smaller lengths is less than the longest length (e.g., 2, 3, 6). **Answer:** Examples provided for circles touching at a point and not intersecting when triangle inequality is not satisfied. > Common mistake: Taking lengths that satisfy the triangle inequality for non-intersecting cases. ### Question 3 *3 marks · Short answer* Frame a complete procedure that can be used to check the existence of a triangle. **Solution** 1. Identify the three given side lengths of the proposed triangle. 2. Check whether each length is strictly smaller than the sum of the other two lengths (the triangle inequality). 3. If the inequality holds for all three comparisons, a triangle exists; if it fails for even one comparison, a triangle does not exist. **Answer:** Compare each side length with the sum of the other two lengths; a triangle exists only if each side is less than the sum of the other two. > Common mistake: Checking only one inequality instead of all three comparisons. ## Figure it Out (Page 159) ### Question 1 *3 marks · Short answer* 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) $5, 10, 12$ **Part (a)** 1. Check the triangle inequality: $1 < 100 + 100$ and $100 < 1 + 100$. 2. Since each side is smaller than the sum of the other two sides, the lengths satisfy the triangle inequality. 3. Therefore, a triangle exists. Answer (a): A triangle exists. **Part (b)** 1. Check the triangle inequality by taking the sum of the two smaller lengths: $3 + 6 = 9$. 2. The sum of the two smaller lengths is equal to the longest length ($9 = 9$), which does not satisfy the strict triangle inequality. 3. Therefore, a triangle cannot exist. Answer (b): A triangle cannot exist. **Part (c)** 1. Check the triangle inequality: sum of the two smaller lengths is $1 + 1 = 2$. 2. Since $2 < 5$, the sum of the two shorter sides is less than the third side. 3. Therefore, a triangle cannot be constructed. Answer (c): A triangle cannot be constructed. **Part (d)** 1. Check the triangle inequality: $5 < 10 + 12$, $10 < 5 + 12$, and $12 < 5 + 10$. 2. Since each length is smaller than the sum of the other two lengths, the triangle inequality is satisfied. 3. Therefore, a triangle exists. Answer (d): A triangle exists. **Answer:** A triangle exists for sets (a) and (d), and does not exist for sets (b) and (c). > Common mistake: Forgetting to check all three conditions of the triangle inequality or confusing equality with existence. ### Question 2 *3 marks · Short answer* 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. **Solution** 1. For sides $50, 50, 50$, each length is smaller than the sum of the other two lengths ($50 < 50 + 50$). 2. Therefore, an equilateral triangle with sides $50, 50, 50$ exists. 3. In general, for any given side length $a$, we always have $a < a + a$, so an equilateral triangle of any side length always exists. **Answer:** Yes, an equilateral triangle with sides $50, 50, 50$ exists, and in general, an equilateral triangle of any side length always exists because it always satisfies the triangle inequality. > Common mistake: Assuming that very large or arbitrary side lengths cannot form an equilateral triangle. ### Question 3 *3 marks · Short answer* 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) $3, 7$ **Part (a)** 1. Let the third length be $x$. By the triangle inequality, the third side must be greater than the difference and less than the sum of the other two sides. 2. The difference is $100 - 1 = 99$ and the sum is $100 + 1 = 101$, so $99 < x < 101$. 3. Five possible values are $99.5, 100, 100.7, 100.5, 99.4$. Answer (a): $99.5, 100, 100.7, 100.5, 99.4$ **Part (b)** 1. For two sides of length $5$ and $5$, the third side $x$ must satisfy $5 - 5 < x < 5 + 5$, which gives $0 < x < 10$. 2. Five possible values within this range are $5, 4, 3, 4.9, 1$. Answer (b): $5, 4, 3, 4.9, 1$ **Part (c)** 1. For two sides of length $3$ and $7$, the third side $x$ must satisfy $7 - 3 < x < 7 + 3$, which gives $4 < x < 10$. 2. Five possible values within this range are $5, 8, 7, 6.4, 6$. Answer (c): $5, 8, 7, 6.4, 6$ **Answer:** Possible values are given for each part based on the triangle inequality. > Common mistake: Choosing values outside the valid range given by the difference and sum of the two given sides. ## In-text questions (Page 160) ### Question 1 *3 marks · Short answer* See if you can describe all possible lengths of the third side in each case, so that a triangle exists with those sidelengths. For example, in case (a), all numbers strictly between 99 and 101 would be possible. **Part (a)** 1. Given two sides of lengths $1\text{ cm}$ and $100\text{ cm}$. 2. The difference between the two lengths is $100 - 1 = 99\text{ cm}$ and their sum is $100 + 1 = 101\text{ cm}$. 3. For a triangle to exist, the third side must be strictly between $99\text{ cm}$ and $101\text{ cm}$. Answer (a): Strictly between $99\text{ cm}$ and $101\text{ cm}$ **Answer:** The possible lengths for the third side are strictly between the difference and the sum of the two given sides. > Common mistake: Including the boundary values 99 and 101, which form a degenerate straight line instead of a triangle. ### Question 2 *3 marks · Short answer* Construct a triangle $ABC$ with $AB = 5\text{ cm}$, $AC = 4\text{ cm}$ and $\angle A = 45^\circ$. **Solution** 1. Construct a line segment $AB$ of length $5\text{ cm}$ using a ruler. 