CBSE Class 9 Mathematics Herons Formula Assignment Set 04

Question 1. If \(ax^3 + bx^2 + x - 6\) has \(x + 2\) as a factor and remainder 4 when divided by \((x - 2)\). Find the values of 'a' and 'b'.
Answer: Let \(P(x) = ax^3 + bx^2 + x - 6\).
Since \((x + 2)\) is a factor of \(P(x)\), by the Factor Theorem, \(P(-2) = 0\):
\(a(-2)^3 + b(-2)^2 + (-2) - 6 = 0\)
\(-8a + 4b - 2 - 6 = 0\)
\(-8a + 4b - 8 = 0\)
Divide by 4:
\(-2a + b - 2 = 0 \implies 2a - b = -2\) --- (Equation 1)
When \(P(x)\) is divided by \((x - 2)\), the remainder is 4. By the Remainder Theorem, \(P(2) = 4\):
\(a(2)^3 + b(2)^2 + (2) - 6 = 4\)
\(8a + 4b + 2 - 6 = 4\)
\(8a + 4b - 4 = 4\)
\(8a + 4b = 8\)
Divide by 4:
\(2a + b = 2\) --- (Equation 2)
Adding Equation 1 and Equation 2:
\((2a - b) + (2a + b) = -2 + 2\)
\(4a = 0 \implies a = 0\)
Substitute \(a = 0\) into Equation 2:
\(2(0) + b = 2 \implies b = 2\)
Therefore, \(a = 0\) and \(b = 2\).
In simple words: We used the factor theorem to set \(P(-2) = 0\) and the remainder theorem to set \(P(2) = 4\), giving two simultaneous equations to solve for a and b.
Exam Tip: Remember that if \((x - c)\) is a factor, \(P(c) = 0\). If it leaves a remainder \(r\), then \(P(c) = r\).

Question 2. If \(a + b + c = 9, ab + bc + ca = 40\), find \(a^2 + b^2 + c^2\).
Answer: Using the algebraic identity for the square of a trinomial:
\((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
Substitute the given values into the identity:
\((9)^2 = a^2 + b^2 + c^2 + 2(40)\)
\(81 = a^2 + b^2 + c^2 + 80\)
\(a^2 + b^2 + c^2 = 81 - 80\)
\(a^2 + b^2 + c^2 = 1\)
In simple words: Expand the formula for \((a + b + c)^2\) and plug in the known values to find the sum of squares.
Exam Tip: Memorize standard algebraic identities such as \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\).

Question 3. Find the area of rhombus whose perimeter is 80m and one of whose diagonal is 24m.
Answer: Perimeter of the rhombus = 80 m
Side of the rhombus \((s) = \frac{80}{4} = 20\) m
Let the two diagonals be \(d_1\) and \(d_2\). Given \(d_1 = 24\) m.
The diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Half of the first diagonal = \(\frac{24}{2} = 12\) m
Using the right-angled triangle formed by half-diagonals and the side of the rhombus:
\(\text{Side}^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2\)
\(20^2 = 12^2 + \left(\frac{d_2}{2}\right)^2\)
\(400 = 144 + \left(\frac{d_2}{2}\right)^2\)
\(\left(\frac{d_2}{2}\right)^2 = 400 - 144 = 256\)
\(\frac{d_2}{2} = \sqrt{256} = 16\) m
\(d_2 = 16 \times 2 = 32\) m
Area of the rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
\(\text{Area} = \frac{1}{2} \times 24 \times 32 = 12 \times 32 = 384\) \(m^2$
In simple words: Find the side length from the perimeter, then use the property that diagonals bisect perpendicularly to find the second diagonal using Pythagoras theorem. Finally, calculate the area.
Exam Tip: Diagonals of a rhombus divide it into 4 congruent right-angled triangles where the hypotenuse is the side of the rhombus.

