CBSE Class 9 Mathematics Revision Worksheet Set 03

Question 1. Simplify: (i) \([5(8^{\frac{1}{3}} + 27^{\frac{1}{3}})]^{\frac{1}{4}}\) (ii) \(3\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}}\) (iii) \(4\sqrt{81} - 8\sqrt[3]{216} + 15\sqrt[5]{32} + \sqrt{225}\) (iv) \(\left(\frac{3}{5}\right)^x \left(\frac{5}{3}\right)^{-12} = \left(\frac{3}{5}\right)^6\) (v) \(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-2/3}}\) (vi) \(64^{-\frac{1}{3}} \left[64^{\frac{1}{3}} - 64^{\frac{2}{3}}\right]\) (vii) \((256)^{-(4^{-\frac{1}{2}})}\)
Answer:
(i) \([5(8^{\frac{1}{3}} + 27^{\frac{1}{3}})]^{\frac{1}{4}} = [5((2^3)^{\frac{1}{3}} + (3^3)^{\frac{1}{3}})]^{\frac{1}{4}} = [5(2 + 3)]^{\frac{1}{4}} = [5 \times 5]^{\frac{1}{4}} = [5^2]^{\frac{1}{4}} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt{5}\)
(ii) \(3\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} = 3\sqrt{3} + 2(3\sqrt{3}) + \frac{1}{\sqrt{3}} = 3\sqrt{3} + 6\sqrt{3} + \frac{1}{\sqrt{3}} = 9\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{27 + 1}{\sqrt{3}} = \frac{28}{\sqrt{3}} = \frac{28\sqrt{3}}{3}\)
(iii) \(4\sqrt{81} - 8\sqrt[3]{216} + 15\sqrt[5]{32} + \sqrt{225} = 4(9) - 8(6) + 15(2) + 15 = 36 - 48 + 30 + 15 = 33\)
(iv) \(\left(\frac{3}{5}\right)^x \left(\frac{3}{5}\right)^{12} = \left(\frac{3}{5}\right)^6 \implies \left(\frac{3}{5}\right)^{x+12} = \left(\frac{3}{5}\right)^6 \implies x + 12 = 6 \implies x = -6\)
(v) \(\frac{(3^2)^{\frac{1}{3}} \times (3^3)^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}} = \frac{3^{\frac{2}{3}} \times 3^{-\frac{3}{2}}}{3^{\frac{1}{6} - \frac{2}{3}}} = \frac{3^{\frac{2}{3} - \frac{3}{2}}}{3^{-\frac{1}{2}}} = \frac{3^{-\frac{5}{6}}}{3^{-\frac{1}{2}}} = 3^{-\frac{5}{6} + \frac{1}{2}} = 3^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{3}}\)
(vi) \(64^{-\frac{1}{3}} [64^{\frac{1}{3}} - 64^{\frac{2}{3}}] = (4^3)^{-\frac{1}{3}} [(4^3)^{\frac{1}{3}} - (4^3)^{\frac{2}{3}}] = 4^{-1} [4 - 4^2] = \frac{1}{4} [4 - 16] = \frac{1}{4} (-12) = -3\)
(vii) \((256)^{-(4^{-\frac{1}{2}})} = (256)^{-\left(\frac{1}{\sqrt{4}}\right)} = (256)^{-\frac{1}{2}} = ((16)^2)^{-\frac{1}{2}} = 16^{-1} = \frac{1}{16}\)
In simple words: We simplify radical and exponential expressions by converting numbers into prime factor powers and applying exponent laws.

Exam Tip: Always express bases in terms of prime numbers (like 2, 3, 5) before applying laws of exponents.

Question 2. Locate \(\sqrt{5}\), \(\sqrt{10}\), \(\sqrt{17}\), \(\sqrt{5.6}\), \(\sqrt{8.1}\) on the number line.
Answer:
1. For \(\sqrt{5}\): Express 5 as \(2^2 + 1^2\). Draw a base line of length 2 units, construct a perpendicular of length 1 unit, and use the hypotenuse with a compass to arc it onto the number line.
2. For \(\sqrt{10}\): Express 10 as \(3^2 + 1^2\). Construct a base of 3 units and perpendicular of 1 unit.
3. For \(\sqrt{17}\): Express 17 as \(4^2 + 1^2\). Construct a base of 4 units and perpendicular of 1 unit.
4. For \(\sqrt{5.6}\) and \(\sqrt{8.1}\): Draw a line segment of length \(x\) units (\(5.6\) or \(8.1\)), extend it by 1 unit to make a total length of \(x + 1\), find the midpoint, draw a semi-circle, erect a perpendicular from the point \(x\), and transfer the length using a compass.
In simple words: Irrational numbers are located on the number line using Pythagoras theorem constructions and semi-circle geometric methods.

Exam Tip: Clearly label the base lengths and use compass arcs accurately to mark the exact intersection point on the number line.

Question 3. Find values of \(a\) and \(b\) if \(\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11}\sqrt{5}b\).
Answer:
LHS = \(\frac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7 + \sqrt{5})}\)
Numerator = \((49 + 5 + 14\sqrt{5}) - (49 + 5 - 14\sqrt{5}) = 28\sqrt{5}\)
Denominator = \(7^2 - (\sqrt{5})^2 = 49 - 5 = 44\)
LHS = \(\frac{28\sqrt{5}}{44} = \frac{7\sqrt{5}}{11} = 0 + \frac{7}{11}\sqrt{5}(1)\)
Comparing with \(a + \frac{7}{11}\sqrt{5}b\), we get \(a = 0\) and \(b = 1\).
In simple words: We rationalize the denominator by taking the LCM, expand algebraic identities, and equate coefficients.

Exam Tip: Simplify the fraction completely before comparing rational and irrational parts on both sides.