2. Draw an angle of $45^\circ$ at vertex $A$ using a protractor and draw the ray forming the arm. 3. Mark a point $C$ on the ray at a distance of $4\text{ cm}$ from $A$ using a compass. 4. Join $B$ and $C$ to obtain the required triangle $ABC$. **Answer:** A triangle $ABC$ with $AB = 5\text{ cm}$, $AC = 4\text{ cm}$, and $\angle A = 45^\circ$ > Common mistake: Measuring the angle from the wrong baseline or marking the side length on the wrong ray. ## Figure it Out (Page 161) ### Question 1 *3 marks · Short answer* Construct triangles for the following measurements where the angle is included between the sides: (a) $3\text{ cm}, 75^\circ, 7\text{ cm}$ (b) $6\text{ cm}, 25^\circ, 3\text{ cm}$ (c) $3\text{ cm}, 120^\circ, 8\text{ cm}$ **Part (a)** 1. Draw a base line segment of length 7 cm. 2. Construct an angle of $75^\circ$ at one end of the base line segment. 3. Mark a point at a distance of 3 cm on the other arm of the angle and join it to the other end of the base. Answer (a): A triangle with sides $3\text{ cm}, 7\text{ cm}$ and included angle $75^\circ$. **Part (b)** 1. Draw a base line segment of length 6 cm. 2. Construct an angle of $25^\circ$ at one end of the base line segment. 3. Mark a point at a distance of 3 cm on the other arm of the angle and join it to the other end of the base. Answer (b): A triangle with sides $6\text{ cm}, 3\text{ cm}$ and included angle $25^\circ$. **Part (c)** 1. Draw a base line segment of length 8 cm. 2. Construct an angle of $120^\circ$ at one end of the base line segment. 3. Mark a point at a distance of 3 cm on the other arm of the angle and join it to the other end of the base. Answer (c): A triangle with sides $3\text{ cm}, 8\text{ cm}$ and included angle $120^\circ$. **Answer:** Triangles constructed using the two sides and the included angle. > Common mistake: Measuring the included angle from the wrong vertex or confusing the side lengths. ## In-text questions (Page 161) ### Question 1 *3 marks · Short answer* We have seen that triangles do not exist for all sets of sidelengths. Is there a combination of measurements in the case of two sides and the included angle where a triangle is not possible? Justify your answer using what you observe during construction. **Solution** 1. Yes, a triangle is not possible if the included angle is greater than or equal to $180^\circ$. 2. When constructing such a triangle, after drawing the base and the given angle, the other two arms of the angles either become parallel or diverge away from each other. 3. Since the two lines never meet to form the third vertex, no triangle can be constructed. **Answer:** A triangle is not possible when the included angle is greater than or equal to $180^\circ$, as the lines do not meet. > Common mistake: Thinking that a triangle can be formed with any measure of the included angle. ### Question 2 *3 marks · Short answer* Construct a triangle $ABC$ where $AB = 5\text{ cm}$, $\angle A = 45^\circ$ and $\angle B = 80^\circ$. **Solution** 1. Draw the base $AB$ of length $5\text{ cm}$. 2. Draw an angle of $45^\circ$ at vertex $A$ and an angle of $80^\circ$ at vertex $B$ using a protractor. 3. Mark the point of intersection of the two new line segments as the third vertex $C$ to get the required triangle $ABC$. **Answer:** Triangle $ABC$ constructed with $AB = 5\text{ cm}$, $\angle A = 45^\circ$, and $\angle B = 80^\circ$. > Common mistake: Constructing the angles on the wrong side or misreading the protractor scale. ## Figure it Out (Page 162) ### Question 1 *3 marks · Short answer* Construct triangles for the following measurements: (a) $75^\circ, 5\text{ cm}, 75^\circ$ (b) $25^\circ, 3\text{ cm}, 60^\circ$ (c) $120^\circ, 6\text{ cm}, 30^\circ$ **Part (a)** 1. Draw a base line segment of length $5\text{ cm}$ and label its ends as base vertices. 2. Construct angles of $75^\circ$ at both ends of the base line segment using a protractor. 3. Extend the arms of the two angles until they intersect to form the third vertex and complete the triangle. Answer (a): A triangle with angles $75^\circ, 75^\circ$ and included side $5\text{ cm}$ is constructed. **Part (b)** 1. Draw a base line segment of length $3\text{ cm}$ and label its ends as base vertices. 2. Construct an angle of $25^\circ$ at one end and an angle of $60^\circ$ at the other end using a protractor. 3. Extend the arms of the two angles until they intersect to form the third vertex and complete the triangle. Answer (b): A triangle with angles $25^\circ, 60^\circ$ and included side $3\text{ cm}$ is constructed. **Part (c)** 1. Draw a base line segment of length $6\text{ cm}$ and label its ends as base vertices. 2. Construct an angle of $120^\circ$ at one end and an angle of $30^\circ$ at the other end using a protractor. 