CBSE Class 8 Mathematics Practice Worksheet Set 11

WORK SHEET

CLASS: 8th | SUBJECT: Mathematics | LESSON: Squares and Square Roots

Question 1. Find the square root of the following numbers by prime factorisation: 361, 729, 4489, 3025, 7056, 8281
Answer:
(i) For 361:
\(361 = 19 \times 19\)
Square root = \(\sqrt{361} = 19\)
(ii) For 729:
\(729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6\)
Square root = \(\sqrt{729} = 3 \times 3 \times 3 = 27\)
(iii) For 4489:
\(4489 = 67 \times 67\)
Square root = \(\sqrt{4489} = 67\)
(iv) For 3025:
\(3025 = 5 \times 5 \times 11 \times 11\)
Square root = \(\sqrt{3025} = 5 \times 11 = 55\)
(v) For 7056:
\(7056 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7\)
Square root = \(\sqrt{7056} = 2 \times 2 \times 3 \times 7 = 84\)
(vi) For 8281:
\(8281 = 91 \times 91\)
Square root = \(\sqrt{8281} = 91\)
In simple words: Break each number down into its prime factors, group them in pairs, and take one factor from each pair to find the square root.

Exam Tip: Always verify your prime factors by multiplying them back before taking the square root.

Question 2. Find the smallest number by which 2475 must be multiplied to get a perfect square.
Answer:
Find prime factors of 2475:
\(2475 = 3 \times 3 \times 5 \times 5 \times 11\)
Pairing the prime factors: \((3 \times 3) \times (5 \times 5) \times 11\)
The prime factor 11 is left unpaired.
Therefore, 2475 must be multiplied by 11 to make it a perfect square.
New number = \(2475 \times 11 = 27225\), and its square root is \(3 \times 5 \times 11 = 165\).
In simple words: Find the prime factors, make pairs of identical numbers, and the number that lacks a pair is the multiplier needed.

Exam Tip: For "multiplied to get a perfect square", find the prime factor with an odd exponent and multiply by it.

Question 3. Find the smallest number by which 3528 must be divided to get a perfect square.
Answer:
Find prime factors of 3528:
\(3528 = 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7\)
Pairing the prime factors: \((2 \times 2) \times (3 \times 3) \times (7 \times 7) \times 2\)
The prime factor 2 is left unpaired.
Therefore, 3528 must be divided by 2 to make it a perfect square.
New number = \(3528 \div 2 = 1764\), and its square root is \(2 \times 3 \times 7 = 42\).
In simple words: Divide the number by the leftover prime factor that does not form a pair.

Exam Tip: Both multiplication and division questions rely on complete prime factorisation.

Question 4. Find the least perfect square number exactly divisible by each one of the numbers 4, 5 and 10.
Answer:
First, find the LCM of 4, 5, and 10:
\(\text{LCM}(4, 5, 10) = 2 \times 2 \times 5 = 20\)
Prime factors of 20 = \(2 \times 2 \times 5\)
To make it a perfect square, 20 must be multiplied by 5 (since 5 is unpaired).
Required least perfect square = \(20 \times 5 = 100\).
In simple words: First find the LCM of the given numbers, then find the smallest number to multiply it with so all prime factors have pairs.

Exam Tip: Always find the LCM first for questions involving numbers that must divide the target number.

Question 5. Find the square root of each of the following by long division method: 961, 1089, 7396, 6889, 5929, 10201
Answer:
(i) For 961: Group digits as 09 61. Square root = 31.
(ii) For 1089: Group digits as 10 89. Square root = 33.
(iii) For 7396: Group digits as 73 96. Square root = 86.
(iv) For 6889: Group digits as 68 89. Square root = 83.
(v) For 5929: Group digits as 59 29. Square root = 77.
(vi) For 10201: Group digits as 01 02 01. Square root = 101.
In simple words: Pair digits from right to left, find the largest square fitting the first pair, bring down the next pair, double the quotient, and continue.

Question 6. Find the least number which must be subtracted from 2361 to make it a perfect square.
Answer:
Using long division for 2361:
Pair digits: 23 61
\(4 \times 4 = 16\) (Remainder = 7)
Bring down 61 to get 761.
Double 4 to get 8_. Trial: \(88 \times 8 = 704\).
Remainder = \(761 - 704 = 57\).
Thus, subtracting 57 from 2361 gives \(2361 - 57 = 2304\), which is \(48^2\).
Required least number to subtract = 57.
In simple words: Perform long division and the final remainder is the amount that must be subtracted.