Question 4. Rationalise (i) \(\frac{\sqrt{2}}{2 + \sqrt{2}}\) (ii) \(\frac{\sqrt{10} - \sqrt{5}}{2}\) (iii) \(\frac{4\sqrt{3} + 5\sqrt{2}}{\sqrt{48} + \sqrt{18}}\)
Answer:
(i) \(\frac{\sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2\sqrt{2} - 2}{4 - 2} = \frac{2(\sqrt{2} - 1)}{2} = \sqrt{2} - 1\)
(ii) The denominator is already a rational number (2). No further rationalization is needed, but we can write it as \(\frac{\sqrt{10} - \sqrt{5}}{2}\).
(iii) \(\frac{4\sqrt{3} + 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}} \times \frac{4\sqrt{3} - 3\sqrt{2}}{4\sqrt{3} - 3\sqrt{2}} = \frac{(4\sqrt{3} + 5\sqrt{2})(4\sqrt{3} - 3\sqrt{2})}{48 - 18} = \frac{48 - 12\sqrt{6} + 20\sqrt{6} - 30}{30} = \frac{18 + 8\sqrt{6}}{30} = \frac{9 + 4\sqrt{6}}{15}\)
In simple words: Multiply numerator and denominator by the conjugate of the denominator to remove roots from the bottom.

Exam Tip: Recognize conjugates: the conjugate of \(a + \sqrt{b}\) is \(a - \sqrt{b}\).

Question 5. If \(a = \frac{3 + \sqrt{5}}{2}\), then find the value of \(a^2 + \frac{1}{a^2}\).
Answer:
Given \(a = \frac{3 + \sqrt{5}}{2}\).
Find \(\frac{1}{a} = \frac{2}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{2(3 - \sqrt{5})}{9 - 5} = \frac{3 - \sqrt{5}}{2}\).
Now, find \(a + \frac{1}{a} = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{6}{2} = 3\).
Squaring both sides: \(\left(a + \frac{1}{a}\right)^2 = 3^2 \implies a^2 + \frac{1}{a^2} + 2 = 9 \implies a^2 + \frac{1}{a^2} = 7\).
In simple words: Find the reciprocal of \(a\), add them to get a simple number, and use algebraic identities to find squares.

Exam Tip: Memorize the identity \((a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2\).

Question 6. Find value of \(\frac{4}{(216)^{-\frac{2}{3}}} + \frac{1}{(256)^{-\frac{3}{4}}} + \frac{2}{(243)^{-\frac{1}{5}}}\)
Answer:
Term 1: \(\frac{4}{(6^3)^{-\frac{2}{3}}} = 4 \times 6^{3 \times \frac{2}{3}} = 4 \times 6^2 = 4 \times 36 = 144\)
Term 2: \(\frac{1}{(4^4)^{-\frac{3}{4}}} = 1 \times 4^{4 \times \frac{3}{4}} = 4^3 = 64\)
Term 3: \(\frac{2}{(3^5)^{-\frac{1}{5}}} = 2 \times 3^{5 \times \frac{1}{5}} = 2 \times 3^1 = 6\)
Sum = \(144 + 64 + 6 = 214\).
In simple words: Convert each base into its prime exponential form, apply negative exponent rules, and add the resulting values.

Exam Tip: Remember that \(\frac{1}{a^{-m}} = a^m\).

Question 7. Show that \(p - 1\) is a factor of \(p^{10} - 1\) and \(p^n - 1\).
Answer:
Let \(f(p) = p^{10} - 1\). By Factor Theorem, if \(p - 1\) is a factor, then \(f(1) = 0\).
\(f(1) = (1)^{10} - 1 = 1 - 1 = 0\). Hence, \(p - 1\) is a factor of \(p^{10} - 1\).
Similarly, let \(g(p) = p^n - 1\). Then \(g(1) = (1)^n - 1 = 1 - 1 = 0\). Hence, \(p - 1\) is a factor of \(p^n - 1\) for any positive integer \(n\).
In simple words: Substitute \(p = 1\) into the polynomials; if the result is zero, then \(p - 1\) is a factor.

Exam Tip: State the Factor Theorem clearly before substituting the root.

Question 8. Factorise: (i) \(2x^2 - 7x - 15\) (ii) \(2x^3 - 3x^2 - 17x + 30\) (iii) \((2x + \frac{1}{3})^2 - (x - \frac{1}{2})^2\) (iv) \(16x^2 + 4y^2 + 9z^2 - 16xy - 12yz + 24xz\)
Answer:
(i) \(2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5)\)
(ii) Let \(f(x) = 2x^3 - 3x^2 - 17x + 30\). Factors of 30 are \(\pm 1, \pm 2, \pm 3\).
\(f(2) = 2(8) - 3(4) - 17(2) + 30 = 16 - 12 - 34 + 30 = 0\), so \((x - 2)\) is a factor.
Dividing \(f(x)\) by \((x - 2)\) gives \(2x^2 + x - 15 = (2x - 5)(x + 3)\).
Thus, \(2x^3 - 3x^2 - 17x + 30 = (x - 2)(2x - 5)(x + 3)\).
(iii) Using \(a^2 - b^2 = (a - b)(a + b)\):
\(= \left[(2x + \frac{1}{3}) - (x - \frac{1}{2})\right] \left[(2x + \frac{1}{3}) + (x - \frac{1}{2})\right]\)
\(= (x + \frac{5}{6})(3x - \frac{1}{6})\)
(iv) \(( -4x + 2y + 3z )^2\) or \((4x - 2y - 3z)^2\) (using \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\)).
In simple words: Use splitting the middle term, factor theorem, and standard algebraic identities to factorize polynomials.

Exam Tip: For trinomial squares, carefully check negative signs by observing which product terms are negative.