3. Extend the arms of the two angles until they intersect to form the third vertex and complete the triangle. Answer (c): A triangle with angles $120^\circ, 30^\circ$ and included side $6\text{ cm}$ is constructed. **Answer:** Triangles constructed using the two angles and the included side for each measurement. > Common mistake: Measuring the angles from the wrong reference line or using the incorrect vertex as the base. ## In-text questions (Page 162) ### Question 1 *3 marks · Short answer* Do triangles exist for every combination of two angles and their included side? Explore. **Solution** 1. State that triangles do not exist for every combination of two angles and their included side. 2. Consider the sum of the two given angles adjacent to the included side. 3. A triangle is possible only when the sum of the two given angles is strictly less than $180^\circ$. **Answer:** Triangles do not always exist; they are formed only when the sum of the two angles is less than $180^\circ$. > Common mistake: Assuming that any two angles can form a triangle with a side. ### Question 2 *3 marks · Short answer* Find examples of measurements of two angles with the included side where a triangle is not possible. **Solution** 1. State an example where a triangle is not possible using two angles and an included side. 2. Consider two angles whose sum is equal to or greater than $180^\circ$. 3. For example, taking angles $90^\circ$ and $90^\circ$, or $100^\circ$ and $90^\circ$ with any side length, the lines will never meet to form a triangle. **Answer:** Examples of measurements where a triangle is not possible are $90^\circ, 5\text{ cm}, 90^\circ$ or $100^\circ, 4\text{ cm}, 90^\circ$. > Common mistake: Choosing angles whose sum is less than $180^\circ$. ### Question 3 *3 marks · Case-based* (a) Try to find a possible $\angle B$ (marked in the figure) for this to happen. (b) What could be smallest value of $\angle B$ for the lines to not meet? **Part (a)** 1. Observe the figure in the textbook (page 162) where one base angle is $40^\circ$. 2. If $\angle B$ is sufficiently large, the lines do not meet. 3. Thus, a possible value for $\angle B$ for this to happen is $145^\circ$ (any angle $\ge 140^\circ$). Answer (a): A possible value for $\angle B$ is $145^\circ$. **Part (b)** 1. Consider the marginal case when the line from $B$ is parallel to the line $l$ passing through $A$. 2. When two lines are parallel, the internal angles on the same side of the transversal $AB$ add up to $180^\circ$. 3. Subtract the given angle from $180^\circ$: $180^\circ - 40^\circ = 140^\circ$, so the smallest value of $\angle B$ is $140^\circ$. Answer (b): The smallest value of $\angle B$ is $140^\circ$. **Answer:** (a) A possible value is $145^\circ$. (b) The smallest value is $140^\circ$. > Common mistake: Confusing the parallel line condition sum of $180^\circ$ with $90^\circ$. ## In-text questions (Page 163) ### Question 1 *3 marks · Short answer* Can you tell the actual value of $\angle B$ be in this case? [Hint: Note that $AB$ is the transversal.] **Solution** 1. Note that line $l$ is parallel to line $m$, and $AB$ acts as a transversal intersecting both parallel lines. 2. The interior angles on the same side of the transversal $AB$ are supplementary. 3. Therefore, $\angle B = 180^\circ - 40^\circ = 140^\circ$. **Answer:** $140^\circ$ > Common mistake: Adding the angles to $180^\circ$ incorrectly or confusing alternate interior angles with interior angles on the same side of the transversal. ## Figure it Out (Page 163) ### Question 1 *3 marks · Short answer* 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) $30^\circ$ (b) $70^\circ$ (c) $54^\circ$ (d) $144^\circ$ **Part (a)** 1. For $30^\circ$, a triangle is possible if the second angle is less than $150^\circ$, for example, $50^\circ$ or $90^\circ$. 2. A triangle is not possible if the second angle is greater than or equal to $150^\circ$, for example, $150^\circ$ or $170^\circ$. Answer (a): Possible: $50^\circ, 90^\circ$; Not possible: $150^\circ, 170^\circ$ **Part (b)** 1. For $70^\circ$, a triangle is possible if the second angle is less than $110^\circ$, for example, $60^\circ$ or $80^\circ$. 2. A triangle is not possible if the second angle is greater than or equal to $110^\circ$, for example, $120^\circ$ or $140^\circ$. Answer (b): Possible: $60^\circ, 80^\circ$; Not possible: $120^\circ, 140^\circ$ **Part (c)** 1. For $54^\circ$, a triangle is possible if the second angle is less than $126^\circ$, for example, $64^\circ$ or $90^\circ$. 2. A triangle is not possible if the second angle is greater than or equal to $126^\circ$, for example, $134^\circ$ or $154^\circ$. Answer (c): Possible: $64^\circ, 90^\circ$; Not possible: $134^\circ, 154^\circ$ **Part (d)** 1. For $144^\circ$, a triangle is possible if the second angle is less than $36^\circ$, for example, $20^\circ$ or $35^\circ$. 2. A triangle is not possible if the second angle is greater than or equal to $36^\circ$, for example, $36^\circ$ or $90^\circ$. Answer (d): Possible: $20^\circ, 35^\circ$; Not possible: $36^\circ, 90^\circ$ **Answer:** Possible and not possible angles found for each given angle using the rule that the sum of two angles must be less than $180^\circ$. > Common mistake: Confusing the condition for triangle existence where the sum of two angles must be strictly less than $180^\circ$. ### Question 2 *3 marks · Short answer* Determine which of the following pairs can be the angles of a triangle and which cannot: (a) $35^\circ, 150^\circ$ (b) $70^\circ, 30^\circ$ (c) $90^\circ, 85^\circ$ (d) $50^\circ, 150^\circ$ **Part (a)** 1. Sum of the given angles $= 35^\circ + 150^\circ = 185^\circ$. 2. Since the sum is greater than $180^\circ$, a triangle cannot be formed. Answer (a): Cannot be the angles of a triangle. **Part (b)** 1. Sum of the given angles $= 70^\circ + 30^\circ = 100^\circ$. 2. Since the sum is less than $180^\circ$, a triangle can be formed. Answer (b): Can be the angles of a triangle. **Part (c)** 1. Sum of the given angles $= 90^\circ + 85^\circ = 175^\circ$. 