CBSE Class 7 Mathematics Triangle And Its Properties Worksheet Set 01

Draw a rough figure using a scale and a pencil where ever required.

Question 1. An exterior angle of a triangle DEF is of measure \(70^{\circ}\) and one of its interior opposite angles is of measure \(25^{\circ}\). Find the measure of the other interior opposite angle.
Answer: Let the measure of the unknown interior opposite angle be \(x\).
By the exterior angle property of a triangle, the measure of an exterior angle is equal to the sum of the lengths of its interior opposite angles.
\(70^{\circ} = 25^{\circ} + x\)
\(x = 70^{\circ} - 25^{\circ} = 45^{\circ}\)
Thus, the measure of the other interior opposite angle is \(45^{\circ}\).
In simple words: Subtract the given interior angle from the exterior angle to find the missing angle.

Exam Tip: Always remember that the exterior angle equals the sum of the two interior opposite angles.

Question 2. The two interior opposite angles of an exterior angle of a triangle LMN are \(60^{\circ}\) and \(80^{\circ}\). Find the measure of the exterior angle.
Answer: Let the measure of the exterior angle be \(y\).
By the exterior angle property, the exterior angle is equal to the sum of the two interior opposite angles.
\(y = 60^{\circ} + 80^{\circ}\)
\(y = 140^{\circ}\)
Thus, the measure of the exterior angle is \(140^{\circ}\).
In simple words: Add the two interior opposite angles together to get the exterior angle.

Exam Tip: Be careful with addition and make sure to include the degree symbol in your final answer.

Question 3. Two angles of a triangle XYZ are \(30^{\circ}\) and \(80^{\circ}\). Find the third angle.
Answer: Let the third angle be \(z\).
By the angle sum property of a triangle, the sum of all interior angles in a triangle is \(180^{\circ}\).
\(30^{\circ} + 80^{\circ} + z = 180^{\circ}\)
\(110^{\circ} + z = 180^{\circ}\)
\(z = 180^{\circ} - 110^{\circ}\)
\(z = 70^{\circ}\)
Thus, the third angle is \(70^{\circ}\).
In simple words: Add the two given angles and subtract the sum from \(180^{\circ}\) to find the remaining angle.

Exam Tip: State the angle sum property clearly before setting up your equation.

Question 4. One of the angles of a triangle PQR is \(80^{\circ}\) and the other two angles are equal. Find the measure of each of the equal angles.
Answer: Let each of the two equal angles be \(x\).
The third angle is given as \(80^{\circ}\).
By the angle sum property of a triangle:
\(x + x + 80^{\circ} = 180^{\circ}\)
\(2x + 80^{\circ} = 180^{\circ}\)
\(2x = 180^{\circ} - 80^{\circ}\)
\(2x = 100^{\circ}\)
\(x = \frac{100^{\circ}}{2} = 50^{\circ}\)
Thus, the measure of each of the equal angles is \(50^{\circ}\).
In simple words: Subtract the given angle from \(180^{\circ}\) and divide the remaining sum by two since the other two angles are equal.

Exam Tip: Use variables like \(x\) for unknown equal quantities to set up a linear equation easily.

Question 4 (Second). The three angles of a triangle ABC are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Answer: Let the angles be \(1x\), \(2x\), and \(1x\).
By the angle sum property of a triangle:
\(1x + 2x + 1x = 180^{\circ}\)
\(4x = 180^{\circ}\)
\(x = \frac{180^{\circ}}{4} = 45^{\circ}\)
Therefore, the three angles are:
\(1 \times 45^{\circ} = 45^{\circ}\)
\(2 \times 45^{\circ} = 90^{\circ}\)
\(1 \times 45^{\circ} = 45^{\circ}\)
The angles are \(45^{\circ}\), \(90^{\circ}\), and \(45^{\circ}\).
Classification 1 (By angles): Right-angled triangle (since one angle is \(90^{\circ}\)).
Classification 2 (By sides): Isosceles triangle (since two angles measuring \(45^{\circ}\) are equal, meaning their opposite sides are also equal).
In simple words: Use the ratio parts with a variable \(x\), solve for \(x\) using \(180^{\circ}\), then find the angles and classify the triangle based on its side lengths and angle measures.

Exam Tip: When given a ratio, always introduce a common multiplier variable such as \(x\).

Question 5. Complete the sentence
(a) A triangle cannot have more than 1 right angle.
(b) A triangle cannot have more than 1 obtuse angle.
(c) A triangle can have 1 obtuse angle and 2 acute angles.
(d) A triangle can have 1 right angle and 2 acute angles. The sum of the acute angles should be \(90^{\circ}\).
Answer:
(a) 1
(b) 1
(c) 1, 2
(d) 1, 2, \(90^{\circ}\)
In simple words: The sum of all angles in any triangle is always \(180^{\circ}\), which limits how many large angles (\(90^{\circ}\) or more) a triangle can contain.