Exam Tip: The remainder of the long division process directly gives the number to be subtracted.

Question 7. Find the least number which must be added to 5678 to make it a perfect square.
Answer:
Using long division for 5678:
Pair digits: 56 78
\(7 \times 7 = 49\) (Remainder = 7)
Bring down 78 to get 778.
Double 7 to get 14_. Trial: \(145 \times 5 = 725\).
Remainder = \(778 - 725 = 53\).
Since \(75^2 < 5678 < 76^2\), the next perfect square is \(76^2 = 5776\).
Number to be added = \(5776 - 5678 = 98\).
In simple words: Find the next integer's square and subtract the given number from it.

Exam Tip: For addition problems, calculate \(( \text{quotient} + 1 )^2\) and subtract the original number.

Question 8. Find the greatest three-digit number which is a perfect square.
Answer:
The greatest 3-digit number is 999.
Using long division on 999, we find \(31^2 = 961\) and \(32^2 = 1024\).
Therefore, the greatest 3-digit perfect square is \(31^2 = 961\).
In simple words: Test numbers near the root of 1000 to find the largest 3-digit square.

Exam Tip: The square of 31 is 961, and the next square 1024 is already a 4-digit number.

Question 9. Find the square root of each of the following: (i) 39.69 (ii) 44.89 (iii) 0.0625 (iv) 0.01 (v) \(6\frac{1}{4}\)
Answer:
(i) \(\sqrt{39.69} = 6.3\)
(ii) \(\sqrt{44.89} = 6.7\)
(iii) \(\sqrt{0.0625} = 0.25\)
(iv) \(\sqrt{0.01} = 0.1\)
(v) \(6\frac{1}{4} = \frac{25}{4}\), so \(\sqrt{\frac{25}{4}} = \frac{5}{2} = 2.5\)
In simple words: Count decimal places or convert fractions before computing the square root.

Exam Tip: The number of decimal places in the square root is half the number of decimal places in the original decimal number.

Question 10. Find the length of the side of a square field of area 7569 sq m. Hence find its perimeter.
Answer:
Area of square = \(\text{side}^2 = 7569\)
Side = \(\sqrt{7569} = 87\text{ m}\)
Perimeter of square = \(4 \times \text{side} = 4 \times 87 = 348\text{ m}\)
In simple words: Take the square root of the area to get the side length, then multiply by 4 to get the perimeter.

Exam Tip: Remember that area uses square units (\(\text{m}^2\)) while perimeter uses linear units (\(\text{m}\)).

Question 11. Find the least four digit perfect square number.
Answer:
The smallest 4-digit number is 1000.
We know \(31^2 = 961\) (3-digit) and \(32^2 = 1024\) (4-digit).
Therefore, the least 4-digit perfect square is 1024.
In simple words: Find the square of the number immediately following the square root of 1000.

Question 12. Find (i) \(\sqrt{18} \times \sqrt{8}\) (ii) \(\sqrt{75} \times \sqrt{12}\) (iii) \(\sqrt{14.4} \times \sqrt{0.1}\) (iv) \(\sqrt{12} \times \sqrt{27}\)
Answer:
(i) \(\sqrt{18} \times \sqrt{8} = \sqrt{18 \times 8} = \sqrt{144} = 12\)
(ii) \(\sqrt{75} \times \sqrt{12} = \sqrt{75 \times 12} = \sqrt{900} = 30\)
(iii) \(\sqrt{14.4} \times \sqrt{0.1} = \sqrt{14.4 \times 0.1} = \sqrt{1.44} = 1.2\)
(iv) \(\sqrt{12} \times \sqrt{27} = \sqrt{12 \times 27} = \sqrt{324} = 18\)
In simple words: Combine terms under a single square root using \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\) and then simplify.

Exam Tip: Multiplying square roots is often easier when you combine them under one radical before simplifying.

Question 13. Using the identity \(a^2 - b^2 = (a+b)(a-b)\), find (i) \(25^2 - 24^2\) (ii) \(5^2 - 4^2\) (iii) \(3.5^2 - 2.5^2\)
Answer:
(i) \(25^2 - 24^2 = (25 + 24)(25 - 24) = (49)(1) = 49\)
(ii) \(5^2 - 4^2 = (5 + 4)(5 - 4) = (9)(1) = 9\)
(iii) \(3.5^2 - 2.5^2 = (3.5 + 2.5)(3.5 - 2.5) = (6.0)(1.0) = 6\)
In simple words: Apply the difference of squares formula by adding and subtracting the two base numbers.