Question 4. If two parallel sides of a trapezium are 60cm and 77cm and other two sides are 25cm and 26cm, find the area of trapezium.
Answer: Let the parallel sides be \(a = 77\) cm and \(b = 60\) cm, and the non-parallel sides be \(c = 25\) cm and \(d = 26\) cm.
Draw a line through one vertex parallel to one of the non-parallel sides to form a triangle and a parallelogram.
The base of the newly formed triangle is \(a - b = 77 - 60 = 17\) cm, and its sides are 25 cm and 26 cm.
Let us find the area of this triangle using Heron's formula with sides \(a_t = 17\), \(b_t = 25\), \(c_t = 26\):
Semi-perimeter \((s) = \frac{17 + 25 + 26}{2} = \frac{68}{2} = 34\) cm
Area of triangle = \(\sqrt{s(s - a_t)(s - b_t)(s - c_t)}\)
\(\text{Area} = \sqrt{34(34 - 17)(34 - 25)(34 - 26)}\)
\(\text{Area} = \sqrt{34 \times 17 \times 9 \times 8}\)
\(\text{Area} = \sqrt{(17 \times 2) \times 17 \times 9 \times (4 \times 2)}\)
\(\text{Area} = \sqrt{17^2 \times 3^2 \times 4^2 \times 2^2} = 17 \times 3 \times 4 \times 2 = 408\) \(cm^2$
The area of the triangle is also given by \(\frac{1}{2} \times \text{base} \times \text{height}\):
\(408 = \frac{1}{2} \times 17 \times h\)
\(17h = 816 \implies h = \frac{816}{17} = 48\) cm
Now, the area of the trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times (77 + 60) \times 48\)
\(\text{Area} = \frac{1}{2} \times 137 \times 48 = 137 \times 24 = 3288\) \(cm^2$
In simple words: Convert the trapezium into a triangle and a parallelogram by shifting a side. Find the height of the triangle using Heron's formula, which is also the height of the trapezium.
Exam Tip: For any trapezium with non-parallel sides, creating a parallel triangle helps easily determine the height.

Question 5. Find the percentage increase in the area of a triangle if its each side is doubled.
Answer: Let the original sides of the triangle be \(a\), \(b\), and \(c\).
Original semi-perimeter \((s) = \frac{a + b + c}{2}\)
Original area \((A_1) = \sqrt{s(s - a)(s - b)(s - c)}\)
If each side is doubled, the new sides are \(2a\), \(2b\), and \(2c\).
New semi-perimeter \((s') = \frac{2a + 2b + 2c}{2} = 2\left(\frac{a + b + c}{2}\right) = 2s\)
New area \((A_2) = \sqrt{s'(s' - 2a)(s' - 2b)(s' - 2c)}\)
\(A_2 = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}\)
\(A_2 = \sqrt{2s \times 2(s - a) \times 2(s - b) \times 2(s - c)}\)
\(A_2 = \sqrt{16 \cdot s(s - a)(s - b)(s - c)}\)
\(A_2 = 4 \sqrt{s(s - a)(s - b)(s - c)} = 4A_1\)
Increase in area = \(A_2 - A_1 = 4A_1 - A_1 = 3A_1\)
Percentage increase = \(\frac{3A_1}{A_1} \times 100\% = 300\%\)
In simple words: When all sides are doubled, the semi-perimeter doubles and the new area becomes 4 times the original area, leading to a 300% increase.
Exam Tip: Whenever linear dimensions are doubled, area increases by a factor of \(2^2 = 4\), which means a 300% net increase.

