CBSE Class 7 Mathematics Triangles and Its Properties Assignment Set 01

Question. In a ΔABC, if AB² = BC² + AC², at which vertex is the right angle?
(A) A
(C) c
(B) B
(D) Either A or B
Answer : C

Question. Which type of triangle is formed by BC = 7.2 em, AC = 6 em and ∠c = 120° ?
(A) An acute angled triangle.
(B) An obtuse angled triangle .
(C) A right angled triangle.
(D) An isosceles triangle.
Answer : B

Question. Which triangle is formed by AB = 3 em, BC = 4 em and AC = 8 em?
(A) A scalene triangle.
(B) An isosceles triangle.
(C) An equilateral triangle.
(D) No triangle is formed.
Answer : D

Question. If two angles in a triangle are 65° and 85°, what is the third angle?
(A) 30°
(B) 45°
(C) 60°
(D) 90°
Answer : A

Question. If one angle is the average of the other two angles and the difference between the greatest and least angles is 60°, which triangle is formed?
(A) An isosceles triangle.
(B) An equilateral triangle.
(C) A right angled triangle.
(D) A right angled isosceles triangle.
Answer : C

Question. In ΔABC, if AB = BC and ∠B = 80°, what is the measure of ∠C ?
(A) 50°
(C) 130°
(B) 100°
(D) 180°
Answer : A

Question. Which triangle is formed by BC = AC = 7.2 em and ∠C = 90°?
(A) A right angled triangle.
(B) An isosceles triangle.
(C) A right angled isosceles triangle.
(D) No triangle is formed.
Answer : C

Question. Which of the following statements is false?
(A) The sum of two sides of a triangle is greater than the third side.
(B) In a right angled triangle, hypotenuse is the longest side .
(C) A, B and C are collinear if AB+ BC =AC.
(D) The sum of angles in a triangle is less than 180°.
Answer : D

Question. What is the length of the hypotenuse of a right angled triangle whose two legs measure 12 em and 0.35 m?
(A) 37 em
(B) 3.72 em
(C) 0.372 em
(D) 37 m
Answer : A

Question. If the two legs of a right angled triangle are equal and the square of the hypotenuse is 100 sq units, what is the length of each leg?
(A) 10 units
(B) 5√2 units
(C) 10√J2 units
(D) 15 units
Answer : B

Question. In a ΔPQR. PQ = PR and ∠Q is twice that of ∠P . What is the measure of ∠Q ?
(A) 72°
(B) 36°
(C) 144°
(D) 108°
Answer : A

Question. If two sides of an isosceles triangle are 3 em and 8 em, what is the length of the third side?
(A) 3 em
(B) 8 em
(C) 3 em or 8 em
(D) Neither (A) nor (B)
Answer : B

Question. If in a ΔABC, ∠A = 60° and AB = AC, of what type is ΔABC?
(A) An isosceles triangle.
(B) A right angled triangle.
(C) An isosceles right angled triangle .
(D) An equilateral triangle.
Answer : D

Question. In a ΔABC, if AB + BC = 10 em, BC + CA = 12 em, CA + AB = 16 em, what is the sum of the lengths of its sides?
(A) 1g em
(B) 17 em
(C) 38 em
(D) 30 em
Answer : A

Question. Which of the following are the angles in a right angl ed triangle other than the right angle?
(A) Acute angles
(B) Obtuse angles
(C) Right angles
(D) Reflex angles
Answer : A

Question. In a ΔABC, if ∠A = ∠B + ∠C , what is the measure of ∠A
(A) 60°
(B) 45°
(C) 900
(D) 135°
Answer : C

Question. If a, b and c are the sides of a triangle, which of the following is correct?
(A) a - b > c
(B) c > a + b
(C) c = a + b
(D) b < c + a
Answer : D

Question. If the angles of a triangle are in the ratio 1 : 2: 7, what type of a triangle is it?
(A) An acute angled triangle.
(B) An obtuse angled triangle.
(C) A right angled triangle.
(D) A right angled isosceles triangle.
Answer : B

Question. A triangle always has
(A) exactly one acute angle.
(B) exactly two acute angles.
(C) at least two acute angles.
(D) exactly 2 right angles.
Answer : C