2. Since the sum is less than $180^\circ$, a triangle can be formed. Answer (c): Can be the angles of a triangle. **Part (d)** 1. Sum of the given angles $= 50^\circ + 150^\circ = 200^\circ$. 2. Since the sum is greater than $180^\circ$, a triangle cannot be formed. Answer (d): Cannot be the angles of a triangle. **Answer:** Determined which angle pairs can form a triangle based on the rule that the sum of two angles must be less than $180^\circ$. > Common mistake: Stating that a triangle can exist when the sum of the angles is equal to or greater than $180^\circ$. ## In-text questions (Page 163) ### Question 1 *3 marks · Short answer* Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible? **Solution** 1. A triangle with two given angles is possible if and only if the sum of the two angles is strictly less than $180^\circ$. 2. If the sum of the two angles is greater than or equal to $180^\circ$, the non-included sides (or lines drawn from the base vertices) will never meet to form a third vertex. 3. Thus, the rule describing the two angles $A$ and $B$ for a triangle to be possible is $0^\circ < A + B < 180^\circ$. **Answer:** $0^\circ < A + B < 180^\circ$ > Common mistake: Including $180^\circ$ or $0^\circ$ as valid sums for triangle formation. ### Question 2 *3 marks · Short answer* Can the sum of the two angles be used for framing this rule? **Solution** 1. Yes, the sum of the two given angles can be directly used for framing this rule. 2. From the property of parallel lines and transversals, the lines drawn from the base vertices will only intersect to form the third vertex if their sum is less than $180^\circ$. 3. Therefore, the sum of the two angles must be strictly less than $180^\circ$ for a triangle to exist. **Answer:** Yes, the sum of the two angles being less than $180^\circ$ is used to frame the rule. > Common mistake: Stating that the sum must equal $180^\circ$ instead of being less than $180^\circ$. ## In-text questions (Page 164) ### Question 1 *3 marks · Short answer* What could the measure of the third angle be? Does this measure change if the base length is changed to some other value, say $7\text{ cm}$? Construct and find out. **Solution** 1. Let the two given angles be $60^\circ$ and $70^\circ$. 2. By the angle sum property of triangles, the third angle is $180^\circ - (60^\circ + 70^\circ) = 50^\circ$. 3. This measure does not change if the base length is changed to some other value, say $7\text{ cm}$. 4. The third angle is $50^\circ$. **Answer:** $50^\circ$, and it does not change when the base length is changed. > Common mistake: Thinking that changing the base length changes the angles of the triangle. ### Question 2 *3 marks · Short answer* 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. **Solution** 1. Construct triangles with the same two angles but different lengths for the included side. 2. Measure the third angle in each constructed triangle. 3. We observe that the third angle remains the same for a fixed set of two angles. 4. Thus, the third angle does not depend on the included sidelength. **Answer:** No, the third angle does not depend on the included sidelength once the two angles are fixed. > Common mistake: Assuming that larger sides must have larger angles directly without geometric constraints. ### Question 3 *3 marks · Short answer* Try experimenting with different triangles to see if there is a relation between any two angles and the third one. To find this relation, what data will you keep track of and how will you organise the data you collect? **Solution** 1. Keep track of the measures of all three angles of different triangles in a table. 2. Organise the collected data by recording $\angle A$, $\angle B$, and $\angle C$ for each triangle. 3. Calculate the sum of the three angles in each case to find the relation. 4. The sum of the three angles is always found to be $180^\circ$. **Answer:** Keep track of all three angle measures in a table and find their sum for different triangles. > Common mistake: Not recording enough data points to spot the constant sum of $180^\circ$. ### Question 4 *3 marks · Short answer* Consider a triangle $ABC$ with $\angle B = 50^\circ$ and $\angle C = 70^\circ$. Let us see how we can find $\angle A$ without construction. **Solution** 1. Construct a line $XY$ parallel to $BC$ passing through the vertex $A$ as shown in Fig. 7.6. 2. Use the parallel lines property where alternate interior angles are equal with transversals $AB$ and $AC$. 3. Write $\angle XAB = \angle B = 50^\circ$ and $\angle YAC = \angle C = 70^\circ$. 4. Since the angles on a straight line add up to $180^\circ$, $\angle A = 180^\circ - (50^\circ + 70^\circ) = 60^\circ$. **Answer:** $\angle A = 60^\circ$ > Common mistake: Forgetting to draw the parallel line through the vertex to relate the angles. ### Question 5 *3 marks · Short answer* We can see new angles being formed here: $\angle XAB$, and $\angle YAC$. What are their values? **Solution** 1. Given that line $XY$ is parallel to $BC$, and $AB$ and $AC$ act as transversals. 2. The angle $\angle XAB$ is equal to $\angle B$ because they are alternate interior angles. 3. The angle $\angle YAC$ is equal to $\angle C$ because they are alternate interior angles. 