Question 6. Let \(R_1\) and \(R_2\) are the remainders when the polynomials \(x^3 + 2x^2 - 5ax - 7\) and \(x^3 + ax^2 - 12x + 6\) are divided by \(x + 1\) and \(x - 2\) respectively. If \(2R_1 + R_2 = 6\), find the value of 'a'.
Answer: Let \(P(x) = x^3 + 2x^2 - 5ax - 7\) and \(Q(x) = x^3 + ax^2 - 12x + 6\).
\(R_1\) is the remainder when \(P(x)\) is divided by \((x + 1)\). By Remainder Theorem:
\(R_1 = P(-1) = (-1)^3 + 2(-1)^2 - 5a(-1) - 7\)
\(R_1 = -1 + 2(1) + 5a - 7\)
\(R_1 = -1 + 2 + 5a - 7 = 5a - 6\)
\(R_2\) is the remainder when \(Q(x)\) is divided by \((x - 2)\). By Remainder Theorem:
\(R_2 = Q(2) = (2)^3 + a(2)^2 - 12(2) + 6\)
\(R_2 = 8 + 4a - 24 + 6\)
\(R_2 = 4a - 10\)
Given equation: \(2R_1 + R_2 = 6\)
\(2(5a - 6) + (4a - 10) = 6\)
\(10a - 12 + 4a - 10 = 6\)
\(14a - 22 = 6\)
\(14a = 6 + 22 = 28\)
\(a = \frac{28}{14} = 2\)
Therefore, \(a = 2\).
In simple words: Find expressions for remainders \(R_1\) and \(R_2\) using the remainder theorem, substitute them into the given condition, and solve for a.
Exam Tip: Be careful with signs when substituting negative numbers into powers and coefficients.

Question 7. Factorise: \(2x^3 - 3x^2 - 17x + 30\).
Answer: Let \(P(x) = 2x^3 - 3x^2 - 17x + 30\).
Factors of the constant term 30 are \(\pm 1, \pm 2, \pm 3, \pm 5, \dots\)
Let us test \(x = 2\):
\(P(2) = 2(2)^3 - 3(2)^2 - 17(2) + 30\)
\(P(2) = 2(8) - 3(4) - 34 + 30\)
\(P(2) = 16 - 12 - 34 + 30 = 4 - 34 + 30 = 0\)
Since \(P(2) = 0\), by the Factor Theorem, \((x - 2)\) is a factor of \(P(x)\).
Now, perform polynomial long division or synthetic division of \(P(x)\) by \((x - 2)\):
\(2x^3 - 3x^2 - 17x + 30 = (x - 2)(2x^2 + x - 15)\)
Now factorize the quadratic polynomial \(2x^2 + x - 15\) by splitting the middle term:
\(2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)\)
Thus, the complete factorization is:
\(2x^3 - 3x^2 - 17x + 30 = (x - 2)(2x - 5)(x + 3)\)
In simple words: Find one factor by trial and error using the constant term, then divide the polynomial and factorize the remaining quadratic expression.
Exam Tip: Always check factors of the constant term for trial values. Usually, \(\pm 1, \pm 2, \pm 3\) yield zero for school-level problems.

Question 8. If the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144cm. Find the area of triangle and height to the longest side.
Answer: Let the sides of the triangle be \(3x\), \(4x\), and \(5x\).
Perimeter = \(3x + 4x + 5x = 144\)
\(12x = 144 \implies x = 12\) cm
The sides are:
\(a = 3 \times 12 = 36\) cm
\(b = 4 \times 12 = 48\) cm
\(c = 5 \times 12 = 60\) cm
Notice that \(36^2 + 48^2 = 1296 + 2304 = 3600 = 60^2\), so this is a right-angled triangle with the longest side (60 cm) as the hypotenuse.
Area of the right-angled triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times 36 \times 48 = 18 \times 48 = 864\) \(cm^2$
Alternatively, using Heron's formula: semi-perimeter \(s = \frac{144}{2} = 72\),
\(\text{Area} = \sqrt{72(72 - 36)(72 - 48)(72 - 60)} = \sqrt{72 \times 36 \times 24 \times 12} = 864\) \(cm^2$
Now, to find the height \(h\) corresponding to the longest side (hypotenuse = 60 cm):
\(\text{Area} = \frac{1}{2} \times \text{Longest Side} \times h\)
\(864 = \frac{1}{2} \times 60 \times h\)
\(864 = 30h \implies h = \frac{864}{30} = 28.8\) cm
In simple words: Find the actual side lengths from the ratio, recognize it is a right-angled triangle, calculate the area, and use the area formula again with the longest side as base to find the height.
Exam Tip: Checking for Pythagorean triples (\(3:4:5\) ratio) quickly reveals if a triangle is right-angled, saving time on Heron's formula calculations.

