CBSE Class 7 Mathematics A Tale of Three Intersecting Lines Worksheet Set 01

Long Answer Questions

Question 1. An equilateral triangle ABC of side 5.5 cm has to be drawn.
(i) Explain why drawing it with a marked ruler alone is difficult.
(ii) Write the complete steps of construction using a ruler and a compass.
(iii) Justify why the point where the two arcs cross is exactly the third vertex C.
Answer:
(i) Why a ruler alone is hard. With a ruler we can easily draw AB = 5.5 cm and also mark a point C with AC = 5.5 cm. But we cannot make sure that BC also comes out as 5.5 cm at the same time. We would have to move C again and again by trial and error, which is slow and never exact.
(ii) Steps of construction.
Step 1: Draw a line segment AB = 5.5 cm with a ruler.
Step 2: Open the compass to 5.5 cm using the ruler. With A as centre, draw a long arc above AB.
Step 3: With the same opening and B as centre, draw another arc cutting the first one. Mark the crossing point as C.
Step 4: Join AC and BC. △ABC is the required equilateral triangle.
(iii) Justification. The first arc holds every point that is 5.5 cm from A; the second arc holds every point that is 5.5 cm from B. C lies on both arcs, so AC = 5.5 cm and BC = 5.5 cm together. With AB = 5.5 cm, all three sides are equal.
In simple words: A ruler cannot fix two distances at once, but two equal arcs from A and B meet at the exact point C.

Teacher's Note:
1. A compass draws all points at a fixed distance in one sweep.
2. Keep the compass opening unchanged for both arcs.
3. The crossing point satisfies both length conditions together.

Question 2. A triangle ABC is to be constructed with AB = 7 cm, BC = 5 cm and CA = 6 cm.
(i) First check, by comparison, that such a triangle exists.
(ii) Write the steps of construction, taking AB as the base.
(iii) Classify the triangle by its sides, and explain what would go wrong if the given lengths were 7 cm, 2 cm and 4 cm instead.
Answer:
(i) Checking existence. 7 < 5 + 6 = 11, 6 < 7 + 5 = 12, 5 < 7 + 6 = 13. The triangle inequality is satisfied, so the triangle exists.
(ii) Steps of construction.
Step 1: Draw the base AB = 7 cm.
Step 2: With A as centre and radius 6 cm (the length CA), draw a long enough arc above AB.
Step 3: With B as centre and radius 5 cm (the length BC), draw another arc cutting the first one at C.
Step 4: Join AC and BC to get △ABC. C is on both arcs, so it is 6 cm from A and 5 cm from B, as required.
(iii) The sides 7 cm, 5 cm and 6 cm are all different, so it is a scalene triangle. For 7, 2 and 4: 2 + 4 = 6 < 7, so the triangle inequality fails. With 7 cm as the base, arcs of radii 2 cm and 4 cm would stay apart and never cross, so no third vertex and no triangle would be obtained.
In simple words: 7, 5, 6 passes the test and gives a scalene triangle, but 7, 2, 4 fails because 2 + 4 is less than 7.

Teacher's Note:
1. Use the correct radius at each end: CA from A, BC from B.
2. Always check existence before constructing.
3. Arcs that never cross mean no triangle.

Teacher's Note:
1. Every triangle has one name from each grouping.
2. Check each angle example adds up to 180°.
3. Side examples must satisfy the triangle inequality.

Question 4. Altitudes are an important set of measurements of a triangle.
(i) Define an altitude of a triangle and state how many altitudes a triangle has.
(ii) Write the steps for drawing the altitude from A to BC in a triangle ABC using a ruler and a set square.
(iii) Describe where the foot of this altitude lies when the triangle is acute-angled, when ∠B = 90°, and when ∠B is obtuse.
Answer:
(i) Meaning. An altitude is the perpendicular line segment from a vertex to the opposite side, or to that side extended. Its length is the height of that vertex from the chosen base. A triangle has three vertices, so it has three altitudes.
(ii) Steps using a ruler and a set square.
Step 1: Place the ruler along the base BC and hold it firmly.
Step 2: Place the set square on the ruler so that one edge of its right angle lies flat along the ruler.
Step 3: Slide the set square along the ruler until its upright edge just touches the vertex A.
Step 4: Draw a line along that upright edge from A down to BC. Name its foot D. AD is the required altitude, with ∠ADB = 90°.
(iii) Position of the foot. In an acute-angled triangle the foot D lies inside the base BC. If ∠B = 90°, the altitude from A is the side AB itself and the foot is the vertex B. If ∠B is obtuse, D falls outside the base, so BC must first be extended beyond B and the perpendicular dropped onto that extended line.
In simple words: An altitude is the 90° line from a corner to the opposite side; its foot is inside, at B, or outside, as ∠B is acute, right or obtuse.

