A Tale of Three Intersecting Lines Class 7 Mathematics NCERT Solutions

Q1Are Triangles Possible for any Lengths?

Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle? Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle is possible for these sidelengths.

Solution

Case 1: Sidelengths 3 cm, 4 cm, and 8 cm

Attempted Construction:

Draw a line segment AB of length 8 cm.
With A as the center and a radius of 3 cm, draw an arc.
With B as the center and a radius of 4 cm, draw another arc.

Observation: The two arcs do not intersect. The sum of the radii (3 cm + 4 cm = 7 cm) is less than the distance between the centers (8 cm). Therefore, it is impossible to find a third point C that is 3 cm from A and 4 cm from B simultaneously.

Conclusion: A triangle with sidelengths 3 cm, 4 cm, and 8 cm cannot be constructed.

Case 2: Sidelengths 2 cm, 3 cm, and 6 cm

Attempted Construction:

Draw a line segment PQ of length 6 cm.
With P as the center and a radius of 2 cm, draw an arc.
With Q as the center and a radius of 3 cm, draw another arc.

Observation: The two arcs do not intersect. The sum of the radii (2 cm + 3 cm = 5 cm) is less than the distance between the centers (6 cm).

Conclusion: A triangle with sidelengths 2 cm, 3 cm, and 6 cm is not possible.

Q1Construct

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

Solution

To Construct: Triangles with given side lengths.

(a) Sidelengths: 4 cm, 4 cm, 6 cm Steps of Construction:

Draw a line segment BC of length 6 cm.
With B as the center and a radius of 4 cm, draw an arc.
With C as the center and a radius of 4 cm, draw another arc to intersect the first arc at point A.
Join AB and AC. Result: $\triangle ABC$ is the required isosceles triangle with sides 4 cm, 4 cm, and 6 cm.

(b) Sidelengths: 3 cm, 4 cm, 5 cm Steps of Construction:

Draw a line segment BC of length 4 cm.
With B as the center and a radius of 3 cm, draw an arc.
With C as the center and a radius of 5 cm, draw another arc to intersect the first arc at point A.
Join AB and AC. Result: $\triangle ABC$ is the required scalene (and right-angled) triangle with sides 3 cm, 4 cm, and 5 cm.

(c) Sidelengths: 1 cm, 5 cm, 5 cm Steps of Construction:

Draw a line segment BC of length 1 cm.
With B as the center and a radius of 5 cm, draw an arc.
With C as the center and a radius of 5 cm, draw another arc to intersect the first arc at point A.
Join AB and AC. Result: $\triangle ABC$ is the required isosceles triangle with sides 1 cm, 5 cm, and 5 cm.

(d) Sidelengths: 4 cm, 6 cm, 8 cm Steps of Construction:

Draw a line segment BC of length 8 cm.
With B as the center and a radius of 6 cm, draw an arc.
With C as the center and a radius of 4 cm, draw another arc to intersect the first arc at point A.
Join AB and AC. Result: $\triangle ABC$ is the required scalene triangle with sides 4 cm, 6 cm, and 8 cm.

(e) Sidelengths: 3.5 cm, 3.5 cm, 3.5 cm Steps of Construction:

Draw a line segment BC of length 3.5 cm.
With B as the center and a radius of 3.5 cm, draw an arc.
With C as the center and a radius of 3.5 cm, draw another arc to intersect the first arc at point A.
Join AB and AC. Result: $\triangle ABC$ is the required equilateral triangle with all sides of length 3.5 cm.

Q1Do triangles always exist? (ASA)

Find examples of measurements of two angles with the included side where a triangle is not possible.

Solution

A triangle is not possible if the sum of the two given angles is greater than or equal to 180°. This is because the sum of all three angles in a triangle must be exactly 180°. If two angles already sum up to 180° or more, the third angle would have to be 0° or negative, which is impossible.

Examples:

Angles: 90°, 90°; Included side: 5 cm
- Sum of angles = $90° + 90° = 180°$ .
- If we construct a 90° angle at both ends of the 5 cm base, the resulting lines will be parallel and will never intersect to form a third vertex.
- Therefore, a triangle is not possible.
Angles: 120°, 70°; Included side: 8 cm
- Sum of angles = $120° + 70° = 190°$ .
- Since $190° > 180°$ , a triangle is not possible. The lines drawn from the base will diverge and never meet.
Angles: 100°, 80°; Included side: 4 cm
- Sum of angles = $100° + 80° = 180°$ .
- The lines would be parallel and never intersect. A triangle is not possible.

Q2Do triangles always exist? (ASA)

Now we make one of the base angles an acute angle, say 40°. What are the possible values that the other angle should take so that the lines don't meet?

(a)

Try to find a possible ∠B for this to happen.

(b)

What could be smallest value of ∠B for the lines to not meet?

Solution

Given: A line segment AB (base) and an angle at A,

\angle A = 40°

. We need to find the values of

\angle B

such that the lines drawn from A and B do not meet to form a triangle.