Question 9. Find the area of the shaded region if AD = 12cm, AC = 52cm, BC = 48cm, BD = 16cm and ADB is a right angled triangle.

Answer: The figure consists of two triangles: \(\triangle ADB\) and \(\triangle ABC\).
Step 1: Find the area of right-angled \(\triangle ADB\).
Given \(AD = 12\) cm, \(BD = 16\) cm, and \(\angle ADB = 90^\circ\).
\(\text{Area of } \triangle ADB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BD \times AD\)
\(\text{Area of } \triangle ADB = \frac{1}{2} \times 16 \times 12 = 8 \times 12 = 96\) \(cm^2$
Using Pythagoras theorem in \(\triangle ADB\), find side \(AB\):
\(AB^2 = AD^2 + BD^2 = 12^2 + 16^2 = 144 + 256 = 400\)
\(AB = \sqrt{400} = 20\) cm
Step 2: Find the area of \(\triangle ABC\) using Heron's formula.
The sides of \(\triangle ABC\) are \(AB = 20\) cm, \(BC = 48\) cm, and \(AC = 52\) cm.
Notice that \(20^2 + 48^2 = 400 + 2304 = 2704\) and \(52^2 = 2704\), which means \(\triangle ABC\) is also a right-angled triangle at \(B\).
\(\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 20 \times 48 = 10 \times 48 = 480\) \(cm^2$
Step 3: Find the area of the shaded region.
From the diagram, the shaded region is the area of \(\triangle ABC\) minus the area of \(\triangle ADB\):
\(\text{Shaded Area} = \text{Area of } \triangle ABC - \text{Area of } \triangle ADB\)
\(\text{Shaded Area} = 480 - 96 = 384\) \(cm^2$
In simple words: Calculate the areas of both right-angled triangles separately and subtract the inner triangle from the outer triangle to get the shaded region.

Exam Tip: Always check if triangles are right-angled using Pythagoras theorem to simplify area calculations without needing full semi-perimeter formulas.

Question 10. Find the value of \(64x^3 - 125z^3\), if \(4x - 5z = 16\) and \(xz = 12\).
Answer: We need to find the value of \(64x^3 - 125z^3\), which can be written as \((4x)^3 - (5z)^3\).
Using the algebraic identity: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) or \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\).
Let \(a = 4x\) and \(b = 5z\). Then \(a - b = 4x - 5z = 16\).
Cube both sides of \((4x - 5z) = 16\):
\((4x - 5z)^3 = 16^3\)
Using the expansion \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\):
\((4x)^3 - (5z)^3 - 3(4x)(5z)(4x - 5z) = 4096\)
\(64x^3 - 125z^3 - 60xz(4x - 5z) = 4096\)
Substitute the given values \(xz = 12\) and \((4x - 5z) = 16\):
\(64x^3 - 125z^3 - 60(12)(16) = 4096\)
Calculate \(60 \times 12 \times 16\):
\(60 \times 192 = 11520\)
\(64x^3 - 125z^3 - 11520 = 4096\)
\(64x^3 - 125z^3 = 4096 + 11520\)
\(64x^3 - 125z^3 = 15616\)
In simple words: Cube the given binomial expression \((4x - 5z)\) and substitute the values of \(xz\) and \((4x - 5z)\) into the expanded identity to solve for the target expression.
Exam Tip: Memorize cubic identities like \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\) as they frequently appear in polynomial manipulation problems.