RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles

Exercise 15A

Question 1. Find the measure of angle C in a triangle ABC where angle A is 72° and angle B is 63°.
Answer: The total of all angles in a triangle is 180°.

\( \angle A + \angle B + \angle C = 180° \)
\( 72° + 63° + \angle C = 180° \)
\( \angle C = 45° \)

Therefore, angle C measures 45°.
In simple words: Add the two angles you know, then subtract from 180° to find the missing angle.

Exam Tip: Always remember that the sum of the three interior angles of any triangle must equal 180°. Use this property to find any missing angle when two are given.

Question 2. In triangle DEF, angle D is 105° and angle E is 40°. Find angle F.
Answer: The total of all angles in a triangle is 180°.

In triangle DEF:
\( \angle D + \angle E + \angle F = 180° \)
\( 105° + 40° + \angle F = 180° \)
\( \angle F = 180° - (105° + 40°) \)
\( \angle F = 35° \)

Therefore, angle F measures 35°.
In simple words: Add the two given angles and subtract the total from 180° to find the third angle.

Exam Tip: Verify your answer by checking that all three angles sum to 180°. This quick check helps catch arithmetic errors.

Question 3. In triangle XYZ, angle X is 90°, angle Y is 48°. Find angle Z.
Answer: The total of all angles in a triangle is 180°.

In triangle XYZ:
\( \angle X + \angle Y + \angle Z = 180° \)
\( 90° + 48° + \angle Z = 180° \)
\( \angle Z = 180° - 138° = 42° \)

Therefore, angle Z measures 42°.
In simple words: Combine the two known angles, then subtract their sum from 180° to get the third angle.

Exam Tip: When one angle is already 90° (a right angle), finding the other two angles becomes straightforward—they must sum to 90°.

Question 4. The three angles of a triangle are (4x)°, (3x)° and (2x)°. Find the measure of each angle.
Answer: Let the three angles be represented as (4x)°, (3x)°, and (2x)°.

Since the total of all angles in a triangle equals 180°:
\( 4x + 3x + 2x = 180 \)
\( 9x = 180 \)
\( x = 20 \)

Therefore, the three angles measure (4 × 20)° = 80°, (3 × 20)° = 60°, and (2 × 20)° = 40°.
In simple words: Combine all the expressions, solve for x, then substitute back to find each angle's actual measurement.

Exam Tip: When angles are given in terms of an unknown variable, always set their sum equal to 180° and solve for the variable first.

Question 5. A right-angled triangle has one angle of 36°. Find the other acute angle.
Answer: Since it is a right-angled triangle, one angle is 90°. Another angle is given as 36°.

Using the angle sum property:
\( 36° + 90° + x = 180° \)
\( x = 54° \)

Therefore, the other acute angle measures 54°.
In simple words: In a right triangle, one angle is always 90°. Add the given angle and 90°, then subtract from 180° to find the third angle.

Exam Tip: In a right-angled triangle, the two acute angles must always sum to 90°. This property provides a quick way to check your answer.

Question 6. The two acute angles of a right-angled triangle are (2x)° and (x)°. Find the measure of each angle.
Answer: Let the two acute angles measure (2x)° and (x)°.

The total of all angles in a triangle is 180°:
\( 2x + 90 = 180 \)
\( (3x) = 180 - 90 \)
\( (3x) = 90 \)
\( x = 30 \)

Therefore, the angles measure (2 × 30)° = 60° and 30°. The three angles of the triangle are 60°, 30°, and 90°.
In simple words: The two acute angles add up to 90°. Set their sum equal to 90°, solve for x, then find each angle.

Exam Tip: In a right triangle, the two non-right angles always complement each other (add to 90°). Use this to set up your equation quickly.

Question 7. One angle of an isosceles triangle is 100°. Find the other two angles.
Answer: The other two angles are equal. Let each of these angles measure x°.

Since the total of all angles in a triangle equals 180°:
\( x + x + 100 = 180 \)
\( 2x = 80 \)
\( x = 40 \)

Therefore, the equal angles of the triangle each measure 40°.
In simple words: In an isosceles triangle, two angles are always the same size. If one angle is 100°, the remaining two must share the leftover degrees equally.

Exam Tip: Always use the property that an isosceles triangle has two equal angles. This cuts the work in half since you only need to find one unknown value.

Question 8. In an isosceles triangle, the third angle measures x°. The two equal angles each measure (2x)°. Find all three angles.
Answer: Let the third angle be x°. The two equal angles each measure (2x)°.

Since the total of all angles in a triangle equals 180°:
\( 2x + 2x + x = 180 \)
\( 5x = 180 \)
\( x = 36 \)

Therefore, the three angles measure 36°, (2 × 36)° = 72°, and (2 × 36)° = 72°. The angles of the triangle are 36°, 72°, and 72°.
In simple words: Set up an equation with all three angles expressed in terms of x. Combine like terms and solve for x, then substitute back to find each angle.

Exam Tip: When angles are given as algebraic expressions in an isosceles triangle, express all angles in terms of the same variable and use the sum property to solve.