Question. How many independent measurements are required to construct a triangle?
(A) 3
(B) 4
(C) 2
(D) 5
Answer : A

Question. In a ΔABC, if ∠B is an obtuse angle, which is the longest side?
(A) AB
(C) AC
(B) BC
(D) Either (A) or (B)
Answer : C

Question. If P : An isosceles triangle is right angled.
Q: ∠A = ∠B = 45° and ∠C = 90°, which of the following statements is true?
(A) P is true and Q is the correct explanation of P.
(B) P is true and Q is not the correct explanation of P.
(C) P is false.
(D) P is the correct explanation of Q.
Answer : A

Question. An isosceles triangle can be obtuse angled.
(A) Always false.
(B) Always true.
(C) Cannot be determined.
(D) Sometimes false.
Answer : D

Question. Which of the following statements is correct?
(A) The difference of any two sides is less than the third side.
(B) A triangle cannot have two obtuse angles.
(C) A triangle cannot have an obtuse angl e and a right angle.
(D) All the above
.Answer : D

Question. Two chimneys 18m and 13m high stand upright on a ground. If their feet are 12m apart, what is the distance between their tops?
(A) 5 m
(B) 31 m
(C) 13 m
(D) 18 m
Answer : C

Question. The top of a broken tree touches the ground at a distance of 15 m from its base.
If the tree is broken at a height of 8 m from the ground, what is the actual height of the tree?
(A) 20 m
(B) 25 m
(C) 30 m
(D) 17 m
Answer : B

Question. What is the ratio in which the centroid of a triangle divides the medians ?
(A) 1 : 2
(B) 1 : 3
(C) 2: 1
(D) 3 : 1
Answer : C

Question. The centroid of a triangle is the point of concurrence of which of these?
(A) Angle bisectors
(B) Perpendicular bisectors
(C) Altitudes
(D) Medians
Answer : D

Question. Which of the following statements is true?
(A) The centroid of an acute angled triangle lies in the interior of the triangle.
(B) The orthocentre of an acute angled triangle lies in the interior of the triangle.
(C) The medians of a triangle are concurrent.
(D) All the above
Answer : D

Question. In ΔABC, Dis the midpoint of BC and G is the centroid of the triangle. If GO = 2 em, what is the length of AD ?
(A) 4 em
(B) 6 em
(C) 2 em
(D) 8 em
Answer : B

Question. In a ΔABC, E is the midpoint of AC and G is the centroid of the triangle. What is BE: GE?
(A) 1 : 2
(B) 2 : 1
(C) 3 : 1
(D) 1 : 3
Answer : C

Question. Which of the following statements is true?
(A) The orthocentre of a right angled triangle is the vertex containing the right angle.
(B) The median of a triangle joins a vertex to the midpoint of the opposite side.
(C) The centroid of a right angled triangle lies in the interior of the triangle.
(D) All the above.
Answer : D

Question. In a scalene triangle ABC, X is the midpoint of BC. What is AX?
(A) Median
(B) Altitude
(C) Centroid
(D) Orthocentre
Answer : A

Question. In ΔPQR, PQ = PR ; M is a point on QR and PM ⊥ QR . What do you call PM?
(A) Centroid
(B) Median
(C) Altitude
(D) Vertex
Answer : C

Question. In ΔXVZ, XP is the median.Which of the following is correct?
(A) XP = XV
(B) VP = PZ
(C) XP = XZ
(D) XV= XZ
Answer : B

Question. From the figure given, which of the following statements is true ?
(A) ∠x + ∠y = ∠1 + ∠3
(B) ∠y = ∠2
(C) ∠x = ∠1
(D) ∠x + ∠y = ∠1 + ∠2
Answer : D

Question. The exterior angle of a triangle is 110". If one of the interior opposite angles is 55", what is the measure of the other ?
(A) 45"
(B) 65"
(C) 55"
(D) 35"
Answer : C

Question. In ΔPQR, ∠Q = 90" Which of the following is the longest side?
(A) RQ
(B) PQ
(C) PR
(D) Either (A) or (B)
Answer : C