Exam Tip: This identity saves time compared to calculating large squares individually.

Question 14. Solve for \(x\): (i) \(x^2 - 1 = 24\) (ii) \(x^2 + 1 = 65\) (iii) \(2x^2 - 1 = 17\) (iv) \(x^2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15\)
Answer:
(i) \(x^2 - 1 = 24\)
\(x^2 = 25\)
\(x = \pm 5\)
(ii) \(x^2 + 1 = 65\)
\(x^2 = 64\)
\(x = \pm 8\)
(iii) \(2x^2 - 1 = 17\)
\(2x^2 = 18\)
\(x^2 = 9\)
\(x = \pm 3\)
(iv) \(x^2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15\) (Sum of first 8 odd numbers is \(8^2 = 64\))
\(x^2 = 64\)
\(x = \pm 8\)
In simple words: Isolate \(x^2\) on one side of the equation and take the square root of both sides.

Exam Tip: Always remember to include both positive and negative roots when solving \(x^2 = k\).

Question 15. Express 100 as the sum of ten odd numbers.
Answer:
The sum of the first \(n\) odd numbers is \(n^2\). Here, \(10^2 = 100\).
Therefore, 100 can be expressed as the sum of the first 10 odd numbers:
\(100 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19\)
In simple words: Use the property that the sum of the first \(n\) odd numbers equals \(n^2\).

Exam Tip: List consecutive odd numbers starting from 1 until you reach the desired count.

Question 16. How many numbers are there between the squares of (i) 12 and 13 (ii) 3 and 4
Answer:
The number of non-square numbers between \(n^2\) and \((n+1)^2\) is \(2n\).
(i) For 12 and 13: \(n = 12\), so number of numbers = \(2 \times 12 = 24\).
(ii) For 3 and 4: \(n = 3\), so number of numbers = \(2 \times 3 = 6\).
In simple words: Multiply the smaller number by 2 to find how many non-square numbers lie between two consecutive squares.

Exam Tip: Use the shortcut formula \(2n\) where \(n\) is the smaller of the two consecutive numbers.

Question 17. Find the other two members of the Pythagorean triplet, one of whose members is: (i) 10 (ii) 14 (iii) 20 (iv) 26
Answer:
Pythagorean triplets are given in the form \(2m, m^2 - 1, m^2 + 1\).
(i) If \(2m = 10\), then \(m = 5\).
Other members: \(m^2 - 1 = 25 - 1 = 24\), and \(m^2 + 1 = 25 + 1 = 26\).
Triplet: (10, 24, 26).
(ii) If \(2m = 14\), then \(m = 7\).
Other members: \(m^2 - 1 = 49 - 1 = 48\), and \(m^2 + 1 = 49 + 1 = 50\).
Triplet: (14, 48, 50).
(iii) If \(2m = 20\), then \(m = 10\).
Other members: \(m^2 - 1 = 100 - 1 = 99\), and \(m^2 + 1 = 100 + 1 = 101\).
Triplet: (20, 99, 101).
(iv) If \(2m = 26\), then \(m = 13\).
Other members: \(m^2 - 1 = 169 - 1 = 168\), and \(m^2 + 1 = 169 + 1 = 170\).
Triplet: (26, 168, 170).
In simple words: Equate the given even number to \(2m\) to find \(m\), then calculate \(m^2 - 1\) and \(m^2 + 1\).

Exam Tip: Always assume the given even number is \(2m\) when generating standard Pythagorean triplets.

Question 18. The length and breadth of a rectangle are respectively 15 cm and 8 cm. Find its diagonal.
Answer:
Length \(l = 15\text{ cm}\), Breadth \(b = 8\text{ cm}\)
Diagonal \(d = \sqrt{l^2 + b^2}\)
\(d = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ cm}\)
In simple words: Use Pythagoras theorem on the rectangle's sides to find the diagonal length.

Exam Tip: The diagonal divides a rectangle into two right-angled triangles.

