CBSE Class 7 Mathematics Triangles and Its Properties Assignment Set 02

2 mark questions

Directions: Completely solve all the questions. All the calculations are to be shown in the solution. Also draw the diagrams when you are solving.

Question 1. Find the value of the unknown exterior angle x in the following diagrams:
Answer:
Using the Exterior Angle Property of a triangle, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
(i) \(x = 50^{\circ} + 70^{\circ} = 120^{\circ}\)
(ii) \(x = 65^{\circ} + 45^{\circ} = 110^{\circ}\)
(iii) \(x = 30^{\circ} + 40^{\circ} = 70^{\circ}\)
(iv) \(x = 60^{\circ} + 60^{\circ} = 120^{\circ}\)
(v) \(x = 50^{\circ} + 50^{\circ} = 100^{\circ}\)
(vi) \(x = 30^{\circ} + 60^{\circ} = 90^{\circ}\)
In simple words: An exterior angle always equals the sum of the two opposite interior angles.

Exam Tip: Always identify the two non-adjacent interior angles before adding them up.

Question 2. Find the value of the unknown x in the following diagrams:
Answer:
(i) Using the angle sum property of a triangle, the sum of all interior angles is \(180^{\circ}\):
\(50^{\circ} + 60^{\circ} + x = 180^{\circ}\)
\(110^{\circ} + x = 180^{\circ}\)
\(x = 180^{\circ} - 110^{\circ} = 70^{\circ}\)
(ii) Using the angle sum property in a right-angled triangle:
\(90^{\circ} + 30^{\circ} + x = 180^{\circ}\)
\(120^{\circ} + x = 180^{\circ}\)
\(x = 180^{\circ} - 120^{\circ} = 60^{\circ}\)
(iii) Using linear pair and angle sum property:
\(\angle Y = 180^{\circ} - 110^{\circ} = 70^{\circ}\)
In triangle XYZ, \(30^{\circ} + 70^{\circ} + x = 180^{\circ}\)
\(100^{\circ} + x = 180^{\circ}\)
\(x = 80^{\circ}\)
(iv) Since two angles are equal to \(x\):
\(50^{\circ} + x + x = 180^{\circ}\)
\(2x = 180^{\circ} - 50^{\circ} = 130^{\circ}\)
\(x = 65^{\circ}\)
(v) Since all three angles are equal to \(x\):
\(x + x + x = 180^{\circ}\)
\(3x = 180^{\circ}\)
\(x = 60^{\circ}\)
(vi) Using the right-angled triangle property:
\(x + 2x = 90^{\circ}\)
\(3x = 90^{\circ}\)
\(x = 30^{\circ}\)
In simple words: The three angles of any triangle always add up to \(180^{\circ}\).

Exam Tip: For equilateral triangles, immediately divide \(180^{\circ}\) by 3 to get \(60^{\circ}\).

3 mark questions

Directions: Completely solve all the questions. All the calculations are to be shown in the solution. Also draw the diagram if it is given in the question.

Question 3. Find the value of x in the following diagrams:
Answer:
(i) Using the exterior angle property:
\(50^{\circ} + x = 115^{\circ}\)
\(x = 115^{\circ} - 50^{\circ} = 65^{\circ}\)
(ii) Using the exterior angle property:
\(70^{\circ} + x = 100^{\circ}\)
\(x = 100^{\circ} - 70^{\circ} = 30^{\circ}\)
(iii) Using the exterior angle property in a right-angled triangle:
\(90^{\circ} + x = 125^{\circ}\)
\(x = 125^{\circ} - 90^{\circ} = 35^{\circ}\)
(iv) Using the exterior angle property:
\(60^{\circ} + x = 120^{\circ}\)
\(x = 120^{\circ} - 60^{\circ} = 60^{\circ}\)
(v) Using the exterior angle property:
\(30^{\circ} + x = 80^{\circ}\)
\(x = 80^{\circ} - 30^{\circ} = 50^{\circ}\)
(vi) Using the exterior angle property:
\(35^{\circ} + x = 75^{\circ}\)
\(x = 75^{\circ} - 35^{\circ} = 40^{\circ}\)
In simple words: Exterior angle minus one interior opposite angle gives the other interior opposite angle.

Exam Tip: Set up the equation as Interior 1 + Interior 2 = Exterior angle.