For the lines not to meet (or to be parallel), the sum of the interior angles on the same side of the transversal AB must be 180°. Let the line from A be

l

. The line from B, say

m

, will not meet

l

if it is parallel to

l

or diverges from it.

(a) A possible $\angle B$ for the lines not to meet:

If the sum of $\angle A$ and $\angle B$ is 180°, the lines will be parallel. So, $\angle B = 180° - 40° = 140°$ .
If the sum is greater than 180°, the lines will diverge. So, any angle $\angle B > 140°$ will also work.
For example, if $\angle B = 150°$ , the sum is $40° + 150° = 190° > 180°$ , so the lines will not meet.
A possible value for $\angle B$ is 150°.

(b) Smallest value of $\angle B$ for the lines to not meet:

The lines will meet and form a triangle as long as the sum of the angles is less than 180°.
The boundary case where they stop meeting (on one side) is when they become parallel. This happens when the sum of the angles is exactly 180°.
Given $\angle A = 40°$ , the parallel case occurs when $\angle B = 180° - 40° = 140°$ .
If $\angle B$ is any value less than 140°, the sum will be less than 180°, and the lines will intersect to form a triangle.
Therefore, the smallest value of $\angle B$ for which the lines do not meet to form a triangle is 140°.

Q1End of Chapter Problem

There is a spider in a corner of a box. It wants to reach the farthest opposite corner (marked in the figure). Since it cannot fly, it can reach the opposite point only by walking on the surfaces of the box. What is the shortest path it can take?

Solution

Problem: To find the shortest path for a spider to travel from one corner of a rectangular box to the diagonally opposite corner by walking on the surfaces.

Concept: The shortest distance between two points is a straight line. Since the spider is confined to the surfaces of the box, we can find the shortest path by 'unfolding' the box to create a flat surface. The straight line drawn between the start and end corners on this flat surface represents the shortest path.

Let the dimensions of the box be:

Length = $l$
Width = $w$
Height = $h$

Let the spider start at corner A and want to reach the opposite corner G. There are three possible ways to unfold the box to create a path between A and G.

Case 1: Unfolding the front face and top face.

The spider travels across the front face and then the top face.
When unfolded, this forms a flat rectangle with dimensions $(l+h) \times w$ .
The distance is the hypotenuse of a right-angled triangle with sides $(l+h)$ and $w$ .
Path length, $d_1 = \sqrt{(l+h)^2 + w^2}$

Case 2: Unfolding the front face and right side face.

The spider travels across the front face and then the side face.
When unfolded, this forms a flat rectangle with dimensions $(l+w) \times h$ .
The distance is the hypotenuse of a right-angled triangle with sides $(l+w)$ and $h$ .
Path length, $d_2 = \sqrt{(l+w)^2 + h^2}$

Case 3: Unfolding the bottom face and right side face.

The spider travels across the bottom face and then the side face.
When unfolded, this forms a flat rectangle with dimensions $(h+w) \times l$ .
The distance is the hypotenuse of a right-angled triangle with sides $(h+w)$ and $l$ .
Path length, $d_3 = \sqrt{(h+w)^2 + l^2}$

Conclusion: The shortest path is the minimum of these three possible distances (

d_1, d_2, d_3

). To find the actual shortest path, one must substitute the values of length, width, and height of the box and calculate the three distances. The smallest of the three values will be the length of the shortest path.

Example: If a box has dimensions

l=30

cm,

w=12

cm,

h=12

cm.

$d_1 = \sqrt{(30+12)^2 + 12^2} = \sqrt{42^2 + 12^2} = \sqrt{1764 + 144} = \sqrt{1908} \approx 43.68$ cm
$d_2 = \sqrt{(30+12)^2 + 12^2} = \sqrt{42^2 + 12^2} = \sqrt{1908} \approx 43.68$ cm
$d_3 = \sqrt{(12+12)^2 + 30^2} = \sqrt{24^2 + 30^2} = \sqrt{576 + 900} = \sqrt{1476} \approx 38.42$ cm

In this example, the shortest path is approximately 38.42 cm.

Q1Exterior Angles

Find $\angle ACD$ , if $\angle A=50°$ , and $\angle B=60°$ .

Solution

Given: In

\triangle ABC

,

\angle A = 50°

and

\angle B = 60°

. The side BC is extended to a point D, forming an exterior angle

\angle ACD

.

To Find: The measure of the exterior angle

\angle ACD

.

Method 1: Using Angle Sum Property

Step 1: Find the interior angle $\angle ACB$ . Using the angle sum property of a triangle:

\angle A + \angle B + \angle ACB = 180°

50° + 60° + \angle ACB = 180°

110° + \angle ACB = 180°

\angle ACB = 180° - 110° = 70°

Step 2: Find the exterior angle $\angle ACD$ . Angles

\angle ACB

and

\angle ACD

form a linear pair on the straight line BCD. Therefore, their sum is 180°.

\angle ACB + \angle ACD = 180°

70° + \angle ACD = 180°

\angle ACD = 180° - 70° = 110°

Method 2: Using Exterior Angle Theorem

Exterior Angle Theorem: The measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

In this case,

\angle ACD

is the exterior angle, and its interior opposite angles are

\angle A

and

\angle B

.

