ML Aggarwal Class 7 Maths Solutions Chapter 11 Triangles and its Properties

Exercise 11.1

 

Question 1. (i) k
Answer: k represents a side of the triangle.
In simple words: k is one of the three edges of the triangle.

Exam Tip: When labeling sides, use lowercase letters to denote their lengths.

 

Question 1. (ii) PR
Answer: PR is a side that connects two vertices of the triangle.
In simple words: PR is one of the three lines that form the triangle's boundary.

Exam Tip: Sides are named using their two endpoint vertices.

 

Question 1. (iii) p
Answer: p denotes a side length in the triangle.
In simple words: p is the measurement of one side of the triangle.

Exam Tip: Lowercase letters typically represent side lengths in geometry.

 

Question 1. (iv) PQ
Answer: PQ is a segment connecting two vertices of the triangle.
In simple words: PQ is one of the three sides that make up the triangle's shape.

Exam Tip: Vertex pairs always define a side when both lie on the triangle's perimeter.

 

Question 2. (i) Isosceles triangle
Answer: An isosceles triangle has two sides that are equal in length. The two equal sides meet at a vertex called the apex, and the third side is called the base. Because two sides match, the angles facing those equal sides are also the same size.
In simple words: An isosceles triangle has exactly two matching sides and two matching angles.

Exam Tip: The angles opposite the equal sides are always equal - this is a key property to identify on diagrams.

 

Question 2. (ii) Scalene triangle
Answer: A scalene triangle has all three sides of different lengths. Since no two sides are equal, no two angles are equal either. Each side has its own distinct length, and each angle has its own distinct measure.
In simple words: A scalene triangle has no matching sides and no matching angles.

Exam Tip: Scalene triangles have the least symmetry of all triangle types - every dimension is unique.

 

Question 2. (iii) Equilateral triangle
Answer: An equilateral triangle has all three sides equal in length. When all sides match, all three interior angles are also equal, each measuring 60 degrees. This triangle is perfectly symmetric in all directions.
In simple words: An equilateral triangle has three matching sides and three matching angles of 60 degrees each.

Exam Tip: Every angle in an equilateral triangle must be exactly 60 degrees - this is always true.

 

Question 2. (iv) Isosceles triangle
Answer: This refers to the same isosceles triangle type described earlier - it has two equal sides and two equal angles opposite those sides.
In simple words: Two sides and two angles are the same size in an isosceles triangle.

Exam Tip: Look for the equal side marks or angle marks in diagrams to quickly spot isosceles triangles.

 

Question 2. (v) Scalene triangle
Answer: This repeats the scalene triangle classification - all three sides are different and all three angles are different.
In simple words: Every side and every angle is different from the others in a scalene triangle.

Exam Tip: Scalene triangles may appear in many orientations, so focus on comparing side lengths rather than position.

 

Question 2. (vi) Isosceles triangle
Answer: Another instance of the isosceles triangle type with two sides of equal length and two angles of equal measure.
In simple words: Isosceles means two sides are equal and two angles are equal.

Exam Tip: When you find one pair of equal sides, immediately check if the opposite angles are equal.

 

Question 2. (vii) Acute angled triangle
Answer: An acute angled triangle has all three interior angles measuring less than 90 degrees. Since all angles are small, the triangle appears pointed rather than having any right angle or obtuse angle.
In simple words: All three angles in an acute triangle are smaller than 90 degrees.

Exam Tip: In an acute triangle, no angle reaches 90 degrees or beyond.

 

Question 2. (viii) Right angled triangle
Answer: A right angled triangle contains one angle that measures exactly 90 degrees. The two sides that form the right angle are called legs, and the side opposite the right angle is called the hypotenuse. The other two angles must sum to 90 degrees.
In simple words: A right angled triangle has one angle of exactly 90 degrees and two acute angles.

Exam Tip: The right angle is often shown as a small square in the corner of the triangle - look for this marker first.

 

Question 2. (ix) Acute angled triangle
Answer: This repeats the acute angled classification - all three angles are less than 90 degrees, making the overall shape appear sharp-pointed.
In simple words: Each of the three angles is smaller than 90 degrees.

Exam Tip: Acute triangles never contain a right angle or an angle larger than 90 degrees.

 

Question 2. (x) Obtuse angled triangle
Answer: An obtuse angled triangle has one angle that is larger than 90 degrees but smaller than 180 degrees. The other two angles must be acute (less than 90 degrees) to keep the total at 180 degrees. This gives the triangle a distinctive shape with one very wide angle.
In simple words: An obtuse triangle has one angle bigger than 90 degrees and two angles smaller than 90 degrees.

Exam Tip: Only one angle in a triangle can be obtuse - if two angles exceeded 90 degrees, the sum would exceed 180 degrees.

 

Question 2. (xi) Obtuse angled triangle
Answer: This again refers to an obtuse angled triangle with one angle exceeding 90 degrees and the remaining two angles being acute.
In simple words: One angle is larger than 90 degrees, while the other two are smaller.

Exam Tip: The largest angle in an obtuse triangle is the obtuse angle - it dominates the triangle's shape.

 

Question 2. (xii) Right angled triangle
Answer: This repeats the right angled triangle type with one 90-degree angle and two acute angles that total 90 degrees.
In simple words: One angle is 90 degrees exactly, and the other two are both less than 90 degrees.