Teacher's Note:
1. The set square gives the right angle; the ruler keeps it on the base line.
2. Extend the base on the side of the obtuse angle.
3. Every triangle has exactly three altitudes.

Question 5. Construct △PQR in which PQ = 6 cm, ∠P = 50° and PR = 4 cm.
(i) Write the steps of construction.
(ii) Without measuring, find the sum of ∠Q and ∠R and give the property you used.
(iii) For a set of measurements of the form “side, included angle, side”, explain for which included angles no triangle can be drawn.
Answer:
(i) Steps of construction.
Step 1: Draw PQ = 6 cm as the base.
Step 2: At P, use a protractor to draw a ray PX making ∠QPX = 50° with PQ.
Step 3: On the ray PX, mark the point R such that PR = 4 cm.
Step 4: Join QR. △PQR is the required triangle.
(ii) By the angle sum property, ∠P + ∠Q + ∠R = 180°, so ∠Q + ∠R = 180° − 50° = 130°, with no measuring needed.
(iii) The two arms of the included angle open out from the vertex and are then joined. This works for every included angle less than 180°. If the included angle is exactly 180°, the two sides lie along one straight line, the three points become collinear and no closed shape forms. An angle greater than 180° is not a valid angle of a triangle at all. So no triangle can be drawn when the included angle is 180° or more; it must lie strictly between 0° and 180°.
In simple words: Draw PQ, make 50° at P, mark R 4 cm along it and join QR; the other two angles total 130°.

Teacher's Note:
1. The included angle is at the vertex shared by the two given sides.
2. A 180° 'angle' is just a straight line.
3. Angle sum gives ∠Q + ∠R without measuring.

Question 6. Construct △XYZ in which XY = 6 cm, ∠X = 60° and ∠Y = 45°.
(i) Write the steps of construction and say why XY is called the included side.
(ii) Find ∠Z without measuring, and classify the triangle by its angles.
(iii) A classmate tries the same construction with ∠X = 60° and ∠Y = 125° on the same 6 cm side. Explain what he will observe and why.
Answer:
(i) Steps of construction.
Step 1: Draw the base XY = 6 cm.
Step 2: At X, draw a ray making 60° with XY.
Step 3: At Y, draw a ray making 45° with YX, on the same side of XY.
Step 4: Mark the point where the two rays cross as Z. △XYZ is the required triangle.
XY is called the included side because it is an arm of both the given angles, ∠X and ∠Y.
(ii) By the angle sum property, ∠Z = 180° − (60° + 45°) = 180° − 105° = 75°. The angles 60°, 45° and 75° are all less than 90°, so the triangle is acute-angled.
(iii) Here 60° + 125° = 185°, which is more than 180°. The two rays bend away from each other, so however far they are extended they never meet and no third vertex is found. A triangle exists only when the two given angles add up to less than 180°, and the length of the included side makes no difference.
In simple words: The 60° and 45° rays meet at Z with ∠Z = 75°, but 60° and 125° add to 185°, so those rays never meet.

Teacher's Note:
1. Draw both rays on the same side of the base.
2. Check the angle sum before constructing.
3. A longer base would not help the classmate.

Question 7. The figure shows △ABC with a line XY drawn through A parallel to BC. The three angles made at A are marked 1, 2 and 3.
(i) Using alternate angles, explain how this figure proves that the three angles of a triangle add up to 180°.
(ii) In a triangle ∠A = ∠B and ∠C = 80°. Find ∠A and ∠B.
(iii) Describe a paper-folding activity by which a student can check the same property for a triangle cut out of paper.
Answer:
(i) The proof. XY is drawn through A parallel to BC.
AB is a transversal of the parallel lines XY and BC, so angle 1 (∠XAB) and ∠B are alternate angles: angle 1 = ∠B.
AC is also a transversal, so angle 3 (∠YAC) and ∠C are alternate angles: angle 3 = ∠C.
Angle 2 is ∠BAC, that is, ∠A.
Angles 1, 2 and 3 lie together along the straight line XY, so angle 1 + angle 2 + angle 3 = 180°, that is, ∠B + ∠A + ∠C = 180°. This is the angle sum property; the idea of the parallel line through the top vertex comes from Euclid’s Elements (about 300 BCE).
(ii) ∠A + ∠B + 80° = 180°, so ∠A + ∠B = 100°. As ∠A = ∠B, each is 100° ÷ 2 = 50°.
(iii) Paper folding. Cut out a paper triangle. Fold the top vertex down so that it lands on the base, with the crease parallel to the base. Then fold the two base corners inwards so that they meet at the same point. The three corners now lie side by side along the base and together form a straight line, showing that their measures add up to 180°.
In simple words: The parallel line XY copies ∠B and ∠C to the top, where with ∠A they fill a straight angle of 180°.

