NCERT Solutions for Class 7 Maths Chapter 7 a Tale of Three Intersecting Lines (PDF) – 2027

No triangle is formed at all. The figure collapses into a single line segment.

Use a compass — a ruler alone would need many trials.

1. Draw AB = 4 cm with a ruler.
2. With A as centre and radius 4 cm, draw a long arc above AB.
3. With B as centre and the same radius 4 cm, draw another arc cutting the first one at C.
4. Join AC and BC. ∆ABC is the required equilateral triangle.

Yes, it can be done with only a marked ruler and a pencil — but it takes many trials.

Replace the guessing with two compass arcs.

Because C lies on both arcs at once.

The same two-arc method works — only the two radii are now different.

1. Choose any one of the three lengths as the base and draw it.
2. From one end, draw an arc whose radius is the second length.
3. From the other end, draw an arc whose radius is the third length.
4. Join the crossing point to both ends.

Take AB = 4 cm as the base, AC = 5 cm and BC = 6 cm.

1. Step 1: Draw AB = 4 cm.
2. Step 2: With A as centre, draw a sufficiently long arc of radius 5 cm.
3. Step 3: With B as centre, draw an arc of radius 6 cm so that it cuts the first arc.
4. Step 4: Call the crossing point C. Join AC and BC to get ∆ABC.

Do not draw the full circles — two short arcs are enough.

In every case draw the longest side as the base, then cut two arcs of the other two lengths.

Take the centre O and any two points P and Q on the circle. ∆OPQ is isosceles.

Each circle passes through the centre of the other, so in both figures AB = radius. In the first figure, call the two crossing points P and Q.

No. The two arcs never meet, so the third vertex simply does not exist.

Not possible. The same thing happens, only worse.

Here are five impossible sets, and the pattern behind all of them.

The pattern: a triangle is impossible exactly when

No such triangle exists.

The red path — the straight line from the tent to the tree — is shorter.

Yes. Suppose such a ∆ABC existed, with BC = 10 cm, AB = 15 cm and CA = 30 cm. Test all three pairs of vertices.

No triangle exists.

Verifying by construction: draw AB = 7 cm, then an arc of radius 3 cm from A and another of radius 3 cm from B. The two arcs stop 1 cm apart and never cut each other, so there is no third vertex.

No. Rearranging the labels changes nothing.

Yes — one comparison settles each case.

None of the three sets gives a triangle.

Yes, always. At most one comparison can ever fail.

Reason: write the lengths in increasing order. Each of the two smaller lengths is already no bigger than the largest one, so it is certainly less than the largest one plus another length. Only the largest length can fail its comparison.

Yes. Put the three lengths in increasing order — the two smaller ones pass automatically.

Compare the longest length with the sum of the other two. That is the only check needed.

When each length is smaller than the sum of the other two, the lengths are said to satisfy the triangle inequality.

Because the other two comparisons are already guaranteed by the size order.

Take AB = 8 cm as the base. Draw the circle of radius 4 cm about A and the circle of radius 5 cm about B — they overlap in a lens-shaped region.

BX = 4 cm. Yes — it shows at once that the two circles must cross.

Compare the longest length with the sum of the other two in each set.

Yes — and we can be sure by looking at how the two circles can possibly sit.

Take the base AB = the longest length, and draw circles at A and B whose radii are the two smaller lengths. Exactly three things can happen:

The two radii together must fall short of AB.

In the figure, circle A ends before circle B begins, so a gap is left on AB between the two circles and they have no point in common.

Yes — the analysis closes the argument completely.

Conclusion: if a given set of three lengths satisfies the triangle inequality, a triangle exists having those as sidelengths. If it does not, no such triangle exists.

They land in Case 1 or Case 2.

(a) Circles touch at a point — sum of the two smaller lengths = longest length:

(b) Circles do not intersect — sum of the two smaller lengths < longest length:

A three-step routine that never fails.

1. Step 1: Make sure all three lengths are in the same unit.
2. Step 2: Pick out the largest length. Add the other two.
3. Step 3: Compare.
   • largest < sum of the other two → the triangle exists
   • largest = sum of the other two → the three points are collinear, no triangle
   • largest > sum of the other two → no triangle

Compare the largest length with the sum of the other two.

Yes to both.

For any sidelength s (a positive number):

The third length must be more than the difference and less than the sum of the two given lengths.

In every case the answer is an open range: difference < third length < sum.

Draw one given side as the base, build the given angle at one end, and cut off the second side along the new arm.

1. Draw the first given side as the base.
2. At the end where the angle sits, use a protractor to draw the second arm of the angle.
3. Along that arm, mark off the second given length.
4. Join the marked point to the far end of the base.

Take AB as the base, since the given angle sits at A.

1. Step 1: Construct a side AB of length 5 cm.
2. Step 2: Construct ∠A = 45° by drawing the other arm of the angle.
3. Step 3: Mark the point C on the other arm such that AC = 4 cm.
4. Step 4: Join BC to get the required triangle.

In each case draw the longer given side as the base, make the angle at one end, and mark the other side along the new arm.

For (a), for example:

1. Draw PQ = 7 cm.
2. At P draw an arm making 75° with PQ.
3. Mark R on that arm with PR = 3 cm.
4. Join QR. ∆PQR is the required triangle.

