RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions

Exercise 17A

Question 1. Construct a perpendicular to a given line AB at a given point P on it.
Answer: Steps of construction:
1. Draw a line AB.
2. Take a point Q on AB and a point P outside AB, and join PQ.
3. With Q as the centre and any radius, draw an arc to cut AB at X and PQ at Z.
4. With P as the centre and the same radius, draw an arc cutting QP at Y.
5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.
6. Join PE and produce it on both the sides to get the required line.

Exam Tip: Ensure that all arc radii are kept equal where specified, and verify that the final line PE is perpendicular to AB by checking that the arcs intersect accurately.

Question 2. Draw a line parallel to a given line AB at a distance of 3.5 cm from it.
Answer: Steps for construction:
1. Let AB be the given line.
2. Take any two points P and Q on AB.
3. Construct \( \angle BPE = 90° \) and \( \angle BQF = 90° \)
4. With P as the centre and the radius equal to 3.5 cm, cut PE at R.
5. With Q as the centre and the radius equal to 3.5 cm, cut QF at S.
6. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3.5 cm from it.

Exam Tip: Keep the perpendicular distances equal at both points P and Q to ensure the constructed line is truly parallel to the original line.

Question 3. Draw a line parallel to a given line l at a distance of 4.3 cm from it.
Answer: Steps of construction:
1. Let l be the given line.
2. Take any two points A and B on line l.
3. Construct \( \angle BAE = 90° \) and \( \angle ABF = 90° \)
4. With A as the centre and the radius equal to 4.3 cm, cut AE at C.
5. With B as the centre and the radius equal to 4.3 cm, cut BF at D.
6. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4.3 cm from it.

Exam Tip: Double-check that the perpendicular distances from both A and B are exactly equal to maintain parallelism accurately.

Exercise 17B

Question 2. Construct the bisector of angle P where \( \angle P = 60° \) and \( \angle Q = 53° \) and PQ = 6 cm.
Answer: Steps of construction:
1. Draw a line segment QR of length 6 cm.
2. Draw arcs of 4.4 cm and 5.3 cm from Q and R, respectively. They intersect at P.
3. Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively.
4. With S as the centre and the radius more than half of ST, draw an arc.
5. With T as the centre and the same radius, draw another arc cutting the previously drawn arc at X.
6. Join P and X. Then, PX is the bisector of \( \angle P \).

Exam Tip: Ensure that the arcs drawn from S and T intersect clearly above or below the angle to create a distinct bisector line.

Question 4. Construct a triangle ABC where BC = 5.3 cm, \( \angle B = 60° \), and \( \angle C = 50° \).
Answer: Steps of construction:
1. Draw BC = 5.3 cm
2. Draw an arc of radius 4.8 cm from the centre B.
3. Draw another arc of radius 4.8 cm from the centre C.
4. Both of these arcs intersect at A.
5. Join AB and AC.
6. With A as the centre and any radius, draw an arc cutting BC at M and N.
7. With M as the centre and the radius more than half of MN, draw an arc.
8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at D.
9. Join AP, cutting BC at D. Then, \( AD \perp BC \)

Exam Tip: Verify that the angle measurements at B and C sum with the third angle to equal 180°, confirming the triangle is properly constructed.

Question 5. Construct a triangle ABC such that AB = 3.8 cm, \( \angle BAZ = 60° \), and AC = 5 cm.
Answer: Steps of construction:
1. Draw AB of length 3.8 cm.
2. Draw \( \angle BAZ = 60° \)
3. With the centre as A, cut ray AZ at 5 cm at C.
4. Join BC. Then, ABC is the required triangle.

Exam Tip: Ensure that the angle is measured accurately from AB, and the arc from A cuts the ray at exactly 5 cm to form a valid triangle.

Question 6. Construct a triangle ABC where AC = 6 cm, \( \angle ACB = 45° \), and AB = 4.3 cm at point B.
Answer: Steps of construction:
1. Draw AC = 6 cm
2. Draw \( \angle ACZ = 45° \)
3. With C as the centre, cut ray BZ at 4.3 cm at point B.
4. Join AB. Then, ABC is the required triangle.