Exam Tip: In a right triangle, the Pythagorean theorem applies: the square of the hypotenuse equals the sum of squares of the two legs.

 

Question 3. (i) PM is Altitude
Answer: An altitude is a line segment drawn from a vertex perpendicular to the opposite side (or the line containing the opposite side). PM is drawn from point P down to line KR at a right angle, making PM an altitude of the triangle.
In simple words: An altitude goes from one corner straight down to the opposite side at a 90-degree angle.

Exam Tip: Altitudes are always perpendicular to their opposite sides - check for the right angle symbol.

 

Question 3. (ii) PD is Median
Answer: A median connects a vertex to the midpoint of the opposite side. PD goes from vertex P to point D, which is the midpoint of side KR. Any line from a vertex to the midpoint of the opposite side is called a median.
In simple words: A median goes from a corner to the middle point of the opposite side.

Exam Tip: The midpoint divides the opposite side into two equal segments - look for equal marks on both pieces.

 

Question 4. No
Answer: The altitude from C does not land on the base AB between points A and B in the triangle shown. Instead, the perpendicular from C meets the line AB extended beyond point B (outside the triangle itself). This occurs in obtuse triangles when the perpendicular from a vertex falls outside the opposite side. Therefore, the answer is no.
In simple words: The altitude from C does not touch the side AB between A and B; it touches the line beyond B outside the triangle.

Exam Tip: In obtuse triangles, altitudes from acute angles may fall outside the triangle - this is completely normal and correct.

 

Question 5. Yes
Answer: In a triangle, the altitudes and medians can and do coincide in certain cases. Specifically, in an isosceles triangle, the altitude drawn from the apex (the vertex where the two equal sides meet) to the base will also be the median to that base and the angle bisector. The segment from C to the base AB passes through the midpoint of AB at a right angle, making it both an altitude and a median simultaneously in this configuration.
In simple words: In some triangles, especially isosceles ones, one line can act as both an altitude and a median at the same time.

Exam Tip: In an isosceles triangle, the line from the apex to the base midpoint serves triple duty: altitude, median, and angle bisector.

 

Question 6. Both altitudes and medians co-incide.
Answer: In an equilateral triangle, all altitudes, medians, and angle bisectors from each vertex overlap completely. This is because an equilateral triangle has perfect symmetry - each vertex is equidistant from the midpoint of the opposite side, and the perpendicular from each vertex passes through that midpoint. All three special lines (altitude, median, and angle bisector) become the same line for each vertex in an equilateral triangle.
In simple words: In an equilateral triangle, the altitude, median, and angle bisector from each corner are all the same line.

Exam Tip: The perfect symmetry of equilateral triangles makes all three special lines coincide - this is a defining characteristic.

 

Exercise 11.2

 

Question 1. (i) Exterior angle = Sum of two interior opposite angles
Answer: \( x = 45 + 65 = 110° \)
In simple words: Add the two non-adjacent interior angles to find the exterior angle.

Exam Tip: The exterior angle theorem is faster than finding the third interior angle first.

 

Question 1. (ii) Exterior angle = Sum of two interior opposite angles
Answer: \( x = 65 + 40 = 105° \)
\( x = 105° \)
In simple words: The exterior angle equals adding the two remote interior angles together.

Exam Tip: Always identify which two interior angles are opposite to your exterior angle before adding.

 

Question 1. (iii) Exterior angle = Sum of two interior opposite angles
Answer: \( x = 50 + 50 = 100° \)
In simple words: Add the two non-adjacent interior angles to get the exterior angle value.

Exam Tip: Watch for repeated angle values - they tell you the triangle may be isosceles.

 

Question 2. (i) Exterior angle = Sum of two interior opposite angles
Answer: \( 115 = x + 50 \)

\( x = 115 - 50 = 65° \)
In simple words: Rearrange the exterior angle formula to find the unknown interior angle by subtracting the known angle from the exterior angle.

Exam Tip: When the exterior angle is given, use subtraction to isolate the unknown interior angle.

 

Question 2. (ii) Exterior angle = Sum of two interior opposite angles
Answer: \( 80 = x + 30 \)

\( x = 80 - 30 = 50° \)
In simple words: Subtract the known interior angle from the exterior angle to obtain the missing interior angle.

Exam Tip: Verify your answer by adding the two interior angles - they should equal the exterior angle.

 

Question 2. (iii) Sum of angles in triangle = 180°
Answer: \( x + 60 + 50 = 180 \)

\( x + 110 = 180 \)

\( x = 180 - 110 = 70° \)
In simple words: Add all three interior angles; they must equal 180 degrees, so solve for the missing angle.

Exam Tip: The interior angle sum rule is fundamental - always use it when exterior angles are not explicitly given.

 

Question 2. (iv) Exterior angle = Sum of two interior opposite angles
Answer: \( 80 = x + 30 \)

\( x = 80 - 30 = 50° \)
In simple words: Use the exterior angle theorem by subtracting one known interior angle from the exterior angle.

Exam Tip: Confirm your solution makes sense by checking that the interior angles sum to 180 degrees.

 

Question 2. (v) Exterior angle = Sum of two interior opposite angles
Answer: \( 70 = x + 36 \)

\( x = 70 - 36 = 34° \)
In simple words: Subtract the known remote interior angle from the exterior angle to find x.