Question 4. Find the values of x and y in the following diagrams:
Answer:
(i) Using linear pair at the base: \(y = 180^{\circ} - 120^{\circ} = 60^{\circ}\).
Now using the angle sum property of the triangle:
\(50^{\circ} + y + x = 180^{\circ}\)
\(50^{\circ} + 60^{\circ} + x = 180^{\circ}\)
\(110^{\circ} + x = 180^{\circ}\)
\(x = 70^{\circ}\).
(ii) Using vertically opposite angles or linear pair: \(y = 180^{\circ} - 80^{\circ} = 100^{\circ}\) (or exterior angle property). Let us use exterior angle property: \(50^{\circ} + x = 80^{\circ}\), so \(x = 30^{\circ}\).
Then, using linear pair for \(y\): \(x + y = 180^{\circ}\) (adjacent angles on straight line) or using triangle sum: \(50^{\circ} + y = 180^{\circ}\) is incorrect, wait. Let us re-examine diagram (ii): exterior angle is \(80^{\circ}\), interior opposite angles are \(50^{\circ}\) and \(x\). So \(50^{\circ} + x = 80^{\circ} \implies x = 30^{\circ}\).
For angle \(y\), it forms a linear pair with interior angle adjacent to it, or looking at the triangle containing \(y\): interior angles are \(50^{\circ}\) and \(x = 30^{\circ}\), and \(y\) is exterior to the right triangle, or \(y\) is adjacent to \(80^{\circ}\). Thus \(y = 180^{\circ} - 80^{\circ} = 100^{\circ}\).
(iii) In the triangle with angles \(50^{\circ}\), \(x\), and \(60^{\circ}\):
\(50^{\circ} + x + 60^{\circ} = 180^{\circ}\)
\(110^{\circ} + x = 180^{\circ}\)
\(x = 70^{\circ}\)
For \(y\), using linear pair with \(x\) or straight line: \(x + y = 180^{\circ} \implies 70^{\circ} + y = 180^{\circ} \implies y = 110^{\circ}\).
In simple words: Combine linear pairs and angle sum properties step by step.

Exam Tip: Solve for the unknown that has direct linear or triangle relations first.

Question 5. PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Answer:
Using Pythagoras theorem in right-angled triangle PQR:
\(QR^2 = PQ^2 + PR^2\)
\(QR^2 = (10)^2 + (24)^2\)
\(QR^2 = 100 + 576\)
\(QR^2 = 676\)
\(QR = \sqrt{676} = 26 \text{ cm}\)
In simple words: In a right triangle, the square of the hypotenuse equals the sum of squares of the other two sides.

Exam Tip: Remember common Pythagorean triplets like (10, 24, 26) to verify your answer.

Question 6. ABC is a triangle right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Answer:
Using Pythagoras theorem in right-angled triangle ABC:
\(AB^2 = AC^2 + BC^2\)
\((25)^2 = (7)^2 + BC^2\)
\(625 = 49 + BC^2\)
\(BC^2 = 625 - 49\)
\(BC^2 = 576\)
\(BC = \sqrt{576} = 24 \text{ cm}\)
In simple words: Subtract the square of the given leg from the square of the hypotenuse to find the missing leg.

Exam Tip: Always make sure you subtract from the hypotenuse squared, not the other way around.

Question 7. In the given figure, AD = CD and AB = CB.
(i) State the three pairs of equal parts in \(\Delta ABD\) and \(\Delta CBD\).
(ii) Is \(\Delta ABD \cong \Delta CBD\)? Why or why not?
(iii) Does BD bisect \(\angle ABC\)? Give reasons.
Answer:
(i) The three pairs of equal parts are:
1. \(AD = CD\) (Given)
2. \(AB = CB\) (Given)
3. \(BD = BD\) (Common side)
(ii) Yes, \(\Delta ABD \cong \Delta CBD\) by SSS (Side-Side-Side) congruence criterion because all three corresponding sides of both triangles are equal.
(iii) Yes, BD bisects \(\angle ABC\) because corresponding parts of congruent triangles are equal (\(\angle ABD = \angle CBD\) by CPCT).
In simple words: If all three sides of one triangle match another, the triangles are identical and their angles match too.

Question 8. In \(\Delta ABC\), \(\angle A = 30^{\circ}\), \(\angle B = 40^{\circ}\) and \(\angle C = 110^{\circ}\)
In \(\Delta PQR\), \(\angle P = 30^{\circ}\), \(\angle Q = 40^{\circ}\) and \(\angle R = 110^{\circ}\)
A student says that \(\Delta ABC \cong \Delta PQR\) by AAA congruence criterion. Is he justified? Why or why not?
Answer:
No, the student is not justified.
Reason: There is no AAA (Angle-Angle-Angle) congruence criterion in geometry. Equal corresponding angles only ensure that the two triangles are similar (same shape), not necessarily congruent (same size and shape). One triangle could be larger than the other while having the exact same angles.
In simple words: AAA proves similarity, not congruence. You always need at least one side length to prove congruence.