Exam Tip: Verify the angle measurement at C and confirm that the distance AB equals 4.3 cm to validate the construction.

Question 7. Construct a triangle ABC where AB = 5.2 cm, \( \angle BAX = 120° \), and AC = 5.3 cm at point C.
Answer: Steps of construction:
1. Draw AB = 5.2 cm
2. Draw \( \angle BAX = 120° \)
3. With A as the centre, cut the ray AX at 5.3 cm at point C.
4. Join BC.
5. With A as the centre and any radius, draw an arc cutting BC at M and N.
6. With M as the centre and the radius more than half of MN, draw an arc.
7. With N as the centre and the same radius as before, draw another arc cutting the previously drawn arc at P.

8. Join AP meeting BC at D.
\( \therefore AD \perp BC \)

Exam Tip: The perpendicular from A to BC is an important feature - ensure the arcs from M and N intersect clearly to establish this perpendicularity.

Question 8. Construct a triangle ABC where BC = 6.2 cm, \( \angle BCX = 45° \), and \( \angle CBY = 60° \).
Answer: Steps of construction:
1. Draw BC = 6.2 cm
2. Draw \( \angle BCX = 45° \)
3. Draw \( \angle CBY = 60° \)
4. The ray CX and BY intersect at A. Then, ABC is the required triangle.

Exam Tip: Ensure both angles are drawn on the same side of BC, and extend the rays until they meet to form the third vertex A.

Question 9. 
Answer: By angle sum property:
\( \angle B = 180° - \angle A - \angle C \)
\( = 180° - 45° - 75° \)
\( = 60° \)
Steps of construction:
1. Draw AB = 7 cm
2. Draw \( \angle BAX = 45° \)
3. Draw \( \angle ABY = 60° \)
4. The ray AX and BY intersect at C. Then, ABC is the required triangle.

Exam Tip: Always compute the third angle using the angle sum property before starting construction to avoid errors in the geometric drawing.

Question 10. Construct a triangle ABC where BC = 5.8 cm, \( \angle BCY = 30° \), and \( \angle CBX = 30° \).
Answer: Steps of construction:
1. Draw BC = 5.8 cm
2. Draw \( \angle BCY = 30° \)
3. Draw \( \angle CBX = 30° \)
4. The ray BX and CY intersect at A. Then, ABC is the required triangle.
On measuring AB and AC:
\( AB = AC = 3.4 \) cm

Exam Tip: When both base angles are equal, the triangle is isosceles, so the two sides opposite to these angles will be equal in length.

Question 11. Construct a right-angled triangle ABC where BC = 4.8 cm and \( \angle C = 90° \).
Answer: Steps of construction:
1. Draw BC = 4.8 cm
2. Draw a perpendicular on C such that \( \angle C \) is equal to \( 90° \).
3. Draw an arc of radius 6.3 cm from the centre B.
4. Join AB.

Exam Tip: The perpendicular at C ensures a 90° angle; verify by checking that the angle formed at C is a right angle before measuring the hypotenuse.

Question 12. 
Answer: Steps of construction:
1. Draw AB = 3.5 cm
2. Construct \( \angle ABX = 90° \)
3. With centre A, draw an arc of radius 6 cm cutting BX at C.
4. Join AC. Then, ABC is the required triangle.

Exam Tip: Use a set square or compass to ensure the 90° angle at B is accurate, as this determines the right-angled nature of the triangle.

Question 13. Construct a triangle ABC where \( \angle A = 30° \), \( \angle C = 90° \), and the hypotenuse AB = 5.6 cm.
Answer: Here, \( \angle A = 30° \) and \( \angle C = 90° \)
By angle sum property:
\( \angle B = 60° \)
1. Draw the hypotenuse AB of length 5.6 cm.
2. Draw \( \angle BAX = 30° \) and \( \angle ABY = 60° \)
3. The ray AX and BY intersect at C. Then, ABC is the required triangle.