Question 10. In triangle ABC, angle A equals angle B and angle C is 90°. Given that 2∠A = 3∠B = 6∠C = x. Find all three angles.
Answer: Given that 2∠A = 3∠B = 6∠C = x.

From this relationship:
\( \angle A = \left(\frac{x}{2}\right)° \)
\( \angle B = \left(\frac{x}{3}\right) \) and \( \angle C = \left(\frac{x}{6}\right) \)

Since the total of all angles in a triangle equals 180°:
\( \angle A + \angle B + \angle C = 180° \)
\( \frac{x}{2} + \frac{x}{3} + \frac{x}{6} = 180° \)
\( \frac{3x + 2x + x}{6} = 180° \)
\( \frac{6x}{6} = 180° \)
\( x = 180 \)

Therefore:
\( \angle A = \left(\frac{180}{2}\right)° = 90° \)
\( \angle B = \left(\frac{180}{3}\right)° = 60° \)
\( \angle C = \left(\frac{180}{6}\right)° = 30° \)

The three angles measure 90°, 60°, and 30°.
In simple words: Express each angle using the given ratio, add them together and set equal to 180°. Solve for x, then calculate each angle by substituting x back.

Exam Tip: When angles are related by a common variable through a ratio, find a common denominator for fractions before solving. This eliminates arithmetic errors.

Question 11. An equilateral triangle has all angles equal. Find the measure of each angle.
Answer: We know that all angles of an equilateral triangle are equal. Let the measure of each angle be x°.

\( x + x + x = 180 \)
\( 3x = 180 \)
\( x = 60 \)

Therefore, the measure of each angle of an equilateral triangle is 60°.
In simple words: All three angles are identical in an equilateral triangle. Since they must sum to 180°, divide 180° by 3 to find each angle.

Exam Tip: The 60° angle measure in an equilateral triangle is a fundamental property to memorize—it appears frequently in geometry problems.

Question 12. In triangle ABC, line DE is parallel to BC. (i) If ∠ABC = ∠ADE = 55°, what can you conclude about angles? (ii) Find ∠ACB. (iii) Show that ∠AED = ∠ACB.
Answer: (i) Since DE || BC, the angles ∠ABC and ∠ADE are corresponding angles, so ∠ABC = ∠ADE = 55°.

(ii) The total of all angles in a triangle is 180°.

\( \angle A + \angle B + \angle C = 180° \)
\( \angle C = 180° - (65° + 55°) = 60° \)

Therefore, ∠ACB = 60°.

Since DE || BC:
\( \angle AED = \angle ACB = 60° \)

(iii) We have found in part (ii) that ∠C = 60°. Since ∠AED = ∠ACB = 60° (corresponding angles when DE || BC).
In simple words: When one line is parallel to another, matching angles are equal. Use this fact along with the angle sum property to find missing angles and verify relationships.

Exam Tip: When parallel lines and a transversal are involved, identify which angles are corresponding, alternate, or co-interior—these relationships speed up the solution.

Question 14. (i) Can a triangle have all angles less than 60°? (ii) Can a right triangle have all sides of different lengths? (iii) Can an equilateral triangle be right-angled? (iv) Can a triangle have two right angles? (v) Can a triangle have two obtuse angles? (vi) Can a triangle with one obtuse angle of 120° and two other angles of 30° each be an equilateral triangle?
Answer: (i) No. This is not possible because the total of all angles is always 180°. If all three angles were less than 60°, their sum would be less than 180°, which violates the angle sum property.

(ii) Yes. A right triangle can have all sides of different measurements. For instance, a triangle with sides 3, 4, and 5 units is a right triangle where all sides differ.

(iii) No. An equilateral triangle cannot be a right-angled triangle since the square of the hypotenuse cannot equal the sum of the squares of the other two sides in an equilateral triangle.

(iv) No. The total of all angles cannot exceed 180°. If two angles are right angles (each 90°), their sum alone equals 180°, leaving no room for a third angle.

(v) No. An obtuse angle is larger than 90°. If a triangle had two obtuse angles, their sum would exceed 180°, which is impossible since one angle has to be more than 60° as the total of all angles is always 180°.

(vi) Yes. A triangle with an obtuse angle of 120° and two other angles of 30° each will be an isosceles triangle since two angles are equal.
In simple words: Test each statement against the angle sum property (angles total 180°), properties of special triangles (equilateral, right-angled, isosceles), and the definitions of acute and obtuse angles.

Exam Tip: These true-or-false questions test deep understanding of triangle properties. Always justify your answer using a specific property or by showing a contradiction.

Question 15. Fill in the blanks. (i) In a right triangle, the angle opposite the hypotenuse is ___. (ii) Each acute angle in a right triangle is ___. (iii) Each acute angle in a right triangle equals ___. (iv) In a right triangle, the third angle measures ___. (v) The longest side of a right triangle is called the ___. (vi) The perimeter of a triangle is ___.
Answer: (i) obtuse (since the sum of the other two angles of the right triangle is 90°)

(ii) equal to the sum of 90°

(iii) 45° (since their sum is equal to 90°)

(iv) 60°

(v) a hypotenuse

(vi) perimeter
In simple words: Review each property of right triangles and special triangles to complete each statement accurately.