Yes — but only when the given angle cannot be an angle of a triangle, that is, when it is 180° or more (or 0°). For any angle between 0° and 180°, a triangle is formed whatever the two lengths are.

Take the given side AB as the base and grow both angles from it.

1. Step 1: Draw the base AB of length 5 cm.
2. Step 2: Draw ∠A and ∠B of measures 45° and 80° respectively, on the same side of AB.
3. Step 3: The point of intersection of the two new line segments is the third vertex C.

Draw the given side as the base and set the two given angles at its two ends.

No. If the two given angles are too big, the two arms never meet.

Any pair whose sum reaches 180° will do, whatever the length of the included side.

With ∠A = 40°, the arm from B must bend far enough to the right.

(a) ∠B = 150° works — so do 145°, 160°, 170°. With any of these the line from B never meets the line l drawn from A.

(b) The smallest such value is ∠B = 140°.

∠B = 140°.

A triangle does not exist when ∠B ≥ 140°, and the length AB plays no part at all.

In general, for base angles ∠A and ∠B:

The other angle must be less than 180° minus the given angle for a triangle to be possible.

Add the pair and compare with 180°.

Yes. The rule uses exactly the sum of the two angles.

• If the sum is less than 180°, a triangle with these two angles exists.
• If the sum is 180° or more, no such triangle exists.

The third angle is 50°, and it does not change when the base changes.

No. The third angle depends only on the two given angles.

Keep a three-column table of the three angles, and add a fourth column for their sum.

The relation:

∠XAB = 50°, ∠YAC = 70°, and therefore ∠BAC = 60°.

Verification: construct BC of any length with ∠B = 50° and ∠C = 70°; on measuring, ∠A = 60°.

Draw XY through A parallel to BC. The two base angles reappear at A as alternate angles, so the three angles at A fill a straight angle.

No, such a triangle is impossible.

If two of the angles are 70°:

If all three angles are equal:

∠B = ∠C = 65°.

The sum is always 180° — for every triangle, big or small, thin or fat.

This result is called the angle sum property of triangles.

Yes — fold the three corners inwards and they line up along the base.

1. Cut out a paper triangle ABC; take the longest side BC as the base.
2. Fold A down onto BC, keeping the crease parallel to BC.
3. Fold B and C inwards to meet A at the same point.
4. ∠A, ∠B and ∠C now fit side by side along BC, with no gap and no overlap.

∠ACD = 110°.

The exterior angle always equals the sum of the two opposite interior angles.

The two hints prove it in one line:

The height of an object is how far its top is above the ground, measured straight up — that is, perpendicular to the ground.

Drop a perpendicular from A to BC and measure it. That perpendicular is the height.

The other two altitudes are BE and CF — the perpendiculars from B and from C to their own opposite sides.

The foot of the perpendicular falls outside the triangle, so BC has to be extended first.

1. Cut out ∆ABC and take BC as the base.
2. Fold the paper so the crease passes through A and part of BC falls exactly on the rest of BC.
3. Open it out. The crease from A to BC is the altitude.

Why the crease is perpendicular:

Use a set square sliding along a ruler — a ruler alone cannot give an accurate 90°.

1. Step 1: Keep the ruler aligned to the base BC. Place the set square on the ruler so that one of the edges of its right angle touches the ruler.
2. Step 2: Slide the set square along the ruler till the vertical edge of the set square touches the vertex A.
3. Step 3: Draw the altitude to BC through A using the vertical edge of the set square.

Yes — in a right-angled triangle. If ∠B = 90°, then AB is itself the altitude from A to BC.

Triangles having one right angle are called right-angled triangles, or simply right triangles.

Yes. Besides equilateral, isosceles, scalene and right-angled triangles, the chapter also throws up acute-angled and obtuse-angled triangles.

Three types altogether — acute-angled, right-angled and obtuse-angled.

An acute-angled triangle is one in which all three angles are acute. Defining it as “a triangle with one acute angle” would be useless.

First check the triangle inequality, then build it and drop the perpendicular.

1. Draw BC = 5 cm.
2. With B as centre, draw an arc of radius 6 cm (this is AB).
3. With C as centre, draw an arc of radius 5 cm (this is CA), cutting the first arc at A.
4. Join AB and AC.
5. Align a ruler along BC, slide a set square on it until its vertical edge touches A, and draw AD ⟂ BC.

∠R is the included angle between RY and TR, so build the angle at R.

1. Draw RY = 4 cm.
2. At R, draw an arm making ∠R = 140° with RY, on one side of it.
3. Mark T on that arm with TR = 7 cm.
4. Join TY. ∆TRY is the required triangle.
5. For the altitude, extend YR beyond R, then drop the perpendicular from T to that extended line; call its foot D. TD is the altitude.

Infinitely many different triangles fit these two conditions.

How to construct one: draw AC = 5 cm, choose any ∠A between 0° and 90°, make ∠C = 90° – ∠A, and let the two arms meet at B.

Equilateral: neither is possible. Isosceles: both are possible.

To draw each isosceles one: make the angle at B, mark 4 cm along both arms (BA = BC = 4 cm) and join AC.

Flatten the box. Open out two adjoining faces so they lie in one plane — then the shortest path is the straight line joining the two corners on that flat sheet.