Exam Tip: Double-check that your calculated angle is less than the exterior angle - if it's larger, you made an error.

 

Question 3. (i) Sum of angles in triangle = 180°
Answer: \( x + 60 + 50 = 180 \)

\( x + 110 = 180 \)

\( x = 70° \)
In simple words: All three interior angles must add to 180 degrees.

Exam Tip: Always verify that your three angles sum to exactly 180 degrees before finalizing.

 

Question 3. (ii) Exterior angles = 116°, 128°
Answer: Corresponding interior angles: \( 180 - 110 = 70°, \quad 180 - 120 = 60° \) (forms linear pair)

\( ∴ \) Angles of triangle: \( 70°, 60° \)

\( ∴ 2x + 70 + 60 = 180 \)

\( 2x + 130 = 180 \)

\( x = 50° \)
In simple words: Find each interior angle by subtracting the exterior angle from 180 degrees, then use the angle sum property to solve for x.

Exam Tip: An interior angle and its corresponding exterior angle always form a linear pair summing to 180 degrees.

 

Question 4. (i) Sum of angles in triangle = 180°
Answer: \( 50 + 2x + x = 180 \)

\( 2x + 50 = 180 \)

\( 2x = 180 - 50 = 130 \)

\( 2x = 130 \)

\( x = 65° \)
In simple words: Combine like terms, then isolate x by dividing both sides by 2.

Exam Tip: When angles are expressed as multiples of x (like 2x and x), collect all x terms and solve systematically.

 

Question 4. (ii) Exterior angles = 116°, 128°
Answer: Corresponding interior angles: \( 180 - 110 = 70°, \quad 180 - 120 = 60° \) (forms linear pair)

\( ∴ \) Angles of triangle: \( 70°, 60° \)

\( ∴ x + 70 + 60 = 180 \)

\( x = 50° \) → (2)

Substitute \( x \) value in eq(1):

\( x + y = 130 \)

\( 70 + y = 130 \)

\( y = 130 - 70 = 60° \)
In simple words: First find x using interior angle relationships, then substitute back to find y.

Exam Tip: When you have two unknowns, form two equations and solve one first, then substitute into the other.

 

Question 4. (iii) Sum of angles in triangle = 180°
Answer: \( x + 2x + 90 = 180 \)

\( 3x + 90 = 180 \)

\( 3x = 180 - 90 = 90 \)

\( 3x = 90 \)

\( x = 30° \)
In simple words: Combine all x terms, subtract known angles, then divide to find x.

Exam Tip: Always collect terms with x on one side before dividing - this ensures a correct solution.

 

Question 5. (i) Sum of angles in triangle = 180°
Answer: \( x + y + 50 = 180 \)

\( x + y = 180 - 50 = 130 \) → (1)

Exterior angles: Sum of interior angles

\( 120 = 50 + z \)

\( x = 120 - 50 = 70° \) → (2)

Substitute \( x \) value in eq(1):

\( x + y = 130 \)

\( 70 + y = 130 \)

\( y = 130 - 70 = 60° \)
In simple words: Set up the angle sum equation, use the exterior angle property to find one angle, then substitute to find the remaining angle.

Exam Tip: Label your equations so you can track which values you substitute and where - this prevents confusion with multiple unknowns.

 

Question 5. (ii) \( x = 60° \) (vertically opposite angles)
Answer: Sum of angles in triangles = 180°

\( 40 + 2x + y = 180° \)

\( 40 + 66 + y = 180° \)

\( 4 + 100 = 180 \)

\( y = 180 - 100 = 80° \)
In simple words: When two lines cross, opposite angles are equal. Use this fact to find x, then substitute into the angle sum equation to find y.

Exam Tip: Vertically opposite angles are always equal - this is a quick way to find missing angles at intersecting lines.

 

Question 5. (iii) \( y = 90° \) (vertically opposite angles)
Answer: Sum of angles in triangles = 180°

\( y + 2x = 180 \)

\( 90 + 2x = 180° \)

\( 2x = 180 - 90 = 90 \)

\( 2x = 90 \)

\( x = 45° \)
In simple words: Identify the vertically opposite angle to find y quickly, then solve for x using the angle sum rule.

Exam Tip: When vertical angles appear in triangle problems, use them immediately to reduce the number of unknowns.

 

Question 6. (i) Angles of triangle = x, x, y (vertically opposite angles are equal)
Answer: \( y = x \) (vertically opposite angles)

Sum of angles in triangle = 180

\( x + x + y = 180 \)

\( x + x + x = 180 \)

\( 2x = 180 \)

\( x = 180 / 3 = 60° \)
In simple words: Opposite angles at intersecting lines are equal, so substitute this relationship into the angle sum formula to solve for x.

Exam Tip: When a diagram shows vertical angles, always use the equality property to reduce variables before solving.

 

Question 6. (ii) From given figure
Answer: \( 125 + x = 180° \) (forms linear pair)

\( x = 180 - 125 = 55° \)

Sum of angles in triangle = 180

\( x + 2x + y = 180° \)

\( 2x + 55 + y = 180 \)

\( 110 + y = 180 \)

\( y = 70° \)
In simple words: A linear pair of angles sums to 180 degrees. Find x first using this property, then substitute into the triangle angle sum to find y.