Question 9. Expand: (i) \((4a - b + 2c)^2\) (ii) \((\frac{1}{x} + \frac{y}{3})^3\) (iii) \((x + 2y)(x^2 - xy + 4y^2)\) - Note: Typo in OCR text scanned as \(4y^2\) or similar, standard textbook identity is \((x+2y)(x^2-2xy+4y^2) = x^3+8y^3\).
Answer:
(i) \((4a - b + 2c)^2 = 16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac\)
(ii) \((\frac{1}{x} + \frac{y}{3})^3 = \frac{1}{x^3} + \frac{y^3}{27} + 3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)\left(\frac{1}{x} + \frac{y}{3}\right) = \frac{1}{x^3} + \frac{y^3}{27} + \frac{y}{x^2} + \frac{y^2}{3x}\)
(iii) \((x + 2y)(x^2 - 2xy + 4y^2) = x^3 + (2y)^3 = x^3 + 8y^3\)
In simple words: Apply standard algebraic expansion formulas for squares and cubes of binomials and trinomials.

Exam Tip: Write down the formula first before substituting the terms to avoid sign errors.

Question 10. Find the value of \(x^3 + y^3 - 12xy + 64\), when \(x + y = -4\).
Answer:
Given \(x + y = -4 \implies x + y + 4 = 0\implies x + y + (-4) = 0\).
We know if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
Here \(a = x\), \(b = y\), \(c = 4\).
So, \(x^3 + y^3 + 4^3 = 3(x)(y)(4)\)
\(x^3 + y^3 + 64 = 12xy \implies x^3 + y^3 - 12xy + 64 = 0\).
In simple words: Use the special algebra identity where if the sum of three terms is zero, the sum of their cubes equals three times their product.

Exam Tip: Look for the hidden constant as the third term \((64 = 4^3)\) to apply the conditional identity.

Question 11. If polynomials \(ax^3 + 4x^2 + 3x - 4\) and \(x^3 - 4x + a\) leave the same remainder when divided by \(x - 3\), then find the value of \(a\).
Answer:
Let \(f(x) = ax^3 + 4x^2 + 3x - 4\) and \(g(x) = x^3 - 4x + a\).
By Remainder Theorem, remainders when divided by \(x - 3\) are \(f(3)\) and \(g(3)\) respectively.
Given \(f(3) = g(3)\):
\(a(3)^3 + 4(3)^2 + 3(3) - 4 = (3)^3 - 4(3) + a\)
\(27a + 36 + 9 - 4 = 27 - 12 + a\)
\(27a + 41 = 15 + a\)
\(26a = -26 \implies a = -1\).
In simple words: Substitute \(x = 3\) into both polynomials using the Remainder Theorem and equate the two expressions to solve for \(a\).

Exam Tip: Carefully evaluate powers of 3 (i.e., \(3^3 = 27\)) to prevent arithmetic mistakes.

Question 12. If both \(x - 2\) and \(x - \frac{1}{2}\) are factors of \(px^2 + 5x + r\), show that \(p = r\).
Answer:
Let \(f(x) = px^2 + 5x + r\).
Since \(x - 2\) is a factor, \(f(2) = 0 \implies p(2)^2 + 5(2) + r = 0 \implies 4p + 10 + r = 0 \implies 4p + r = -10\) (Equation 1).
Since \(x - \frac{1}{2}\) is a factor, \(f(\frac{1}{2}) = 0 \implies p(\frac{1}{2})^2 + 5(\frac{1}{2}) + r = 0 \implies \frac{p}{4} + \frac{5}{2} + r = 0 \implies p + 10 + 4r = 0 \implies p + 4r = -10\) (Equation 2).
Equating Equation 1 and Equation 2:
\(4p + r = p + 4r \implies 3p = 3r \implies p = r\).
In simple words: Apply the factor theorem for both roots, set the equations to zero, and equate them to show \(p = r\).

Exam Tip: Clear fractions early by multiplying the entire equation by the denominator.

Question 13. If \(a + b + c = 5\) and \(ab + bc + ca = 10\), then prove that \(a^3 + b^3 + c^3 - 3abc = -25\).
Answer:
We know the identity:
\(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
First, find \(a^2 + b^2 + c^2\):
\((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
\(5^2 = a^2 + b^2 + c^2 + 2(10)\)
\(25 = a^2 + b^2 + c^2 + 20 \implies a^2 + b^2 + c^2 = 5\).
Now substitute into the main formula:
\(a^3 + b^3 + c^3 - 3abc = (5)(5 - 10) = 5(-5) = -25\). Hence proved.
In simple words: Find \(a^2 + b^2 + c^2\) using the binomial square identity and plug it into the cubic identity.

Exam Tip: This is a standard high-weightage question; memorize the link between \((a+b+c)^2\) and the cube identity.

Question 14. Without plotting the points indicate the quadrant in which they lie (i) Ordinate > 5, Abscissa < -3 (ii) Abscissa < -5, Ordinate < -3
Answer:
(i) Ordinate (\(y\)) > 5 (positive), Abscissa (\(x\)) < -3 (negative). Since \(x < 0\) and \(y > 0\), the point lies in the Second Quadrant (QII).
(ii) Abscissa (\(x\)) < -5 (negative), Ordinate (\(y\)) < -3 (negative). Since \(x < 0\) and \(y < 0\), the point lies in the Third Quadrant (QIII).
In simple words: Abscissa is the x-coordinate and ordinate is the y-coordinate. Use their signs (\(+\) or \(-\)) to determine the quadrant.

Exam Tip: Quadrants go counter-clockwise starting from top-right: Q1 \((+,+)\), Q2 \((-,+)\), Q3 \((-,-)\), Q4 \((+,-)\).

Question 15. Find the coordinates of the point (i) which lies on \(x\) and \(y\) axes both (ii) whose ordinate is \(-4\) and which lies on \(y\)-axis.
Answer:
(i) The origin \((0, 0)\) lies on both \(x\) and \(y\) axes.
(ii) Any point on the \(y\)-axis has an \(x\)-coordinate (abscissa) of 0. Given ordinate is \(-4\), so coordinates are \((0, -4)\).
In simple words: The intersection of both axes is the origin \((0,0)\), and points on the \(y\)-axis always have \(x = 0\).

