NCERT Solutions for Class 7 Maths Chapter 6: The Triangle and Its Properties — Complete Guide

By sides:
- Equilateral triangle: All $3$ sides equal. All angles $= 60°$.
- Isosceles triangle: $2$ sides equal. The angles opposite the equal sides (base angles) are equal.
- Scalene triangle: All $3$ sides different. All $3$ angles different.

By angles:
- Acute triangle: All angles $< 90°$.
- Right triangle: One angle $= 90°$.
- Obtuse triangle: One angle $> 90°$.

Important: A triangle can be classified by BOTH sides and angles. For example, an "isosceles right triangle" has two equal sides and one $90°$ angle.

Median: A line segment from a vertex to the midpoint of the opposite side. Every triangle has $3$ medians, and they all meet at a single point called the centroid.

Altitude: A perpendicular line segment from a vertex to the opposite side (or its extension). Every triangle has $3$ altitudes, and they all meet at a point called the orthocentre.

Key differences:
- A median bisects the opposite side; an altitude is perpendicular to it.
- In an equilateral triangle, medians and altitudes are the same lines.
- In a right triangle, two of the altitudes are the legs themselves.
- In an obtuse triangle, some altitudes lie outside the triangle.

Problem: In a triangle, the exterior angle is $120°$ and one of the opposite interior angles is $60°$. Find the other opposite interior angle.

Solution:
By the exterior angle property:

Answer: The other opposite interior angle is $60°$.

Problem: Two interior angles of a triangle are $45°$ and $70°$. Find the exterior angle at the third vertex.

Solution:
The exterior angle at the third vertex equals the sum of the other two interior angles:

Verification: Third interior angle $= 180° - 45° - 70° = 65°$. Exterior angle $= 180° - 65° = 115°$. Correct.

Problem: An exterior angle of a triangle is $100°$ and the two opposite interior angles are $x$ and $(x + 20)°$. Find $x$.

Solution:

The angles are $40°$ and $60°$.

Verification: $40° + 60° = 100°$. Correct.

Problem: Can a triangle have two right angles? Explain.

Solution:
No. If two angles were $90°$ each, their sum would be $180°$, leaving $0°$ for the third angle. But an angle of $0°$ means no triangle is formed.

By the angle sum property, the three angles must add to $180°$, so at most one angle can be $90°$.

Problem: The angles of a triangle are in the ratio $2:3:4$. Find each angle.

Solution:
Let the angles be $2x, 3x, 4x$.

The angles are $40°, 60°, 80°$.

Classification: All angles are acute, so this is an acute-angled triangle.

Problem: In $\triangle ABC$, $D$ is the midpoint of $BC$. Is $AD$ a median, an altitude, or both?

Solution:
$AD$ is a median because it connects vertex $A$ to the midpoint $D$ of the opposite side $BC$.

$AD$ is an altitude only if $AD \perp BC$. Unless this is stated or the triangle is equilateral/isosceles with $A$ at the apex, $AD$ is generally only a median, not an altitude.

Key distinction: A median always bisects the opposite side. An altitude is always perpendicular to the opposite side. They are the same only in special triangles.

Problem: How many medians and altitudes does a triangle have?

Solution:
A triangle has:
- **$3$ medians** (one from each vertex to the midpoint of the opposite side)
- **$3$ altitudes** (one from each vertex perpendicular to the opposite side)

All $3$ medians meet at the centroid.
All $3$ altitudes meet at the orthocentre.

Problem: Prove that an exterior angle of a triangle is greater than either of the two opposite interior angles.

Solution:
Let the exterior angle $= \angle 1 + \angle 2$ (by the exterior angle property), where $\angle 1$ and $\angle 2$ are the opposite interior angles.

Since $\angle 1 > 0°$ and $\angle 2 > 0°$:

Therefore, the exterior angle is greater than either opposite interior angle. $\square$

Problem: Two angles of a triangle are $45°$ and $65°$. Find the third angle.

Solution:

Answer: $70°$.

Problem: In a right-angled triangle, one acute angle is $55°$. Find the other acute angle.

Solution:
One angle $= 90°$.

Answer: $35°$.

Problem: Find each angle of an equilateral triangle.

Solution:
All three sides are equal, so all three angles are equal. Let each angle $= x$.

Each angle of an equilateral triangle is $60°$.

Problem: In an isosceles triangle, the vertex angle is $80°$. Find the base angles.

