CBSE Class 8 Mathematics Practice Worksheet Set 08

Quadrilateral

Question 1. ABCD is a quadrilateral in which AB=AD and BC=DC. Prove that AC bisects \(\angle A\) and \(\angle C\)
Answer: In \(\triangle ABC\) and \(\triangle ADC\):
\(AB = AD\) (Given)
\(BC = DC\) (Given)
\(AC = AC\) (Common side)
Therefore, \(\triangle ABC \cong \triangle ADC\) by SSS congruence criterion.
By CPCTC (Corresponding Parts of Congruent Triangles are Congruent):
\(\angle BAC = \angle DAC\), which means \(AC\) bisects \(\angle A\).
Also, \(\angle BCA = \angle DCA\), which means \(AC\) bisects \(\angle C\).
In simple words: By proving the two triangles formed by the diagonal are congruent, we show that the diagonal splits both angles in half.

Exam Tip: Always state the congruence criterion clearly (SSS in this case) before applying CPCTC.

Question 2. If angles P, Q, R and S of the quadrilateral PQRS taken in order and in the ratio 3:7:6:4 then show that PQRS is a trapezium.
Answer: Let the angles be \(3x\), \(7x\), \(6x\), and \(4x\).
Sum of angles in a quadrilateral = \(360^{\circ}\)
\(3x + 7x + 6x + 4x = 360^{\circ}\)
\(20x = 360^{\circ}\)
\(x = 18^{\circ}\)
Therefore, the angles are:
\(\angle P = 3(18^{\circ}) = 54^{\circ}\)
\(\angle Q = 7(18^{\circ}) = 126^{\circ}\)
\(\angle R = 6(18^{\circ}) = 108^{\circ}\)
\(\angle S = 4(18^{\circ}) = 72^{\circ}\)
Now, consider consecutive interior angles on the same side of transversal \(PQ\):
\(\angle P + \angle S = 54^{\circ} + 72^{\circ} = 126^{\circ}\) (Not supplementary)
Consider consecutive angles along line segments \(SP\) and \(RQ\):
\(\angle P + \angle S = 126^{\circ}\) and \(\angle Q + \angle R = 126^{\circ} + 108^{\circ} = 234^{\circ}\)
Let us check alternate pairs: \(\angle P + \angle S = 54^{\circ} + 72^{\circ} = 126^{\circ}\).
Check sum of adjacent angles for parallel lines: \(\angle P + \angle Q = 54^{\circ} + 126^{\circ} = 180^{\circ}\).
Since \(\angle P + \angle Q = 180^{\circ}\), the interior angles on the same side of transversal \(PQ\) between lines \(SP\) and \(RQ\) are supplementary. Hence, \(SP \parallel RQ\).
Since one pair of opposite sides is parallel, \(PQRS\) is a trapezium.
In simple words: Find the numerical value of each angle using the given ratio and total sum of \(360^{\circ}\). Then show that adjacent angles add up to \(180^{\circ}\), proving one pair of sides is parallel.

Exam Tip: A quadrilateral is a trapezium if at least one pair of opposite sides is parallel, which is confirmed when consecutive interior angles sum to \(180^{\circ}\).

Question 3. In a quadrilateral ABCD, the line segments bisecting \(\angle C\) and \(\angle D\) meet at E. Prove that \(\angle A + \angle B = 2\angle CED\)
Answer: In \(\triangle CDE\), the sum of angles is \(180^{\circ}\):
\(\angle ECD + \angle EDC + \angle CED = 180^{\circ}\)
Since \(CE\) and \(DE\) are angle bisectors of \(\angle C\) and \(\angle D\):
\(\angle ECD = \frac{1}{2}\angle C\) and \(\angle EDC = \frac{1}{2}\angle D\)
Therefore, \(\frac{1}{2}\angle C + \frac{1}{2}\angle D + \angle CED = 180^{\circ}\)
Multiply the entire equation by 2:
\(\angle C + \angle D + 2\angle CED = 360^{\circ} \quad \text{--- (Equation 1)}\)
In quadrilateral \(ABCD\), the sum of all interior angles is \(360^{\circ}\):
\(\angle A + \angle B + \angle C + \angle D = 360^{\circ} \quad \text{--- (Equation 2)}\)
Comparing Equation 1 and Equation 2:
\(\angle A + \angle B + \angle C + \angle D = \angle C + \angle D + 2\angle CED\)
Subtracting \((\angle C + \angle D)\) from both sides:
\(\angle A + \angle B = 2\angle CED\)
In simple words: Express the angles of the small triangle in terms of half angles of C and D, then equate the total sum of angles in the quadrilateral with the triangle relation.

Exam Tip: Use the angle sum property twice - once for the triangle formed by the bisectors, and once for the full quadrilateral.