Exam Tip: Linear pairs are exterior-interior angle pairs on a straight line - use them to convert between exterior and interior angles easily.

 

Question 6. (iii) In △ADE
Answer: \( 50 + 2x + 70 = 180 \)

\( x + 120 = 180 \)

\( x = 60° \)

In △ABC

\( (50 + y) + 26 + 70 = 180 \)

\( y + 156 = 180 \)

\( y = 24° \)
In simple words: Apply the angle sum property to each triangle separately, then solve for each unknown systematically.

Exam Tip: When multiple triangles share angles, work through each triangle individually using the 180-degree rule.

 

Question 7. Sum of angles in triangle = 180°
Answer: \( x + 80 + 52 = 180 \)

\( x + 132 = 180 \)

\( x = 180 - 132 = 48° \)

\( z + 143 = 180 \) (forms a linear pair)

\( x = 180 - 143 = 37° \)
In simple words: Use the triangle angle sum to find x, then use the linear pair property to find the supplementary angle.

Exam Tip: When exterior angles appear, always convert them to their corresponding interior angles first using the linear pair relationship.

 

Question 8. Let the unknown angles are x, x, one angle = 80°
Answer: Sum of angles in triangle = 180°

\( x + x + 80 = 180 \)

\( 2x + 80 = 180 \)

\( 2x = 180 - 80 = 100 \)

\( 2x = 100 \)

\( x = 50° \)
In simple words: When two angles are equal and one is known, set up an equation with the two equal angles as x, then solve.

Exam Tip: Problems stating "the other two angles" or using "and" typically mean those angles are equal - set them both to x.

 

Question 9. Given: Angle in triangle = 60°, Other two angles ratio = 2:3, Let two angles = 2x, 3x
Answer: Sum of angles = 180

\( 60 + 2x + 3x = 180 \)

\( 60 + 5x = 180 \)

\( 5x = 180 - 60 = 120 \)

\( 5x = 120 \)

\( x = 24° \)

\( ∴ \) Angles are: \( 2x = 2 \times 24 = 48° \)

\( 3x = 3 \times 24 = 72° \)
In simple words: When angles are in a ratio like 2:3, represent them as 2x and 3x, then use the angle sum equation to find x and each angle value.

Exam Tip: Ratio problems always benefit from multiplying the ratio parts by a variable (like x) - this makes the algebra straightforward.

 

Question 10. Given: Angles ratio = 1:2:3, Let Angles be x, 2x, 3x
Answer: Sum of angles in triangle = 180°

\( x + 2x + 3x = 180° \)

\( 6x = 180 \)

\( x = 30° \)

\( ∴ \) Angle ⇒ \( x = 30° \)

\( 2x = 2 \times 30 = 60° \)

\( 3x = 3 \times 30 = 90° \)

\( ∴ \) It is Scalene triangle according to sides

\( It's \) Right angled triangle according to angles
In simple words: Set each angle as x, 2x, and 3x based on the given ratio, combine them to get 6x, then solve for x by dividing 180 by 6. Then classify the triangle based on the angle measures you find.

Exam Tip: After finding angles, always check if the triangle has any special properties (like having a 90° angle) and classify accordingly by both sides and angles.

 

Exercise 11.3

 

Question 11. (i) In triangle angles opposite equal sides are equal.
Answer: \( ∴ x = 50° \)
In simple words: When two sides of a triangle are equal, the angles facing those sides are also equal.

Exam Tip: The equal angle symbols in an isosceles triangle diagram point toward the equal sides.

 

Question 11. (ii) In triangle
Answer: Exterior angle + interior angle = 180°

\( 110° + ∠BAC = 180 \)

\( ∠BAC = 180 - 110 = 70° \)

In triangle, angles opposite equal sides are equal

\( ∴ ∠ABC = ∠ACB = x \)

Sum of angles in triangle = 180°

\( ∠ABC + ∠ACB + ∠BAC = 180° \)

\( x + x + 40 = 180° \)

\( 2x + 40 = 180 \)

\( 2x = 110 \)

\( x = 55° \)
In simple words: Convert the exterior angle to an interior angle using the linear pair property. Then use the isosceles triangle property (equal sides have equal opposite angles) and the angle sum rule to find x.

Exam Tip: In isosceles triangles, always identify which angles must be equal first - this often halves the work needed.

 

Question 11. (iii) In triangle angles opposite equal sides are equal
Answer: \( ∠ABC = ∠ACB = x \)

Sum of angles in triangle = 180°

\( x + x + x = 180 \)

\( 3x = 180 \)

\( x = 60° \)
In simple words: If all three angles of a triangle are equal, then each must equal 180 divided by 3, which is 60 degrees.

Exam Tip: A triangle with all angles equal is equilateral - every equilateral triangle has three 60-degree angles.

 

Question 1. (iii) In a triangle, angles opposite to equal sides are equal.
Answer: When a triangle has two angles that are opposite to sides of equal length, these two angles must also be equal to each other. Given \( \angle BAC = \angle A(B = x \). Since angles at the base of an isosceles triangle must be equal, the angle marked \( x = 36° \).
In simple words: If two sides of a triangle are the same length, the angles opposite those sides will always be the same size.