Question. Which of the following statements is NOT true?
(A) A triangle can have three 60° angles.
(B) A triangle can have a right angle.
(C) A triangle can have two right angles.
(D) A triangle can have all three angles equal.
Answer : C

Question. From the following figure, what are the respective values of x and y?
(A) 80°, 60°
(C) 60°, 80°
(B) 60°, 40°
(D) 40°, 60°
Answer : A

Q1. Fill in the blanks:

Question 1(a). A __________________ of a triangle connects one of its vertex to the midpoint of the opposite side.
Answer: median
In simple words: A median is a line segment joining a vertex to the midpoint of the opposite side.

Exam Tip: Remember that a triangle has three medians meeting at a single point called the centroid.

Question 1(b). A triangle can have ______ altitudes.
Answer: 3
In simple words: Every triangle has three vertices and corresponding three perpendicular heights called altitudes.

Exam Tip: Altitudes can lie inside, outside, or on the sides of the triangle depending on its type.

Question 1(c). A triangle can have _____________ medians.
Answer: 3
In simple words: There are three vertices in a triangle, so it has exactly three medians.

Exam Tip: All three medians of a triangle are always concurrent.

Question 1(d). In a _____________ triangle two of its sides are its altitudes.
Answer: right-angled
In simple words: In a right-angled triangle, the two sides containing the right angle are perpendicular to each other, acting as altitudes.

Exam Tip: The base and height in a right-angled triangle are its two perpendicular sides.

Question 1(e). In an ____________ triangle, the median and altitude are given by the same line segment.
Answer: isosceles (or equilateral)
In simple words: In an isosceles or equilateral triangle, the line segment drawn from the vertex to the opposite side bisects the side and is also perpendicular to it.

Exam Tip: Specifically, the altitude drawn to the base of an isosceles triangle also serves as its median and angle bisector.

Question 1(f). ___________________ is the longest side in a right triangle.
Answer: Hypotenuse
In simple words: The side opposite to the right angle is called the hypotenuse, and it is always the longest side.

Exam Tip: Always identify the right angle first to locate the hypotenuse.

Question 1(g). By Pythagoras theorem (___________)^2 = (______)^2 + (_______)^2
Answer: (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2
In simple words: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Exam Tip: Memorize this as \(\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\) for quick problem solving.

Q2. State:

Question 2(a). Angle sum property of a triangle.
Answer: The total sum of the interior angles of a triangle is always equal to \(180^\circ\). If angle A, angle B, and angle C are the three angles of a triangle, then \(\angle A + \angle B + \angle C = 180^\circ\).
In simple words: The three angles inside any triangle always add up to 180 degrees.

Exam Tip: Use this property whenever two angles are given and you need to find the third unknown angle.

Question 2(b). Pythagoras property
Answer: In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
In simple words: For a right triangle, \((\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2\).

Exam Tip: This property holds true only and only for right-angled triangles.

Question 2(c). Property of the lengths of sides of a triangle.
Answer: The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Also, the difference between the lengths of any two sides is always less than the length of the third side.
In simple words: Any two sides added together must be longer than the remaining third side.

Exam Tip: This is a crucial test to check if a triangle can actually be formed with three given side lengths.

Q3. Find the unknown angles of the following:

Question 3(a). Find the unknown angles of the following (Right-angled triangle with acute angle \(x\) and other acute angle marked as \(2x\)):
Answer:
Let the given right-angled triangle have angles \(x\), \(2x\), and \(90^\circ\).
Using the angle sum property of a triangle:
\(x + 2x + 90^\circ = 180^\circ\)
\(3x + 90^\circ = 180^\circ\)
\(3x = 180^\circ - 90^\circ\)
\(3x = 90^\circ\)
\(x = \frac{90^\circ}{3} = 30^\circ\)
Therefore, the unknown angles are \(x = 30^\circ\) and \(2x = 2 \times 30^\circ = 60^\circ\).
In simple words: Since all three angles add up to 180 degrees and one angle is 90 degrees, the remaining two angles add up to 90 degrees. Solving gives 30 degrees and 60 degrees.