\angle ACD = \angle A + \angle B

\angle ACD = 50° + 60°

\angle ACD = 110°

Final Answer:

\angle ACD = 110°

.

Q2Exterior Angles

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? [Hint: From angle sum property, we have $\angle A+\angle B+\angle ACB=180°$ . We also have $\angle ACD+\angle ACB=180°$ , since they form a straight angle. What does this show?]

Solution

Let's explore the relation. Consider a triangle

\triangle ABC

. Let the side BC be extended to a point D. The exterior angle is

\angle ACD

. The interior opposite angles are

\angle A

and

\angle B

.

From the angle sum property of a triangle, we have:

\angle A + \angle B + \angle ACB = 180° \quad ...(1)

Since BCD is a straight line,

\angle ACB

and

\angle ACD

form a linear pair. So:

\angle ACD + \angle ACB = 180° \quad ...(2)

From equation (1), we can write:

\angle A + \angle B = 180° - \angle ACB

From equation (2), we can write:

\angle ACD = 180° - \angle ACB

Comparing the two results, we see that the right-hand sides are identical. Therefore, the left-hand sides must be equal:

\angle ACD = \angle A + \angle B

Relation/Conclusion: This shows that the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles. This is known as the Exterior Angle Theorem.

Q1Figure it Out - Altitudes and Triangle Types

Construct a triangle ABC with BC=5 cm, AB=6 cm, CA=5 cm. Construct an altitude from A to BC.

Solution

To Construct: An isosceles triangle ABC and its altitude from A to BC.

Steps of Construction of Triangle:

Draw a line segment AB of length 6 cm.
With A as the center and a radius of 5 cm, draw an arc.
With B as the center and a radius of 5 cm, draw another arc to intersect the first arc at point C. (Note: The question has a typo. Assuming sides are 5cm, 5cm, 6cm. Let's take BC=6cm, AB=5cm, CA=5cm). Let's follow the question text strictly: BC=5 cm, AB=6 cm, CA=5 cm.
Draw a line segment BC of length 5 cm.
With C as the center and a radius of 5 cm, draw an arc.
With B as the center and a radius of 6 cm, draw another arc to intersect the first arc at point A.
Join AB and AC. $\triangle ABC$ is the required triangle.

Steps of Construction of Altitude from A to BC:

Place the ruler along the base BC.
Place a set square on the ruler such that one edge of its right angle is aligned with the ruler.
Slide the set square along the ruler until its other edge (the vertical edge) touches the vertex A.
Draw a line segment from A along this vertical edge of the set square until it meets the base BC. Let the point of intersection be D.

Result: The line segment AD is the required altitude from vertex A to the side BC.

Q2Figure it Out - Altitudes and Triangle Types

Construct a triangle TRY with RY=4 cm, TR=7 cm, $\angle R=140°$ . Construct an altitude from T to RY.

Solution

To Construct: An obtuse triangle TRY and its altitude from T to RY.

Steps of Construction of Triangle:

Draw a line segment RY of length 4 cm.
At point R, construct an angle of 140° using a protractor. Let the ray be RX.
From point R, cut an arc of radius 7 cm on the ray RX. Mark the point as T.
Join TY. Result: $\triangle TRY$ is the required obtuse triangle.

Steps of Construction of Altitude from T to RY: Since

\angle R

is an obtuse angle, the altitude from T will fall outside the triangle on the extension of the side RY.

Extend the line segment RY from the R side.
Place a ruler along the extended line RY.
Place a set square on the ruler with one edge of its right angle aligned with the ruler.
Slide the set square along the ruler until its vertical edge touches the vertex T.
Draw a line segment from T along this vertical edge until it meets the extended line of RY. Let this point be S.

Result: The line segment TS is the required altitude from vertex T to the side RY.

Q3Figure it Out - Altitudes and Triangle Types

Construct a right-angled triangle $\triangle ABC$ with $\angle B=90°$ , AC=5 cm. How many different triangles exist with these measurements?

Solution

Given: A right-angled triangle

\triangle ABC

with

\angle B=90°

and hypotenuse AC = 5 cm.

Construction: One way to construct such a triangle is to use the property that an angle in a semicircle is a right angle.

Draw the hypotenuse AC of length 5 cm.
Find the midpoint of AC, let's call it O.
With O as the center and radius OA (or OC, which is 2.5 cm), draw a semicircle.
Choose any point B on the arc of the semicircle.
Join AB and CB. Result: $\triangle ABC$ will be a right-angled triangle with $\angle B=90°$ and AC = 5 cm.

How many different triangles exist? In the construction above, we can choose any point B on the semicircle. Since there are infinitely many points on the arc of the semicircle, we can draw an infinite number of different triangles that satisfy the given conditions. Each choice of B will result in different lengths for the sides AB and BC, but the hypotenuse AC will always be 5 cm and the angle at B will always be 90°.