Exam Tip: Remember that the right angle is at C (opposite to the hypotenuse AB), and verify this angle measures exactly 90° after construction.

Exercise 17C

Question 1. Find the supplement of 45°.
Answer: (c) 135°

Supplement of \( 45° = 180° - 45° = 135° \)
In simple words: Two angles whose sum is 180° are called supplementary angles. Subtract 45° from 180° to find its supplement, which is 135°.

Exam Tip: Always remember that supplementary angles add up to 180°, so subtract the given angle from 180° to find its supplement.

Question 2. Find the complement of 80°.
Answer: (b) 10°

Complement of \( 80° = 90° - 80° = 10° \)
In simple words: Two angles whose sum is 90° are called complementary angles. Subtract 80° from 90° to find its complement, which is 10°.

Exam Tip: Complementary angles always add up to 90°, so subtract the given angle from 90° to find its complement.

Question 3. An angle is equal to its complement. Find the angle.
Answer: (b) 45°

Suppose the angle is \( x \).
Then, the complement is also \( x \).
Complement of \( x = 90° - x \)
\( \Rightarrow x = 90° - x \)
\( \Rightarrow x + x = 90° \)
\( \Rightarrow 2x° = 90° \)
\( \Rightarrow x = \frac{90}{2} \)
\( \Rightarrow x = 45 \)
In simple words: If an angle equals its own complement, both parts must be the same size, so each is half of 90°, which gives 45°.

Exam Tip: Set up the equation where the angle equals 90° minus itself, then solve to find that the angle must be 45°.

Question 4. 
Answer: Suppose the angle is \( x \).
\( x = \frac{180 - 2}{5} \)
\( \Rightarrow 5x = 180 - x \)
\( \Rightarrow 5x + x = 180 \)
\( \Rightarrow x = \frac{180}{6} \)
\( \Rightarrow x = 30° \)

Exam Tip: When an angle is expressed as a fraction of its supplement, set up an algebraic equation and solve to find the exact angle measurement.

Question 5. An angle is 24° more than its supplement. Find the angle.
Answer: (b) 57°

Suppose the angle is x.
\( x = 90 - x + 24 \)
\( \Rightarrow x + x = 114 \)
\( \Rightarrow 2x = 114 \)
\( \Rightarrow x = \frac{114}{2} \)
\( \Rightarrow x = 57° \)
In simple words: Set up an equation where the angle is 24° more than its supplement. Add the angle and its supplement to get 180°, then solve for the angle.

Exam Tip: Write the relationship as an equation: angle = supplement + 24°, then use the fact that angle + supplement = 180° to solve.

Question 6. An angle is 32° less than its supplement. Find the angle.
Answer: (b) 74°

Suppose the angle is x.
\( x = 180 - x - 32 \)
\( \Rightarrow x + x = 148 \)
\( \Rightarrow 2x = 148 \)
\( \Rightarrow x = \frac{148}{2} \)
\( \Rightarrow x = 74° \)
In simple words: If an angle is 32° less than its supplement, then the angle plus 32° gives the supplement. Since angle + supplement = 180°, you can solve for the angle.

Exam Tip: Express the relationship as angle = supplement - 32°, then combine with angle + supplement = 180° to find the solution.

Question 7. Two supplementary angles differ by 36°. Find the angles.
Answer: (c) 72°

Supplementary angles:
\( 3x + 2x = 180 \)
\( \Rightarrow x = \frac{180}{5} \)
\( \Rightarrow x = 36° \)
Smaller angle = \( (2 \times 36°) \)
\( = 72° \)
In simple words: Let the two angles differ by a certain amount and add up to 180°. Solve both conditions together to get the individual angles.

Exam Tip: Use two equations: one for the sum (180°) and one for the difference (36°). Solve the system to find both angles.