4. Therefore, $\angle XAB = 50^\circ$ and $\angle YAC = 70^\circ$. **Answer:** $\angle XAB = 50^\circ$ and $\angle YAC = 70^\circ$ > Common mistake: Confusing alternate interior angles with corresponding or interior angles on the same side. ## In-text questions (Page 165) ### Question 1 *3 marks · Short answer* Therefore, $\angle XAB = 50^\circ$, and $\angle YAC = 70^\circ$. Can we find $\angle BAC$ from this? **Solution** 1. We know that the line $XY$ is parallel to $BC$ and the angles $\angle XAB$, $\angle BAC$, and $\angle YAC$ together lie on a straight line. 2. The sum of these three angles forming a straight angle is equal to $180^\circ$. 3. Substituting the given values, $50^\circ + \angle BAC + 70^\circ = 180^\circ$, which gives $\angle BAC = 60^\circ$. **Answer:** $60^\circ$ > Common mistake: Forgetting that angles on a straight line add up to $180^\circ$. ### Question 2 *3 marks · Short answer* Now construct a triangle (taking $BC$ to be of any suitable length) and verify if this is indeed the case. **Solution** 1. Draw a base $BC$ of any suitable length, for example $5\text{ cm}$. 2. Construct angles of $50^\circ$ at $B$ and $70^\circ$ at $C$ such that their arms intersect at vertex $A$. 3. Measure the angle $\angle BAC$ at the third vertex, which is found to be $60^\circ$, verifying the calculated value. **Answer:** The triangle is successfully constructed and verified. > Common mistake: Errors in measuring angles using a protractor during construction. ## Figure it Out (Page 165) ### Question 1 *3 marks · Short answer* Find the third angle of a triangle (using a parallel line) when two of the angles are: (a) $36^\circ, 72^\circ$ (b) $150^\circ, 15^\circ$ (c) $90^\circ, 30^\circ$ (d) $75^\circ, 45^\circ$ **Part (a)** 1. Consider a triangle $\text{ABC}$ with $\angle B = 36^\circ$ and $\angle C = 72^\circ$. 2. Construct a line $XY$ parallel to $BC$ through vertex $A$, making $AB$ and $AC$ transversals. 3. Using alternate angles, $\angle XAB = \angle B = 36^\circ$ and $\angle YAC = \angle C = 72^\circ$. 4. On the straight line $XY$, the sum of angles is $\angle XAB + \angle BAC + \angle YAC = 180^\circ$. 5. Substitute the given angle values: $36^\circ + \angle BAC + 72^\circ = 180^\circ$, giving $\angle BAC = 72^\circ$. Answer (a): $72^\circ$ **Part (b)** 1. The sum of the two given angles is $150^\circ + 15^\circ = 165^\circ$. 2. Using the straight line angle sum $180^\circ$, subtract the sum from $180^\circ$. 3. $\angle BAC = 180^\circ - 165^\circ = 15^\circ$. Answer (b): $15^\circ$ **Part (c)** 1. The sum of the two given angles is $90^\circ + 30^\circ = 120^\circ$. 2. Subtract this sum from $180^\circ$ to find the third angle. 3. $\angle BAC = 180^\circ - 120^\circ = 60^\circ$. Answer (c): $60^\circ$ **Part (d)** 1. The sum of the two given angles is $75^\circ + 45^\circ = 120^\circ$. 2. Subtract this sum from $180^\circ$ to find the third angle. 3. $\angle BAC = 180^\circ - 120^\circ = 60^\circ$. Answer (d): $60^\circ$ **Answer:** The third angles are (a) $72^\circ$, (b) $15^\circ$, (c) $60^\circ$, and (d) $60^\circ$. > Common mistake: Adding the angles incorrectly before subtracting from $180^\circ$. ### Question 2 *3 marks · Short answer* 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. **Part (i)** 1. Calculate the sum of all three angles if they were each $70^\circ$: $70^\circ + 70^\circ + 70^\circ = 210^\circ$. 2. Since the sum of angles in a triangle must be exactly $180^\circ$, a triangle with all angles equal to $70^\circ$ is not possible. Answer (i): No, such a triangle is not possible. **Part (ii)** 1. Let the two given angles be $70^\circ$ and $70^\circ$. 2. Subtract their sum from $180^\circ$: $180^\circ - (70^\circ + 70^\circ) = 180^\circ - 140^\circ = 40^\circ$. Answer (ii): $40^\circ$ **Part (iii)** 1. Let each of the three equal angles of the triangle be $x$. 2. Using the angle sum property, $x + x + x = 180^\circ$, which gives $3x = 180^\circ$. 3. Divide by 3 to get $x = 60^\circ$. Answer (iii): $60^\circ$ **Answer:** A triangle with all angles equal to $70^\circ$ is not possible; if two angles are $70^\circ$, the third is $40^\circ$; if all angles are equal, each must measure $60^\circ$. > Common mistake: Forgetting that the sum of all angles in a triangle must always equal $180^\circ$. ### Question 3 *3 marks · Short answer* Here is a triangle in which we know $\angle B = \angle C$ and $\angle A = 50^\circ$. Can you find $\angle B$ and $\angle C$? **Solution** 1. Given that $\angle A = 50^\circ$ and $\angle B = \angle C$. 2. Using the angle sum property of triangles, the sum of all three angles is $\angle A + \angle B + \angle C = 180^\circ$. 3. Substitute the given value and replace $\angle C$ with $\angle B$: $50^\circ + \angle B + \angle B = 180^\circ$. 4. Simplify the equation: $2 \angle B = 180^\circ - 50^\circ = 130^\circ$. 5. Divide by 2 to find the measure: $\angle B = \frac{130^\circ}{2} = 65^\circ$. 6. Since $\angle B = \angle C$, both angles are equal to $65^\circ$. **Answer:** $\angle B = \angle C = 65^\circ$ > Common mistake: Dividing $180^\circ$ by 2 directly without subtracting $\angle A$ first. ## In-text questions (Page 165) ### Question 1 *3 marks · Short answer* What can we say about the sum of the angles of any triangle? **Solution** 1. Consider a triangle ABC and construct a line XY parallel to BC passing through the vertex A. 2. Since XY is parallel to BC and AB and AC act as transversals, the alternate interior angles give $\angle XAB = \angle B$ and $\angle YAC = \angle C$. 