Question 20. In an auditorium, the number of rows is equal to the number of chairs in each row. If there are 1764 chairs in the auditorium, find the number of chairs in each row.
Answer:
Let the number of rows be \(x\). Then chairs per row = \(x\).
Total chairs = \(x \times x = x^2\)
\(x^2 = 1764\)
\(x = \sqrt{1764} = 42\)
Number of chairs in each row = 42.
In simple words: Take the square root of the total number of chairs when rows and columns are equal.

Exam Tip: Whenever rows and items per row are equal, the total count is always a square number.

Question 21. A ladder reached a window 16m high from the ground on placing it against a wall at a distance of 12 m. Find the length of the ladder (\(l\)).
Answer:
Height of window (\(a\)) = 16 m
Distance from wall (\(b\)) = 12 m
Length of ladder (\(l\)) = \(\sqrt{16^2 + 12^2}\)
\(l = \sqrt{256 + 144} = \sqrt{400} = 20\text{ m}\)
In simple words: Apply Pythagoras theorem where the ladder is the hypotenuse of the right triangle formed by the wall and ground.

Exam Tip: This is a classic 3-4-5 right triangle scaled up by a factor of 4 (\(3\times4=12, 4\times4=16, 5\times4=20\)).

Question 22. A tree is cut off at a height of 5m from the ground and its top touches the ground at a distance of 12 m. Find the original height of the tree.
Answer:
Height of remaining stump = 5 m
Distance from base to top on ground = 12 m
Length of fallen part (hypotenuse) = \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ m}\)
Original height of tree = Stump height + Fallen part = \(5 + 13 = 18\text{ m}\)
In simple words: Find the length of the fallen top using Pythagoras theorem and add it to the height of the remaining stump.

Exam Tip: Do not forget to add the stump height back to the hypotenuse to get the total original height.

Question 23. Find the area of a triangle if area of \(\Delta = \sqrt{s(s-a)(s-b)(s-c)}\) given \(a = 9\text{ cm}, b = 12\text{ cm}, c = 15\text{ cm}\) and \(s = 18\text{ cm}\).
Answer:
Using Heron's formula:
\(s = 18\)
\(s - a = 18 - 9 = 9\)
\(s - b = 18 - 12 = 6\)
\(s - c = 18 - 15 = 3\)
Area = \(\sqrt{18 \times 9 \times 6 \times 3}\)
\(= \sqrt{(9 \times 2) \times 9 \times (3 \times 2) \times 3}\)
\(= \sqrt{9^2 \times 3^2 \times 2^2} = 9 \times 3 \times 2 = 54\text{ cm}^2\)
In simple words: Substitute the side lengths and semi-perimeter into Heron's formula and simplify the square root.

Exam Tip: Group identical factors under the radical to easily extract them without multiplying large numbers.

WORK SHEET

CLASS: 8th | SUBJECT: Mathematics | LESSON: Understanding Quadrilaterals

Question 1. Three angles of a quadrilateral are \(100^\circ, 92^\circ\) and \(48^\circ\). Find the fourth angle.
Answer:
Sum of all angles of a quadrilateral = \(360^\circ\)
Fourth angle = \(360^\circ - (100^\circ + 92^\circ + 48^\circ)\)
\(= 360^\circ - 240^\circ = 120^\circ\)
In simple words: Subtract the sum of the three given angles from \(360^\circ\).

Exam Tip: Angle sum property of any quadrilateral is always \(360^\circ\).

Question 2. The four angles of a quadrilateral are in the ratio \(3:4:5:6\). Find their measures.
Answer:
Let the angles be \(3x, 4x, 5x,\) and \(6x\).
Sum of angles = \(3x + 4x + 5x + 6x = 18x\)
\(18x = 360^\circ\)
\(x = \frac{360^\circ}{18} = 20^\circ\)
Angles are:
\(3(20^\circ) = 60^\circ\)
\(4(20^\circ) = 80^\circ\)
\(5(20^\circ) = 100^\circ\)
\(6(20^\circ) = 120^\circ\)
In simple words: Assign a variable to the ratio parts, equate their sum to \(360^\circ\), solve for \(x\), and find each angle.

Exam Tip: Always check that the sum of your calculated angles equals \(360^\circ\).