Teacher's Note:
1. Alternate angles between parallel lines are equal.
2. Angles on a straight line add up to 180°.
3. Paper folding gives a quick hands-on check.

Question 8. Two sides of a triangle measure 6 cm and 10 cm.
(i) Find the range of lengths the third side can take, and list five possible values.
(ii) Describe what happens to the two construction circles if the third side is taken as 4 cm, and what happens if it is taken as 3 cm.
(iii) Between what two values must the perimeter of such a triangle lie?
Answer:
(i) Let the third side be x cm. The triangle inequality gives x < 6 + 10 = 16 and 10 < 6 + x, that is, x > 4. So x lies strictly between 4 cm and 16 cm. Five possible values: 5 cm, 7.5 cm, 10 cm, 12 cm and 15.5 cm.
(ii) Take the longest length as the base and the other two as radii. If x = 4, the lengths are 4, 6, 10 and 4 + 6 = 10: the two circles just touch at one point on the base, so no triangle forms. If x = 3, the lengths are 3, 6, 10 and 3 + 6 = 9 < 10: the circles stay apart and do not meet. Again no triangle forms.
(iii) Perimeter = 6 + 10 + x = 16 + x. Since 4 < x < 16, we get 16 + 4 < perimeter < 16 + 16, so the perimeter lies strictly between 20 cm and 32 cm.
In simple words: The third side must be between 4 cm and 16 cm, so the perimeter is between 20 cm and 32 cm.

Teacher's Note:
1. Third side: more than the difference, less than the sum.
2. At exactly 4 cm the circles touch; below it they separate.
3. Add 16 to each end of the range to get the perimeter range.

Question 9. A right-angled triangle ABC has ∠B = 90°, AB = 3 cm and BC = 4 cm.
(i) Write the steps of construction and state what the measured length of AC should be, to the nearest half centimetre.
(ii) Name the two altitudes of this triangle that are already sides of it, and describe how the third altitude is drawn.
(iii) How many different triangles can be drawn with ∠B = 90° and AC = 5 cm only? Justify your answer using the angle sum property.
Answer:
(i) Steps of construction.
Step 1: Draw BC = 4 cm.
Step 2: At B, draw a ray perpendicular to BC with a set square or protractor, so that ∠B = 90°.
Step 3: On this ray mark A with BA = 3 cm.
Step 4: Join AC to complete △ABC. On measuring, AC comes out as 5 cm.
(ii) Since ∠B = 90°, AB is perpendicular to BC, so AB is the altitude from A; similarly CB is the altitude from C. For the third altitude, place the ruler along AC, slide a set square along it until its upright edge touches B, and draw the line from B to AC. This perpendicular BD is the third altitude, and it lies inside the triangle.
(iii) Infinitely many. With only ∠B = 90° and AC = 5 cm fixed, the angle sum property gives ∠A + ∠C = 180° − 90° = 90°. So ∠A can be any value strictly between 0° and 90°, with ∠C = 90° − ∠A: for example 30° and 60°, 45° and 45°, or 20° and 70°. Each choice gives a different triangle on the same 5 cm side, so these measurements do not fix one triangle.
In simple words: The 3-4 right triangle has AC = 5 cm; but fixing only the right angle and AC leaves endless choices for the other angles.

Teacher's Note:
1. 3, 4, 5 is a famous right-triangle set.
2. The two arms of the right angle are altitudes.
3. Two measurements alone usually do not fix a triangle.