3. The angles $\angle XAB$, $\angle BAC$, and $\angle YAC$ together lie on a straight line XY, so their sum is $180^\circ$ (${\angle XAB + \angle BAC + \angle YAC = 180^\circ}$). 4. Substituting the equal alternate angles, we get $\angle A + \angle B + \angle C = 180^\circ$. 5. Thus, the sum of the three angles of any triangle is always $180^\circ$, which is known as the angle sum property of triangles. **Answer:** The sum of the three angles of any triangle is always $180^\circ$. > Common mistake: Forgetting to mention the straight line property or incorrectly using angle relationships without establishing parallel lines. ## In-text questions (Page 166) ### Question 1 *3 marks · Short answer* 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 $180^\circ$? **Solution** 1. Take a triangular paper cut-out and fold the top vertex down so that it touches the opposite base. 2. Make a crease parallel to the base such that the three corners (angles) of the triangle meet at a single point along a straight line on the base. 3. Since the three angles together form a straight angle on the line, their sum is $180^\circ$, verifying the angle sum property. **Answer:** The three angles of the triangle meet along a straight line when folded, forming a straight angle of $180^\circ$. > Common mistake: Not aligning the folded vertex properly onto the base line. ## In-text questions (Page 167) ### Question 1 *3 marks · Short answer* Find $\angle ACD$, if $\angle A = 50^\circ$, and $\angle B = 60^\circ$. **Solution** 1. From the angle sum property, in triangle ABC, $\angle A + \angle B + \angle ACB = 180^\circ$. 2. Substitute the given values: $50^\circ + 60^\circ + \angle ACB = 180^\circ$, which gives $\angle ACB = 70^\circ$. 3. Since $\angle ACB$ and $\angle ACD$ form a linear pair on a straight line, $\angle ACD = 180^\circ - 70^\circ = 110^\circ$. **Answer:** $\angle ACD = 110^\circ$ > Common mistake: Subtracting the angles directly from $180^\circ$ without finding the interior adjacent angle first. ### Question 2 *3 marks · Short answer* Find the exterior angle for different measures of $\angle A$ and $\angle B$. Do you see any relation between the exterior angle and these two angles? **Solution** 1. Using the angle sum property for any triangle, $\angle A + \angle B + \angle ACB = 180^\circ$, so $\angle ACB = 180^\circ - (\angle A + \angle B)$. 2. The exterior angle $\angle ACD$ and $\angle ACB$ form a straight angle, so $\angle ACD = 180^\circ - \angle ACB$. 3. Substituting the value of $\angle ACB$, we get $\angle ACD = 180^\circ - (180^\circ - (\angle A + \angle B)) = \angle A + \angle B$. **Answer:** The exterior angle of a triangle is equal to the sum of its two interior opposite angles. > Common mistake: Confusing interior opposite angles with the adjacent interior angle. ### Question 3 *3 marks · Short answer* What does this show? **Solution** 1. We have $\angle ACD + \angle ACB = 180^\circ$ since they form a linear pair on a straight line. 2. We also know from the angle sum property that $\angle A + \angle B + \angle ACB = 180^\circ$. 3. Comparing both equations shows that the exterior angle $\angle ACD$ is equal to the sum of the interior opposite angles $\angle A$ and $\angle B$. **Answer:** This shows that the exterior angle of a triangle is always equal to the sum of the two interior opposite angles. > Common mistake: Stating that the exterior angle equals all three interior angles combined. ### Question 4 *3 marks · Short answer* What would the altitude from $A$ to $BC$ be in this triangle? **Solution** 1. In an obtuse-angled triangle where the perpendicular from a vertex falls outside the base, the side must be extended. 2. We extend the line segment $BC$ outwards. 3. Then we drop a perpendicular from vertex $A$ to this extended line segment $BC$, which gives the required altitude. **Answer:** The altitude is the perpendicular segment dropped from $A$ to the extended line $BC$. > Common mistake: Drawing the altitude inside the obtuse triangle directly from $A$ to $BC$. ## In-text questions (Page 168) ### Question 1 *3 marks · Short answer* 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. **Solution** 1. Fix one of the sides of the paper triangle as the base and hold the opposite vertex. 2. Fold the paper such that the top vertex falls directly on itself and the two parts of the base fall exactly on top of each other. 3. The crease formed passes through the top vertex and meets the base at a $90^\circ$ angle because the two halves of the base fold symmetrically upon each other, making the crease perpendicular to the base. **Answer:** The crease formed is perpendicular to the base because folding the base halves onto each other makes the angles on either side of the crease equal to $90^\circ$. > Common mistake: Stating the crease is perpendicular without explaining the symmetry of the fold. ### Question 2 *3 marks · Short answer* Construct the altitude from $A$ to $BC$ **Solution** 1. Align the straight edge of the ruler along the side $BC$ to act as the base line. 2. Place the set square on the ruler such that one of the edges forming the right angle rests against the ruler. 