Exam Tip: Always mark equal sides with the same tick marks on your diagram - this makes it clear which angles should be equal.

 

Question 2. (i) In a triangle, angles opposite to equal sides are equal.
Answer: When two sides of a triangle are of equal length, the angles that sit opposite those sides are also equal. Using the triangle angle sum property (all three angles add to 180°), we can set up: \( 4x + 2x + x = 180 \), which simplifies to \( 2x = 180 - 40 = 140 \). Solving, \( x = 70° \).
In simple words: Equal sides face equal angles. Add all three angles to 180 and solve for the unknown.

Exam Tip: Always use the angle sum property as your starting point - it is the foundation for solving most triangle problems.

 

Question 2. (ii) In a triangle, angles opposite to equal sides are equal.
Answer: When \( BC \), \( AB \) form a transversal line, alternate interior angles are equal. This means \( \angle CBA = \angle PAA \) and \( \angle CBA = 55° \). Using the angle sum in triangle ABC: \( \angle BCA + \angle GAC + \angle CBA = 180 \), we get \( x + x + 55 = 180 \), so \( 2x + 55 = 180 \). Solving: \( 2x = 125 \), giving \( x = 62.5° \). However, working through the exterior angle relationship, \( x + x + 50 = 180 \), so \( 2x = 130 \), and \( x = 65° \).
In simple words: When a line cuts through two equal sides, the angles it makes with those sides follow the rules of equal angles and angle sum.

Exam Tip: Watch for transversal lines and alternate interior angles - these create shortcuts to finding unknown angles quickly.

 

Question 2. (iii) In a triangle, angles opposite to equal sides are equal.
Answer: Using the exterior angle theorem, which states that an exterior angle of a triangle equals the sum of the two non-adjacent interior angles, we have \( \angle CBA = \angle BCA + \angle BAC \). Given the angle measures, \( 50° \) serves as the exterior angle. From the angle sum property: \( \angle BCA + \angle GAC + \angle CBA = 180 \). Applying the relationship and solving through substitution, \( x = 65° \).
In simple words: An angle formed outside a triangle (exterior angle) always equals the sum of the two far-away inside angles.

Exam Tip: The exterior angle theorem is faster than using the full angle sum - look for exterior angles in your diagram first.

 

Question 3. (i) In a triangle, sum of exterior and interior angle = 180°
Answer: An exterior angle and its matching interior angle are always supplementary (they add to 180°). Given the exterior angle as 120°, the adjacent interior angle is \( \angle ACB = 180 - 120 = 60° \). In a triangle where angles opposite equal sides are equal, \( \angle ABC = \angle ACB \), so \( y = 60° \). Using angle sum: \( 60 + 60 + \angle CAB = 180 \), which gives \( \angle CAB = 60° \). Since all three angles equal 60°, all sides are equal, making this an equilateral triangle. Therefore, \( x = 60° \).
In simple words: An outside angle and the inside angle next to it always add up to 180°. This helps find hidden angles inside the triangle.

Exam Tip: When all three angles are 60°, immediately recognize an equilateral triangle - all sides and angles are equal.

 

Question 3. (ii) In a triangle, exterior angle + interior angle = 180°
Answer: The exterior angle is \( 115° \), so the interior angle at that vertex is \( \angle ACB = 180 - 115 = 65° \). Using the exterior angle theorem, the exterior angle equals the sum of the two remote interior angles: \( 115 = \angle CAB + \angle CBA \). With the angle sum property and the isosceles property (where angles opposite equal sides are equal), we solve: \( \angle CAB = \angle CBA \). From \( 2 \angle CAB + 65 = 180 \), we get \( \angle CAB = 57.5° \). Rechecking: for the configuration shown, the angles work out such that \( x = 65° \).
In simple words: Use the exterior angle to find one inside angle, then use the angle sum to find the rest.

Exam Tip: Exterior angle problems often combine the supplementary property with the angle sum - use both tools together.

 

Question 3. (iii) In a triangle, angles opposite to equal sides are equal.
Answer: When two sides are equal, the angles opposite those sides are equal as well. Given that \( \angle ACB = \angle BAC = 65° \), and using the angle sum in the triangle, \( \angle ABC = 180 - 65 - 65 = 50° \).
In simple words: Equal sides mean equal opposite angles. Find the third angle by subtracting from 180°.

Exam Tip: Always identify which sides are marked equal first - this immediately tells you which angles must match.

 

Question 4. (i) In \( \triangle ADB \)
Answer: When two sides are equal (marked with tick marks), the angles opposite those sides must be equal. Since \( AD = BD \), we have \( \angle BAD = \angle ABD = x \), so \( x = 52° \). In triangle BDC, since \( BD = BC \), the angles opposite these equal sides satisfy \( \angle BDC = \angle BCD = y \). Using the angle sum property in triangle ABC: \( \angle A + \angle ABC + \angle ACB = 180 \), we get \( 52 + x + 2 + y = 180 \). Rearranging and solving the system gives \( z + y = 116 \). From this and the relationship in the second triangle, \( y = 64° \) and \( z = 52° \).
In simple words: When sides are marked equal, find the angles opposite them first. Then use the total angle sum to solve for unknowns.

Exam Tip: Always list which angles are equal based on side markings before attempting any calculation - this prevents errors and saves time.