Question 10. Construct △TRY in which RY = 4 cm, TR = 7 cm and ∠R = 140°.
(i) Write the steps of construction and classify the triangle by its angles.
(ii) Find the sum of ∠T and ∠Y, and explain why both of them have to be acute.
(iii) Explain why the side RY must be extended before the altitude from T can be drawn, and write the steps of drawing that altitude.
Answer:
(i) Steps of construction.
Step 1: Draw RY = 4 cm.
Step 2: At R, use a protractor to draw a ray making ∠YRT = 140° with RY.
Step 3: On this ray mark T with RT = 7 cm.
Step 4: Join TY to complete △TRY.
Since ∠R = 140° is more than 90°, it is an obtuse-angled triangle.
(ii) By the angle sum property, ∠T + ∠Y = 180° − 140° = 40°. Both angles are positive, so each is less than 40°, well below 90°. Hence both are acute. This also shows a triangle can never have two obtuse angles.
(iii) The altitude from T must meet the line RY at a right angle. Because ∠R is obtuse, the sides RT and RY open out wide and the foot of the perpendicular from T falls outside the segment RY, beyond R. So the base must first be extended.
Step 1: Extend YR beyond R to a point M.
Step 2: Place the ruler along the line YM and rest a set square on it.
Step 3: Slide the set square until its upright edge touches T, and draw the line from T meeting YM at N.
Then TN is the required altitude, with its foot N on the extended base beyond R, not on RY itself.
In simple words: The 140° angle makes the triangle obtuse, leaves only 40° for the other two, and pushes the altitude from T outside the base.

Teacher's Note:
1. An obtuse angle leaves less than 90° for the other two together.
2. Extend the base on the side of the obtuse angle.
3. The foot of the altitude is then on the extended line.

Case Study Based Questions

Question 11. A triangular park in a colony.
A colony has a small park PQR in the shape of a triangle. Its three boundary walls measure 30 m, 40 m and 50 m, as shown in the plan below. The residents’ committee wants to put a wire fence all along the boundary and to paint a scale drawing of the park on the notice board. One member suggests that the longest wall should be made 75 m long while keeping the other two walls as they are, so that more children can play inside.
(i) Check, using the triangle inequality, whether a park with these three wall lengths can really exist.
(ii) Find the total length of wire needed for the fence.
(iii) Classify the park by the lengths of its sides and give a reason.
(iv) Can the member’s suggestion of walls 30 m, 40 m and 75 m be carried out? Show the comparison that decides it.
Answer:
(i) Compare each wall with the sum of the other two: 50 < 30 + 40 = 70, 40 < 30 + 50 = 80 and 30 < 40 + 50 = 90. All three hold, so the triangle inequality is satisfied and such a park can exist.
(ii) Length of wire = perimeter = 30 + 40 + 50 = 120 m.
(iii) The walls 30 m, 40 m and 50 m are all different, so the park is a scalene triangle.
(iv) No. Here 30 + 40 = 70, and 70 < 75. The longest wall would be more than the sum of the other two, so the triangle inequality fails and the two shorter walls could never meet. The suggestion cannot be carried out.
In simple words: The 30-40-50 park exists and needs 120 m of wire, but a 75 m wall is longer than 30 + 40 and cannot close the shape.

Teacher's Note:
1. Perimeter means the total length of the boundary.
2. Making one side too long breaks the triangle inequality.
3. No two walls are equal, so the park is scalene.

Question 12. A rangoli of triangles.
For a school festival, Meera draws a rangoli made of many small triangles. Every small triangle has all its three sides of length 6 cm. She first draws one such triangle carefully on a sheet of paper with a ruler and a compass, cuts it out, and uses it as a template to trace all the other triangles. Her friend tells her that if all the three angles of a triangle are equal to one another, then the measure of each of those angles is fixed and can be found without any measurement.
(i) What is this type of triangle called when it is classified by its sides?
(ii) Does the set 6, 6, 6 satisfy the triangle inequality? Show the comparison.
(iii) Write the steps Meera follows to draw one such triangle with a ruler and a compass.
(iv) If all three angles of a triangle are equal, find the measure of each angle.
Answer:
(i) All three sides are equal, so each small triangle is an equilateral triangle.
(ii) Yes. 6 < 6 + 6 = 12, and the same comparison holds for each side. The triangle inequality is satisfied, so the triangle can be drawn.
(iii) Step 1: Draw a base AB = 6 cm. Step 2: Open the compass to 6 cm; with A as centre draw an arc above AB. Step 3: With the same opening and B as centre, draw another arc cutting the first at C. Step 4: Join AC and BC; since C lies on both arcs, AC = BC = 6 cm.
(iv) Let each angle be x. Then x + x + x = 180°, so 3x = 180° and x = 60°. Each angle measures 60°.
In simple words: Meera's triangles are equilateral; 6, 6, 6 passes the test, and each angle is 180° ÷ 3 = 60°.