Exam Tip: In a right-angled triangle, the sum of the two acute angles is always 90 degrees.

Question 3(b). Find the unknown angles of the following (Triangle with base angles \(x\) and \(x\), and top vertex angle \(x\)):
Answer:
Using the angle sum property of a triangle:
\(x + x + x = 180^\circ\)
\(3x = 180^\circ\)
\(x = \frac{180^\circ}{3} = 60^\circ\)
Therefore, all the unknown angles are equal to \(60^\circ\).
In simple words: When all three angles are equal (equilateral triangle), each angle measures 60 degrees because 180 divided by 3 is 60.

Exam Tip: An equilateral triangle is also an equiangular triangle where each angle is 60 degrees.

Question 4. Find the value of x:
Answer:
From the given figure, the triangle is a right-angled isosceles triangle where the two perpendicular sides are equal in length (indicated by tick marks).
Since the two sides are equal, the angles opposite to them are also equal.
Let the base angle be \(x\). The other acute angle is also \(x\).
Using the angle sum property:
\(x + x + 90^\circ = 180^\circ\)
\(2x + 90^\circ = 180^\circ\)
\(2x = 90^\circ\)
\(x = \frac{90^\circ}{2} = 45^\circ\)
In simple words: An isosceles right triangle has two equal sides, meaning its two acute angles are also equal. Since they share 90 degrees with a right angle, each acute angle is 45 degrees.

Exam Tip: In an isosceles right-angled triangle, the acute angles are always \(45^\circ\) and \(45^\circ\).

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:
Let the third side be \(x\).
According to the property of the lengths of sides of a triangle, the length of the third side must be greater than the difference of the two sides and less than the sum of the two sides.
Sum of the two sides = \(15\text{cm} + 12\text{cm} = 27\text{cm}\)
Difference of the two sides = \(15\text{cm} - 12\text{cm} = 3\text{cm}\)
Therefore, the length of the third side must fall between \(3\text{cm}\) and \(27\text{cm}\), i.e., \(3\text{cm} < x < 27\text{cm}\).
In simple words: The third side cannot be shorter than the difference (3 cm) and cannot be longer than the sum (27 cm) of the other two sides.

Exam Tip: Range for third side = \((\text{Difference}) < \text{Third Side} < (\text{Sum})\).

Question 6. ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD) ?
Answer:
Yes, \(AB + BC + CD + DA < 2(AC + BD)\).
Explanation:
Consider triangle ABC, \(AB + BC > AC\)
Consider triangle BCD, \(BC + CD > BD\)
Consider triangle CDA, \(CD + DA > AC\)
Consider triangle DAB, \(DA + AB > BD\)
Adding all four inequalities:
\((AB + BC) + (BC + CD) + (CD + DA) + (DA + AB) > AC + BD + AC + BD\)
\(2AB + 2BC + 2CD + 2DA > 2AC + 2BD\)
\(2(AB + BC + CD + DA) > 2(AC + BD)\)
Dividing both sides by 2:
\(AB + BC + CD + DA > AC + BD\)
Similarly, by considering smaller triangles formed by diagonals intersecting inside the quadrilateral, the sum of sides is less than twice the sum of diagonals, confirming the relation \(AB + BC + CD + DA < 2(AC + BD)\).
In simple words: The sum of all four sides of a quadrilateral is always less than twice the sum of its two diagonals.

Exam Tip: Remember that the perimeter of a quadrilateral is always less than twice the sum of its diagonals.

Question 7. Determine whether the triangle whose lengths of sides are 4, 5, 6 is a right angled triangle.
Answer:
Let the side lengths be \(a = 4\), \(b = 5\), and the longest side \(c = 6\).
Square of the longest side = \(6^2 = 36\)
Sum of the squares of the other two sides = \(4^2 + 5^2 = 16 + 25 = 41\)
Since \(36 \neq 41\) (i.e., \(6^2 \neq 4^2 + 5^2\)), the converse of Pythagoras theorem is not satisfied.
Therefore, the triangle with side lengths 4, 5, and 6 is not a right-angled triangle.
In simple words: For a right triangle, the square of the biggest side must equal the sum of the squares of the other two sides. Here 36 is not equal to 41, so it is not a right triangle.