Exam Tip: Memorize key triangle terminology (hypotenuse, acute angle, obtuse angle) and properties to answer fill-in-the-blank questions confidently.

Exercise 15B

Question 1. In triangle ABC, angle ACD is an exterior angle where angle CAB = 75° and angle CBA = 45°. Find angle ACD.
Answer: We know that an exterior angle of a triangle equals the sum of the two interior opposite angles.

\( \angle ACD = \angle CAB + \angle CBA \)
\( \angle ACD = 75° + 45° = 120° \)

Therefore, angle ACD measures 120°.
In simple words: An exterior angle at any vertex of a triangle always equals the sum of the two interior angles that are not adjacent to it.

Exam Tip: The exterior angle property is a powerful shortcut - you don't need to find the third interior angle first. Use it directly to find exterior angles quickly.

Question 2. In triangle ABC, the exterior angle at C measures 130°. If angle BAC = 62° and angle ABC = 68°, find angle ACB.
Answer: We know that an exterior angle of a triangle equals the sum of the two interior opposite angles.

\( \angle BAC + \angle ABC = \angle ACD \)
\( x + 68 = 130 \)
\( x = 62 \)

The total of the angles in a triangle is 180°.

\( \angle BAC + \angle ABC + \angle ACB = 180° \)
\( 62 + 68 + y = 180 \)
\( y = 50 \)

Therefore, angle ACB measures 50°.
In simple words: First use the exterior angle property to confirm the interior opposite angles sum to the exterior angle. Then use the angle sum property to find the remaining interior angle.

Exam Tip: Set up two equations - one using the exterior angle property and another using the 180° angle sum rule. Both must be satisfied for a valid answer.

Question 3. The exterior angle at vertex C of triangle ABC is 65°. If angle BAC = 32° and angle ABC = y, find the values of x and y.
Answer: We know that an exterior angle of a triangle equals the sum of the two interior opposite angles.

\( \angle BAC + \angle CBA = \angle ACD \)
\( 32 + x = 65 \)
\( x = 33 \)

Also, the total of the angles in a triangle is 180°.

\( \angle BAC + \angle CBA + \angle ACB = 180° \)
\( 32 + 33 + y = 180 \)
\( y = 115 \)

Therefore, x = 33 and y = 115.
In simple words: Use the exterior angle theorem to find one angle, then apply the angle sum property to find the second angle.

Exam Tip: Notice that an interior angle and its adjacent exterior angle are supplementary (sum to 180°). This provides another way to verify your answers.

Question 4. The two interior opposite angles of an exterior angle measuring 110° are (2x)° and (3x)°. Find all three angles of the triangle and the value of x.
Answer: Suppose the two interior opposite angles are (2x)° and (3x)°. We know that an exterior angle of a triangle equals the sum of the interior opposite angles.

\( 3x + 2x = 110 \)
\( 5x = 110 \)
\( x = 22 \)

The interior opposite angles are (2 × 22)° = 44° and (3 × 22)° = 66°. Let the third angle of the triangle be y°.

The total of the angles in a triangle is 180°:

\( 44 + 66 + y = 180 \)
\( y = 70 \)

Therefore, the angles of the triangle are 44°, 66°, and 70°.
In simple words: The exterior angle equals the sum of the two non-adjacent interior angles. Set this sum equal to the exterior angle value and solve for x.

Exam Tip: When two interior angles are expressed in terms of a variable, setting their sum equal to the exterior angle gives you a direct equation to solve - this is faster than finding all three interior angles first.

Question 5. The two interior opposite angles of an exterior angle measuring 100° are x° and x°. Find the three angles of the triangle.
Answer: Suppose the two interior opposite angles of an exterior angle measuring 100° are x° and x°. We know that an exterior angle of a triangle equals the sum of the interior opposite angles.

\( x + x = 100 \)
\( 2x = 100 \)
\( x = 50 \)

Also, the total of the angles in a triangle is 180°. Let the measure of the third angle be y°.

\( x + x + y = 180 \)
\( 50 + 50 + y = 180 \)
\( y = 80 \)

Therefore, the angles measure 50°, 50°, and 80°.
In simple words: When the two interior opposite angles are equal, set their sum equal to the exterior angle. Then find the third angle using the 180° property.

Exam Tip: When you find two equal interior angles, you immediately know the triangle is isosceles. Use this insight to check your work.

Question 6. In triangle ABC, an exterior angle at C is formed. In triangle ABC, angle ACD = 70°. In triangle ECD, angle AED = 70° + 40° = 110°. Find angle ACD in triangle ABC and angle AED in triangle ECD.
Answer: We know that the exterior angle of a triangle equals the sum of the interior opposite angles.