Question 8. Two angles form a linear pair. If one angle is 132°, find the other.
Answer: (b) 48°

\( \angle AOC + \angle BOC = 180° \) (linear pair)
\( \angle AOC = 180° - \angle BOC \)
\( = 180° - 132° \)
\( = 48° \)
In simple words: When two angles form a linear pair, they add up to 180°. Subtract the known angle from 180° to get the other angle.

Exam Tip: Linear pair angles are supplementary, so always remember that their sum equals 180°.

Question 9. Two angles form a linear pair. If one angle is 68°, find the other.
Answer: (x) 112

\( \angle AOC + \angle AOR = 180° \) (linear pair)
\( 68° + x° = 180° \)
\( \Rightarrow x° = 180° - 68° \)
\( \Rightarrow x° = 112° \)
In simple words: Linear pair angles always sum to 180°. Subtract 68° from 180° to find the missing angle, which is 112°.

Exam Tip: For linear pair problems, always apply the property that the two angles sum to a straight angle (180°).

Question 10. One angle is 5 times the other. Find the two angles if they form a linear pair.
Answer: (c) x = 35

\( (2x - 10) + (3x + 15) = 180 \)
\( \Rightarrow 2x - 10 + 3x + 15 = 180 \)
\( \Rightarrow 5x + 5 = 180 \)
\( \Rightarrow 5x = 180 - 5 \)
\( \Rightarrow 5x = 175 \)
\( \Rightarrow x = \frac{175}{5} \)
\( \Rightarrow x = 35 \)
In simple words: If one angle is 5 times another and they form a linear pair (sum to 180°), set up an equation with the two angle expressions and solve for the variable.

Exam Tip: Always verify your answer by checking that the two computed angles indeed sum to 180°.

Question 11. Find x if (2x - 10)° and (3x + 15)° form a linear pair.
Answer: (d) x = 80

\( x + 55 + 45 = 180 \) (linear pair)
\( \Rightarrow x = 180 - 55 - 45 \)
\( \Rightarrow x = 180 - 100 \)
\( \Rightarrow x = 80 \)
In simple words: Since two angles form a linear pair, they sum to 180°. Subtract the known angles from 180° to find x.

Exam Tip: Always arrange the angles on one side and the constant on the other when solving linear pair equations.

Question 12. Find x if x and 5x form a linear pair.
Answer: (a) 100

\( x + y = 180 \) (linear pair)
\( \Rightarrow x + \frac{5}{9} x = 180° \)
\( \Rightarrow 9x - 5 \times 180 \)
\( \Rightarrow x = 100 \)
In simple words: Two angles in a linear pair add to 180°. If one is x and the other is related to x, form the equation and solve for x.

Exam Tip: Check that the two angles computed from x actually sum to 180° to verify your solution.

Question 13. The angles \( \angle AOC \) and \( \angle BOD \) are vertically opposite. If \( \angle AOC = 50° \), find \( \angle BOD \).
Answer: (b) 60°

Here, \( \angle AOC \) and \( \angle BOD \) are vertically opposite angles.
\( \therefore \angle AOC = \angle BOD \)
Given, \( \angle AOC = 50° \)
\( \therefore \angle BOD = 50° \)
In simple words: Vertically opposite angles are always equal. Since angle AOC is 50°, angle BOD is also 50°.

Exam Tip: Remember that vertically opposite angles formed by two intersecting lines are always equal to each other.

Question 14. One angle is 8° more than another. The angles form a linear pair. Find x.
Answer: (a) 32

\( (3x - 8)° + (x + 10)° + 50° = 180° \) (linear pair)
\( \Rightarrow 3x + 52° = 180° \)
\( \Rightarrow 3x + 52° = 180° \)
\( \Rightarrow 3x + 62 = 180° \)
\( \Rightarrow 4x° = 128° \)
\( \Rightarrow x° = 32° \)
\( \therefore x = 32 \)
In simple words: When angles form a linear pair, their sum is 180°. Set up the equation using the given angle expressions and solve for x.

Exam Tip: Always combine like terms and isolate the variable x on one side to solve these linear pair problems.

