Class 7 Mathematics · Part 1 · Chapter 7

7. A Tale of Three Intersecting Lines

Answers to all Figure it Out and in-text questions on constructing triangles, triangle inequality, angle sum property and altitudes. CBSE / NCERT Class 7 Ganita Prakash.

?What happens when the three vertices lie on a straight line?

No triangle is formed. The three points just make a line segment.

7.1 Equilateral Triangles

?Construct a triangle with all sides 4 cm. Why are AC and BC both 4 cm?

Step 1: Draw AB = 4 cm.

Step 2: With A as centre and radius 4 cm, draw an arc.

Step 3: With B as centre and radius 4 cm, draw another arc cutting the first arc at C.

Step 4: Join AC and BC.

C lies on the arc with centre A and radius 4 cm, so AC = 4 cm. C also lies on the arc with centre B and radius 4 cm, so BC = 4 cm.

7.2 Constructing a Triangle When its Sides are Given

?Construct triangles with the following sidelengths (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

For each set, draw the longest side as the base. From one end, draw an arc with the second length. From the other end, draw an arc with the third length. Join the meeting point to both ends.

Example (a): Draw AB = 6 cm. Draw arcs of radius 4 cm from A and from B. They meet at C. ∆ABC is isosceles.

All five triangles can be constructed. (a) and (c) are isosceles; (e) is equilateral.

FIOFigure it Out

1. Use the points on the circle and/or the centre to form isosceles triangles.

Join the centre O to any two points P and Q on the circle. OP = OQ (both are radii), so ∆OPQ is isosceles.

2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles.

Two circles with centres A and B: let C and D be the points where the circles meet. AC = AB = BC (all are radii), so ∆ABC and ∆ABD are equilateral.

Take any point P on the circle with centre B. BA = BP, so ∆ABP is isosceles.

Three circles with centres A, B, C: ∆ABC is equilateral, since each side is a radius. Any centre with two points on its circle gives an isosceles triangle.

?Construct a triangle with sides 3 cm, 4 cm and 8 cm, and with 2 cm, 3 cm and 6 cm. What happens?

The arcs do not meet, so the triangles cannot be constructed.

More such sets: 1, 2, 5; 2, 4, 7; 3, 5, 10. In each, the two smaller lengths add up to less than the longest.

?Can a triangle have sides 3 cm, 3 cm and 7 cm? Can the lengths 10, 15, 30 be rearranged so that all direct paths are shorter?

No. 3 + 3 = 6 is less than 7. The arcs of 3 cm will not meet.

No. Whichever side is 30 cm, the roundabout path is 10 + 15 = 25 cm, which is shorter than 30 cm.

FIOFigure it Out

1. Could you have found that 3, 4, 8 and 2, 3, 6 give no triangle without constructing?

Yes. 3 + 4 = 7 is less than 8, and 2 + 3 = 5 is less than 6. The direct path is longer than the roundabout path.

2. Does a triangle exist?

(a) 10 + 10 = 20 < 25 → No.

(b) 5 + 10 = 15 < 20 → No.

3. Will there always be at least two comparisons where a length is less than the sum of the other two?

Yes. The two smaller lengths are each less than the longest length, so each is less than the sum of the other two.

So, we only need to check if the longest length is less than the sum of the other two.

?Does a triangle exist with sides 4 cm, 5 cm and 8 cm? Why do we not need to check the other two sides?

8 < 4 + 5 = 9, so the triangle inequality is satisfied.

4 and 5 are smaller than 8, so they are surely less than the sum of the other two.

With AB = 8 cm, the circle of radius 4 cm from A reaches X, where BX = 4 cm. The circle from B has radius 5 cm, which is more than 4 cm. So, the circles meet at two points and the triangle exists.

FIOFigure it Out: Which lengths can be the sidelengths of a triangle?

(a) 2, 2, 5: 2 + 2 = 4 < 5 → No

(b) 3, 4, 6: 3 + 4 = 7 > 6 → Yes

(d) 5, 5, 8: 5 + 5 = 10 > 8 → Yes

(e) 10, 20, 25: 10 + 20 = 30 > 25 → Yes

(f) 10, 20, 35: 10 + 20 = 30 < 35 → No

(g) 24, 26, 28: 24 + 26 = 50 > 28 → Yes

?Find 3 sets of lengths for which the circles (a) touch each other at a point, (b) do not intersect. Frame a procedure to check for a triangle.

(a) Touch: 2, 3, 5; 4, 4, 8; 1, 6, 7 (sum of the two smaller lengths = the longest).

(b) Do not intersect: 2, 3, 6; 1, 1, 5; 3, 4, 10 (sum of the two smaller lengths < the longest).

Procedure: find the longest length. Add the other two. If this sum is greater than the longest length, a triangle exists. Otherwise, it does not.

FIOFigure it Out

1. Check if a triangle exists.

(a) 1, 100, 100: 1 + 100 > 100 → exists

(b) 3, 6, 9: 3 + 6 = 9, not greater → does not exist

(d) 5, 10, 12: 5 + 10 > 12 → exists

2. Does an equilateral triangle with sides 50, 50, 50 exist? Does one exist for any sidelength?

Yes. 50 < 50 + 50. For any side a, a < a + a, so an equilateral triangle exists for every sidelength.

3. Give 5 possible values for the third side.

(a) 1, 100: any length between 99 and 101 → 99.5, 99.9, 100, 100.4, 100.8

(b) 5, 5: any length between 0 and 10 → 1, 3, 5, 7, 9.5

7.3 Construction of Triangles When Some Sides and Angles are Given

FIOFigure it Out: Construct triangles where the angle is included between the sides.