Question 3. In a parallelogram ABCD, \(\angle A = (3x - 10)^\circ\) and \(\angle B = (5x + 30)^\circ\), find \(x\) and hence find all the angles of the parallelogram.
Answer:
Adjacent angles of a parallelogram are supplementary:
\(\angle A + \angle B = 180^\circ\)
\((3x - 10) + (5x + 30) = 180\)
\(8x + 20 = 180\)
\(8x = 160\)
\(x = 20\)
Substituting \(x = 20\):
\(\angle A = 3(20) - 10 = 50^\circ\)
\(\angle B = 5(20) + 30 = 130^\circ\)
Opposite angles are equal, so \(\angle C = \angle A = 50^\circ\) and \(\angle D = \angle B = 130^\circ\).
Angles are \(50^\circ, 130^\circ, 50^\circ, 130^\circ\).
In simple words: Set the sum of adjacent angles to \(180^\circ\) to find \(x\), then use opposite angle properties for the rest.

Exam Tip: Remember that opposite angles in a parallelogram are equal, and adjacent angles add up to \(180^\circ\).

Question 4. In the figure, ABCD is a parallelogram. If \(\angle DCE = 65^\circ\), find \(\angle A\) and \(\angle B\).
Answer:
Line DC is extended to E, forming a linear pair at vertex C:
\(\angle DCB + \angle DCE = 180^\circ\)
\(\angle DCB + 65^\circ = 180^\circ\)
\(\angle DCB = 180^\circ - 65^\circ = 115^\circ\)
Since ABCD is a parallelogram, opposite angles are equal:
\(\angle A = \angle DCB = 115^\circ\)
Adjacent angles are supplementary, so:
\(\angle B = 180^\circ - \angle A = 180^\circ - 115^\circ = 65^\circ\)
In simple words: Use the linear pair property to find interior angle C, then use parallelogram properties to find angles A and B.

Exam Tip: Exterior angles on a straight line always form a linear pair with the adjacent interior angle.

Question 5. ABCD is an isosceles trapezium with \(AB \parallel DC\) and \(AD = BC\). If \(\angle A = 130^\circ\), find \(\angle B\), \(\angle C\) and \(\angle D\).
Answer:
In a trapezium with \(AB \parallel DC\), adjacent angles on the same side of a transversal are supplementary:
\(\angle A + \angle D = 180^\circ \implies 130^\circ + \angle D = 180^\circ \implies \angle D = 50^\circ\)
Similarly, \(\angle B + \angle C = 180^\circ\).
In an isosceles trapezium, base angles are equal: \(\angle B = \angle A = 130^\circ\) and \(\angle C = \angle D = 50^\circ\).
Thus, \(\angle B = 130^\circ, \angle C = 50^\circ, \angle D = 50^\circ\).
In simple words: Base angles of an isosceles trapezium are equal, and consecutive interior angles along the legs add up to \(180^\circ\).

Question 6. In a kite ABCD, \(AB = AD\) and \(BC = DC\). If \(\angle A = 80^\circ\) and \(\angle C = 40^\circ\), find \(\angle B\) and \(\angle D\).
Answer:
In a kite, the non-vertex angles are equal: \(\angle B = \angle D\).
Sum of all angles in quadrilateral ABCD = \(360^\circ\)
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
\(80^\circ + \angle B + 40^\circ + \angle B = 360^\circ\)
\(2\angle B + 120^\circ = 360^\circ\)
\(2\angle B = 240^\circ\)
\(\angle B = 120^\circ\)
Therefore, \(\angle B = 120^\circ\) and \(\angle D = 120^\circ\).
In simple words: Use the property that the opposite angles between unequal sides of a kite are equal.

Exam Tip: The diagonal connecting the vertex angles of a kite bisects the non-vertex angles.

Question 7. In the figure, ABCD is a parallelogram. If \(\angle CAB = 32^\circ\), \(\angle ABC = 110^\circ\), find \(x^\circ, y^\circ\) and \(z^\circ\).
Answer:
In triangle ABC:
\(\angle CAB = 32^\circ\)
\(\angle ABC = 110^\circ\)
\(\angle ACB = 180^\circ - (110^\circ + 32^\circ) = 180^\circ - 142^\circ = 38^\circ\)
Since \(AB \parallel DC\) and AC is a transversal, alternate interior angles are equal:
\(x^\circ = \angle ACB = 38^\circ\)
Also, alternate interior angles for parallel lines AD and BC give:
\(y^\circ = \angle CAB = 32^\circ\)
For angle \(z^\circ\), using triangle properties or linear pairs depending on internal diagonals: \(z^\circ = 180^\circ - (110^\circ + 32^\circ)\) or matching alternate angles. Given standard parallel line geometry, \(z^\circ = 38^\circ\).
In simple words: Use angle sum property of triangles and alternate interior angles formed by parallel lines and transversals.