Question 15. 
Answer: (a) 32

\( (3x - 8)° + (x + 10)° + 50° = 180° \) (linear pair)
\( \Rightarrow 3x + 52° = 180° \)
\( \Rightarrow 3x + 62 = 180° \)
\( \Rightarrow 4x° = 128° \)
\( \Rightarrow x° = 32° \)
\( \therefore x = 32 \)
In simple words: Combine all angle expressions, set their sum equal to 180° (the straight angle), and solve for the unknown variable.

Exam Tip: Verify your answer by substituting x back into the original angle expressions to confirm their sum equals 180°.

Question 16. 
Answer: (a) 32

\( (3x - 8)° + (x + 10)° + 50° = 180° \) (linear pair)
\( \Rightarrow 3x + 52° = 180° \)
\( \Rightarrow 3x + 62 = 180° \)
\( \Rightarrow 4x° = 128° \)
\( \Rightarrow x° = 32° \)
\( \therefore x = 32 \)
In simple words: Add the given angle expressions together. Since they form a linear pair or complete angle, their sum must equal 180° or 360°. Solve for x using this relationship.

Exam Tip: Identify whether the angles form a linear pair (sum to 180°) or a complete angle (sum to 360°) before setting up your equation.

Question 17. Find the angle if its exterior angle is 120°.
Answer: (b) 60°

\( \angle BCA = 180° - 120° \) (linear pair)
\( = 60° \)
\( \angle BAC = 180° - (60° + 70°) \) (angle sum property of triangles)
\( = 50° \)
In simple words: The exterior angle and the interior angle at the same vertex are supplementary (add up to 180°). So subtract 120° from 180° to get 60°.

Exam Tip: Remember that an interior angle and its adjacent exterior angle always form a linear pair, so they sum to 180°.

Question 18. Find x if the sum of angles in the figure is 360°.
Answer: (c) 150°

\( x° + 70° + 50° + 90° = 360° \) (complete angle)
\( \Rightarrow x° = 360° - 210° \)
\( = 150° \)
In simple words: All the angles around a point sum to 360°. Add up the known angles and subtract from 360° to find x.

Exam Tip: When dealing with angles around a point, always remember that their total is 360°, regardless of how many angles there are.

Question 19. Find \( \angle ACB \) if \( \angle BCE = 60° \) and \( \angle BAC = 60° \).
Answer: (c) 70°

Here, \( \angle ACE = \angle BAC = 60° \) [alternate angles]
\( \angle ACB + \angle ACE + \angle DCE = 180° \) (linear pair)
\( \angle ACB = 180° - (50° + 60°) \)
\( = 180° - 110° \)
\( = 70° \)
In simple words: Alternate angles are equal when a transversal crosses two parallel lines. Use this property and the linear pair relationship to solve for angle ACB.

Exam Tip: Identify parallel lines and transversals to apply angle properties like alternate angles, corresponding angles, and co-interior angles.

Question 20. Find angle B in the given figure.
Answer: Since the angles of a triangle sum to 180°, we have \( \angle A + \angle B + \angle C = 180° \). Substituting the known values: \( \angle B = 180° - (65° + 85°) = 180° - 150° = 30° \).

Exam Tip: Always remember the triangle angle sum property - the three interior angles of any triangle add up to exactly 180 degrees.

Question 21. What is the sum of all angles of a quadrilateral?
Answer: The sum of all interior angles of a quadrilateral is 1800°.

Exam Tip: This is a fundamental property - memorize that any quadrilateral's four interior angles always total 1800 degrees.

Question 22. Find the sum of all interior angles of a polygon with 10 sides.
Answer: The sum of all interior angles is 360°.

Exam Tip: Use the polygon angle formula: sum = (n - 2) × 180°, where n is the number of sides.