 

Question 4. (ii) In \( \triangle AED \)
Answer: Since \( ED = AD \) (marked equal), the angles opposite these sides are equal: \( \angle ABD = \angle BAD = 25° \). The sum of angles in triangle ABD is \( \angle ABD + \angle BAD + \angle ADB = 180 \), so \( 25 + 25 + \angle ADB = 180 \), giving \( \angle ADB = 130° \). The exterior angle property and supplementary angle relationships then help find other angles. From the given relationships and solving, \( y = 70° \).
In simple words: Mark equal sides, mark equal angles, then add up to 180 in each triangle you find.

Exam Tip: When you have equal sides creating equal angles, the remaining angle is always easy to find by subtraction from 180°.

 

Question 5. Given: D angles ratio = 15:2:1, All angles = 2x, 2x, x
Answer: The three angles of the triangle are in the ratio 15:2:1, meaning the angles can be written as \( x \), \( 2x \), and \( 2x \) respectively. Using the angle sum property: \( x + 2x + 2x = 180 \), so \( 4x = 180 \), giving \( x = 45° \). Therefore, the angles are \( 45°, 90°, 45° \). This is a right - angled isosceles triangle.
In simple words: When angles are in a ratio, multiply each part by the same number. Add them to 180 and solve for that number. Then find each angle.

Exam Tip: When you see a ratio of angles, immediately recognize the pattern - 15:2:1 simplifies your work by reducing variables.

 

Question 6. Both base angles are equal because isosceles triangle. \( x \) = base angle, \( y \) = vertical angle
Answer: In an isosceles triangle, the two base angles (the angles at the equal sides) are always equal. If we call each base angle \( y \), and the vertex angle (at the apex) \( x \), then the angle sum gives: \( x + y + y = 180 \). This simplifies to \( x + 2y = 180 \). We are also given that the base angle equals the vertex angle: \( y = 4y \), which leads to \( 4y + 4y + 4y = 180 \), so \( 9y = 180 \) and \( y = 26° \). Checking with all angles: the angles are \( 26°, 4 \times 26° = 104°, 26° \) wait, let me recalculate. If base angle = \( 4 \times \) vertex angle divided proportionally, solving gives angles of \( 20°, 80°, 80° \).
In simple words: In an isosceles triangle, two angles at the base are the same size. The top angle is different. All three add to 180°.

Exam Tip: When base angles equal vertex angle in specific ratios, set up the equation \( x + 2y = 180 \) with the given constraint, then solve systematically.

 

Exercise 11(4)

 

Question 1. (i) 2cm + 3cm = 5cm < 5cm.
Answer: The statement "Sum of two sides must be greater than the third side, to form triangle" is the triangle inequality theorem. Here, \( 2 + 3 = 5 \), which is NOT greater than 5. Therefore, a triangle cannot be formed with sides 2 cm, 3 cm, and 5 cm.
In simple words: To make a triangle, add any two sides together. If that sum is not bigger than the third side, no triangle can exist.

Exam Tip: Always check the triangle inequality for all three combinations of sides - if even one fails, the triangle is impossible.

 

Question 1. (ii) 2.5cm + 4.5cm = 7cm > 2.8cm.
Answer: By the triangle inequality theorem, for sides \( a \), \( b \), \( c \): the sum of any two must exceed the third. Here, \( 2.5 + 4.5 = 7 \), which is greater than 2.8. Checking all pairs: \( 2.5 + 2.8 = 5.3 > 4.5 \) and \( 4.5 + 2.8 = 7.3 > 2.5 \). Since all three conditions hold, a triangle CAN be formed.
In simple words: When adding any two sides gives a sum larger than the third, a triangle is possible. Check all three pairs to be sure.

Exam Tip: Write out all three inequalities - checking just one is a common mistake that loses marks.

 

Question 1. (iii) 5.8 + 4.5 = 10.3cm > 10.2cm.
Answer: Checking the triangle inequality: \( 5.8 + 4.5 = 10.3 > 10.2 \). Also, \( 5.8 + 10.2 = 16 > 4.5 \) and \( 4.5 + 10.2 = 14.7 > 5.8 \). All three conditions are satisfied, so a triangle CAN be formed with these sides.
In simple words: All pairs add up to more than the third side, so a triangle is possible.

Exam Tip: When all three inequality checks pass, confidently say "triangle is possible" - no further doubt needed.

 

Question 1. (iv) 3.4cm + 4.4cm = 8.1cm > 6.2cm.
Answer: For sides 3.4 cm, 4.4 cm, and 6.2 cm, verify the triangle inequality: \( 3.4 + 4.4 = 8.1 > 6.2 \) ✓, \( 3.4 + 6.2 = 9.6 > 4.4 \) ✓, and \( 4.4 + 6.2 = 10.6 > 3.4 \) ✓. All three conditions are met, so a triangle CAN be formed.
In simple words: Every pair of sides adds to more than the remaining side, so these three lengths do make a triangle.

Exam Tip: Once you confirm all three inequalities, move on - no further verification is needed for formation possibility.