3. Slide the set square along the ruler until the other perpendicular edge touches the vertex $A$, and draw a line segment from $A$ perpendicular to $BC$. **Answer:** The line segment drawn from vertex $A$ perpendicular to $BC$ using a ruler and set square is the required altitude. > Common mistake: Using only a ruler to draw a perpendicular line without a set square. ### Question 3 *2 marks · Very short answer* Can you see how to do this? **Solution** 1. Yes, we can draw the perpendicular by placing a set square on a ruler aligned with the base. 2. Sliding the set square to the vertex allows us to draw a precise $90^\circ$ altitude line. **Answer:** Yes, by placing a set square on a ruler aligned to the base and sliding it until it touches the vertex, we can construct the altitude. > Common mistake: Not mentioning the use of both ruler and set square together. ## In-text questions (Page 169) ### Question 1 *3 marks · Short answer* Does there exist a triangle in which a side is also an altitude? **Solution** 1. Visualise a triangle where one of the sides forms a perpendicular height from a vertex to the opposite base. 2. We see that this happens when one of the angles of the triangle is a right angle ($90^\circ$). 3. Thus, a side is also an altitude in right-angled triangles (or right triangles). **Answer:** Yes, such a triangle exists and it is called a right-angled triangle. > Common mistake: Thinking that an altitude can only lie strictly inside the interior of a triangle. ## In-text questions (Page 170) ### Question 1 *3 marks · Short answer* Did you spot any other type of triangle? **Solution** 1. Triangles are classified by their sides into equilateral, isosceles, and scalene triangles. 2. Triangles are also classified by their angles into right-angled triangles. 3. Thus, right-angled triangles are another type of triangle based on angle measures. **Answer:** Right-angled triangles. > Common mistake: Confusing side-based classification with angle-based classification. ### Question 2 *3 marks · Short answer* Can a similar classification be done based on equality of angles? Is there any relation between these two classifications? We will answer these questions in a later chapter. **Solution** 1. Yes, a similar classification of triangles can be done based on the equality of angles. 2. The relation between these two classifications will be answered in a later chapter. **Answer:** Yes, triangles can be classified based on the equality of angles. > Common mistake: Stating that no relation exists without reading the textbook text. ### Question 3 *3 marks · Short answer* What are the other types of triangles based on angle measures? **Solution** 1. Triangles based on angle measures are classified into acute-angled triangles, right-angled triangles, and obtuse-angled triangles. **Answer:** Acute-angled, right-angled, and obtuse-angled triangles. > Common mistake: Listing side-based types instead of angle-based types. ### Question 4 *3 marks · Short answer* What could an acute-angled triangle be? Can we define it as a triangle with one acute angle? Why not? **Solution** 1. An acute-angled triangle is a triangle in which all three angles are acute angles. 2. We cannot define it as a triangle with just one acute angle because every triangle has at least two acute angles. **Answer:** A triangle in which all three angles are acute angles. It cannot be defined by having one acute angle because every triangle has at least two acute angles. > Common mistake: Defining an acute-angled triangle as having only one acute angle. ## Figure it Out (Page 170-171) ### Question 1 *3 marks · Short answer* Construct a triangle $ABC$ with $BC = 5\text{ cm}$, $AB = 6\text{ cm}$, $CA = 5\text{ cm}$. Construct an altitude from $A$ to $BC$. **Solution** 1. Construct the base BC of length 5 cm using a ruler. 2. With B as centre and radius 6 cm, draw an arc. With C as centre and radius 5 cm, draw another arc intersecting the first arc at point A. 3. Join AB and AC to form the isosceles triangle ABC with AB = 6 cm and AC = 5 cm. 4. Place a set square with one of its right-angle edges along BC, and slide it until the vertical edge passes through vertex A, then draw the perpendicular AD to BC. **Answer:** An isosceles triangle ABC with sides 5 cm, 6 cm, 5 cm and altitude from A to BC constructed. > Common mistake: Drawing the altitude from the wrong vertex or not using a set square for an accurate 90° angle. ### Question 2 *3 marks · Short answer* Construct a triangle $TRY$ with $RY = 4\text{ cm}$, $TR = 7\text{ cm}$, $\angle R = 140^\circ$. Construct an altitude from $T$ to $RY$. **Solution** 1. Given: $RY = 4\text{ cm}$, $TR = 7\text{ cm}$, $\angle R = 140^\circ$. 2. Construct the base segment $RY$ of length $4\text{ cm}$ using a ruler. 3. At vertex $R$, draw an angle of $140^\circ$ using a protractor and extend the arm. 4. Mark the point $T$ on this arm at a distance of $7\text{ cm}$ from $R$ using a compass and join $TY$ to complete $\Delta TRY$. 5. Since $\angle R = 140^\circ$ is obtuse, extend the base line segment $RY$ past $R$, and use a set square and ruler to drop a perpendicular from $T$ onto the extended base line at point $D$. **Answer:** Triangle $TRY$ constructed with an altitude from $T$ meeting the extended base $RY$ at $D$. > Common mistake: Trying to drop the perpendicular from $T$ directly onto the side $RY$ without extending it, even though $\angle R$ is obtuse. ### Question 3 *3 marks · Short answer* Construct a right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$, $AC = 5\text{ cm}$. How many different triangles exist with these measurements? [Hint: Note that the other measurements can take any values. Take $AC$ as the base. What values can $\angle A$ and $\angle C$ take so that the other angle is $90^\circ$] **Solution** 1. Given: A right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$ and hypotenuse $AC = 5\text{ cm}$. 