Exam Tip: Look for Z-shapes (alternate interior angles) when parallel lines are cut by a transversal.

Question 8. ABCD is a trapezium with \(AD \parallel BC\), \(\angle A = (3x + 1)^\circ\) and \(\angle B = (2x + 4)^\circ\). Find \(x\).
Answer:
Since \(AD \parallel BC\), angles A and B are consecutive interior angles on the transversal AB, so they are supplementary:
\(\angle A + \angle B = 180^\circ\)
\((3x + 1) + (2x + 4) = 180\)
\(5x + 5 = 180\)
\(5x = 175\)
\(x = 35\)
In simple words: Add angles A and B to equal \(180^\circ\) because they lie between parallel lines on the same side.

Exam Tip: Co-interior angles between parallel lines always sum to \(180^\circ\).

Question 9. In a trapezium ABCD, \(AB \parallel DC\), \(\angle A : \angle D = 3 : 2\) and \(\angle B : \angle C = 4 : 5\), find all the angles of the trapezium.
Answer:
Since \(AB \parallel DC\), angles A and D are supplementary: \(\angle A + \angle D = 180^\circ\).
Given ratio \(\angle A : \angle D = 3 : 2\), let \(\angle A = 3x\) and \(\angle D = 2x\).
\(3x + 2x = 180^\circ \implies 5x = 180^\circ \implies x = 36^\circ\).
\(\angle A = 3(36^\circ) = 108^\circ\), \(\angle D = 2(36^\circ) = 72^\circ\).
Similarly, for B and C: \(\angle B + \angle C = 180^\circ\). Let \(\angle B = 4y\) and \(\angle C = 5y\).
\(4y + 5y = 180^\circ \implies 9y = 180^\circ \implies y = 20^\circ\).
\(\angle B = 4(20^\circ) = 80^\circ\), \(\angle C = 5(20^\circ) = 100^\circ\).
Angles are \(\angle A = 108^\circ, \angle B = 80^\circ, \angle C = 100^\circ, \angle D = 72^\circ\).
In simple words: Divide \(180^\circ\) according to the given ratio for each pair of adjacent angles.

Exam Tip: Remember that adjacent angles on both legs of a trapezium are supplementary independently.

Question 10. In a parallelogram PQRS, \(\angle P\) is \(30^\circ\) more than \(\angle Q\). Find the measures of all the angles.
Answer:
Let \(\angle Q = x\). Then \(\angle P = x + 30^\circ\).
Adjacent angles in a parallelogram are supplementary:
\(\angle P + \angle Q = 180^\circ\)
\((x + 30) + x = 180\)
\(2x + 30 = 180\)
\(2x = 150 \implies x = 75^\circ\)
\(\angle Q = 75^\circ\), and \(\angle P = 75^\circ + 30^\circ = 105^\circ\).
By opposite angle properties: \(\angle R = \angle P = 105^\circ\) and \(\angle S = \angle Q = 75^\circ\).
Angles are \(105^\circ, 75^\circ, 105^\circ, 75^\circ\).
In simple words: Set up an equation where adjacent angles sum to \(180^\circ\), solve for \(x\), and find all four angles.

Question 11. The adjacent angles of a parallelogram are in the ratio \(2 : 7\). Find their measures.
Answer:
Let the adjacent angles be \(2x\) and \(7x\).
Their sum is \(180^\circ\) (supplementary adjacent angles):
\(2x + 7x = 180^\circ\)
\(9x = 180^\circ \implies x = 20^\circ\)
Angles are \(2(20^\circ) = 40^\circ\) and \(7(20^\circ) = 140^\circ\).
All four angles of the parallelogram are \(40^\circ, 140^\circ, 40^\circ, 140^\circ\).
In simple words: Divide \(180^\circ\) in the ratio \(2:7\) to find the adjacent angles.

Exam Tip: Ratio problems are solved by attaching a variable \(x\) to each ratio term.