Question 23. In the given figure, AB and CD are two straight lines intersecting at O. Find angle AOC.
Answer: Draw a parallel line through O and produce AB and CD on R and P, respectively. By the alternate angles property, \( \angle OCD = \angle COQ = 120° \) (alternate angles). Next, \( \angle COS = 180° - 120° = 60° \) (linear pair). Similarly, \( \angle AOQ = \angle BAO = 150° \) (alternate angles). Further, \( \angle AOS = 180° - 150° = 30° \) (linear pair). Therefore, \( \angle AOC = \angle AOS + \angle COS = 30° + 60° = 90° \).

Exam Tip: When lines intersect, use alternate angle properties and linear pair relationships systematically to find unknown angles.

Question 24. In the given figure, AB || CD. Find angle BAC.
Answer: Since \( \angle PAC = \angle ACS = 100° \) (alternate angles), we have \( \angle PAB + \angle BAC = 100° \). Also, \( \angle PAB + \angle BAC = 100° \), so \( \angle BAC = 100° - 60° = 40° \).

Exam Tip: When two lines are parallel, use the alternate angles theorem to relate angles on opposite sides of a transversal.

Question 25. Find the value of x in the given figure.
Answer: Since the angles within the triangle must sum to 180°, and based on the configuration shown, the calculation reveals that \( x = 30 \).

Exam Tip: Always check that your answer makes sense by verifying the angle sum equals 180 degrees for triangles or 360 degrees for quadrilaterals.

Question 26. Which of the following can be a side of a triangle if two sides are 5 cm and 8 cm?
Answer: The third side must satisfy the triangle inequality: the sum of any two sides must exceed the third side. Given sides of 5 cm and 8 cm, the third side must be greater than 3 cm (8 - 5) and less than 13 cm (8 + 5). So the third side must be greater than the difference of the other two sides.

Exam Tip: Apply the triangle inequality theorem - the sum of any two sides must always be greater than the third side.

Question 27. Which statement about a rhombus is correct?
Answer: The diagonals of a rhombus always intersect each other at right angles.

Exam Tip: Remember key properties of quadrilaterals - a rhombus has perpendicular diagonals that bisect each other.

Question 28. In a right angle triangle, two sides are 5 cm and 13 cm. Find the third side.
Answer: Using the Pythagorean theorem in a right triangle: \( AC^2 = AB^2 + BC^2 \). With \( BC^2 = 13^2 - 5^2 = 169 - 25 = 144 \), we get \( BC = \sqrt{144} = 12 \) cm. The length cannot be negative, so BC = 12 cm.

Exam Tip: Apply the Pythagorean theorem correctly - identify which side is the hypotenuse (longest side opposite the right angle).

Question 29. In triangle ABC, angle A = 37° and angle B = 29°. Find angle C.
Answer: The sum of angles in a triangle equals 180°. So \( \angle A + \angle B + \angle C = 180° \), which gives us \( \angle C = 180° - (37° + 29°) = 180° - 66° = 114° \).

Exam Tip: Always use the angle sum property of triangles: add the two known angles and subtract from 180 degrees.

Question 30. Find the value of angle A in the given figure.
Answer: Based on the geometric configuration and properties shown, angle A equals 105°.

Exam Tip: When working with complex figures, identify all parallel lines and transversals to apply angle relationships systematically.

Question 31. If 2 times angle A equals 3 times angle B, and 3 times angle B equals 6 times angle C, find angle B.
Answer: Given the relationships: \( 2\angle A = 3\angle B \) and \( 3\angle B = 6\angle C \). From the second equation, \( \angle C = \frac{\angle B}{2} \). From the first, \( \angle A = \frac{3\angle B}{2} \). In triangle ABC: \( \angle A + \angle B + \angle C = 180° \), which gives us \( \frac{3\angle B}{2} + \angle B + \frac{\angle B}{2} = 180° \). Simplifying: \( \frac{6\angle B}{2} = 180° \), so \( \angle B = 60° \).

Exam Tip: When angles are related by ratios, express all angles in terms of one variable and use the angle sum to solve.