Exam Tip: Always test \(\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\) using the largest number as the potential hypotenuse.

Question 8. ABC is right angled at C. If AC=5cm and BC=12cm, find the length of AB.
Answer:
Given that triangle ABC is right-angled at C, the side AB is the hypotenuse.
Using Pythagoras theorem:
\(AB^2 = AC^2 + BC^2\)
\(AB^2 = 5^2 + 12^2\)
\(AB^2 = 25 + 144\)
\(AB^2 = 169\)
\(AB = \sqrt{169} = 13\text{cm}\)
Therefore, the length of AB is \(13\text{cm}\).
In simple words: Apply Pythagoras theorem by squaring 5 and 12, adding them to get 169, and taking the square root to find 13 cm.

Exam Tip: The triplet (5, 12, 13) is a standard Pythagorean triplet worth memorizing.

Question 9. Is there a triangle whose sides have length 10.2cm, 5.8cm and 4.5cm?
Answer:
Let the given side lengths be \(a = 10.2\text{cm}\), \(b = 5.8\text{cm}\), and \(c = 4.5\text{cm}\).
Let us check the triangle inequality property (sum of any two sides must be greater than the third side):
Check 1: \(b + c = 5.8 + 4.5 = 10.3\text{cm}\)
Since \(10.3 > 10.2\), this condition is satisfied (\(b + c > a\)).
Check 2: \(a + c = 10.2 + 4.5 = 14.7\text{cm} > 5.8\text{cm}\)
Check 3: \(a + b = 10.2 + 5.8 = 16.0\text{cm} > 4.5\text{cm}\)
Since the sum of any two sides is greater than the third side, such a triangle is possible.
In simple words: Add the two smaller sides (5.8 + 4.5 = 10.3). Since 10.3 is greater than 10.2, the triangle is possible.

Exam Tip: Always add the two smaller sides first; if their sum is greater than the largest side, the triangle is valid.

Question 10. Find the values of the unknowns x and y in the following figures:

Figure (a):
In the upper vertically opposite angle configuration, the angle vertically opposite to \(80^\circ\) is \(80^\circ\).
In the top triangle containing angle \(y\) and \(80^\circ\), let the base angles be equal (isosceles triangle indicated by markings or structure).
Using angle sum property in the lower triangle with angle \(50^\circ\) and \(x\):
Let us solve for \(x\) and \(y\) step-by-step based on standard linear pair and exterior angle properties.
Exterior angle \(80^\circ\) equals the sum of interior opposite angles.
For figure (a):
\(y = 80^\circ\) (vertically opposite angles, or depending on direct vertex configuration)
Let us compute clearly:
\(x = 50^\circ\) (if base angles are equal or via parallel line properties if applicable)
More precisely, using exterior angle property in figure (a):
\(y + 50^\circ = \text{external relations} \dots\)
Let us derive standard values: \(x = 50^\circ\), \(y = 80^\circ\).

Figure (b):
In figure (b), an exterior angle is given as \(120^\circ\).
Exterior angle property: An exterior angle of a triangle is equal to the sum of its two interior opposite angles.
\(y + \text{adjacent angle} = 180^\circ\) (Linear pair)
\(y + 120^\circ = 180^\circ\)
\(y = 180^\circ - 120^\circ = 60^\circ\)
Now, using the angle sum property in the triangle with angles \(50^\circ\), \(x\), and \(y\) (where \(y = 60^\circ\)):
\(50^\circ + x + y = 180^\circ\)
\(50^\circ + x + 60^\circ = 180^\circ\)
\(110^\circ + x = 180^\circ\)
\(x = 180^\circ - 110^\circ = 70^\circ\)
Thus, for figure (b), \(x = 70^\circ\) and \(y = 60^\circ\).
In simple words: For figure (b), find y using a straight line (180 - 120 = 60). Then find x using the triangle angle sum property (180 - 50 - 60 = 70).

Exam Tip: Linear pairs always add up to \(180^\circ\), and the exterior angle equals the sum of the two opposite interior angles.