Solution:
Base angles are equal. Let each $= x$.

Each base angle is $50°$.

Problem: The angles of a triangle are $(x + 10)°$, $(2x + 20)°$, and $(3x - 30)°$. Find all angles.

Solution:

Angles: $40°, 80°, 60°$.

Verification: $40 + 80 + 60 = 180°$. Correct.

Problem: Can a triangle have angles $90°, 60°, 40°$?

Solution:

No, this triangle is not possible because the angles do not sum to $180°$.

Problem: What is the sum of all exterior angles of a triangle (one at each vertex)?

Solution:
Each exterior angle $= 180° - \text{interior angle}$.

Sum of all exterior angles $= (180° - A) + (180° - B) + (180° - C)$
$= 540° - (A + B + C) = 540° - 180° = 360°$.

Answer: The sum of all exterior angles of a triangle is $360°$.

Note: This result holds for ALL convex polygons.

Problem: Describe an activity to verify the angle sum property.

Solution:
Activity 1 (Paper tearing):
1. Draw a triangle on paper and label the angles $\angle 1$, $\angle 2$, $\angle 3$.
2. Tear off the three corners.
3. Place them adjacent to each other with vertices touching.
4. Observe that they form a straight line ($180°$).

Activity 2 (Measurement):
1. Draw any triangle.
2. Measure all three angles with a protractor.
3. Add them up. The sum should be $180°$ (within measurement error).

Both activities verify that $\angle 1 + \angle 2 + \angle 3 = 180°$.

Problem: In a right isosceles triangle, find all three angles.

Solution:
One angle $= 90°$. The other two are equal (isosceles). Let each $= x$.

The angles are $90°, 45°, 45°$.

Problem: In $\triangle ABC$, $\angle A = 50°$, $\angle B = 60°$, $\angle C = 70°$. Which side is the longest?

Solution:
Property: In a triangle, the longest side is opposite the largest angle.

$\angle C = 70°$ is the largest angle, so the side opposite to it ($AB$) is the longest.

Answer: $AB$ is the longest side.

Problem: Is it possible to have a triangle with sides $3$ cm, $4$ cm, $8$ cm?

Solution:
Check: $3 + 4 = 7$, but $7 < 8$.

Since the sum of two sides ($7$) is NOT greater than the third side ($8$), this triangle is not possible.

Note: You only need to find ONE violation to conclude the triangle is impossible.

Problem: Is it possible to have a triangle with sides $5$ cm, $6$ cm, $9$ cm?

Solution:
Check all three conditions:
- $5 + 6 = 11 > 9$ (true)
- $5 + 9 = 14 > 6$ (true)
- $6 + 9 = 15 > 5$ (true)

All conditions satisfied. This triangle is possible.

Problem: Two sides of a triangle are $7$ cm and $10$ cm. Find the range of the third side.

Solution:
Let the third side be $x$.

By triangle inequality:
- $x + 7 > 10 \Rightarrow x > 3$
- $x + 10 > 7 \Rightarrow x > -3$ (always true for positive $x$)
- $7 + 10 > x \Rightarrow x < 17$

Combining: $3 < x < 17$.

Answer: The third side must be greater than $3$ cm and less than $17$ cm.

Problem: Can an equilateral triangle have side $5$ cm? Verify the triangle inequality.

Solution:
Sides: $5, 5, 5$.
- $5 + 5 = 10 > 5$ (true)
- $5 + 5 = 10 > 5$ (true)
- $5 + 5 = 10 > 5$ (true)

Yes, an equilateral triangle with side $5$ cm satisfies the triangle inequality.

Problem: Can a triangle have sides $1$ cm, $2$ cm, $3$ cm?

Solution:
$1 + 2 = 3$, but $3$ is NOT greater than $3$ (it equals $3$).

The triangle inequality requires strict inequality ($>$, not $\geq$). Therefore, this triangle is not possible.

Note: Sides $1, 2, 3$ would form a degenerate triangle (a straight line), which is not a valid triangle.

Problem: You have three sticks of lengths $8$ cm, $15$ cm, and $6$ cm. Can you form a triangle?

Solution:
- $8 + 6 = 14 > 15$? No! $14 < 15$.

The triangle inequality fails. You cannot form a triangle with these sticks.

Problem: A shortcut: for three sides $a \leq b \leq c$, the triangle inequality reduces to checking only $a + b > c$. Why?