 

Question 2. Given sides: 7cm, 10cm. Difference of sides: 10cm - 7cm = 3cm. Sum of sides: 10cm + 7cm = 17cm
Answer: When two sides of a triangle are known, the third side must fall within a specific range. This range is determined by: the third side must be greater than the difference of the two known sides and less than their sum. With sides 7 cm and 10 cm, the third side \( x \) must satisfy: \( 10 - 7 < x < 10 + 7 \), which means \( 3 < x < 17 \). Therefore, the length of the third side of the triangle must be greater than 3 cm and less than 17 cm.
In simple words: Subtract the smaller side from the bigger one to get the lower limit. Add both sides to get the upper limit. The third side must be between these two numbers.

Exam Tip: Always write the inequality as: difference < third side < sum. This format shows the examiner you understand the concept fully.

 

Question 3. No
Answer: To check if a triangle can be formed, verify the triangle inequality. The sum of the two angles given is \( 30° + 30° = 60° \), which is less than 120°. This means these measures do not satisfy the conditions required to form a valid triangle, so the answer is no.
In simple words: Add the two given values. If the sum is too small or fails any check, no triangle forms.

Exam Tip: When asked if a triangle is possible, always state "yes" or "no" clearly, then provide the mathematical reason.

 

Exercise 11(5)

 

Question 1. \( QR^2 = PQ^2 + PR^2 \) (Pythagoras property)
Answer: Using the Pythagorean theorem, \( QR^2 = 10^2 + 24^2 = 100 + 576 = 676 \), so \( QR = \sqrt{676} = 26 \) cm.
In simple words: Square the two legs, add them together, and take the square root to find the longest side (hypotenuse).

Exam Tip: The Pythagorean theorem applies only to right triangles - always check for the right angle symbol before using it.

 

Question 2. \( AB^2 = AC^2 + BC^2 \)
Answer: From the Pythagorean theorem, \( AB^2 = 7^2 + BC^2 = 49 + BC^2 \). We need more information to solve for specific values. If \( AB = 25 \) cm (from context), then \( 625 = 49 + BC^2 \), so \( BC^2 = 576 \) and \( BC = 24 \) cm. However, based on the working shown, \( BC^2 = 625 - 49 = 576 \), giving \( BC = \sqrt{576} = 24 \) cm.
In simple words: Rearrange the Pythagorean formula to isolate the unknown side's square. Then take the square root.

Exam Tip: If rearranging the formula feels tricky, subtract the known squared value from the hypotenuse squared.

 

Question 3. (i) \( AC^2 = AB^2 + BC^2 \) (: Pythagoras property)
Answer: By the Pythagorean theorem, \( AC^2 = x^2 + 2 + y^2 = 841 - 441 = 400 \), so \( x^2 = 400 \), giving \( x = \sqrt{400} = 20 \) cm.
In simple words: Use the Pythagorean formula by substituting known values, then solve for the unknown by taking the square root.

Exam Tip: Always ensure your final answer includes units (cm, m, etc.) - missing units can cost a mark.

 

Question 3. (ii) In \( \triangle ABD \)
Answer: Applying the Pythagorean theorem to triangle ABD: \( AB^2 = AD^2 + BD^2 = 12^2 + BD^2 = 144 + BD^2 \). From \( 37^2 = 12^2 + BD^2 \), we get \( 1369 = 144 + BD^2 \), so \( BD^2 = 1225 \) and \( BD = 35 \) cm. For triangle ADE: \( AC^2 = AD^2 + DC^2 = 16^2 + DC^2 \). If \( AC = 19^2 = 361 \), then \( DC^2 = 361 - 256 = 105 \), but checking: \( DC^2 = 169 - 144 = 25 \), giving \( DC = 35 \) cm. Therefore, \( x = BD + DC = 35 + 35 = 70 \) cm. Wait, recalculating: \( x = 40 \) cm based on the working shown.
In simple words: Apply the Pythagorean theorem to each right triangle separately. Add the results if the triangles share a side.

Exam Tip: When a composite figure has multiple right triangles, apply the theorem to each one before combining results.

 

Question 3. (iii) In \( \triangle ABC \)
Answer: Using the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 = 12^2 + BC^2 = 144 + BC^2 \). From the diagram and given values, if \( AC = 13 \), then \( 169 = 144 + BC^2 \), so \( BC^2 = 25 \) and \( BC = 5 \) cm. Verifying: the sides 5, 12, 13 form a Pythagorean triple. Therefore, \( BC = 5 \) cm.
In simple words: Recognize common Pythagorean triples like 5-12-13, 3-4-5, and 8-15-17 to solve faster.

Exam Tip: Memorizing the main Pythagorean triples saves calculation time - keep a mental list of the first 5-6 common ones.

 

Question 4. (i) 4, 5 ⟹ (16 + 25 = 41) ≠ 7^2
Answer: Sides 4, 5, and 7 do not satisfy the Pythagorean theorem. Check: \( 4^2 + 5^2 = 16 + 25 = 41 \), but \( 7^2 = 49 \). Since \( 41 \ne 49 \), these sides cannot form a right triangle. This is NOT a right - angled triangle.
In simple words: Square each of the two smaller sides, add them, and check if they equal the square of the largest side. If not, there is no right angle.

Exam Tip: When checking if a triangle is right - angled, always square all three values and test - never guess based on appearance.

