CBSE Class 7 Mathematics Triangle And Its Properties Worksheet Set 05

Question 1. Fill in the blanks:
a) A __________________ of a triangle connects one of its vertex to the midpoint of the opposite side.
b) A triangle can have ______ altitudes.
c) A triangle can have _____________ medians.
d) In a _____________ triangle two of its sides are its altitudes.
e) In an ____________ triangle, the median and altitude are given by the same line segment.
f) ___________________ is the longest side in a right triangle.
g) By Pythagoras theorem \( (\_\_\_\_\_\_\_)^2 = (\_\_\_\_\_\_\_)^2 + (\_\_\_\_\_\_\_)^2 \)
Answer:
a) median
b) three
c) three
d) right-angled
e) isosceles / equilateral
f) hypotenuse
g) \( (\mathbf{hypotenuse})^2 = (\mathbf{base})^2 + (\mathbf{altitude/height})^2 \)
In simple words: These blanks teach us about standard rules for triangles. A median connects a corner to the middle of the opposite side, altitudes are height lines, and the hypotenuse is always the longest side in a right-angled triangle.

Exam Tip: Remember that in an equilateral triangle, all three medians are also altitudes because all sides are equal.

Question 2. State:
a) Angle sum property of a triangle.
b) Pythagoras property
c) Property of the lengths of sides of a triangle.
Answer:
a) Angle sum property of a triangle: The total sum of all three internal angles of a triangle is always exactly \( 180^\circ \).
b) Pythagoras property: In a right-angled triangle, the square of the longest side (hypotenuse) is equal to the sum of the squares of the other two sides (legs).
c) Property of the lengths of sides of a triangle: The sum of the lengths of any two sides of a triangle must always be strictly greater than the length of its third side.
In simple words: First, the inside corners of any triangle always add up to \( 180^\circ \). Second, in a square-corner triangle, squaring the two short sides and adding them gives the square of the long side. Third, you cannot make a triangle unless any two sides together are longer than the third side.

Exam Tip: When writing the property for side lengths, remember that the sum of two sides must be strictly "greater than" the third side, not "greater than or equal to".

Question 3. Find the unknown angles of the following:
a)
b)
Answer:
a) In the first figure, we have a right-angled triangle. Using the angle sum property:
\( \implies x + 2x + 90^\circ = 180^\circ \)
\( \implies 3x + 90^\circ = 180^\circ \)
\( \implies 3x = 180^\circ - 90^\circ \)
\( \implies 3x = 90^\circ \)
\( \implies x = 30^\circ \)
The unknown angles are:
- \( x = 30^\circ \)
- \( 2x = 2 \times 30^\circ = 60^\circ \)

b) In the second figure, all three angles are equal to \( x \). Since the sum of angles of a triangle is \( 180^\circ \):
\( \implies x + x + x = 180^\circ \)
\( \implies 3x = 180^\circ \)
\( \implies x = \frac{180^\circ}{3} = 60^\circ \)
Each unknown angle is \( 60^\circ \).
In simple words: For the first triangle, the two sharp corners must add up to \( 90^\circ \), so one is \( 30^\circ \) and the other is \( 60^\circ \). For the second one, since all three corners are identical, we divide \( 180^\circ \) by 3 to get \( 60^\circ \).

Exam Tip: A square-shaped marker in a corner always stands for a right angle (\( 90^\circ \)), even if the text does not state it.

Question 4. Find the value of x:
Answer:
The triangle is a right-angled isosceles triangle, as indicated by the matching double tick marks on the two perpendicular legs.
In an isosceles triangle, the angles opposite the equal sides are also equal.
Therefore, the other acute angle is also \( x \).
Using the angle sum property of a triangle:
\( \implies x + x + 90^\circ = 180^\circ \)
\( \implies 2x = 180^\circ - 90^\circ \)
\( \implies 2x = 90^\circ \)
\( \implies x = 45^\circ \)
The value of \( x \) is \( 45^\circ \).
In simple words: The small double marks on the bottom and vertical lines mean they are equal. This makes the two sharp corners equal. Since the square corner is \( 90^\circ \), the other two corners must split the remaining \( 90^\circ \) equally, which gives us \( 45^\circ \) for \( x \).

Exam Tip: A right-angled isosceles triangle will always have acute angles of exactly \( 45^\circ \) each.

Question 5. The lengths of two sides of a triangle are 12cm and 15cm. Between what two measures should the length of the third side fall?
Answer:
The length of the third side must be less than the sum of the other two sides, and greater than their positive difference.
Let the third side be \( s \).
1. Find the sum of the two given sides:
\( \implies 15\text{ cm} + 12\text{ cm} = 27\text{ cm} \)
Therefore, the third side must be less than 27 cm.
2. Find the difference of the two given sides:
\( \implies 15\text{ cm} - 12\text{ cm} = 3\text{ cm} \)
Therefore, the third side must be greater than 3 cm.
Thus, the length of the third side should fall between 3 cm and 27 cm.
In simple words: The third side of the triangle has to be longer than the difference of the other two sides (which is 3 cm) but shorter than their total sum (which is 27 cm).

Exam Tip: To solve this quickly, remember that the third side \( s \) always satisfies: Difference of sides < \( s \) < Sum of sides.

