Chapter 6, The Triangle and Its Properties (त्रिभुज और उसके गुण), explores the basic elements and fundamental properties of triangles, including medians, altitudes, and important angle theorems.
Table of Contents
Exercise 6.1: Medians and Altitudes
This exercise focuses on the different lines associated with a triangle.
| Q. No. | Type | Example/Concept | Key Concept |
| 1 | Identification | In $\triangle PQR$, $D$ is the midpoint of $QR$. (a) $PD$ is a $\_\_\_$. (b) $PM$ is an $\_\_\_$. (a) $PD$ is a Median. (b) $PM$ is an Altitude (since $PM \perp QR$). | Median: Line segment joining a vertex to the midpoint of the opposite side. Altitude: Line segment from a vertex perpendicular to the opposite side. |
| 2 | Altitude Location | Draw rough sketches of altitudes for: (a) An acute-angled triangle. (b) An obtuse-angled triangle. (a) Altitudes lie inside the triangle. (b) Two altitudes lie outside the triangle, and one lies inside. | Altitudes can lie inside, outside, or on the boundary (for right triangles). |
| 3 | Check Property | Can the altitude and median be the same for an isosceles triangle? Yes, in an isosceles triangle, the median drawn to the unequal side is also the altitude. | Property of Isosceles and Equilateral triangles. |
Exercise 6.2: Exterior Angle Property
This exercise applies the theorem related to the exterior angle of a triangle.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Finding Exterior Angle | In $\triangle ABC$, the interior opposite angles are $50^\circ$ and $70^\circ$. Find the exterior angle $x$. Exterior Angle $x = \text{Sum of Interior Opposite Angles}$ $x = 50^\circ + 70^\circ = \mathbf{120^\circ}$ | Exterior Angle Property: An exterior angle of a triangle is equal to the sum of its two interior opposite angles. |
| 2 | Finding Interior Angle | The exterior angle is $110^\circ$. One interior opposite angle is $45^\circ$. Find the other interior opposite angle $y$. $110^\circ = 45^\circ + y$ $y = 110^\circ – 45^\circ = \mathbf{65^\circ}$ | Applying the property to find an unknown interior angle. |
Exercise 6.3: Angle Sum Property
This exercise uses the fundamental theorem about the sum of a triangle’s interior angles.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Finding Unknown Angle | In a triangle, two angles are $60^\circ$ and $70^\circ$. Find the third angle $x$. Sum of angles $= 180^\circ$ $x + 60^\circ + 70^\circ = 180^\circ$ $x + 130^\circ = 180^\circ$ $x = 180^\circ – 130^\circ = \mathbf{50^\circ}$ | Angle Sum Property: The sum of the interior angles of any triangle is always $180^\circ$. |
| 2 | Complex Problem | In a figure, find $x$ and $y$, where $x$ is an interior angle and $y$ is the exterior angle ($y = 120^\circ$). The other interior angles are $50^\circ$ and $x$. 1. Find $x$ (Exterior Angle Property): $120^\circ = 50^\circ + x \implies x = 70^\circ$. 2. (Alternate Method using Linear Pair): $x + 50^\circ + (\text{angle adjacent to } y) = 180^\circ$ | Combining Angle Sum Property and Exterior Angle Property. |
Exercise 6.4 & 6.5: Pythagorean Property
This focuses on the relationship between the sides of a right-angled triangle.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Finding Hypotenuse | The lengths of the legs of a right triangle are $3 \text{ cm}$ and $4 \text{ cm}$. Find the length of the hypotenuse $c$. Pythagorean Property: $a^2 + b^2 = c^2$ $3^2 + 4^2 = c^2$ $9 + 16 = 25 \implies c = \sqrt{25} = \mathbf{5 \text{ cm}}$ | Pythagorean Property: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides ($a^2 + b^2 = c^2$). |
| 2 | Checking for Right Triangle | Can $6 \text{ cm}, 8 \text{ cm}, 10 \text{ cm}$ be the sides of a right triangle? Check if $6^2 + 8^2 = 10^2$ $36 + 64 = 100 \implies 100 = 100$. Yes, they form a right triangle. | The property holds if and only if the triangle is a right triangle. |
| 3 | Finding a Leg | A ladder $13 \text{ m}$ long is placed against a wall. The foot of the ladder is $5 \text{ m}$ from the wall. Find the height $h$ up to which the ladder reaches. $5^2 + h^2 = 13^2$ $25 + h^2 = 169$ $h^2 = 144 \implies h = \mathbf{12 \text{ m}}$ |