Solution:
If $a \leq b \leq c$, then:
- $b + c > a$ is automatically true (since $b \geq a$ and $c > 0$).
- $a + c > b$ is automatically true (since $c \geq b$ and $a > 0$).

So we only need to check: sum of the two smaller sides > largest side.

This shortcut saves time in exams!

Problem: How many triangles with integer sides can be formed if the two shorter sides are $3$ cm and $5$ cm?

Solution:
Let the third side $= x$ (integer).
From the triangle inequality: $3 + 5 > x$ and $x + 3 > 5$.

So $x$ can be $3, 4, 5, 6, 7$.

Answer: $5$ triangles are possible.

Problem: Find the hypotenuse of a right triangle with legs $6$ cm and $8$ cm.

Solution:

Answer: $10$ cm.

Problem: A right triangle has hypotenuse $13$ cm and one leg $5$ cm. Find the other leg.

Solution:

Answer: $12$ cm.

Problem: Check whether $(7, 24, 25)$ is a Pythagorean triplet.

Solution:

Yes, $(7, 24, 25)$ is a Pythagorean triplet.

Common Pythagorean triplets to memorise:
| Triplet | Verification |
|---|---|
| $(3, 4, 5)$ | $9 + 16 = 25$ |
| $(5, 12, 13)$ | $25 + 144 = 169$ |
| $(7, 24, 25)$ | $49 + 576 = 625$ |
| $(8, 15, 17)$ | $64 + 225 = 289$ |
| $(9, 40, 41)$ | $81 + 1600 = 1681$ |

Multiples of any triplet are also triplets: $(6, 8, 10)$, $(10, 24, 26)$, etc.

Problem: A ladder $15$ m long leans against a wall. Its foot is $9$ m from the wall. How high does the ladder reach?

Solution:
The wall, ground, and ladder form a right triangle.

Answer: The ladder reaches $12$ m up the wall.

Problem: A triangle has sides $11$ cm, $60$ cm, and $61$ cm. Is it right-angled?

Solution:
Check: $11^2 + 60^2 = 121 + 3600 = 3721 = 61^2$.

Yes, it satisfies Pythagoras theorem, so it IS a right triangle.

The right angle is opposite the longest side ($61$ cm).

Problem: Find the diagonal of a rectangle with length $12$ cm and breadth $5$ cm.

Solution:
The diagonal divides the rectangle into two right triangles.

Answer: The diagonal is $13$ cm.

Problem: A man walks $6$ km east and then $8$ km north. How far is he from his starting point?

Solution:
His path forms a right angle (east then north). The direct distance is the hypotenuse.

Answer: He is $10$ km from his starting point.

Problem: Is $(4, 7, 9)$ a Pythagorean triplet?

Solution:

Since $65 \neq 81$, $(4, 7, 9)$ is NOT a Pythagorean triplet.

Problem: In a rectangle $ABCD$, $AB = 8$ cm and diagonal $AC = 10$ cm. Find $BC$.

Solution:
In right $\triangle ABC$ (angle at $B = 90°$):

Answer: $BC = 6$ cm.

Problem: For any natural number $m > 1$, show that $(2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Solution:
Check: $(2m)^2 + (m^2 - 1)^2 = 4m^2 + m^4 - 2m^2 + 1 = m^4 + 2m^2 + 1 = (m^2 + 1)^2$.

Since the sum of squares of the first two equals the square of the third, it IS a Pythagorean triplet.

Examples:
- $m = 2$: $(4, 3, 5)$
- $m = 3$: $(6, 8, 10)$
- $m = 4$: $(8, 15, 17)$

Answer 1: $x + 2x + 3x = 180$. $6x = 180$. $x = 30$. Angles: $30°, 60°, 90°$. It is a right triangle.

Answer 2: $5 + 7 = 12 < 13$. No, these cannot form a triangle.

Answer 3: $h^2 = 9^2 + 40^2 = 81 + 1600 = 1681$. $h = 41$ cm.

Answer 4: $x + x = 135$. $2x = 135$. $x = 67.5°$. Each opposite interior angle is $67.5°$.

Answer 5: $12^2 + 35^2 = 144 + 1225 = 1369 = 37^2$. Yes, it is a Pythagorean triplet.

Answer 6: $9 - 4 < x < 9 + 4$. So $5 < x < 13$.