Question 6. ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD) ?
Answer:
Yes, this statement is correct. Here is the mathematical proof:
Let the diagonals AC and BD intersect at point O.
According to the triangle inequality property, the sum of any two sides of a triangle is greater than the third side. Applying this to the four interior triangles:
1. In \(\Delta AOB\):
\( \implies AB < OA + OB \)
2. In \(\Delta BOC\):
\( \implies BC < OB + OC \)
3. In \(\Delta COD\):
\( \implies CD < OC + OD \)
4. In \(\Delta DOA\):
\( \implies DA < OD + OA \)
Now, add these four inequalities together:
\( \implies AB + BC + CD + DA < (OA + OB) + (OB + OC) + (OC + OD) + (OD + OA) \)
\( \implies AB + BC + CD + DA < 2(OA + OC) + 2(OB + OD) \)
Since \( OA + OC = AC \) and \( OB + OD = BD \), we can substitute these values:
\( \implies AB + BC + CD + DA < 2(AC + BD) \)
This proves the inequality is indeed true.
In simple words: Yes, it is true. If we look at the four small triangles inside the shape where the diagonals cross, the outer boundary lines are always shorter than the lines pointing to the middle. When we add them all up, the total perimeter is less than double the length of the diagonals.

Exam Tip: Be careful with the inequality sign. The sum of the outer sides is greater than the sum of the diagonals, but less than twice the sum of the diagonals.

Question 7. Determine whether the triangle whose lengths of sides are 4, 5, 6 is a right angled triangle.
Answer:
To find out if the triangle is right-angled, we check if the sides satisfy the Pythagoras property. The square of the longest side must equal the sum of the squares of the two shorter sides.
Let the sides be \( a = 4 \), \( b = 5 \), and \( c = 6 \).
1. Calculate the sum of the squares of the two shorter sides:
\( \implies 4^2 + 5^2 = 16 + 25 = 41 \)
2. Calculate the square of the longest side:
\( \implies 6^2 = 36 \)
Since \( 41 \neq 36 \), the Pythagoras property does not hold true.
Therefore, this is not a right-angled triangle.
In simple words: No, it is not a right-angled triangle. When we square the two smaller sides (\( 16 \) and \( 25 \)) and add them, we get \( 41 \). This does not match the square of the longest side, which is \( 36 \).

Exam Tip: Always make sure to use the largest number as the potential hypotenuse when checking if a triangle is right-angled.

Question 8. ABC is right angled at C. If AC=5cm and BC=12cm, find the length of AB.
Answer:
Since the triangle is right-angled at C, the side AB is the hypotenuse.
Using the Pythagoras property:
\( \implies AB^2 = AC^2 + BC^2 \)
\( \implies AB^2 = 5^2 + 12^2 \)
\( \implies AB^2 = 25 + 144 \)
\( \implies AB^2 = 169 \)
\( \implies AB = \sqrt{169} = 13\text{ cm} \)
The length of AB is 13 cm.
In simple words: Since we have a square corner, we can square the two short sides to get \( 25 \) and \( 144 \). Adding them gives \( 169 \). Since \( 13 \times 13 = 169 \), the long side AB is 13 cm.

Exam Tip: 5 - 12 - 13 is a very common Pythagorean triple. Remembering it helps you quickly double-check your work.

Question 9. Is there a triangle whose sides have length 10.2cm, 5.8cm and 4.5cm?
Answer:
According to the triangle inequality property, a triangle can only exist if the sum of any two sides is strictly greater than the third side.
Let's test the three combinations of sides: 10.2 cm, 5.8 cm, and 4.5 cm.
- Case 1: \( 10.2 + 5.8 = 16.0 > 4.5 \) (True)
- Case 2: \( 5.8 + 4.5 = 10.3 > 10.2 \) (True)
- Case 3: \( 10.2 + 4.5 = 14.7 > 5.8 \) (True)
Since the sum of any two sides is indeed greater than the third side in all cases, a triangle with these side lengths can exist.
In simple words: Yes, this triangle can exist. No matter which two sides we choose to add together, their total length is always longer than the remaining third side.

Exam Tip: To save time, you only need to verify if the sum of the two smallest sides is greater than the single largest side.

Question 10 Find the values of the unknowns x and y in the following figures
a)
b)
Answer:
a) In the first figure:
- The angle \( y \) is vertically opposite to the given \( 80^\circ \) angle. Since vertically opposite angles are equal:
\( \implies y = 80^\circ \)
- Now, apply the angle sum property of a triangle:
\( \implies 50^\circ + x + y = 180^\circ \)
\( \implies 50^\circ + x + 80^\circ = 180^\circ \)
\( \implies x + 130^\circ = 180^\circ \)
\( \implies x = 180^\circ - 130^\circ \)
\( \implies x = 50^\circ \)
So, \( x = 50^\circ \) and \( y = 80^\circ \).

b) In the second figure:
- The interior angle \( y \) and the exterior angle \( 120^\circ \) form a linear pair on a straight line:
\( \implies y + 120^\circ = 180^\circ \)
\( \implies y = 180^\circ - 120^\circ \)
\( \implies y = 60^\circ \)
- Using the exterior angle property of a triangle, the exterior angle equals the sum of its two opposite interior angles:
\( \implies x + 50^\circ = 120^\circ \)
\( \implies x = 120^\circ - 50^\circ \)
\( \implies x = 70^\circ \)
So, \( x = 70^\circ \) and \( y = 60^\circ \).
In simple words: In the first drawing, the top inside angle \( y \) is \( 80^\circ \) because it is directly opposite the other \( 80^\circ \) angle. We then find \( x = 50^\circ \) because all three inside angles must add up to \( 180^\circ \). In the second drawing, the straight line at the bottom helps us find \( y = 60^\circ \), and the outside corner rule helps us find \( x = 70^\circ \).

Exam Tip: You can always verify your answers by checking if the calculated values of \( x \), \( y \), and the third angle sum up to \( 180^\circ \) inside the triangle.