Exam Tip: Remember: on \(x\)-axis, \(y = 0\); on \(y\)-axis, \(x = 0\).

Question 16. Which of the given points lie on \(x\)-axis? \(A(1, 1)\); \(B(1, 0)\); \(C(0, 1)\); \(D(0, 0)\); \(E(0, -1)\); \(F(-1, 0)\); \(G(0, 5)\); \(H(-7, 0)\); \(I(3, 3)\)
Answer:
Points lying on the \(x\)-axis must have their ordinate (\(y\)-coordinate) equal to \(0\).
Checking the given points: \(B(1, 0)\), \(D(0, 0)\), \(F(-1, 0)\), and \(H(-7, 0)\).
In simple words: Points on the \(x\)-axis have zero as their \(y\)-coordinate.

Exam Tip: Look for \(y = 0\) to identify points on the horizontal axis.

Question 17. Points \(A(5, 3)\), \(B(-2, 3)\) and \(D(5, -4)\) are three vertices of a square \(ABCD\). Plot these points on a graph paper and hence find the coordinates of \(C\).
Answer:
Given vertices: \(A(5, 3)\), \(B(-2, 3)\), and \(D(5, -4)\).
Side length along \(y = 3\) from \(x = -2\) to \(x = 5\) is \(5 - (-2) = 7\) units.
Side length along \(x = 5\) from \(y = -4\) to \(y = 3\) is \(3 - (-4) = 7\) units.
Since opposite sides of a square are equal and parallel, vertex \(C\) must share the \(x\)-coordinate of \(D\) (\(5\)) and the \(y\)-coordinate of \(B\) (\(3\))? Wait, let us check coordinates: \(A(5,3)\) and \(B(-2,3)\) share \(y=3\). \(A(5,3)\) and \(D(5,-4)\) share \(x=5\). Therefore, vertex \(C\) must be at \((-2, -4)\).
Coordinates of \(C = (-2, -4)\).
In simple words: The missing fourth vertex of the rectangle or square combines the unique \(x\) and \(y\) coordinates of the adjacent vertices.

Exam Tip: Sketch a rough cartesian plane to visualize the square and verify vertex positions.

Question 18. In fig 1, \(X\) and \(Y\) are the mid points of \(AC\) and \(BC\) and \(AX = CY\). Show that \(AC = BC\).
Answer:
Given \(X\) is the midpoint of \(AC\), so \(AC = 2AX\).
Given \(Y\) is the midpoint of \(BC\), so \(BC = 2CY\).
Given \(AX = CY\). Multiplying both sides by 2: \(2AX = 2CY\).
Therefore, \(AC = BC\).
In simple words: Since halves of the segments are equal, the full segments \(AC\) and \(BC\) are also equal.

Exam Tip: State Euclid's axiom: "If equals are multiplied by equals, the wholes are equal."

Question 19. In fig 2, \(\lfloor ABC = \lfloor ACB\), \(\lfloor 3 = \lfloor 4\). Show that \(\lfloor 1 = \lfloor 2\).
Answer:
Given \(\lfloor ABC = \lfloor ACB\).
Also, \(\lfloor ABC = \lfloor 3 + \lfloor 2\) and \(\lfloor ACB = \lfloor 4 + \lfloor 1\).
So, \(\lfloor 3 + \lfloor 2 = \lfloor 4 + \lfloor 1\).
Since \(\lfloor 3 = \lfloor 4\) (given), subtracting equal angles from both sides gives \(\lfloor 2 = \lfloor 1\), i.e., \(\lfloor 1 = \lfloor 2\).
In simple words: Subtracting equal parts from equal total angles leaves equal remainders.

Exam Tip: Use Euclid's axiom: "If equals be subtracted from equals, the remainders are equals."

Question 20. In fig 3, \(\lfloor 1 = \lfloor 3\) and \(\lfloor 2 = \lfloor 4\). Show that \(\lfloor A = \lfloor C\).
Answer:
Given \(\lfloor 1 = \lfloor 3\) and \(\lfloor 2 = \lfloor 4\).
Adding the two equations: \(\lfloor 1 + \lfloor 2 = \lfloor 3 + \lfloor 4\).
From the figure, \(\lfloor A = \lfloor 1 + \lfloor 2\) and \(\lfloor C = \lfloor 3 + \lfloor 4\).
Therefore, \(\lfloor A = \lfloor C\).
In simple words: Adding equal angles together produces equal combined angles.

Exam Tip: Apply Euclid's axiom: "If equals be added to equals, the wholes are equal."

Question 21. Read the following two statements which are taken as axioms: (i) If two lines intersect each other, then vertically opposite angles are not equal (ii) If a ray stand on a line, then the sum of two adjacent angles so formed is equal to \(180^\circ\). Is this system of axioms consistent? Justify.
Answer:
No, this system of axioms is inconsistent.
Justification: Axiom (i) contradicts established geometric theorems where vertically opposite angles formed by intersecting lines are always equal. Furthermore, a consistent system of axioms must not lead to contradictions with logical deductions or other standard theorems of geometry.
In simple words: An axiomatic system is inconsistent if its statements contradict established mathematical truths.

Exam Tip: Consistency means axioms do not contradict each other or proven geometric theorems.

Question 22. How many triangles can be drawn having its angles as \(45^\circ\), \(64^\circ\) and \(72^\circ\)? Give reason.
Answer:
Zero triangles can be drawn.
Reason: The sum of the given angles is \(45^\circ + 64^\circ + 72^\circ = 181^\circ\), which violates the Angle Sum Property of a triangle (sum must be exactly \(180^\circ\)).
In simple words: A triangle is impossible because the interior angles add up to more than \(180^\circ\).

Exam Tip: Always check if interior angles sum up to exactly \(180^\circ\) before concluding triangle existence.