Exam Tip: Remember that if a triangle had two right angles (\(90^{\circ} + 90^{\circ} = 180^{\circ}\)), the third angle would be \(0^{\circ}\), which is impossible.

Question 6. If all the three angle of a triangle LMN are equal then find the measure of each angle.
Answer: Let each of the three equal angles be \(x\).
By the angle sum property of a triangle:
\(x + x + x = 180^{\circ}\)
\(3x = 180^{\circ}\)
\(x = \frac{180^{\circ}}{3} = 60^{\circ}\)
Thus, the measure of each angle is \(60^{\circ}\).
In simple words: Divide \(180^{\circ}\) by 3 since all three angles are equal in an equilateral triangle.

Exam Tip: A triangle with all equal angles is called an equilateral triangle, and each interior angle is always \(60^{\circ}\).

Question 7. One of the acute angles of a right angled triangle is \(35^{\circ}\). Find the measure of the other acute angle of the triangle.
Answer: In a right-angled triangle, one angle is always \(90^{\circ}\).
Let the other acute angle be \(x\).
By the angle sum property of a triangle:
\(90^{\circ} + 35^{\circ} + x = 180^{\circ}\)
\(125^{\circ} + x = 180^{\circ}\)
\(x = 180^{\circ} - 125^{\circ}\)
\(x = 55^{\circ}\)
Alternatively, the two acute angles of a right-angled triangle are complementary (\(90^{\circ} - 35^{\circ} = 55^{\circ}\)).
Thus, the measure of the other acute angle is \(55^{\circ}\).
In simple words: Subtract the given acute angle from \(90^{\circ}\) to find the remaining acute angle.

Exam Tip: For right-angled triangles, you can directly subtract the given acute angle from \(90^{\circ}\) to save time.

Question 8. Two angles of a triangle are equal and its third angle is double of the equal angle. Find the measures of the three angles of the triangle.
Answer: Let each of the two equal angles be \(x\).
Then the third angle is \(2x\).
By the angle sum property of a triangle:
\(x + x + 2x = 180^{\circ}\)
\(4x = 180^{\circ}\)
\(x = \frac{180^{\circ}}{4} = 45^{\circ}\)
Therefore, the three angles are:
First angle = \(45^{\circ}\)
Second angle = \(45^{\circ}\)
Third angle = \(2 \times 45^{\circ} = 90^{\circ}\)
Thus, the measures of the three angles are \(45^{\circ}\), \(45^{\circ}\), and \(90^{\circ}\).
In simple words: Set up the equation with two equal angles (\(x\) and \(x\)) and double the angle (\(2x\)), then solve for \(x\).

Exam Tip: Double-check your final values by adding them up: \(45^{\circ} + 45^{\circ} + 90^{\circ} = 180^{\circ}\).

Question 9. If one angle of a triangle is \(100^{\circ}\) and the other angles are in the ratio of 2:3, find the angles.
Answer: Let one of the remaining angles be \(2x\) and the other be \(3x\).
The sum of all interior angles of a triangle is \(180^{\circ}\).
Given one angle is \(100^{\circ}\).
\(100^{\circ} + 2x + 3x = 180^{\circ}\)
\(100^{\circ} + 5x = 180^{\circ}\)
\(5x = 180^{\circ} - 100^{\circ}\)
\(5x = 80^{\circ}\)
\(x = \frac{80^{\circ}}{5} = 16^{\circ}\)
Therefore, the two unknown angles are:
\(2x = 2 \times 16^{\circ} = 32^{\circ}\)
\(3x = 3 \times 16^{\circ} = 48^{\circ}\)
Thus, the three angles of the triangle are \(100^{\circ}\), \(32^{\circ}\), and \(48^{\circ}\).
In simple words: Subtract the known angle from \(180^{\circ}\), divide the remainder according to the given ratio, and find the individual angles.

Exam Tip: Always verify that the sum of your calculated angles equals \(180^{\circ}\).

Question 10. Which of the following triplets can be the angles of a triangle
(a) \(80^{\circ}, 73^{\circ}, 27^{\circ}\)
(b) \(90^{\circ}, 80^{\circ}, 30^{\circ}\)
(c) \(57^{\circ}, 61^{\circ}, 62^{\circ}\)
(d) \(30^{\circ}, 142^{\circ}, 20^{\circ}\)
Answer: (c) \(57^{\circ}, 61^{\circ}, 62^{\circ}\)
Let us check the sum of angles for each option using the angle sum property (\(180^{\circ}\)):
(a) \(80^{\circ} + 73^{\circ} + 27^{\circ} = 180^{\circ}\) (Wait, let us recalculate: \(80 + 73 + 27 = 180\)? No, \(73 + 27 = 100\), \(100 + 80 = 180\). Let us re-verify: \(80 + 73 + 27 = 180\). Let us check options carefully. Let us sum all options properly):
Let us check option (a): \(80 + 73 + 27 = 180^{\circ}\).
Let us check option (b): \(90 + 80 + 30 = 200^{\circ} \neq 180^{\circ}\).
Let us check option (c): \(57 + 61 + 62 = 180^{\circ}\).
Let us check option (d): \(30 + 142 + 20 = 192^{\circ} \neq 180^{\circ}\).
Note: Both (a) and (c) sum up to \(180^{\circ}\). Let us verify option (a) again: \(80 + 73 + 27 = 180\). Both (a) and (c) are valid triplets.
In simple words: A set of angles can form a triangle only if their exact sum equals \(180^{\circ}\).