Question 12. P is a point in the interior of \(\angle AOB\). \(PM \perp OA\) and \(PN \perp OB\). If \(\angle AOB = 60^\circ\), find \(\angle P\).
Answer:
In quadrilateral PMON (where O is the vertex of angle AOB, M is on OA, and N is on OB):
\(\angle PMO = 90^\circ\) (since \(PM \perp OA\))
\(\angle PNO = 90^\circ\) (since \(PN \perp OB\))
\(\angle MON = \angle AOB = 60^\circ\)
Sum of angles in quadrilateral PMON = \(360^\circ\)
\(\angle P + \angle PMO + \angle MON + \angle PNO = 360^\circ\)
\(\angle P + 90^\circ + 60^\circ + 90^\circ = 360^\circ\)
\(\angle P + 240^\circ = 360^\circ \implies \angle P = 120^\circ\)
In simple words: Use the angle sum property of quadrilateral PMON, knowing that perpendiculars create \(90^\circ\) angles.

Exam Tip: Perpendicular lines always form right angles (\(90^\circ\)).

Question 13. One of the diagonals of a rhombus is equal to one of its sides. Find the angles of the rhombus.
Answer:
Let ABCD be a rhombus where diagonal BD equals side AB. Then triangle ABD is an equilateral triangle.
Therefore, \(\angle A = 60^\circ\), and \(\angle ABD = \angle ADB = 60^\circ\).
Since opposite angles of a rhombus are equal, \(\angle C = \angle A = 60^\circ\).
Adjacent angles are supplementary: \(\angle B = 180^\circ - 60^\circ = 120^\circ\), and \(\angle D = 120^\circ\).
The angles of the rhombus are \(60^\circ, 120^\circ, 60^\circ, 120^\circ\).
In simple words: An equilateral triangle is formed inside the rhombus, giving \(60^\circ\) and \(120^\circ\) as its angles.

Exam Tip: A rhombus with a diagonal equal to its side is composed of two equilateral triangles.

Question 14. Find the perimeter of a rhombus whose diagonals are 6 cm and 8 cm.
Answer:
Diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Half diagonals are \(d_1/2 = 6/2 = 3\text{ cm}\) and \(d_2/2 = 8/2 = 4\text{ cm}\).
Using Pythagoras theorem in one of the four right triangles formed at the center:
\(\text{Side}^2 = 3^2 + 4^2 = 9 + 16 = 25\)
\(\text{Side} = \sqrt{25} = 5\text{ cm}\)
Perimeter of rhombus = \(4 \times \text{Side} = 4 \times 5 = 20\text{ cm}\)
In simple words: Half the diagonals form a right-angled triangle with the side of the rhombus as the hypotenuse.

Exam Tip: Diagonals of a rhombus bisect each perpendicularly, creating four congruent right triangles.

Question 15. Lengths of two sides of a parallelogram are in the ratio \(2:3\). Find the sides of the parallelogram if its perimeter = 120 cm.
Answer:
Let the sides be \(2x\) and \(3x\).
Perimeter of parallelogram = \(2(\text{side}_1 + \text{side}_2)\)
\(2(2x + 3x) = 120\)
\(2(5x) = 120 \implies 10x = 120 \implies x = 12\text{ cm}\)
Side 1 = \(2(12) = 24\text{ cm}\)
Side 2 = \(3(12) = 36\text{ cm}\)
The sides of the parallelogram are 24 cm, 36 cm, 24 cm, and 36 cm.
In simple words: Use the perimeter formula for a parallelogram and solve for the variable \(x\).

Exam Tip: Perimeter is the total distance around the outside, calculated as twice the sum of adjacent sides.

Question 16. Find the measure of an exterior angle of a regular (i) Hexagon (ii) Octagon (iii) Decagon
Answer:
Formula for exterior angle of a regular polygon of \(n\) sides = \(\frac{360^\circ}{n}\)
(i) Hexagon (\(n = 6\)): Exterior angle = \(\frac{360^\circ}{6} = 60^\circ\)
(ii) Octagon (\(n = 8\)): Exterior angle = \(\frac{360^\circ}{8} = 45^\circ\)
(iii) Decagon (\(n = 10\)): Exterior angle = \(\frac{360^\circ}{10} = 36^\circ\)
In simple words: Divide \(360^\circ\) by the number of sides to get the measure of each exterior angle.

Exam Tip: The sum of all exterior angles of any convex polygon is always \(360^\circ\).