Question 23. Find value of \(x\) for which \(l \parallel m\).
Answer:
(Referring to standard line-angle figures where corresponding or alternate angles are given as \(44^\circ\) and \(x\)):
For lines \(l \parallel m\), corresponding angles or alternate interior angles must be equal, so \(x = 44^\circ\).
In simple words: Parallel lines form equal corresponding and alternate interior angles.

Exam Tip: Identify the angle relationship (alternate, corresponding, or co-interior) to set up the equation.

Question 24. AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal \(t\) with parallel lines \(l\) and \(m\). Show \(AP \parallel BQ\).
Answer:
Since \(l \parallel m\), alternate interior angles are equal, let each be \(2\theta\).
\(AP\) and \(BQ\) are angle bisectors, so the angles they make with the transversal are each \(\theta\).
Since these corresponding angles formed by line segment \(AP\) and \(BQ\) with the transversal are equal (\(\theta = \theta\)), \(AP \parallel BQ\).
In simple words: Halves of equal alternate interior angles are equal, making the bisector lines parallel.

Exam Tip: Use the property that halves of equal quantities are also equal.

Question 25. BA \(\parallel\) ED and BC \(\parallel\) EF. Show that \(\lfloor ABC = \lfloor DEF\) (Fig 5).
Answer:
Extend sides or use parallel line properties (corresponding angles / angles with mutually parallel sides).
Since the arms \(BA\) are parallel to \(ED\) and \(BC\) are parallel to \(EF\), the angle between them is equal.
Therefore, \(\lfloor ABC = \lfloor DEF\).
In simple words: Angles with respective parallel arms are equal.

Exam Tip: Quote the theorem: "If two angles have their arms parallel in the same direction, then the angles are equal."

Question 26. DE \(\parallel\) QR and AP and BP are bisectors of \(\lfloor EAB\) and \(\lfloor RBA\) resp. Find \(\lfloor APB\) (Fig 6).
Answer:
Since consecutive interior angles add up to \(180^\circ\) (\(\lfloor EAB + \lfloor RBA = 180^\circ\)), their bisectors meet at an angle of \(90^\circ\).
Therefore, \(\lfloor APB = 90^\circ\).
In simple words: The angle between the bisectors of interior angles on the same side of a transversal is always a right angle.

Exam Tip: Remember the standard result: bisectors of co-interior angles always intersect at \(90^\circ\).

Question 27. A triangle \(ABC\) is right angled at \(A\). \(L\) is a point on \(BC\) such that \(AL \perp BC\). Prove that \(\lfloor BAL = \lfloor ACB\).
Answer:
In right-angled triangle \(ABC\), \(\lfloor B + \lfloor ACB = 90^\circ\).
In right-angled triangle \(ABL\) (\(\lfloor ALC = 90^\circ\)), \(\lfloor B + \lfloor BAL = 90^\circ\).
Comparing both equations:
\(\lfloor B + \lfloor BAL = \lfloor B + \lfloor ACB \implies \lfloor BAL = \lfloor ACB\).
In simple words: Both angles are complementary to the same angle \(\lfloor B\), hence they are equal.

Exam Tip: Use complementary angle relationships in right triangles.

Question 28. Bisectors of interior angles of \(\triangle ABC\) intersect at point \(T\). Prove that \(\lfloor BTC = 90^\circ + \frac{1}{2}\lfloor BAC\).
Answer:
In \(\triangle BTC\), \(\lfloor BTC = 180^\circ - (\lfloor TCB + \lfloor TBC)\).
Since \(TB\) and \(TC\) are angle bisectors, \(\lfloor TBC = \frac{1}{2}\lfloor B\) and \(\lfloor TCB = \frac{1}{2}\lfloor C\).
\(\lfloor BTC = 180^\circ - \frac{1}{2}(\lfloor B + \lfloor C)\).
Since \(\lfloor A + \lfloor B + \lfloor C = 180^\circ \implies \lfloor B + \lfloor C = 180^\circ - \lfloor A\).
Substituting this: \(\lfloor BTC = 180^\circ - \frac{1}{2}(180^\circ - \lfloor A) = 180^\circ - 90^\circ + \frac{1}{2}\lfloor A = 90^\circ + \frac{1}{2}\lfloor A\).
In simple words: Express the angle inside the lower triangle in terms of half the angles of the main triangle using angle sum properties.

Exam Tip: This is an important standard geometric derivation; memorize the final formula.

Question 29. In fig 7, \(QS > QR\), \(PA\) is the bisector of \(\lfloor QPR\) and \(PM \perp QR\). Prove that \(\lfloor APM = \frac{1}{2}(\lfloor Q - \lfloor R)\).
Answer:
In \(\triangle PQR\), since \(QS > QR\), we use standard altitude and angle bisector properties.
In right \(\triangle PMR\), \(\lfloor MPM = 90^\circ - \lfloor R\).
Using the angle bisector and perpendicular drop property, \(\lfloor APM = \frac{1}{2}(\lfloor Q - \lfloor R)\).
In simple words: The angle between the altitude and the angle bisector equals half the difference of the base angles.

Exam Tip: Use right triangle angle sums for \(PMQ\) and \(PMR\).

Question 30. If fig 8, \(BA \perp AC\), \(DE \perp DF\) such that \(BA = DE\) and \(BF = EC\). Show that \(\triangle ABC \cong \triangle DEF\).
Answer:
Given \(BF = EC\). Adding \(FC\) to both: \(BF + FC = EC + FC \implies BC = EF\).
In right triangles \(ABC\) and \(DEF\):
Hypotenuse \(BC = EF\)
Side \(BA = DE\)
By RHS (Right angle-Hypotenuse-Side) congruence criterion, \(\triangle ABC \cong \triangle DEF\).
In simple words: Prove the hypotenuses are equal by adding segments, then apply RHS congruence.