2. Take the hypotenuse $AC$ of length $5\text{ cm}$ as the base. 3. The vertex $B$ must form a right angle ($\angle B = 90^\circ$), so $B$ lies on the semicircle drawn with $AC$ as diameter. 4. Since the other angles $\angle A$ and $\angle C$ can take any acute values such that their sum is $90^\circ$, infinitely many points $B$ can be chosen on this semicircle. 5. Therefore, infinitely many different right-angled triangles exist with these measurements. **Answer:** Infinitely many different triangles exist. > Common mistake: Concluding that only one right-angled triangle exists because the hypotenuse and right angle are fixed. ### Question 4 *3 marks · Short answer* 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. **Solution** 1. An equilateral triangle has all three angles equal to $60^\circ$, so it cannot be right-angled or obtuse-angled. 2. An isosceles right-angled triangle can be constructed (e.g., with angles $90^\circ, 45^\circ, 45^\circ$). 3. An isosceles obtuse-angled triangle can also be constructed (e.g., with angles $100^\circ, 40^\circ, 40^\circ$). **Answer:** Equilateral triangle cannot be right-angled or obtuse-angled; isosceles triangles can be both right-angled and obtuse-angled. > Common mistake: Assuming an equilateral triangle can have a $90^\circ$ angle, which contradicts the angle sum property of $180^\circ$. ## In-text questions (Page 172) ### Question 1 *3 marks · Short answer* What is the shortest path it can take? **Solution** 1. Unfold the three-dimensional rectangular box into a two-dimensional flat net showing the surfaces. 2. The shortest path between two opposite corners on the surface of a box is a straight line drawn on the unfolded flat net of the adjacent faces. 3. The length of this straight-line path is found by applying Pythagoras theorem to the dimensions of the unfolded rectangular faces. **Answer:** The shortest path is the straight line joining the opposite corners on the unfolded net of the box surfaces. > Common mistake: Thinking the spider can walk through the interior of the box along a direct three-dimensional diagonal. ### Question 2 *3 marks · Short answer* Take a cardboard box and mark the path that you think is the shortest from one corner to its opposite corner. Compare the length of this path with that of the paths made by your friends. **Solution** 1. Take a rectangular cardboard box and mark one starting corner and its farthest opposite corner. 2. Draw different paths along the adjacent outer surfaces of the box connecting the two opposite corners. 3. Measure the length of each path with a string or scale and compare them to verify that the straight line on the unfolded net gives the minimum length. **Answer:** The path corresponding to the straight line on the unfolded net of the box is the shortest among all paths drawn by friends. > Common mistake: Drawing roundabout zig-zag paths instead of unfolding the box surfaces into a single plane. ## Frequently asked questions ### How many total questions are covered in the NCERT solutions for Class 7 Maths Chapter 7? This chapter covers numerous in-text questions, 'Construct' tasks, and 'Figure it Out' sections across pages 146 to 172. SwaVid provides complete step-by-step solutions and a free PDF for all these questions right on this page. ### Which major topics are included in the Class 7 Maths Chapter 7 NCERT questions? The questions cover concepts such as triangles and collinear vertices, construction of equilateral triangles, the Triangle Inequality theorem, circle intersections, triangle construction using side and angle criteria, parallel lines and transversals, the angle sum property, altitudes of triangles, and triangle classifications. You can check SwaVid's free PDF on this page for detailed explanations of each topic. ### What are the hardest question types in this chapter and how should we approach them? Case-based and multi-step construction questions involving the Triangle Inequality, circle intersections, or parallel lines and transversals can be challenging. To approach them, carefully note the given side lengths or angle conditions, apply geometric theorems step-by-step, and verify your constructions before writing the final answer. ### How can I write answers in exams to score full marks for Class 7 Maths Chapter 7? To score full marks, always state the geometric property or theorem being used, such as the triangle inequality or the angle sum property, before applying it in calculations. Present your steps clearly in a logical sequence, and SwaVid's free PDF solutions on this page serve as an ideal reference for proper presentation. ### Is the free PDF for Class 7 Maths Chapter 7 solutions available on SwaVid? Yes, the complete free PDF for NCERT Solutions for Class 7 Maths Chapter 7: A Tale of Three Intersecting Lines is available right on this page. It aligns with the new NCERT book under the NCF 2023 syllabus for the 2026-27 session. ## Related pages - [Class 7 Maths chapters](https://www.swavid.com/maths/class/7) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student