Teacher's Note:
1. Three equal lengths always form a triangle.
2. Equal angles share 180° equally.
3. Use the same compass opening for both arcs.

Question 13. The frame of a kite.
Rehan is making a kite. The upper part of its frame is a triangle made of three thin sticks of lengths 25 cm, 25 cm and 30 cm. He joins the ends of the sticks firmly with thread and then pastes paper over the frame. His younger brother says that the same two 25 cm sticks could also be joined with a longer 55 cm stick to make a wider kite of the same shape.
(i) Name the type of this triangle according to its sides and give the reason.
(ii) Verify that the three sticks of 25 cm, 25 cm and 30 cm can really form a triangle.
(iii) Find the total length of stick used in the upper triangle of Rehan’s kite.
(iv) Is the brother’s idea of 25 cm, 25 cm and 55 cm possible? Give the comparison and the reason.
Answer:
(i) It is an isosceles triangle, because exactly two of its sides, the two 25 cm sticks, are equal.
(ii) Check the longest stick against the sum of the other two: 30 < 25 + 25 = 50. Also 25 < 25 + 30 = 55. The triangle inequality is satisfied, so the frame closes into a triangle.
(iii) Total length of stick = 25 + 25 + 30 = 80 cm.
(iv) No. 25 + 25 = 50, and 50 < 55. The longest stick is more than the sum of the other two, so the two 25 cm sticks can never meet above the long one. The brother’s idea is impossible.
In simple words: 25, 25, 30 makes an isosceles frame using 80 cm of stick, but 25 + 25 is too short to span 55 cm.

Teacher's Note:
1. Two equal sticks make the frame isosceles.
2. A longer base needs longer slanting sticks too.
3. Always compare the longest stick with the other two together.

Teacher's Note:
1. Arcs refusing to cross means the triangle inequality fails.
2. Aarav's 3, 4, 5 also works and is right-angled.
3. 8 > 7 by just 1, but that is enough for Dia.

Question 15. A bridge truss.
A footbridge is supported by steel panels, each shaped like the triangle LTR shown below. In each panel the base LR rests on the deck of the bridge, and the two slanting bars LT and TR meet at the top point T. The engineer has fixed the two base angles as ∠L = 55° and ∠R = 65°. The deck line is later extended beyond R to fit the railing.
(i) Find ∠T, the angle at the top of the panel.
(ii) Classify this panel by its angle measures.
(iii) When the deck line LR is extended beyond R to a point S, find the exterior angle ∠TRS.
(iv) The engineer now wants both base angles to be 95°. Is such a panel possible? Give a reason.
Answer:
(i) By the angle sum property, ∠L + ∠R + ∠T = 180°, so ∠T = 180° − (55° + 65°) = 180° − 120° = 60°.
(ii) The three angles are 55°, 65° and 60°, each less than 90°. So the panel is an acute-angled triangle.
(iii) ∠TRL = 65°, and ∠TRL and ∠TRS together form a straight angle along LS. So ∠TRS = 180° − 65° = 115°. This also equals 55° + 60°, the sum of the two opposite interior angles.
(iv) No. 95° + 95° = 190°, which is more than 180°. Two angles of a triangle must add up to less than 180°, so the two bars would bend away from each other and never meet at a top point.
In simple words: The top angle is 60°, the panel is acute-angled, the exterior angle at R is 115°, and two 95° base angles are impossible.

Teacher's Note:
1. Exterior angle = sum of the two opposite interior angles.
2. Two base angles of 90° or more can never close a triangle.
3. All three angles below 90° means acute-angled.