In triangle ABC:
\( \angle ACD = \angle BAC + \angle ABC = 25° + 45° = 70° \)
\( \angle ACD = 70° \)

(ii) In triangle ECD:
\( \angle AED = \angle ECD + \angle EDC = 70° + 40° = 110° \)
\( \angle AED = 110° \)

Therefore, angle ACD measures 70° and angle AED measures 110°.
In simple words: Apply the exterior angle property to each triangle separately. The exterior angle at any vertex equals the sum of the two remote interior angles.

Exam Tip: When dealing with multiple triangles, carefully identify which angles belong to which triangle to avoid mixing up the exterior angle relationships.

Question 7. (i) In triangle ABC, find angle BAC given angles at B and C. (ii) In triangle ACD, find angle AED and related angles. (iii) In triangle DAB, find angle DAE.
Answer: The total of the angles in a triangle is 180°.

In triangle ABC:
\( \angle BAC + \angle CBA + \angle ACB = 180° \)
\( \angle BAC = 180° - (40° + 100°) \)
\( \angle BAC = 40° \)

We know that the exterior angle of a triangle equals the sum of the interior opposite angles.

\( \angle ACD = \angle BAC + \angle CBA = 40° + 40° = 80° \)

(i) ∠ACD = 80°

(ii) In triangle ACD:
\( \angle CAD + \angle ACD + \angle ADC = 180° \)
\( \angle ADC = 180° - (50° + 80°) \)
\( \angle ADC = 50° \)
\( \angle DAC = 50° \)

(iii) Since BE is a straight line:
\( \angle DAB + \angle DAE = 180° \)
\( \angle DAE = 180° - (\angle DAC + \angle CAB) \)
\( \angle DAE = 180° - (50° + 40°) \)
\( \angle DAE = 90° \)

Therefore, angle DAE measures 90°.
In simple words: Use the angle sum property to find unknown angles in each triangle. Apply the exterior angle property and straight line angle property as needed.

Exam Tip: For complex diagrams with multiple triangles, work systematically through each triangle separately. Track which angles you have found and which remain to be found.

Question 8. In a triangle, the ratio of one angle to another is given by x/y = 3/4. An exterior angle equals 130°. Find all three angles and the values of x, y, and z.
Answer: \( \frac{x}{y} = \frac{3}{4} \)

\( \Rightarrow 3x = 2y \)
\( \Rightarrow x = \frac{2}{3}y \)

We know that the exterior angle of a triangle equals the sum of the interior opposite angles.

\( x + y = 130° \)

\( \Rightarrow \frac{2}{3}y + y = 130 \)
\( \Rightarrow 5y = 130 \times 3 \)
\( \Rightarrow 5y = 390 \)
\( \Rightarrow y = 78 \)
\( \Rightarrow x = \frac{2}{3} \times 78 \)
\( \Rightarrow x = 52 \)

Also, the total of the angles in a triangle is 180°:
\( x + y + z = 180 \)
\( z = 180 - 78 - 52 \)
\( z = 50 \)

Therefore, x = 52, y = 78, and z = 50. The three angles are 52°, 78°, and 50°.
In simple words: Use the given ratio to express one variable in terms of another. Substitute into the exterior angle equation to solve for the unknowns.

Exam Tip: When a ratio is given, express it as an equation and solve the system using substitution. Always verify your answer satisfies both the ratio and the angle sum property.

Exercise 15C

Question 1. Which of the following sets of lengths can form the sides of a triangle? (i) 1, 1, 1 (ii) 2, 3, 4 (iii) 7, 8, 15 (iv) 3.4, 2.1, 5.3 (v) 6, 7, 14
Answer: (i) Consider measurements 1, 1, and 1.

Clearly: 1 + 1 > 1, 1 + 1 > 1, 1 + 1 > 1

The sum of any two sides is greater than the third side. Therefore, a triangle with sides 1 cm, 1 cm, and 1 cm can be formed.

(ii) Clearly: 2 + 3 > 4, 3 + 4 > 2

The sum of any two sides is greater than the third side. Therefore, a triangle with sides 2 cm, 3 cm, and 4 cm can be formed.

(iii) Clearly: 7 + 8 = 15

The sum of these two numbers is not greater than the third number. Therefore, it is not possible to form a triangle with sides 7 cm, 8 cm, and 15 cm.

(iv) Consider measurements 3.4, 2.1, and 5.3.

Clearly: 3.4 + 2.1 > 5.3, 5.3 + 2.1 > 3.4, 5.3 + 3.4 > 2.1

The sum of any two sides is greater than the third side. Therefore, a triangle with sides 3.4 cm, 2.1 cm, and 5.3 cm can be formed.

(v) Consider measurements 6, 7, and 14.

Clearly: 6 + 7 = 13 < 14

The sum of these two numbers is not greater than the third number. Therefore, it is not possible to form a triangle with sides 6 cm, 7 cm, and 14 cm.
In simple words: For any triangle, the sum of any two sides must be greater than the third side. Check this rule for each set of lengths.

Exam Tip: Always verify the triangle inequality for all three possible pairs of sides. If even one pair fails the test, the triangle cannot be formed.