Final Answer: An infinite number of different triangles exist with these measurements.

Q4Figure it Out - Altitudes and Triangle Types

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

Part 1: Equilateral Triangle An equilateral triangle has all three angles equal. By the angle sum property, each angle must be

180° / 3 = 60°

.

(i)

Right-angled equilateral triangle:

A right-angled triangle must have one angle of 90°.
An equilateral triangle must have all angles as 60°.
Since these conditions contradict each other, it is not possible to construct a right-angled equilateral triangle.

(ii)

Obtuse-angled equilateral triangle:

An obtuse-angled triangle must have one angle greater than 90°.
An equilateral triangle must have all angles as 60°.
These conditions also contradict each other. It is not possible to construct an obtuse-angled equilateral triangle.

Part 2: Isosceles Triangle An isosceles triangle has two equal sides and two equal base angles.

(i)

Right-angled isosceles triangle:

Yes, this is possible. Let the right angle be $\angle B = 90°$ . The other two angles, $\angle A$ and $\angle C$ , must be equal and their sum must be $180° - 90° = 90°$ . Therefore, $\angle A = \angle C = 45°$ . The sides opposite to equal angles are equal, so AB = BC.
Construction:
1. Draw a line segment BC of any length, say 4 cm.
2. At B, construct a 90° angle.
3. Cut off BA = 4 cm along the perpendicular line.
4. Join AC. $\triangle ABC$ is a right-angled isosceles triangle.

(ii)

Obtuse-angled isosceles triangle:

Yes, this is possible. Let the obtuse angle be $\angle B = 120°$ . The other two angles, $\angle A$ and $\angle C$ , must be equal and their sum must be $180° - 120° = 60°$ . Therefore, $\angle A = \angle C = 30°$ . The sides AB and BC will be equal.
Construction:
1. Draw a line segment BC of any length, say 5 cm.
2. At B, construct a 120° angle.
3. Cut off BA = 5 cm along the new arm.
4. Join AC. $\triangle ABC$ is an obtuse-angled isosceles triangle.

Q1Figure it Out - Angle Sum Property

For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category:

(a)

30°

(b)

70°

(c)

54°

(d)

144°

Solution

Rule: A triangle is possible if the sum of the two given angles is less than 180°. It is not possible if the sum is greater than or equal to 180°.

(a) Given angle: 30°

(a) Possible: The second angle must be less than $180° - 30° = 150°$ $180° - 30° = 150°$ .
- Example 1: 60° (Sum = 90° < 180°)
- Example 2: 140° (Sum = 170° < 180°)
(b) Not Possible: The second angle must be greater than or equal to 150°.
- Example 1: 150° (Sum = 180°)
- Example 2: 160° (Sum = 190° > 180°)

(b) Given angle: 70°

(a) Possible: The second angle must be less than $180° - 70° = 110°$ $180° - 70° = 110°$ .
- Example 1: 50° (Sum = 120° < 180°)
- Example 2: 100° (Sum = 170° < 180°)
(b) Not Possible: The second angle must be greater than or equal to 110°.
- Example 1: 110° (Sum = 180°)
- Example 2: 120° (Sum = 190° > 180°)

(c) Given angle: 54°

(a) Possible: The second angle must be less than $180° - 54° = 126°$ $180° - 54° = 126°$ .
- Example 1: 90° (Sum = 144° < 180°)
- Example 2: 125° (Sum = 179° < 180°)
(b) Not Possible: The second angle must be greater than or equal to 126°.
- Example 1: 126° (Sum = 180°)
- Example 2: 130° (Sum = 184° > 180°)

(d) Given angle: 144°

(a) Possible: The second angle must be less than $180° - 144° = 36°$ $180° - 144° = 36°$ .
- Example 1: 10° (Sum = 154° < 180°)
- Example 2: 35° (Sum = 179° < 180°)
(b) Not Possible: The second angle must be greater than or equal to 36°.
- Example 1: 36° (Sum = 180°)
- Example 2: 40° (Sum = 184° > 180°)

Q2Figure it Out - Angle Sum Property

Determine which of the following pairs can be the angles of a triangle and which cannot:

(a)

35°, 150°

(b)

70°, 30°

(c)

90°, 85°

(d)

50°, 150°

Solution

Rule: A pair of angles can be angles of a triangle if their sum is less than 180°.

(a) 35°, 150°

Sum: $35° + 150° = 185°$ .
Since $185° > 180°$ , this pair cannot be the angles of a triangle.

(b) 70°, 30°

Sum: $70° + 30° = 100°$ .
Since $100° < 180°$ , this pair can be the angles of a triangle.

(c) 90°, 85°

Sum: $90° + 85° = 175°$ .
Since $175° < 180°$ , this pair can be the angles of a triangle.

(d) 50°, 150°

Sum: $50° + 150° = 200°$ .
Since $200° > 180°$ , this pair cannot be the angles of a triangle.