Question 16. Marking a triangular flower bed.
A gardener has to mark a triangular flower bed ABC in a lawn. From a fixed corner A of the lawn he measures one edge AB = 7 m and a second edge AC = 5 m, and he keeps the angle between these two edges equal to 60°. He then joins B to C with a rope to complete the bed. Before digging, he makes a scale drawing in his notebook taking 1 cm for 1 m.
(i) Which kind of data is given here: three sides, two sides with the included angle, or two angles with the included side?
(ii) Write the steps of the scale drawing in the notebook.
(iii) Find the sum of the other two angles ∠B and ∠C of the bed, and name the property used.
(iv) What would happen if the gardener kept the angle at A equal to 180°? Explain.
Answer:
(i) Two sides and the included angle: the sides AB = 7 m and AC = 5 m, with ∠A = 60° lying between them.
(ii) Step 1: Draw AB = 7 cm (scale 1 cm = 1 m). Step 2: At A, draw a ray making 60° with AB using a protractor. Step 3: On this ray mark C with AC = 5 cm. Step 4: Join BC to complete the drawing of the bed.
(iii) By the angle sum property, ∠B + ∠C = 180° − 60° = 120°.
(iv) With an angle of 180° at A, the edges AB and AC would lie along one straight line. The three corners would be in a single line, no closed figure would form, and there would be no flower bed at all.
In simple words: This is side-angle-side data drawn at 1 cm for 1 m; ∠B + ∠C = 120°, and a 180° angle at A would flatten the bed into a line.

Teacher's Note:
1. Scale drawings keep angles the same and shrink lengths.
2. The included angle must be less than 180°.
3. Angle sum gives ∠B + ∠C without measuring.

Question 17. A roof truss for a shed.
A carpenter is making the triangular front of a shed. The horizontal beam AB at the bottom is 8 m long. He fixes the rafter at A so that it makes an angle of 35° with the beam, and the rafter at B so that it makes an angle of 55° with the beam. The two rafters meet at the ridge point C. He then has to fix a vertical support from C straight down to the beam AB, so that the roof does not sag in the middle.
(i) Which construction data is this, and why is AB called the included side?
(ii) Find ∠C and classify the triangle by its angles.
(iii) Can the carpenter instead fix the two rafters at 90° and 95° with the beam? Give the reason.
(iv) What is the vertical support from C to AB called in geometry, and which instrument will the carpenter use with a ruler to draw it accurately?
Answer:
(i) Two angles and the included side are given. AB is the included side because it is an arm of both the given angles, ∠A and ∠B.
(ii) ∠C = 180° − (35° + 55°) = 180° − 90° = 90°. One angle is exactly 90°, so the front of the shed is a right-angled triangle.
(iii) No. 90° + 95° = 185°, which is more than 180°. Two given angles form a triangle only when their sum is less than 180°, so the rafters would never meet at a ridge point.
(iv) The vertical support is the altitude from C to the base AB. The carpenter will use a set square with the ruler, sliding it along the ruler until its upright edge touches C, so that the line drawn makes exactly 90° with AB.
In simple words: The rafters at 35° and 55° meet at a 90° ridge; 90° and 95° cannot meet, and the support from C is an altitude drawn with a set square.

Teacher's Note:
1. The included side is common to both given angles.
2. 35° + 55° = 90° leaves exactly 90° at the top.
3. A set square gives the right angle for an altitude.

Question 18. Painting a triangular signboard.
A painter has three triangular signboards to paint. For each board he first fixes one side as the base and then measures the height of the opposite corner from that base, because the height tells him how much paint he will need. On the first board, which is acute-angled, the base is 60 cm and the height he measures is 40 cm. The second board is right-angled and the third board is obtuse-angled. He uses a ruler together with a set square to draw each height line.
(i) What is the height line drawn from a corner to the opposite side called, and what angle does it make with that side?
(ii) Write the steps the painter follows with the ruler and the set square to draw this line on the first board.
(iii) On the obtuse-angled board, the foot of the height line does not fall on the base at all. Explain where it falls and what the painter must do.
(iv) On the right-angled board, two of the three height lines need no drawing at all. Explain why.
Answer:
(i) It is called an altitude of the triangle. It makes an angle of 90° with the side taken as the base.
(ii) Step 1: Place a ruler along the base of the board. Step 2: Rest the set square on the ruler so that one edge of its right angle lies along the ruler. Step 3: Slide the set square along the ruler until its upright edge touches the opposite corner. Step 4: Draw the line along that edge from the corner down to the base; this is the required height line of 40 cm.
(iii) When the base is a side next to the obtuse angle, the foot of the altitude from the opposite corner falls outside the base, beyond the obtuse-angled corner. The painter must first extend the base beyond that corner and then drop the perpendicular onto the extended line.
(iv) On a right-angled board the two sides forming the right angle are already perpendicular to each other. So each of them is the altitude from the vertex at its far end, and only the third altitude has to be drawn.
In simple words: The height line is an altitude at 90° to the base; it may need an extended base on an obtuse board, and on a right-angled board two sides already are altitudes.