Exam Tip: Always add up all three numbers in each option to test the angle sum property.

Question 11. Give reason
(a) A triangle cannot have each angle greater than \(60^{\circ}\).
(b) A triangle can have each angle equal to \(60^{\circ}\).
(c) A triangle cannot have each angle less than \(60^{\circ}\).
Answer:
(a) If each angle of a triangle were greater than \(60^{\circ}\), say \(61^{\circ}\), then the sum of the three angles would be greater than \(61^{\circ} + 61^{\circ} + 61^{\circ} = 183^{\circ}\), which violates the angle sum property of a triangle (\(180^{\circ}\)).
(b) If each angle is equal to \(60^{\circ}\), their sum is \(60^{\circ} + 60^{\circ} + 60^{\circ} = 180^{\circ}\), which perfectly satisfies the angle sum property of a triangle.
(c) If each angle of a triangle were less than \(60^{\circ}\), say \(59^{\circ}\), then the sum of the three angles would be less than \(59^{\circ} + 59^{\circ} + 59^{\circ} = 177^{\circ}\), which is less than \(180^{\circ}\) and thus violates the angle sum property of a triangle.
In simple words: The sum of the angles in a triangle must always equal exactly \(180^{\circ}\), so all angles cannot be simultaneously greater than or less than \(60^{\circ}\).

Exam Tip: Use proof by contradiction logic or test boundary values (\(60^{\circ}\)) to explain inequalities involving triangle angles.

Question. Two angles of a triangle are equal and third angle is smaller than others by 15°. Find the angles.
Answer : 65, 65, 50

Question. Two sides of a triangle are 11 cm and 5.5 cm. What will be the minimum length of the third side?
Answer : More than 5.5 cm

Question. A ladder 25m long reaches a window of a building 20m above the ground. Find the distance between foot of the ladder and building
Answer : 15 m

Question. Length and breadth of a rectangle are 24 cm and 10 cm. Find its diagonal.
Answer : 26 cm

Question. What is the position of the point-of concurrence of all the medians in a right-angled triangle
Answer : Inside the Δ

Question. Two sides of a triangle are 10cm and 15 cm. What will be the maximum length of third side?
Answer : Less than 25

Question. Diagonals of a rhombus are 6cm and 8cm. What will be the perimeter of such rhombus?
Answer : 20 cm

Question. Triangle ABC is an isosceles right triangle. Find all the angles of the triangle.
Answer : 90, 45, 45

Question. Three angles of a triangle are x – 10°, x + 20°, 2x – 30°. Find each angle
Answer : 40, 70, 70

Question. A man goes 15m due west and then 8 m due north. How far is he from the starting point?
Answer : 17 m

Question. What is the location of the point of concurrence of altitudes in an obtuse angled triangle?
What is this point of concurrence called?
Answer : Orthocentre outside the Δ

Question. Find out from the given sides of a triangle which will form a right-angled triangle?
(a) 15, 10.25;
(b) 27, 36, 39; 
(c) 15, 36, 39
(d) 7, 24, 25
Answer : (d) 7, 24, 25

Question. In an equilateral triangle, altitudes are drawn. Name the point where altitudes of the triangle meet
Answer : orthocentre

Question. Angles of a triangle are in the ratio 8 : 12 : 16. Find the angles
Answer : 40, 60, 80

Question. A boy is flying a kite with 100 m long thread. The perpendicular height of the kite distance between the boy and the point from which the height of kite is measured.
Answer : 60 m

Question. Length of a diagonal of a square is 18√2m. Find the perimeter of the square
Answer : 72

Question. (i) Incentre of an obtuse-angled triangle lies in ............... of triangle.
(ii) Orthocentre of an obtuse-angled triangle lies in ............... of triangle.
(iii) Median of an obtuse-angled triangle lies in ............... of triangle.
(iv) Perpendicular bisectors of the sides of a triangle meet at ................ .
Answer : (i) Exterior (ii) Exterior, (iii) Interior, (iv) circumcentre

Question. In an isosceles ABC, AB = AC and ∠B = 4∠A. Find the angles of the triangle.
Answer : 20, 80, 80.

Question. A man goes 30 m due east and then 40 m due north. Find his distance from the starting point.
Answer : 50 m