Q3Figure it Out - Angle Sum Property

Find the third angle of a triangle (using a parallel line) when two of the angles are:

(a)

36°, 72°

(b)

150°, 15°

(c)

90°, 30°

(d)

75°, 45°

Solution

Method: The sum of the three angles in any triangle is 180°. If two angles are given, the third angle is 180° minus the sum of the two given angles.

(a) 36°, 72°

Sum of given angles = $36° + 72° = 108°$ .
Third angle = $180° - 108° = 72°$ .

(b) 150°, 15°

Sum of given angles = $150° + 15° = 165°$ .
Third angle = $180° - 165° = 15°$ .

(c) 90°, 30°

Sum of given angles = $90° + 30° = 120°$ .
Third angle = $180° - 120° = 60°$ .

(d) 75°, 45°

Sum of given angles = $75° + 45° = 120°$ .
Third angle = $180° - 120° = 60°$ .

Q4Figure it Out - Angle Sum Property

Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.

Solution

Part 1: Can you construct a triangle with all angles equal to 70°?

If all three angles are 70°, their sum would be $70° + 70° + 70° = 210°$ .
The sum of angles in a triangle must be 180°.
Since $210° \neq 180°$ , it is not possible to construct such a triangle.

Part 2: If two of the angles are 70°, what would the third angle be?

Sum of the two given angles = $70° + 70° = 140°$ .
The third angle = $180° - 140° = 40°$ .

Part 3: If all the angles in a triangle have to be equal, what must its measure be?

Let the measure of each equal angle be $x$ .
According to the angle sum property of a triangle: $x + x + x = 180°$ .
$3x = 180°$
$x = \frac{180°}{3}$
$x = 60°$
Answer: If all angles in a triangle are equal, each angle must measure 60°. Such a triangle is called an equiangular triangle, which is also an equilateral triangle.

Q5Figure it Out - Angle Sum Property

Here is a triangle in which we know $\angle B=\angle C$ and $\angle A=50°$ . Can you find $\angle B$ and $\angle C$ ?

Solution

Given: In a triangle,

\angle A = 50°

and

\angle B = \angle C

.

To Find: The measures of

\angle B

and

\angle C

.

Formula: The sum of the angles in a triangle is 180°.

\angle A + \angle B + \angle C = 180°

Solution: Let

\angle B = \angle C = x

. Substitute the given values into the formula:

50° + x + x = 180°

50° + 2x = 180°

Subtract 50° from both sides:

2x = 180° - 50°

2x = 130°

Divide by 2:

x = \frac{130°}{2}

x = 65°

Therefore,

\angle B = 65°

and

\angle C = 65°

.

Final Answer:

\angle B = 65°

and

\angle C = 65°

.

Q1Figure it Out - Circles and Triangles

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

Solution

Given: A circle with a center and points on its circumference.

To Do: Form isosceles triangles.

Solution: Let the center of the circle be O. Let A and B be any two distinct points on the circumference of the circle.

Join OA and OB. These are radii of the circle, so their lengths are equal, i.e., OA = OB.
Join AB to form a triangle $\triangle OAB$ .
In $\triangle OAB$ , since two sides OA and OB are equal, it is an isosceles triangle.

By choosing any two points on the circle and connecting them to the center, we can form an infinite number of isosceles triangles.

Q2Figure it Out - Circles and Triangles

Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.

Solution

Case 1: Two intersecting circles of the same size

Given: Two circles of the same radius, say

r

, with centers A and B. They intersect at points C and D. The distance between the centers A and B is such that they intersect.

Isosceles Triangles:

Consider $\triangle ABC$ . Sides AC and BC are radii of their respective circles. If the distance AB is not equal to the radius, but AC = r and BC = r, it's not guaranteed to be isosceles based on centers A and B. However, consider $\triangle ACD$ . AC and AD are both radii of the circle with center A, so AC = AD = r. Thus, $\triangle ACD$ is an isosceles triangle.
Similarly, in $\triangle BCD$ , BC and BD are both radii of the circle with center B, so BC = BD = r. Thus, $\triangle BCD$ is an isosceles triangle.
Consider $\triangle ABC$ . AC is the radius of the circle with center A. AB is the distance between centers. BC is the radius of the circle with center B. Since the circles are of the same size, AC = BC = r. Therefore, $\triangle ABC$ is an isosceles triangle.
Similarly, $\triangle ABD$ is also an isosceles triangle because AD = BD = r.

Equilateral Triangles:

An equilateral triangle can be formed if the distance between the centers of the two circles is equal to their radius, i.e., AB = r.
In this case, in $\triangle ABC$ , we have AC = r (radius of first circle), BC = r (radius of second circle), and AB = r (distance between centers). Since all three sides are equal, $\triangle ABC$ is an equilateral triangle.
Similarly, in $\triangle ABD$ , we have AD = r, BD = r, and AB = r. Thus, $\triangle ABD$ is also an equilateral triangle.

Case 2: Three intersecting circles of the same size

Given: Three circles of the same radius, r, with centers A, B, and C. Each circle passes through the centers of the other two.