Question 17. Find the sum of the interior angles of a (i) Heptagon (ii) Nonagon (iii) Pentagon
Answer:
Formula for sum of interior angles = \((n - 2) \times 180^\circ\)
(i) Heptagon (\(n = 7\)): Sum = \((7 - 2) \times 180^\circ = 5 \times 180^\circ = 900^\circ\)
(ii) Nonagon (\(n = 9\)): Sum = \((9 - 2) \times 180^\circ = 7 \times 180^\circ = 1260^\circ\)
(iii) Pentagon (\(n = 5\)): Sum = \((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\)
In simple words: Subtract 2 from the number of sides and multiply the result by \(180^\circ\).

Exam Tip: This formula comes from dividing an \(n\)-sided polygon into \(n-2\) triangles.

Question 18. Each interior angle of a regular polygon is \(140^\circ\). Find the number of sides of the polygon.
Answer:
Exterior angle = \(180^\circ - \text{Interior angle} = 180^\circ - 140^\circ = 40^\circ\)
Number of sides \(n = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{40^\circ} = 9\)
The polygon has 9 sides (nonagon).
In simple words: Find the exterior angle first by subtracting the interior angle from \(180^\circ\), then divide \(360^\circ\) by it.

Exam Tip: Working with exterior angles is usually much faster than using the interior angle formula directly.

Question 19. The measure of an exterior angle of a regular polygon is \(72^\circ\). What is the name of the polygon?
Answer:
Number of sides \(n = \frac{360^\circ}{72^\circ} = 5\)
A 5-sided regular polygon is called a pentagon.
In simple words: Divide \(360^\circ\) by the exterior angle to find the number of sides.

Question 20. Find the measure of an interior angle of a regular hexagon.
Answer:
For a regular hexagon (\(n = 6\)), exterior angle = \(\frac{360^\circ}{6} = 60^\circ\)
Interior angle = \(180^\circ - \text{Exterior angle} = 180^\circ - 60^\circ = 120^\circ\)
In simple words: Find the exterior angle first and subtract it from \(180^\circ\) to get the interior angle.

Exam Tip: Interior and exterior angles at any vertex of a polygon always form a linear pair.

Question 21. The ratio of an interior angle to the exterior angle of a regular polygon is \(5:1\). Find the number of sides.
Answer:
Let interior angle = \(5x\) and exterior angle = \(x\).
Since interior and exterior angles are supplementary:
\(5x + x = 180^\circ \implies 6x = 180^\circ \implies x = 30^\circ\)
Exterior angle = \(30^\circ\)
Number of sides \(n = \frac{360^\circ}{30^\circ} = 12\)
In simple words: Use the ratio to find the exterior angle, then divide \(360^\circ\) by it to find the number of sides.

Exam Tip: Adjacent interior and exterior angles always add up to \(180^\circ\).

Question 22. The sum of the interior angles of a polygon is \(540^\circ\). What is the name of the polygon?
Answer:
Sum of interior angles = \((n - 2) \times 180^\circ = 540^\circ\)
\(n - 2 = \frac{540}{180} = 3\)
\(n = 3 + 2 = 5\)
A 5-sided polygon is a pentagon.
In simple words: Divide the angle sum by \(180^\circ\) and add 2 to find the number of sides.

Exam Tip: A polygon with a sum of \(540^\circ\) has 5 sides and is a pentagon.

Question 23. How many diagonals can we draw in a (i) pentagon (ii) Hexagon (iii) Triangle (iv) Quadrilateral
Answer:
Formula for number of diagonals in an \(n\)-sided polygon = \(\frac{n(n-3)}{2}\)
(i) Pentagon (\(n = 5\)): Diagonals = \(\frac{5(5-3)}{2} = \frac{5 \times 2}{2} = 5\)
(ii) Hexagon (\(n = 6\)): Diagonals = \(\frac{6(6-3)}{2} = \frac{6 \times 3}{2} = 9\)
(iii) Triangle (\(n = 3\)): Diagonals = \(\frac{3(3-3)}{2} = 0\)
(iv) Quadrilateral (\(n = 4\)): Diagonals = \(\frac{4(4-3)}{2} = \frac{4 \times 1}{2} = 2\)
In simple words: Use the formula \(\frac{n(n-3)}{2}\) where \(n\) is the number of sides.

Exam Tip: Triangles have 0 diagonals because no two non-adjacent vertices exist.