Teacher's Note:
1. The ruler holds the base line; the set square gives 90°.
2. Extend the base beyond the obtuse-angled corner.
3. In a right-angled triangle, the legs are altitudes.

Question 19. Choosing rods for a frame.
A workshop has four straight iron rods of lengths 6 m, 8 m, 15 m and 17 m. A triangular support frame has to be welded using exactly three of these rods, joined end to end without cutting or bending any rod. The supervisor wants to know, before any welding is started, which choices of three rods will actually close up into a triangle, so that no rod is wasted.
(i) Write down all the four possible sets of three rods.
(ii) Find which of these sets will form a triangle. Show the comparison for each.
(iii) Find which sets cannot form a triangle, and state the reason in terms of the construction circles.
(iv) The 6 m rod is to be replaced by a new rod of whole-number length so that it forms a triangle with the 8 m and 15 m rods. Find the shortest and the longest such rod.
Answer:
(i) The four sets are (6, 8, 15), (6, 8, 17), (6, 15, 17) and (8, 15, 17).
(ii) (6, 15, 17): 17 < 6 + 15 = 21, 15 < 6 + 17 = 23, 6 < 15 + 17 = 32, so a triangle forms. (8, 15, 17): 17 < 8 + 15 = 23, 15 < 8 + 17 = 25, 8 < 15 + 17 = 32, so a triangle forms. Two sets work: (6, 15, 17) and (8, 15, 17).
(iii) (6, 8, 15): 6 + 8 = 14 < 15. (6, 8, 17): 6 + 8 = 14 < 17. These two sets cannot form a triangle. If the longest rod is the base and circles of the two smaller radii are drawn at its ends, the circles do not reach each other, so no third vertex exists and the frame cannot close.
(iv) Let the new rod be x m. Then x < 8 + 15 = 23 and 15 < 8 + x, that is, x > 7. So x lies strictly between 7 and 23. The shortest whole-number rod is 8 m and the longest is 22 m.
In simple words: Only (6, 15, 17) and (8, 15, 17) close into triangles, and a replacement rod must be between 8 m and 22 m.

Teacher's Note:
1. List the sets in an organised way so none is missed.
2. The 6 m and 8 m rods together are only 14 m, too short for 15 m or 17 m.
3. Third side: between the difference and the sum.

Question 20. The face of a camping tent.
A camping tent has a triangular front face ABC. The ground edge BC is the base, and the two sloping canvas edges AB and AC meet at the top point A. The tent is designed so that the two angles at the ground, ∠B and ∠C, are equal to each other, and the angle at the top, ∠A, is twice each ground angle. A straight pole is fixed from the top point A down to the ground so that it stands upright on BC. The company also makes a larger tent of exactly the same design but with a wider base.
(i) Find the measures of ∠A, ∠B and ∠C.
(ii) Classify this triangular face by its angle measures.
(iii) What is the upright pole called in geometry, and does its foot fall inside the base BC or outside it? Give a reason.
(iv) In the larger tent the base is wider but the design is the same. Do the three angles change? Justify your answer using what the chapter says about the third angle and the included side.
Answer:
(i) Let ∠B = ∠C = x. Then ∠A = 2x, and by the angle sum property 2x + x + x = 180°, so 4x = 180° and x = 45°. Hence ∠B = ∠C = 45° and ∠A = 90°. Check: 90° + 45° + 45° = 180°.
(ii) One angle is exactly 90°, so the front face is a right-angled triangle.
(iii) The upright pole is the altitude from A to the base BC, since it meets BC at 90°. Both base angles are acute (45° each), so its foot falls inside the segment BC, between B and C.
(iv) No, the angles do not change. Once the two angles at the ends of a side are fixed, the third angle is decided by the angle sum property alone and does not depend on the length of that side. So the wider tent still has angles 90°, 45° and 45°.
In simple words: The face has angles 90°, 45°, 45°; the pole is an altitude landing inside BC, and a wider base does not change the angles.

Teacher's Note:
1. Write all angles using one unknown.
2. Acute base angles put the foot of the altitude inside the base.
3. Same angles with a bigger base gives a bigger tent of the same shape.