Step 1: Draw the first side as the base, e.g. AB = 7 cm for (a).

Step 2: At A, draw the given angle with a protractor (75° for (a)).

Step 3: On the new arm, mark C so that AC is the other side (3 cm for (a)).

Step 4: Join BC.

Do the same for (b) 6 cm, 25°, 3 cm and (c) 3 cm, 120°, 8 cm.

?Is there a combination of two sides and the included angle where a triangle is not possible?

Only if the angle is 180° or more. For any angle between 0° and 180°, the arm can always be drawn and the two ends can always be joined.

FIOFigure it Out: Construct triangles for (a) 75°, 5 cm, 75° (b) 25°, 3 cm, 60° (c) 120°, 6 cm, 30°.

Step 1: Draw the given side as the base AB.

Step 2: At A draw the first angle and at B draw the second angle, on the same side of AB.

Step 3: The arms meet at C. ∆ABC is the required triangle.

In (a), the triangle is isosceles because the base angles are equal.

?With ∠A = 40°, for which values of ∠B do the lines not meet? Does AB play any part?

(a) For example, ∠B = 150°.

(b) The smallest such ∠B is when the line from B is parallel to l. Interior angles on the same side add up to 180°, so ∠B = 180° − 40° = 140°.

A triangle does not exist when ∠B ≥ 140°. The length AB plays no part.

FIOFigure it Out

1. Find another angle for which a triangle is possible and not possible.

(a) 30°: possible with 50° or 90°; not possible with 150° or 170°.

(b) 70°: possible with 60° or 100°; not possible with 110° or 130°.

(d) 144°: possible with 20° or 30°; not possible with 36° or 50°.

2. Which pairs can be the angles of a triangle?

(a) 35° + 150° = 185° → cannot

(b) 70° + 30° = 100° → can

(d) 50° + 150° = 200° → cannot

Rule

A triangle is possible when the sum of the two angles is less than 180°.

?For angles 60° and 70°, what is the third angle? Does it change with the base length?

The third angle is 180° − 60° − 70° = 50°. It does not change when the base is 5 cm or 7 cm. The third angle depends only on the two angles.

FIOFigure it Out

1. Find the third angle.

(a) 180° − 36° − 72° = 72°

(b) 180° − 150° − 15° = 15°

(d) 180° − 75° − 45° = 60°

2. Can all angles be 70°? If two angles are 70°, what is the third? If all angles are equal, what is each?

No, because 70° + 70° + 70° = 210°, not 180°.

Third angle = 180° − 140° = 40°.

If all are equal, each is 180° ÷ 3 = 60°.

3. ∠B = ∠C and ∠A = 50°. Find ∠B and ∠C.

∠B + ∠C = 180° − 50° = 130°. So, ∠B = ∠C = 65°.

?How does folding a triangular paper cut-out show that the angles add up to 180°?

When the three corners are folded to meet at one point on the base, the three angles fit together on a straight line. A straight angle is 180°.

?Find the relation between the exterior angle ∠ACD and ∠A and ∠B.

∠A + ∠B + ∠ACB = 180° and ∠ACD + ∠ACB = 180°.

So, ∠ACD = ∠A + ∠B. The exterior angle equals the sum of the two opposite interior angles. Example: 50° + 60° = 110°.

7.4 Constructions Related to Altitudes of Triangles

?Fold a paper triangle so that the crease is an altitude. Why is the crease perpendicular to the base?

Fold so that the base falls exactly on itself and the crease passes through the top vertex.

The two angles at the base on either side of the crease are then equal. Together they make 180°, so each is 90°.

?Does there exist a triangle in which a side is also an altitude?

Yes. In a right-angled triangle, the two sides that make the right angle are also altitudes.

7.5 Types of Triangles

?Can we define an acute-angled triangle as a triangle with one acute angle? Why not?

No. Every triangle has at least two acute angles, even right-angled and obtuse-angled triangles. So, an acute-angled triangle must have all three angles acute.

FIOFigure it Out

1. Construct ∆ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm and the altitude from A to BC.

Draw BC = 5 cm. Draw an arc of 6 cm from B and an arc of 5 cm from C to get A. Then place a ruler along BC and slide a set square along it till its edge touches A. Draw AD perpendicular to BC.

2. Construct ∆TRY with RY = 4 cm, TR = 7 cm, ∠R = 140° and the altitude from T to RY.

Draw RY = 4 cm. At R draw 140° and mark T so that RT = 7 cm. Join TY. ∠R is obtuse, so the altitude falls outside: extend YR beyond R and draw the perpendicular from T to this extended line.

3. Right-angled ∆ABC with ∠B = 90° and AC = 5 cm. How many triangles exist?

Infinitely many. ∠A can be any angle between 0° and 90°, and then ∠C = 90° − ∠A. Each choice gives a different triangle.

4. Equilateral and isosceles triangles that are right-angled or obtuse-angled

An equilateral triangle can be neither. All its angles are 60°.

An isosceles triangle can be right-angled (90°, 45°, 45°) and obtuse-angled (for example 120°, 30°, 30°).

?Shortest Path in a Box: What is the shortest path for the spider?

Open the box flat (make its net). Draw a straight line from the spider’s corner to the opposite corner on the net. This straight line, folded back on the box, is the shortest path. It crosses one edge of the box.

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