Isosceles and Equilateral Triangles:

In this configuration, the distance between any two centers is equal to the radius, i.e., AB = BC = CA = r.
Therefore, the triangle formed by joining the centers, $\triangle ABC$ , is an equilateral triangle.
Many other isosceles and equilateral triangles can be formed using the intersection points as well, similar to Case 1.

Q1Figure it Out - Conclusion Section

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

Solution

We use the triangle inequality rule: the sum of the two smaller sides must be greater than the largest side.

(a) 1, 100, 100

Sum of smaller sides: $1 + 100 = 101$ .
Largest side: 100.
Is $101 > 100$ ? True.
Answer: Yes, a triangle exists.

(b) 3, 6, 9

Sum of smaller sides: $3 + 6 = 9$ .
Largest side: 9.
Is $9 > 9$ ? False. (The sum must be strictly greater).
Answer: No, a triangle does not exist. The points would be collinear.

(c) 1, 1, 5

Sum of smaller sides: $1 + 1 = 2$ .
Largest side: 5.
Is $2 > 5$ ? False.
Answer: No, a triangle does not exist.

(d) 5, 10, 12

Sum of smaller sides: $5 + 10 = 15$ .
Largest side: 12.
Is $15 > 12$ ? True.
Answer: Yes, a triangle exists.

Q2Figure it Out - Conclusion Section

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

Part 1: Equilateral triangle with sides 50, 50, 50

To check: We apply the triangle inequality. Let the sides be

a=50, b=50, c=50

.

$a + b > c \implies 50 + 50 > 50 \implies 100 > 50$ (True) Since all sides are equal, all three checks will be the same. Answer: Yes, an equilateral triangle with sides 50, 50, 50 exists.

Part 2: Equilateral triangle of any sidelength

To check: Let the sidelength be

s

, where

s

is any positive number. The three sides are

s, s, s

. We apply the triangle inequality:

$s + s > s \implies 2s > s$ This inequality is true for any positive value of $s$ .

Justification: Since the triangle inequality (

2s > s

) holds for any positive length

s

, an equilateral triangle can be constructed for any given positive sidelength. Answer: Yes, an equilateral triangle of any positive sidelength exists.

Q3Figure it Out - Conclusion Section

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

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.

Solution

To find the possible range for the third side, we use the property derived from the triangle inequality: If two sides of a triangle are of length

a

and

b

, the third side

c

must satisfy the inequality:

|a - b| < c < a + b

.

(a) Given sides: 1, 100

Difference: $100 - 1 = 99$ .
Sum: $100 + 1 = 101$ .
Range for the third side (c): $99 < c < 101$ .
5 possible values: 99.1, 99.5, 100, 100.5, 100.9.
Description of all possible lengths: The third side must be a length strictly greater than 99 and strictly less than 101.

(b) Given sides: 5, 5

Difference: $5 - 5 = 0$ .
Sum: $5 + 5 = 10$ .
Range for the third side (c): $0 < c < 10$ .
5 possible values: 1, 3, 5, 7.5, 9.9.
Description of all possible lengths: The third side must be a length strictly greater than 0 and strictly less than 10.

(c) Given sides: 3, 7

Difference: $7 - 3 = 4$ .
Sum: $7 + 3 = 10$ .
Range for the third side (c): $4 < c < 10$ .
5 possible values: 4.1, 5, 6.5, 8, 9.5.
Description of all possible lengths: The third side must be a length strictly greater than 4 and strictly less than 10.

Q1Figure it Out - Construction with Angles

Construct triangles for the following measurements where the angle is included between the sides:

(a)

3 cm, 75°, 7 cm

(b)

6 cm, 25°, 3 cm

(c)

3 cm, 120°, 8 cm

Solution

To Construct: Triangles given two sides and the included angle (SAS).

(a) Sides 3 cm, 7 cm and included angle 75° Steps of Construction:

Draw a line segment AB of length 7 cm.
At point A, construct an angle of 75° using a protractor. Let the ray be AX.
From point A, cut an arc of radius 3 cm on the ray AX. Mark the point as C.
Join BC. Result: $\triangle ABC$ is the required triangle.

(b) Sides 6 cm, 3 cm and included angle 25° Steps of Construction:

Draw a line segment PQ of length 6 cm.
At point P, construct an angle of 25° using a protractor. Let the ray be PX.
From point P, cut an arc of radius 3 cm on the ray PX. Mark the point as R.
Join QR. Result: $\triangle PQR$ is the required triangle.

(c) Sides 3 cm, 8 cm and included angle 120° Steps of Construction:

Draw a line segment YZ of length 8 cm.
At point Y, construct an angle of 120° using a protractor. Let the ray be YX.
From point Y, cut an arc of radius 3 cm on the ray YX. Mark the point as X.
Join XZ. Result: $\triangle XYZ$ is the required triangle.

Q2Figure it Out - Construction with Angles

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

Answer: No, a unique triangle is always possible.

Justification: When we are given two sides and the included angle (SAS), the construction process is as follows:

Draw one side as the base.
At one end of the base, construct the given angle.
Along the new arm of the angle, mark the length of the second side.
Join the endpoints to form the triangle.