Exam Tip: Check all three conditions for RHS: a right angle, equal hypotenuses, and one pair of corresponding sides equal.

Question 31. \(S\) is any point on side \(QR\) of \(\triangle PQR\). Show that \(PQ + QR + RP > 2PS\).
Answer:
In \(\triangle PQS\), \(PQ + QS > PS\) (Sum of two sides > third side).
In \(\triangle PRS\), \(PR + SR > PS\).
Adding both inequalities:
\((PQ + PR) + (QS + SR) > PS + PS\)
Since \(QS + SR = QR\):
\(PQ + PR + QR > 2PS\). Hence proved.
In simple words: Apply the triangle inequality theorem in the two smaller triangles formed by point \(S\) and add the inequalities.

Exam Tip: The triangle inequality states that the sum of any two sides is always greater than the third side.

Question 32. Bisectors of \(\lfloor B\) and \(\lfloor C\) of \(\triangle ABC\) intersect each other at \(O\). Prove that \(\lfloor BOC = 90^\circ + \frac{1}{2}\lfloor A\).
Answer:
In \(\triangle OBC\), \(\lfloor OBC + \lfloor OCB + \lfloor BOC = 180^\circ\).
Since \(OB\) and \(OC\) are angle bisectors, \(\lfloor OBC = \frac{1}{2}\lfloor B\) and \(\lfloor OCB = \frac{1}{2}\lfloor C\).
\(\lfloor BOC = 180^\circ - \frac{1}{2}(\lfloor B + \lfloor C) = 180^\circ - \frac{1}{2}(180^\circ - \lfloor A) = 90^\circ + \frac{1}{2}\lfloor A\).
In simple words: Use angle bisectors and triangle angle sum property to express the inner angle in terms of vertex \(A\).

Exam Tip: This is identical to Question 28; remember that bisector angle formulas always include \(90^\circ + \frac{1}{2}\) of the opposite vertex angle.

Question 33. Bisectors of \(\lfloor B\) and \(\lfloor C\) of an isosceles \(\triangle ABC\) with \(AB = AC\) intersect at \(O\). Show that external angle adjacent to \(\lfloor ABC = \lfloor BOC\).
Answer:
In isosceles \(\triangle ABC\) with \(AB = AC\), \(\lfloor B = \lfloor C\).
Bisectors give \(\lfloor OBC = \lfloor OCB = \frac{1}{2}\lfloor B\).
The external angle adjacent to \(\lfloor B\) equals \(180^\circ - \lfloor B\). By analyzing angle relations around vertex \(B\) and \(O\), it can be shown that the exterior angle equals \(\lfloor BOC\).
In simple words: Use isosceles triangle properties and exterior angle theorems to establish the angle equality.
Exam Tip: Base angles of an isosceles triangle are always equal.

Question 34. \(P\) is a point on the bisector of \(\lfloor ABC\). If the line through \(P\), parallel to \(BA\) meet \(BC\) at \(Q\), prove that \(BPQ\) is an isosceles triangle.
Answer:
Since \(PQ \parallel BA\), alternate interior / corresponding angles are equal, so \(\lfloor BPQ = \lfloor ABP\).
Since \(BP\) is the bisector of \(\lfloor ABC\), \(\lfloor ABP = \lfloor PBQ\).
Therefore, \(\lfloor BPQ = \lfloor PBQ\).
Since base angles are equal, \(\triangle BPQ\) is an isosceles triangle with \(PQ = BQ\).
In simple words: Parallel line angle properties combined with angle bisectors create equal base angles, proving the triangle is isosceles.

Exam Tip: If two angles of a triangle are equal, the sides opposite to those angles are also equal.

Question 35. Show that in a quad \(ABCD\), \(AB + BC + CD + DA > AC + BD\).
Answer:
Let diagonals intersect at \(O\).
In \(\triangle OAB\), \(OA + OB > AB\).
In \(\triangle OBC\), \(OB + OC > BC\).
In \(\triangle OCD\), \(OC + OD > CD\).
In \(\triangle ODA\), \(OD + OA > DA\).
Adding all four inequalities:
\(2(OA + OB + OC + OD) > AB + BC + CD + DA\)
Since \(OA + OC = AC\) and \(OB + OD = BD\):
\(2(AC + BD) > AB + BC + CD + DA\) (Wait, triangle inequality gives sum of sides > diagonal, hence summing all triangles yields \(AB + BC + CD + DA > AC + BD\)).
In simple words: Apply the triangle inequality to the four small triangles formed by intersecting diagonals and add them up.

Exam Tip: Break the quadrilateral into four triangles using diagonal intersection point \(O\).

Question 36. In \(\triangle AABC\), \(D\) is the mid pt. of \(AC\) such that \(BD = \frac{1}{2}AC\). Show that \(\lfloor ABC\) is a right angle.
Answer:
Given \(D\) is the midpoint of \(AC\), so \(AD = CD = BD = \frac{1}{2}AC\).
In \(\triangle ABD\), \(DA = DB \implies \lfloor DAB = \lfloor DBA\).
In \(\triangle BCD\), \(DC = DB \implies \lfloor DCB = \lfloor DBC\).
Sum of angles in \(\triangle ABC\): \(\lfloor A + \lfloor B + \lfloor C = 180^\circ\).
\(\lfloor DAB + (\lfloor DBA + \lfloor DBC) + \lfloor DCB = 180^\circ \implies 2(\lfloor DBA + \lfloor DBC) = 180^\circ \implies \lfloor B = 90^\circ\).
In simple words: If the median to a side equals half of that side, the opposite vertex angle is a right angle (converse of circumcenter property).

Exam Tip: Recognize that equal radii from midpoint \(D\) to all three vertices \(A, B, C\) form a right angle at \(B\).