Question 2. Two sides of a triangle measure 5 cm and 9 cm. Find the range of possible lengths for the third side.
Answer: Let the length of the third side be x cm.

The sum of any two sides of a triangle is greater than the third side.

\( 5 + 9 > x \)
\( x < 14 \)

Therefore, the length of the third side must be less than 14 cm.
In simple words: Apply the triangle inequality rule. Add the two known sides and set the sum as an upper limit for the third side.

Exam Tip: The third side of a triangle must satisfy two inequalities simultaneously - it must be less than the sum of the other two sides AND greater than their difference.

Question 3. Use appropriate symbols (<, >, or =) to complete the following. (i) ___ (ii) ___ (iii) ___
Answer: (i) >

(ii) >

(iii) <

The reason for the above three results is that the sum of any two sides of a triangle is greater than the third side.
In simple words: The triangle inequality states that the sum of any pair of sides must exceed the third side. Apply this rule to determine which symbol fits.

Exam Tip: Master the triangle inequality thoroughly - it is one of the most tested properties and appears in many variations.

Question 4. Prove that the sum of any two sides of a triangle is greater than the third side.
Answer: The sum of any two sides of a triangle is greater than the third side.

In triangle AMB:
\( AB + BM > AM \) ... (i)

In triangle AMC:
\( AC + CM > AM \) ... (ii)

Adding the above two equations:
\( AB + BM + AC + CM > AM + AM \)
\( AB + BC + AC > 2AM \)
\( AB + BC + AC > 2AM \)

Therefore, proved.
In simple words: Use auxiliary constructions and apply the triangle inequality to simpler triangles, then combine the results to establish the general property.

Exam Tip: For proof-based questions, clearly state what you are proving and show each logical step. Label any auxiliary constructions (like points M in this case) clearly.

Question 5. Point P lies inside triangle ABD. Prove that AB + AC + BC > 2AP.
Answer: The sum of any two sides of a triangle is greater than the third side.

In triangle APB:
\( AD + BP > AP \)

In triangle APC:
\( AC + PC > AP \)

Adding the corresponding sides:
\( AD + BP + AC + PC > AP + AP \)
\( AD + AC + BC > 2AP \)

Therefore, proved.
In simple words: Apply the triangle inequality to triangles formed by the interior point and the vertices, then add the resulting inequalities.

Exam Tip: When a point lies inside a polygon, divide the figure into triangles using that point and apply the triangle inequality to each smaller triangle.

Question 6. Prove that the sum of the sides AB + BC + CD + DA of quadrilateral ABCD is greater than twice the sum AC + BD of the diagonals.
Answer: The sum of any two sides of a triangle is greater than the third side.

In triangle ABC:
\( AB + BC > AC \)

In triangle ACD:
\( CD + DA > AC \)

Adding the above two inequalities:
\( AB + BC + CD + DA > 2AC \) ... (i)

In triangle ABD:
\( AD + AB > BD \)

In triangle BCD:
\( CD + BC > BD \)

Adding the above two:
\( AB + BC + CD + DA > 2BD \) ... (ii)

Adding equations (i) and (ii):
\( AB + BC + CD + DA > 2(AC + BD) \)
\( AB + BC + CD + DA > AC + BD \)

Therefore, proved.
In simple words: Divide the quadrilateral into triangles using both diagonals. Apply the triangle inequality to each triangle and combine results to reach the conclusion.

Exam Tip: When proving geometric inequalities involving quadrilaterals, use the diagonals as auxiliary lines to create triangles where the inequality can be applied.

Question 7. In quadrilateral OAOB, prove that OA + OB + OC > AB + BC + CA.
Answer: We know that the sum of any two sides of a triangle is greater than the third side.

In triangle AOB:
\( OA + OB > AB \) ... (1)

In triangle BOC:
\( OB + OC > BC \) ... (2)

In triangle AOC:
\( OA + OC > CA \) ... (3)

Adding (1), (2) and (3):
\( OA + OB + OB + OC + OA + OC > AB + BC + CA \)
\( 2(OA + OB + OC) > AB + BC + CA \)

Therefore, proved.
In simple words: Apply the triangle inequality to each of the three triangles formed, then add all three inequalities together.

Exam Tip: When you add multiple inequalities, the result is also a valid inequality. Use this technique to combine several triangle inequality statements into one final proof.

Exercise 15(D)

Question 1. Find the length of the hypotenuse of a right triangle with legs 9 cm and 12 cm.
Answer: Let the hypotenuse measure a cm.

By Pythagoras theorem:
\( a^2 = 9^2 + 12^2 \)
\( \Rightarrow a^2 = 81 + 144 \)
\( \Rightarrow a^2 = 225 \)
\( \Rightarrow a = \sqrt{225} \)
\( \Rightarrow a = 15 \)

The length of the hypotenuse is 15 cm.
In simple words: Square each leg, add them together, then take the square root of the total to find the hypotenuse length.

Exam Tip: Always verify your answer by checking that the sum of the squares of the two shorter sides equals the square of the longest side.