Exam Tip: Memorize valid congruence criteria: SSS, SAS, ASA, and RHS. AAA is never a valid congruence rule.

5 mark questions

Directions: Completely solve all the questions. All the calculations are to be shown in the solution. Also draw the diagram if it is given in the question or is asked in the question.

Question 9. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Answer:
Let the wall, ground, and ladder form a right-angled triangle where:
Length of the ladder (Hypotenuse) = 15 m
Height of the window (Perpendicular) = 12 m
Distance of the foot of the ladder from the wall (Base) = \(a\)
Using Pythagoras theorem:
\(\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\)
\(15^2 = a^2 + 12^2\)
\(225 = a^2 + 144\)
\(a^2 = 225 - 144\)
\(a^2 = 81\)
\(a = \sqrt{81} = 9 \text{ m}\)
The distance of the foot of the ladder from the wall is 9 m.
In simple words: Use the right triangle rule where the ladder is the hypotenuse leaning against the vertical wall.

Exam Tip: Draw a neat rough sketch of a wall, ground, and leaning ladder to clearly identify the right triangle.

Question 10. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Answer:
Let the tree break at point B, height from ground \(AB = 5 \text{ m}\).
The top of the tree touches the ground at point C, distance from base \(AC = 12 \text{ m}\).
The broken part forms the hypotenuse BC of the right-angled triangle ABC.
Using Pythagoras theorem:
\(BC^2 = AB^2 + AC^2\)
\(BC^2 = 5^2 + 12^2\)
\(BC^2 = 25 + 144\)
\(BC^2 = 169\)
\(BC = \sqrt{169} = 13 \text{ m}\)
The original height of the tree is the sum of the standing part and the broken leaning part:
\(\text{Original height} = AB + BC = 5 + 13 = 18 \text{ m}\)
In simple words: The broken top forms the slanted hypotenuse. Add it to the standing stem to get the total tree height.

Exam Tip: Do not forget to add the lower standing portion (5 m) to the calculated hypotenuse (13 m) at the end.

Question 11. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer:
Let the length of the rectangle be \(l = 40 \text{ cm}\) and breadth be \(b\).
The diagonal \(d = 41 \text{ cm}\) divides the rectangle into two right-angled triangles where the diagonal is the hypotenuse.
Using Pythagoras theorem:
\(d^2 = l^2 + b^2\)
\(41^2 = 40^2 + b^2\)
\(1681 = 1600 + b^2\)
\(b^2 = 1681 - 1600\)
\(b^2 = 81\)
\(b = \sqrt{81} = 9 \text{ cm}\)
Now, calculate the perimeter of the rectangle:
\(\text{Perimeter} = 2 \times (l + b)\)
\(\text{Perimeter} = 2 \times (40 + 9)\)
\(\text{Perimeter} = 2 \times 49 = 98 \text{ cm}\)
In simple words: Find the breadth using the diagonal and length via Pythagoras theorem, then plug both into the perimeter formula.

Exam Tip: Perimeter formula for a rectangle is \(2(l + b)\).

Question 12. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Answer:
Let the diagonals of the rhombus be \(d_1 = 16 \text{ cm}\) and \(d_2 = 30 \text{ cm}\).
The diagonals of a rhombus bisect each other at right angles (\(90^{\circ}\)).
Half of the first diagonal = \(\frac{16}{2} = 8 \text{ cm}\)
Half of the second diagonal = \(\frac{30}{2} = 15 \text{ cm}\)
Each side \(s\) of the rhombus forms the hypotenuse of a right-angled triangle with these half-diagonals as its legs.
Using Pythagoras theorem:
\(s^2 = 8^2 + 15^2\)
\(s^2 = 64 + 225\)
\(s^2 = 289\)
\(s = \sqrt{289} = 17 \text{ cm}\)
Since all four sides of a rhombus are equal:
\(\text{Perimeter} = 4 \times s = 4 \times 17 = 68 \text{ cm}\)
In simple words: Diagonals bisect at right angles. Use half the diagonals to find one side, then multiply by 4.

Question 13. Explain, why \(\Delta ABC \cong \Delta FED\)?
Answer:
Based on the given figure markings:
1. Side \(AB = FE\) (indicated by single tick marks)
2. Side \(BC = ED\) (indicated by double tick marks)
3. Since triangle ABC and triangle FED have all three corresponding sides equal (including \(AC = FD\) by third side property or given markers), by the SSS (Side-Side-Side) congruence criterion, \(\Delta ABC \cong \Delta FED\).
In simple words: The tick marks on the sides show that corresponding lengths are equal, proving congruence by SSS.

Exam Tip: Always check the tick markings on sides and angles to identify which congruence rule applies.