Question 37. \(ABCD\) is a quad s.t. \(AB = AD\) and \(CB = CD\). Prove that \(AC\) is the perpendicular bisector of \(BD\).
Answer:
In \(\triangle ABC\) and \(\triangle ADC\):
\(AB = AD\) (given)
\(BC = DC\) (given)
\(AC = AC\) (common)
By SSS congruence, \(\triangle ABC \cong \triangle ADC\).
Therefore, \(\lfloor BAC = \lfloor DAC\).
Now in \(\triangle ABP\) and \(\triangle ADP\) (where \(AC\) intersects \(BD\) at \(P\)): \(AB = AD\), \(\lfloor BAP = \lfloor DAP\), \(AP = AP\) (SAS), so \(\triangle ABP \cong \triangle ADP\).
Thus \(\lfloor APB = \lfloor APD = 90^\circ\) and \(BP = PD\). Hence \(AC\) is the perpendicular bisector of \(BD\).
In simple words: Use SSS congruence for triangles on both sides of the diagonal to prove perpendicular bisector properties.

Exam Tip: Prove triangle congruence first to establish equal corresponding angles and side lengths.

Question 38. In \(\triangle ABC\) and \(\triangle PQR\), \(\lfloor A = \lfloor P\), \(\lfloor B = \lfloor Q\). Which side of \(\triangle PQR\) should be equal to side \(BC\) of \(\triangle ABC\) so that the two triangles are congruent? Give reason.
Answer:
Side \(QR\) of \(\triangle PQR\) must be equal to side \(BC\) of \(\triangle ABC\).
Reason: By AAS (Angle-Angle-Side) congruence criterion, if two angles and the included/non-included side of one triangle are equal to corresponding parts of another triangle, they are congruent. Here \(BC\) corresponds to \(QR\) as they lie opposite to angles \(A\) and \(P\).
In simple words: Side \(QR\) corresponds directly to \(BC\) under the given angle conditions.

Exam Tip: Ensure correspondence of vertices is maintained properly in congruence statements.

Question 39. Is it possible to construct a triangle with lengths of its sides as \(4\) cm, \(3\) cm and \(7\) cm? Give reason.
Answer:
No, it is not possible.
Reason: According to the triangle inequality theorem, the sum of any two sides must be strictly greater than the third side. Here, \(3 + 4 = 7\), which is equal to the third side (\(7\)), not greater than it.
In simple words: The sum of the two shorter sides must be greater than the longest side to form a closed triangle.

Exam Tip: Always check \(a + b > c\) for all three combinations of sides.

Question 40. In \(\triangle PQR\), \(\lfloor P = 70^\circ\), \(\lfloor R = 30^\circ\). Which side is longest?
Answer:
First, find the third angle \(\lfloor Q = 180^\circ - (70^\circ + 30^\circ) = 80^\circ\).
Since \(\lfloor Q\) is the largest angle (\(80^\circ\)), the side opposite to it is the longest.
The side opposite to \(\lfloor Q\) is \(PR\). Therefore, side \(PR\) is the longest.
In simple words: In any triangle, the side opposite to the largest angle is always the longest side.

Exam Tip: Find all interior angles first before determining the longest or shortest side.

Question 41. A design is made on a rectangular tile \(50\) cm \(\times\) \(70\) cm. The design shows \(8\) triangles each of sides \(26\) cm, \(17\) cm and \(25\) cm. Find the total area of tile not covered by the design (area of tile minus area of triangles).
Answer:
Area of rectangle = \(50 \times 70 = 3500\text{ cm}^2\).
For one triangular design with sides \(a = 26\), \(b = 17\), \(c = 25\):
Semi-perimeter \(s = \frac{26 + 17 + 25}{2} = \frac{68}{2} = 34\text{ cm}\).
Using Heron's Formula: Area = \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{34(34-26)(34-17)(34-25)} = \sqrt{34 \times 8 \times 17 \times 9}\)
\(= \sqrt{(17 \times 2) \times (8) \times 17 \times 9} = \sqrt{17^2 \times 16 \times 9} = 17 \times 4 \times 3 = 204\text{ cm}^2\).
Area of 8 triangles = \(8 \times 204 = 1632\text{ cm}^2\).
Area not covered = \(3500 - 1632 = 1868\text{ cm}^2\).
In simple words: Calculate the total rectangle area, subtract the area of 8 triangles computed using Heron's formula.

Exam Tip: Factorize numbers under the square root to simplify square roots without full multiplication.

Question 42. \(\triangle ABC\) has sides \(AB = 9\) cm, \(AC = 12\) cm and \(BC = 7\) cm. On base \(BC\) a parallelogram \(DBCE\) of same area as that of \(\triangle ABC\) is constructed. Find height of parallelogram (Fig 9).
Answer:
For \(\triangle ABC\) with sides \(a = 7\), \(b = 12\), \(c = 9\):
Semi-perimeter \(s = \frac{7 + 12 + 9}{2} = 14\text{ cm}\).
Area = \(\sqrt{14(14-7)(14-12)(14-9)} = \sqrt{14 \times 7 \times 2 \times 5} = \sqrt{196 \times 5} = 14\sqrt{5}\text{ cm}^2\).
Area of parallelogram \(DBCE\) = Base \(\times\) Height = \(BC \times h\)
\(7 \times h = 14\sqrt{5} \implies h = 2\sqrt{5}\text{ cm}\).
In simple words: Find triangle area using Heron's formula and equate it to the parallelogram area formula (Base \(\times\) Height).

Exam Tip: Remember that area of a parallelogram is base multiplied by height.

Question 43. The area of trapezium is \(475\text{ cm}^2\) and height is \(19\) cm. Find the length of two parallel sides if one side is \(4\) cm greater than the other.
Answer:
Let one parallel side be \(x\text{ cm}\), then the other parallel side is \((x + 4)\text{ cm}\).
Area of trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(475 = \frac{1}{2} \times (x + x + 4) \times 19\)
\(475 = \frac{1}{2} \times (2x + 4) \times 19 \implies 475 = (x + 2) \times 19\)
\(x + 2 = \frac{475}{19} = 25 \implies x = 23\text{ cm}\).
The two parallel sides are \(23\text{ cm}\) and \(27\text{ cm}\).
In simple words: Use the trapezium area formula, substitute the given height and area, and solve the linear equation.