This process will always result in a unique third point. The two endpoints of the base and this third point will form a triangle, provided the two given side lengths are positive and the included angle is greater than 0° and less than 180°. If the angle is 0° or 180°, the three vertices would lie on a straight line (a degenerate triangle), not a closed shape.

Therefore, for any two positive lengths and an included angle between 0° and 180°, a unique triangle can always be constructed.

Q3Figure it Out - Construction with Angles

Construct triangles for the following measurements:

(a)

75°, 5 cm, 75°

(b)

25°, 3 cm, 60°

(c)

120°, 6 cm, 30°

Solution

To Construct: Triangles given two angles and the included side (ASA).

(a) Angles 75°, 75° and included side 5 cm Steps of Construction:

Draw a line segment AB of length 5 cm.
At point A, construct an angle of 75° using a protractor. Let the ray be AX.
At point B, construct an angle of 75° using a protractor. Let the ray be BY.
The two rays AX and BY will intersect at a point. Mark this point as C. Result: $\triangle ABC$ is the required triangle.

(b) Angles 25°, 60° and included side 3 cm Steps of Construction:

Draw a line segment PQ of length 3 cm.
At point P, construct an angle of 25°. Let the ray be PX.
At point Q, construct an angle of 60°. Let the ray be QY.
The two rays PX and QY will intersect at a point. Mark this point as R. Result: $\triangle PQR$ is the required triangle.

(c) Angles 120°, 30° and included side 6 cm Steps of Construction:

Draw a line segment YZ of length 6 cm.
At point Y, construct an angle of 120°. Let the ray be YX.
At point Z, construct an angle of 30°. Let the ray be ZX.
The two rays YX and ZX will intersect at a point. Mark this point as X. Result: $\triangle XYZ$ is the required triangle.

Q1Figure it Out - Sidelengths of a Triangle

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

Solution

To check if the lengths can form a triangle, we check if the sum of the two smaller sides is greater than the largest side.

(a) 2, 2, 5

Sum of smaller sides: $2 + 2 = 4$ .
Largest side: 5.
Is $4 > 5$ ? False.
Answer: No, a triangle cannot be formed.

(b) 3, 4, 6

Sum of smaller sides: $3 + 4 = 7$ .
Largest side: 6.
Is $7 > 6$ ? True.
Answer: Yes, a triangle can be formed.

(c) 2, 4, 8

Sum of smaller sides: $2 + 4 = 6$ .
Largest side: 8.
Is $6 > 8$ ? False.
Answer: No, a triangle cannot be formed.

(d) 5, 5, 8

Sum of smaller sides: $5 + 5 = 10$ .
Largest side: 8.
Is $10 > 8$ ? True.
Answer: Yes, a triangle can be formed.

(e) 10, 20, 25

Sum of smaller sides: $10 + 20 = 30$ .
Largest side: 25.
Is $30 > 25$ ? True.
Answer: Yes, a triangle can be formed.

(f) 10, 20, 35

Sum of smaller sides: $10 + 20 = 30$ .
Largest side: 35.
Is $30 > 35$ ? False.
Answer: No, a triangle cannot be formed.

(g) 24, 26, 28

Sum of smaller sides: $24 + 26 = 50$ .
Largest side: 28.
Is $50 > 28$ ? True.
Answer: Yes, a triangle can be formed.

Q1Figure it Out - Triangle Inequality

We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.

Solution

Yes, we can determine if a triangle is possible without construction using the Triangle Inequality Property.

Triangle Inequality Property: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Case 1: Sidelengths 3 cm, 4 cm, 8 cm Let the sides be

a=3

,

b=4

,

c=8

.

$a + b > c \implies 3 + 4 > 8 \implies 7 > 8$ (False)
$a + c > b \implies 3 + 8 > 4 \implies 11 > 4$ (True)
$b + c > a \implies 4 + 8 > 3 \implies 12 > 3$ (True)

Since one of the conditions (

3 + 4 > 8

) is false, a triangle with these sidelengths cannot be formed.

Case 2: Sidelengths 2 cm, 3 cm, 6 cm Let the sides be

a=2

,

b=3

,

c=6

.

$a + b > c \implies 2 + 3 > 6 \implies 5 > 6$ (False)
$a + c > b \implies 2 + 6 > 3 \implies 8 > 3$ (True)
$b + c > a \implies 3 + 6 > 2 \implies 9 > 2$ (True)

Since the condition

2 + 3 > 6

is false, a triangle with these sidelengths cannot be formed.

Q2Figure it Out - Triangle Inequality

Can we say anything about the existence of a triangle for each of the following sets of lengths?

(a)

10 km, 10 km and 25 km

(b)

5 mm, 10 mm and 20 mm

(c)

12 cm, 20 cm and 40 cm

Solution

To check for the existence of a triangle, we use the Triangle Inequality Property. A simplified way to check is to see if the sum of the two shorter sides is greater than the longest side.