Question 2. Find the missing side of a right triangle with hypotenuse 26 cm and one leg 10 cm.
Answer: Let the other side measure a cm.

By Pythagoras theorem:
\( 26^2 = 10^2 + a^2 \)
\( \Rightarrow a^2 = 676 - 100 \)
\( \Rightarrow a^2 = 576 \)
\( \Rightarrow a = \sqrt{576} \)
\( \Rightarrow a = 24 \)

The length of the other side is 24 cm.
In simple words: Subtract the square of the known leg from the square of the hypotenuse, then find the square root to get the missing leg.

Exam Tip: Remember that the hypotenuse is always the longest side in a right triangle - it is the side opposite the right angle.

Question 3. Find the missing side of a right triangle with sides 4.5 cm and 7.5 cm (where 7.5 cm is the hypotenuse).
Answer: Let the other side measure a cm.

By Pythagoras theorem:
\( 4.5^2 + a^2 = 7.5^2 \)
\( \Rightarrow a^2 = 56.25 - 20.25 \)
\( \Rightarrow a^2 = 36 \)
\( \Rightarrow a = \sqrt{36} \)
\( \Rightarrow a = 6 \)

The length of the other side of the triangle is 6 cm.
In simple words: When you know the hypotenuse and one leg, subtract the square of the known leg from the square of the hypotenuse, then take the square root.

Exam Tip: Check that your calculated leg is shorter than the hypotenuse - if it is not, you have made an error.

Question 4. The two legs of a right triangle are equal, and their sum is 50 cm. Find the length of each leg.
Answer: Let the length of each leg be a cm and another leg a cm (since both are equal).

By Pythagoras theorem:
\( a^2 + a^2 = 50 \)
\( \Rightarrow 2a^2 = 50 \)
\( \Rightarrow a^2 = 25 \)
\( \Rightarrow a = \sqrt{25} \)
\( \Rightarrow a = 5 \)

The length of each leg is 5 cm.
In simple words: When both legs are equal and add up to 50, divide by 2 to find that each leg is 25 when squared. Take the square root to get 5 cm.

Exam Tip: In isosceles right triangles, both legs are equal - this simplifies the calculation significantly.

Question 5. Show that a triangle with sides 15 cm, 36 cm, and 39 cm is a right triangle.
Answer: The largest side of the triangle is 39 cm.

\( 15^2 + 36^2 = 225 + 1296 = 1521 \)

Also, \( 39^2 = 1521 \)

\( \therefore 15^2 + 36^2 = 39^2 \)

The sum of the squares of the two sides equals the square of the third side. Therefore, the triangle is right angled (by the Converse of Pythagoras Theorem).
In simple words: When you square the two smaller sides and their sum matches the square of the biggest side, the triangle must have a 90-degree angle.

Exam Tip: Use the Converse of Pythagoras Theorem to prove right angles - always square the sides and compare.

Question 6. Determine whether a triangle with sides 6 cm, 8 cm, and 10 cm is right angled or not (largest side is the hypotenuse).
Answer: We need to check if the triangle with sides 6 cm, 8 cm, and 10 cm satisfies the Pythagoras relation.

(i) When largest side c = 25 cm:

\( a^2 + b^2 = 225 + 400 = 625 \)

Also, \( c^2 = 625 \)

\( \therefore a^2 + b^2 = c^2 \)

The given triangle is right angled using Pythagoras theorem.

(ii) When largest side c = 16 cm:

\( a^2 + b^2 = 81 + 144 = 225 \)

Also, \( c^2 = 256 \)

\( a^2 + b^2 \neq c^2 \)

Therefore, the given triangle is not right angled.
In simple words: For a right triangle, the sum of squares of the two smaller sides must equal the square of the longest side. If this relationship holds, it is right angled; otherwise it is not.

Exam Tip: Always identify the longest side first - it is the potential hypotenuse in a right triangle.

Question 7. Show that a triangle with sides 9 cm and 12 cm (right angle between them) has a hypotenuse of 15 cm.
Answer: Let the hypotenuse be c cm. The two legs are 9 cm and 12 cm.

By Pythagoras theorem:
\( a^2 + b^2 = c^2 \)
\( \Rightarrow 9^2 + 12^2 = c^2 \)
\( \Rightarrow 81 + 144 = c^2 \)
\( \Rightarrow c^2 = 225 \)
\( \Rightarrow c = 15 \)

The hypotenuse is 15 cm. (This is a Pythagorean triple: 9, 12, 15 or 3-4-5 scaled by 3)
In simple words: Add the squares of both legs: 81 plus 144 gives 225. The square root of 225 is 15, which is the hypotenuse.

Exam Tip: Learn common Pythagorean triples like 3-4-5, 5-12-13, and 8-15-17 to check answers quickly.

Question 8. In a right triangle ABC, angle B = 35° and angle C = 55°. Verify that angle A = 90°.
Answer: We are given:\
\( \angle B = 35° \) and \( \angle C = 55° \)

\( \therefore \angle A = 180° - 35° - 55° = 90° \) (since the sum of angles in any triangle is 180°)

We know that the side opposite to the right angle is the hypotenuse.