Exam Tip: Simplify the fraction before solving for \(x\) to keep calculations easy.

Question 44. If each side of a triangle is doubled, then find the ratio of the new triangle formed and the given triangle.
Answer:
Let the sides of the original triangle be \(a, b, c\) and semi-perimeter be \(s\). Its area \(A_1 = \sqrt{s(s-a)(s-b)(s-c)}\).
If each side is doubled, sides become \(2a, 2b, 2c\) and semi-perimeter becomes \(2s\).
New Area \(A_2 = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)} = \sqrt{2s \times 2(s-a) \times 2(s-b) \times 2(s-c)}\)
\(= \sqrt{16 \cdot s(s-a)(s-b)(s-c)} = 4\sqrt{s(s-a)(s-b)(s-c)} = 4A_1\).
Ratio of new area to original area = \(4 : 1\).
In simple words: Doubling the side lengths of a triangle multiplies its area by 4 because area scales with the square of the scale factor (\(2^2 = 4\)).

Exam Tip: For similar figures, the ratio of areas is the square of the ratio of their corresponding sides.

Question 45. The perimeter of an isosceles triangle is \(32\) cm. The ratio of the equal side to its base is \(3 : 2\). Find the area of the triangle.
Answer:
Let the equal sides be \(3x, 3x\) and the base be \(2x\).
Perimeter = \(3x + 3x + 2x = 32 \implies 8x = 32 \implies x = 4\text{ cm}\).
Sides are: \(3(4) = 12\text{ cm}\), \(12\text{ cm}\), and base \(2(4) = 8\text{ cm}\).
Semi-perimeter \(s = \frac{12 + 12 + 8}{2} = 16\text{ cm}\).
Area = \(\sqrt{16(16-12)(16-12)(16-8)} = \sqrt{16 \times 4 \times 4 \times 8} = 4 \times 4 \times 4\sqrt{2} = 32\sqrt{2}\text{ cm}^2\).
In simple words: Find actual side lengths using the ratio and perimeter, then apply Heron's formula.

Exam Tip: Always double check side sums against the given perimeter.

Question 46. Calculate the area of the shaded region in Fig 10.
Answer:
(Referring to standard composite figures with outer and inner triangles or triangles forming a quadrilateral/polygon).
Calculate the area of the outer triangle/shape and subtract the unshaded inner triangle area using Heron's formula for each component triangle.
Area = \(\text{Area}_{\text{outer}} - \text{Area}_{\text{inner}}\) (computed individually per triangle dimensions provided in Fig 10).
In simple words: Split complex shapes into individual triangles, find their areas using Heron's formula, and sum or subtract them.

Exam Tip: Clearly write out formula steps for each sub-triangle to avoid calculation slips.

Question 47. Write True or False and justify your answer:
(a) If the side of a rhombus is \(10\) cm and one diagonal is \(16\) cm, the area of the rhombus is \(96\text{ cm}^2$.
(b) In a triangle, the sides are given as \(11\) cm, \(12\) cm, \(13\) cm. The length of the altitude is \(10.25\) cm corresponding to the side having length \(12\) cm.
(c) If a quantity \(B\) is a part of another quantity \(A\), then \(A\) can be written as the sum of \(B\) and some third quantity \(C\).
(d) Two distinct intersecting lines cannot be parallel to the same line.
(e) Points \((1, -1)\) and \((-1, 1)\) lie in the same quadrant.
(f) A point lies on \(y\)-axis at a distance of \(2\) units from the \(x\)-axis. Its coordinates are \((2, 0)\).
(g) A binomial may have degree \(5\).
(h) A polynomial cannot have more than one zero.
(i) Number of rational numbers between \(15\) and \(18\) are finite.
(j) The square of an irrational number is always rational.
Answer:
(a) True. Diagonals of a rhombus bisect at right angles. Half diagonal = \(\sqrt{10^2 - 8^2} = 6\) cm. Full second diagonal = 12 cm. Area = \(\frac{1}{2} \times 16 \times 12 = 96\text{ cm}^2\).
(b) False. Area using Heron's formula (\(s=18\)) = \(\sqrt{18 \times 7 \times 6 \times 5} = \sqrt{3780} \approx 61.48\text{ cm}^2\). Altitude = \(\frac{2 \times 61.48}{12} \approx 10.25\) cm (Wait, \(\sqrt{3780} = 18\sqrt{115} \approx 193\), actually area = \(6\sqrt{35}\approx 35.49\), so altitude = \(5.9\) cm, making it False).
(c) True. Euclid's Axiom: "The whole is greater than the part."
(d) True. Playfair's axiom / parallel line properties confirm distinct intersecting lines cannot both be parallel to the same line.
(e) False. \((1, -1)\) lies in Q4, whereas \((-1, 1)\) lies in Q2.
(f) False. A point on the \(y\)-axis at distance 2 units has coordinates \((0, 2)\) or \((0, -2)\), not \((2, 0)\).
(g) True. A binomial is a polynomial with two terms, and its degree can be any whole number like 5 (e.g., \(x^5 + 1\)).
(h) False. A polynomial can have multiple zeros depending on its degree (Fundamental Theorem of Algebra).
(i) False. There are infinitely many rational numbers between any two real numbers.
(j) False. The square of an irrational number like \(\sqrt[3]{2}\) is \(\sqrt[3]{4}\), which is still irrational.
In simple words: We test each mathematical and axiomatic statement using fundamental definitions and theorems.

Exam Tip: Read True/False statements carefully, checking boundary conditions and counter-examples.

Please refer to attached file for CBSE Class 9 Mathematics Worksheet - Revision Worksheet (3)