Question 32. If angle A + angle B = 65° and angle B + angle C = 140°, find angle B.
Answer: From the given conditions: \( \angle A + \angle B = 65° \) and \( \angle B + \angle C = 140° \). Subtracting: \( \angle C - \angle A = 75° \). In triangle ABC: \( \angle A + \angle B + \angle C = 180° \). Substituting values of \( \angle A = 65° - \angle B \) and \( \angle C = 140° - \angle B \): we get \( (65° - \angle B) + \angle B + (140° - \angle B) = 180° \). Solving: \( 205° - \angle B = 180° \), so \( \angle B = 25° \).

Exam Tip: When given multiple angle relationships, use substitution carefully to reduce to a single variable equation.

Question 33. Find the angle of the third angle in a triangle where the other two angles are in the ratio 2:3 and their difference is 30°.
Answer: Let the two angles be 2x and 3x. Given their difference is 30°: \( 3x - 2x = 30° \), so \( x = 30° \). The two angles are 60° and 90°. The third angle is \( 180° - 60° - 90° = 30° \).

Exam Tip: When angles are in a given ratio, use the ratio multiplier to set up equations involving the difference or sum.

Question 34. If the angles of a triangle are (3x)°, (2x - 7)°, and (4x - 11)°, find x.
Answer: Since the sum of angles in a triangle is 180°: \( (3x) + (2x - 7) + (4x - 11) = 180° \). Simplifying: \( 9x - 18 = 180° \), so \( 9x = 198° \), giving us \( x = 22° \).

Exam Tip: When angles are given in algebraic form, always combine like terms and solve the resulting equation carefully.

Question 35. In a right angle triangle ABC with right angle at A, AB = 7 cm and AC = 24 cm. Find BC.
Answer: Using the Pythagorean theorem: \( BC^2 = AB^2 + AC^2 = 7^2 + 24^2 = 49 + 576 = 625 \), so \( BC = \sqrt{625} = 25 \) cm. Since length cannot be negative, BC = 25 cm.

Exam Tip: In a right triangle, the hypotenuse is always the longest side - it's opposite the right angle and found using the Pythagorean theorem.

Question 36. A ladder 25 m long is placed against a wall. If the base of the ladder is 20 m away from the wall, find the height of the wall up to which the ladder reaches.
Answer: The ladder, wall, and ground form a right triangle with the ladder as the hypotenuse. Using the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 \). With \( AC = 25 \) m (ladder) and \( AB = 20 \) m (distance from wall): \( 25^2 = 20^2 + BC^2 \), giving us \( BC^2 = 625 - 400 = 225 \), so \( BC = 15 \) m. Since length cannot be negative, the ladder reaches a height of 25 m on the wall.

Exam Tip: Visualize word problems as geometric figures - identify the right angle first, then apply the Pythagorean theorem.

Question 37. Two vertical poles AE and BD stand on level ground. If AB = 12 m, AE = 6 m, and BC = 12 m, find the distance ED between the tops of the poles.
Answer: (a) Since EC = AB = 12 m (because ABCE forms a rectangle) and AE = BC = 6 m (because ABCE is a rectangle), we have DC = BD - AE = 11 - 6 = 5 m. In the right angled triangle ECD: \( ED^2 = EC^2 + DC^2 \) (Pythagorean theorem). So \( ED^2 = 5^2 + 12^2 = 25 + 144 = 160 \), giving us \( ED = \sqrt{160} = 4\sqrt{10} \) m. The length cannot be negative, therefore ED = 13 m.

Exam Tip: For problems involving two structures, identify rectangles formed by equal segments to simplify calculations.

Question 38. In a right isosceles triangle with AC equal to BC and angle C = 90°, where each leg measures 5 cm, find the hypotenuse AB.
Answer: In this right angled isosceles triangle, both equal legs have length 5 cm. Using the Pythagorean theorem: \( AB^2 = AC^2 + BC^2 = 5^2 + 5^2 = 25 + 25 = 50 \), so \( AB = \sqrt{50} = 5\sqrt{2} \) cm.

Exam Tip: For an isosceles right triangle with legs of length a, the hypotenuse always equals \( a\sqrt{2} \) - memorize this shortcut.