By Pythagoras theorem:
\( BC^2 = AB^2 + AC^2 \)

Hence, (iii) is true.
In simple words: Since angles B and C add to 90 degrees, angle A must be 90 degrees. The side opposite angle A (which is BC) becomes the hypotenuse.

Exam Tip: The side opposite a right angle is always the hypotenuse - the longest side in the triangle.

Question 9. A ladder 15 m long leans against a wall. Its foot is at a distance of 9 m from the wall. Find how far the ladder reaches up the wall.
Answer: By Pythagoras theorem in triangle ABC:
\( AB^2 = AC^2 + BC^2 \)
\( 15^2 = x^2 + 9^2 \)
\( 225 = x^2 + 81 \)
\( \Rightarrow x^2 = 225 - 81 = 144 \)
\( \Rightarrow x = 12 \)

The distance of the foot of the ladder from the wall is 9 cm.
In simple words: The ladder, wall, and ground form a right triangle. Use Pythagoras theorem: square the ladder length, subtract the square of the ground distance, then take the square root.

Exam Tip: Always identify the hypotenuse first - in ladder problems, it is the ladder itself since it is longest.

Question 10. A ladder 5 m long is leaning against a wall. The foot of the ladder is 4.8 m from the wall. How far up the wall does the ladder reach?
Answer: Let the foot of the ladder be x m from the wall.

Let the ladder be represented by AB. The height at which it reaches the wall is AC. The distance between the foot of ladder and wall is BC.

By Pythagoras theorem:
\( AB^2 = AC^2 + BC^2 \)
\( \Rightarrow 5^2 = 4.8^2 + x^2 \)
\( \Rightarrow x^2 = 25 - 23.04 \)
\( \Rightarrow x^2 = 1.96 \)
\( \Rightarrow x^2 = \left(1.4\right)^2 \)
\( \Rightarrow x = 1.4 \)

The foot of the ladder is 1.4 m from the wall.
In simple words: The ladder and wall create a right triangle. Square the ladder length, subtract the square of the distance from the wall, then take the square root to find how high it reaches.

Exam Tip: In real-world problems involving ladders, walls, and ground, always form a right angle at the base of the wall.

Question 11. A 15 m tall tree is broken at a point 9 m above the ground. The broken part bends down and its top touches the ground. How far from the base of the tree does the top touch the ground?
Answer: Let BD be the height of the tree broken at point C. Suppose CD takes the position CA.

Now, from the given conditions: AB = 9 m, BC = 12 m

By Pythagoras theorem in triangle ABC:
\( AC^2 = AB^2 + BC^2 \)
\( \Rightarrow AC^2 = 12^2 + 9^2 \)
\( \Rightarrow AC^2 = 144 + 81 \)
\( \Rightarrow AC^2 = 225 \)
\( \Rightarrow AC^2 = 15^2 \)
\( \Rightarrow AC = 15 \)

Length of the tree before it broke = AC + AB = 15 + 9 = 24 m
In simple words: The unbroken part of the tree is 9 m tall. The broken part of length 15 m bends to touch the ground. Use Pythagoras theorem to find the horizontal distance.

Exam Tip: In these problems, the bent part (hypotenuse) length remains the same, and it forms a right triangle with the unbroken part and the ground distance.

Question 12. Two poles 18 m and 13 m tall stand vertically on the ground. The distance between them is 12 m. Find the distance between their tops.
Answer: Let the two poles be AB and CD, having length 18 m and 13 m respectively. Distance between them, BD, is equal to 12 m.

From C, draw CE - AB.

AE = AB - EB
= AB - CD (CD = EB)
= 18 - 13
= 5 cm
EC = BD = 12 m

Now, by Pythagoras theorem in triangle AEC:
\( AC^2 = AE^2 + EC^2 \)
\( \Rightarrow AC^2 = 5^2 + 12^2 \)
\( \Rightarrow AC^2 = 25 + 144 \)
\( \Rightarrow AC^2 = 169 \)
\( \Rightarrow AC^2 = 13^2 \)
\( \Rightarrow AC = 13 \)

The distance between their tops is 13 m.
In simple words: Draw a perpendicular from the top of the shorter pole to the taller pole. This creates a right triangle with height difference 5 m and horizontal distance 12 m. Use Pythagoras theorem to find the distance between tops.

Exam Tip: When poles of different heights are involved, always find the height difference - this becomes one leg of your right triangle.

Question 13. A man starts at point A and walks 35 m west to point B. He then walks 12 m north to point C. Find his distance from the starting point.
Answer: The man starts at point A and travels 35 m towards west, reaching point B. He then travels 12 m north, reaching point C.

We need to find AC.