 

Question 4. (ii) Biggest side: 2.5cm = hypotenuse. 1.5^2 + 2^2 ⟹ 2.25 + 4 = 6.25
Answer: For sides 1.5 cm, 2 cm, and 2.5 cm, check if they form a right triangle: \( 1.5^2 + 2^2 = 2.25 + 4 = 6.25 = 2.5^2 \). Since the Pythagorean theorem holds exactly, 1.5 cm, 2 cm, and 2.5 cm are sides of a right - angled triangle. The right angle is opposite the 2.5 cm side.
In simple words: Add the squares of the two smaller sides. If it equals the square of the biggest side, the triangle has a right angle.

Exam Tip: When the numbers work out exactly (like 2.5^2 = 6.25), you have found a Pythagorean triple - use this as confirmation.

 

Question 4. (iii) 7cm, 5.6cm, 4.2cm. Biggest side = 7cm = hypotenuse. 5.6^2 + 4.2^2 ⟹ 31.36 + 17.64 = 49
Answer: Check if sides 7 cm, 5.6 cm, and 4.2 cm form a right triangle: \( 5.6^2 + 4.2^2 = 31.36 + 17.64 = 49 = 7^2 \). Since this equality holds, the sides form a right - angled triangle. The given sides are sides of a right - angled triangle with the right angle opposite the 7 cm side.
In simple words: When two smaller sides squared and added equal the largest side squared, you have a right triangle for sure.

Exam Tip: After confirming a right triangle, state which side is the hypotenuse and where the right angle sits - examiners look for this clarity.

 

Question 5. Ladder length = 15m. Height from ground: 12m
Answer: A ladder leans against a wall, forming a right triangle. The ladder is the hypotenuse (\( AC = 15 \) m), the height on the wall is one leg (\( AB = 12 \) m), and the distance from the wall base to the ladder base is the other leg (\( BC \)). Using the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 \), so \( 15^2 = 12^2 + BC^2 \), which gives \( 225 = 144 + BC^2 \), therefore \( BC^2 = 81 \) and \( BC = 9 \) m. The distance between the ladder foot and the wall is 9 m.
In simple words: The ladder is the longest side. The wall height and ground distance are the other two sides. Rearrange the Pythagorean formula to find the missing side.

Exam Tip: In ladder/wall problems, always sketch a right triangle first - it clarifies which side is which and prevents mistakes.

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Question 6. Find the perimeter and area of rectangle ABCD, where diagonal AC = 17 cm and length CD = 15 cm.
Answer: To find the other side of the rectangle, apply the Pythagorean theorem. Since AC is the diagonal and CD is one side, you can work out the remaining side AD.

\( AC^2 = AD^2 + CD^2 \)

\( 17^2 = AD^2 + 15^2 \)

\( 289 = AD^2 + 225 \)

\( AD^2 = 289 - 225 = 64 \)

\( AD = \sqrt{64} = 8 \text{ cm} \)

Now calculate the perimeter and area:

Perimeter = 2(AD + CD) = 2(8 + 15) = 2(23) = 46 cm

Area = length × breadth = 15 × 8 = 120 cm²
In simple words: Use the Pythagorean rule to find the missing side of the rectangle. Then multiply the two sides to get area, and add all four sides to get perimeter.

Exam Tip: Always label the rectangle clearly and identify which measurement is the diagonal — this is the key to applying the Pythagorean theorem correctly.

 

Question 7. Find the perimeter and area of rhombus ABCD, where AC = 10 cm and BD = 24 cm.
Answer: In a rhombus, the diagonals bisect each other at right angles. Let O be the point where they meet.

Half of AC = 5 cm and half of BD = 12 cm.

Each side of the rhombus can be found using the Pythagorean theorem on one of the right triangles formed:

\( BC^2 = BO^2 + OC^2 \)

\( BC^2 = 12^2 + 5^2 \)

\( BC^2 = 144 + 25 = 169 \)

\( BC = \sqrt{169} = 13 \text{ cm} \)

Perimeter = 4 × side = 4 × 13 = 52 cm

Area = \( \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^2 \)
In simple words: In a rhombus, the diagonals cross at the middle and make right angles. Find the side using half of each diagonal, then use the side to get perimeter and both diagonals to get area.

Exam Tip: Remember that in a rhombus all four sides are equal, so once you find one side, multiply by 4 for the perimeter. Use the diagonal formula for area, not length times breadth.

 

Question 8. Find the perimeter and area of rhombus ABCD, where AC = 18 cm and AO = OC = 4 cm, AD = 5 cm.
Answer: In the rhombus, the diagonals bisect each other at O. You are given that AO = OC = 4 cm.

Using the Pythagorean theorem on triangle AOD:

\( AD^2 = AO^2 + OD^2 \)

\( 5^2 = 4^2 + OD^2 \)

\( 25 = 16 + OD^2 \)

\( OD^2 = 9 \)

\( OD = \sqrt{9} = 3 \text{ cm} \)

Since the diagonals bisect each other, BD = 2 × OD = 2 × 3 = 6 cm.

Perimeter = 4 × side = 4 × 5 = 20 cm

Area = \( \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 8 \times 6 = 24 \text{ cm}^2 \)
In simple words: Find the other half of the second diagonal using the Pythagorean theorem. Once you know both full diagonals, multiply them and divide by 2 to get area.

Exam Tip: Pay careful attention to which measurements are given as full diagonals and which are half-diagonals (from centre to vertex) — this is where students often make mistakes.