(a) 10 km, 10 km, 25 km

The two shorter sides are 10 km and 10 km. The longest side is 25 km.
Sum of shorter sides: $10 + 10 = 20$ km.
Comparison: Is $20 > 25$ ? No, it is false.
Conclusion: A triangle with these sidelengths cannot exist.

(b) 5 mm, 10 mm, 20 mm

The two shorter sides are 5 mm and 10 mm. The longest side is 20 mm.
Sum of shorter sides: $5 + 10 = 15$ mm.
Comparison: Is $15 > 20$ ? No, it is false.
Conclusion: A triangle with these sidelengths cannot exist.

(c) 12 cm, 20 cm, 40 cm

The two shorter sides are 12 cm and 20 cm. The longest side is 40 cm.
Sum of shorter sides: $12 + 20 = 32$ cm.
Comparison: Is $32 > 40$ ? No, it is false.
Conclusion: A triangle with these sidelengths cannot exist.

Q3Figure it Out - Triangle Inequality

For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are two comparisons where this happens: $10 < 15+30$ $15 < 10+30$ But this doesn't happen for the third length: $30 > 10+15$ . 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. 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.] Given three sidelengths, what do we need to compare to check for the existence of a triangle?

Solution

Part 1: Will there always be at least two true comparisons?

Yes. Let the three lengths be

a, b, c

such that

a \leq b \leq c

.

Comparison for side $a$ : We need to check if $a < b + c$ . Since $b$ and $c$ are positive, and $a \leq b$ , it is always true that $a < b+c$ . For example, $a$ is less than $b$ alone, so it must be less than $b+c$ .
Comparison for side $b$ : We need to check if $b < a + c$ . Since $a$ is positive and $b \leq c$ , it is always true that $b < a+c$ .

So, for any set of three positive lengths, the two smaller lengths will always be less than the sum of the other two. This means at least two comparisons will always be true.

Part 2: Identifying which comparisons are true without calculation?

Yes. If we arrange the lengths in increasing order, say

a \leq b \leq c

, the two smaller sides (

a

and

b

) will always be less than the sum of the other two.

$a < b+c$ is always true because $a \leq b < b+c$ .
$b < a+c$ is always true because $b \leq c < a+c$ .

The only comparison that needs to be checked is whether the longest side is less than the sum of the other two.

Part 3: What do we need to compare?

To check for the existence of a triangle given three sidelengths, we only need to perform one comparison: Check if the sum of the two shorter lengths is greater than the longest length. If this single condition is true, the other two conditions will automatically be true, and a triangle can be formed.

A Tale of Three Intersecting LinesClass 7 Mathematics NCERT Solutions

Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle? Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle is possible for these sidelengths.

Solution

Construct triangles having the following sidelengths (all the units are in cm):

Solution

Find examples of measurements of two angles with the included side where a triangle is not possible.

Solution

Now we make one of the base angles an acute angle, say 40°. What are the possible values that the other angle should take so that the lines don't meet?

Solution

There is a spider in a corner of a box. It wants to reach the farthest opposite corner (marked in the figure). Since it cannot fly, it can reach the opposite point only by walking on the surfaces of the box. What is the shortest path it can take?

Solution

Find ∠ACD\angle ACD∠ACD, if ∠A=50°\angle A=50°∠A=50°, and ∠B=60°\angle B=60°∠B=60°.

Solution

Solution

Construct a triangle ABC with BC=5 cm, AB=6 cm, CA=5 cm. Construct an altitude from A to BC.

Solution

Construct a triangle TRY with RY=4 cm, TR=7 cm, ∠R=140°\angle R=140°∠R=140°. Construct an altitude from T to RY.

Solution

Construct a right-angled triangle △ABC\triangle ABC△ABC with ∠B=90°\angle B=90°∠B=90°, AC=5 cm. How many different triangles exist with these measurements?

Solution

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

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:

Solution

Determine which of the following pairs can be the angles of a triangle and which cannot:

Solution

Find the third angle of a triangle (using a parallel line) when two of the angles are:

Solution

Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.

Solution

Here is a triangle in which we know ∠B=∠C\angle B=\angle C∠B=∠C and ∠A=50°\angle A=50°∠A=50°. Can you find ∠B\angle B∠B and ∠C\angle C∠C?

Solution

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

Solution

Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.

Solution

Check if a triangle exists for each of the following set of lengths:

Solution

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

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):

Solution

Construct triangles for the following measurements where the angle is included between the sides:

Solution

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

Construct triangles for the following measurements:

Solution

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.

Solution

We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.

Solution

Can we say anything about the existence of a triangle for each of the following sets of lengths?

Solution

Solution

Find $\angle ACD$ , if $\angle A=50°$ , and $\angle B=60°$ .

Construct a triangle TRY with RY=4 cm, TR=7 cm, $\angle R=140°$ . Construct an altitude from T to RY.

Construct a right-angled triangle $\triangle ABC$ with $\angle B=90°$ , AC=5 cm. How many different triangles exist with these measurements?

Here is a triangle in which we know $\angle B=\angle C$ and $\angle A=50°$ . Can you find $\angle B$ and $\angle C$ ?