By Pythagoras theorem:
\( AC^2 = BC^2 + AB^2 \)
\( \Rightarrow AC^2 = 35^2 + 12^2 \)
\( \Rightarrow AC^2 = 1225 + 144 \)
\( \Rightarrow AC^2 = 1369 \)
\( \Rightarrow AC^2 = 37^2 \)
\( \Rightarrow AC = 37 \) m

The man is 37 m from the starting point.
In simple words: The man's two walks form the two legs of a right triangle. Square both distances, add them, then take the square root to find the direct distance from the starting point.

Exam Tip: When movement is described in perpendicular directions (north-south and east-west), always form a right triangle to find the direct distance.

Question 14. A man starts at point A and goes 3 km north, reaching B. Then he travels 4 km east to reach C. How far is he from his starting position?
Answer: Suppose the man starts from A and travels 3 km north, arriving at B. He then travels 4 km towards east, arriving at C.

\( \therefore AB = 3 \) km
\( BC = 4 \) km
We have to find AC.

By Pythagoras theorem:
\( \Rightarrow AC^2 = AB^2 + BC^2 \)
\( \Rightarrow AC^2 = 3^2 + 4^2 \)
\( \Rightarrow AC^2 = 25 \)
\( \Rightarrow AC^2 = 5^2 \)
\( \Rightarrow AC = 5 \) km

He is 5 km from the initial position.
In simple words: Two perpendicular journeys of 3 km and 4 km form the legs of a right triangle. Square each distance, add them to get 25, then take the square root to get 5 km as the direct distance.

Exam Tip: The 3-4-5 Pythagorean triple is very common - recognize it immediately to save calculation time.

Question 15. In a rectangle, the sides are 16 cm and 12 cm respectively. Find the length of the diagonal.
Answer: Suppose the sides are x and y of lengths 16 cm and 12 cm, respectively.

Let the diagonal be z cm.

Clearly, the diagonal is the hypotenuse of the right triangle with legs x and y

By Pythagoras theorem:
\( z^2 = x^2 + y^2 \)
\( \Rightarrow z^2 = 16^2 + 12^2 \)
\( \Rightarrow z^2 = 256 + 144 \)
\( \Rightarrow z^2 = 400 \)
\( \Rightarrow z^2 = 20^2 \)
\( \Rightarrow z = 20 \)

The length of the diagonal is 20 cm.
In simple words: A diagonal of a rectangle splits it into two right triangles. The diagonal becomes the hypotenuse. Use Pythagoras theorem with the two side lengths to find the diagonal.

Exam Tip: In any rectangle, both diagonals are equal in length - calculate one and you have both.

Question 16. The diagonals of a rectangle are 41 cm. One side is 40 cm. Find the other side.
Answer:


AB = 40 cm, Diagonal AC = 41 cm

Then, by Pythagoras theorem in right triangle ABC:
\( AC^2 = AB^2 + BC^2 \)
\( \Rightarrow BC^2 = 41^2 - 40^2 \)
\( \Rightarrow BC^2 = 1681 - 1600 \)
\( \Rightarrow BC^2 = 81 \)
\( \Rightarrow BC^2 = 9^2 \)
\( \Rightarrow BC = 9 \) cm
In simple words: The diagonal is the hypotenuse of a right triangle formed with the two sides. Subtract the square of the known side from the square of the diagonal, then take the square root.

Exam Tip: Always verify: 40² + 9² should equal 41² to confirm your answer is correct.

Question 17. The diagonals of a rhombus are 16 cm and 30 cm. Find the perimeter of the rhombus.
Answer: We know that the diagonals of a rhombus bisect each other at right angles. Therefore, in right triangle AOB, we have:
AO = 8 cm
BO = 15 cm

By Pythagoras theorem in triangle AOB:
\( AB^2 = AO^2 + BO^2 \)
\( \Rightarrow AB^2 = 8^2 + 15^2 \)
\( \Rightarrow AB^2 = 64 + 225 \)
\( \Rightarrow AB^2 = 289 \)
\( \Rightarrow AB^2 = 17^2 \)
\( \Rightarrow AB = 17 \) cm

Now, since all sides of a rhombus are equal:
\( \therefore \) Perimeter of the rhombus = 4(side)
= 4(17)
= 68 cm
In simple words: The rhombus diagonals split it into four right triangles. Each triangle has legs measuring half of each diagonal. Use Pythagoras theorem to find one side, then multiply by 4 for the perimeter.

Exam Tip: Remember - diagonals of a rhombus bisect each other at 90 degrees. This creates four congruent right triangles inside the rhombus.

Question 18. Identify which of the following is true.
(i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
(ii) If the square of one side of a triangle is equal to the sum of the squares of the other two sides then the triangle is right angled.
(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest.
Answer: (i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides - TRUE. This is the Pythagorean theorem.

(ii) If the square of one side of a triangle is equal to the sum of the squares of the other two sides then the triangle is right angled - TRUE. This is the converse of Pythagoras theorem.

(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest - TRUE. This is a well-established geometric principle.
In simple words: All three statements are true. Statement (i) is Pythagoras theorem. Statement (ii) is its converse. Statement (iii) is a basic property about perpendicular distances.

Exam Tip: Both Pythagoras theorem and its converse are crucial for identifying and working with right triangles.