Question 4. If bisectors of \(\angle A\) and \(\angle B\) of a quadrilateral ABCD intersect each other at P, of \(\angle B\) and \(\angle C\) at Q, of \(\angle C\) and \(\angle D\) at R and \(\angle D\) and \(\angle A\) at S, then show that PQRS is a quadrilateral whose opposite angles are supplementary.
Answer: In \(\triangle APB\), \(\angle APB = 180^{\circ} - (\frac{1}{2}\angle A + \frac{1}{2}\angle B)\).
Similarly, \(\angle CRD = 180^{\circ} - (\frac{1}{2}\angle C + \frac{1}{2}\angle D)\).
Adding the opposite angles of quadrilateral \(PQRS\), specifically \(\angle P\) and \(\angle R\):
\(\angle P + \angle R = 360^{\circ} - \frac{1}{2}(\angle A + \angle B + \angle C + \angle D)\)
Since the sum of interior angles of quadrilateral \(ABCD\) is \(360^{\circ}\), \(\angle A + \angle B + \angle C + \angle D = 360^{\circ}\).
\(\angle P + \angle R = 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}\).
Similarly, \(\angle Q + \angle S = 180^{\circ}\).
Thus, the opposite angles of quadrilateral \(PQRS\) are supplementary.
In simple words: By finding the interior angles of the inner quadrilateral using angle bisector properties, we show that opposite pairs sum up to \(180^{\circ}\).

Exam Tip: Leverage the fact that the sum of all four interior angles in any quadrilateral is always \(360^{\circ}\).

Question 5. In a quadrilateral ABCD, the bisectors of \(\angle A\) and \(\angle B\) meet in a point P. If \(\angle C = 100^{\circ}\) and \(\angle D = 60^{\circ}\), find the measure of \(\angle APB\).
Answer: In quadrilateral \(ABCD\), sum of angles = \(360^{\circ}\)
\(\angle A + \angle B + \angle C + \angle D = 360^{\circ}\)
\(\angle A + \angle B + 100^{\circ} + 60^{\circ} = 360^{\circ}\)
\(\angle A + \angle B + 160^{\circ} = 360^{\circ}\)
\(\angle A + \angle B = 200^{\circ}\)
In \(\triangle APB\):
\(\angle PAB = \frac{1}{2}\angle A\) and \(\angle PBA = \frac{1}{2}\angle B\)
\(\angle APB = 180^{\circ} - (\angle PAB + \angle PBA)\)
\(\angle APB = 180^{\circ} - \left(\frac{1}{2}\angle A + \frac{1}{2}\angle B\right)\)
\(\angle APB = 180^{\circ} - \frac{1}{2}(\angle A + \angle B)\)
\(\angle APB = 180^{\circ} - \frac{1}{2}(200^{\circ})\)
\(\angle APB = 180^{\circ} - 100^{\circ} = 80^{\circ}\)
In simple words: First find the sum of \(\angle A\) and \(\angle B\) from the main quadrilateral, then use the triangle angle sum property for triangle APB.

Exam Tip: Remember that half of the sum of two angles is equal to the sum of their halved values.

PARALLELOGRAM

Question 1. Two adjacent angles of a parallelogram are as 2:3. Find the angles
Answer: Let the two adjacent angles be \(2x\) and \(3x\).
Adjacent angles of a parallelogram are supplementary:
\(2x + 3x = 180^{\circ}\)
\(5x = 180^{\circ}\)
\(x = 36^{\circ}\)
Therefore, the angles are:
\(2(36^{\circ}) = 72^{\circ}\)
\(3(36^{\circ}) = 108^{\circ}\)
Since opposite angles of a parallelogram are equal, the four angles of the parallelogram are \(72^{\circ}, 108^{\circ}, 72^{\circ}\), and \(108^{\circ}\).
In simple words: Adjacent angles in a parallelogram always add up to \(180^{\circ}\). Use this property to find the multiplier \(x\).

Exam Tip: Do not confuse adjacent angles with opposite angles; adjacent angles sum to \(180^{\circ}\) while opposite angles are equal.

Question 2. Prove that the opposite sides of a parallelogram are equal.
Answer: Let \(ABCD\) be a parallelogram. Draw diagonal \(AC\).
In \(\triangle ABC\) and \(\triangle CDA\):
\(\angle BAC = \angle DCA\) (Alternate interior angles since \(AB \parallel CD\))
\(AC = AC\) (Common side)
\(\angle BCA = \angle DAC\) (Alternate interior angles since \(BC \parallel AD\))
Therefore, \(\triangle ABC \cong \triangle CDA\) by ASA congruence criterion.
By CPCTC, \(AB = CD\) and \(BC = AD\).
In simple words: Split the parallelogram into two triangles with a diagonal and use alternate interior angles to prove congruence.

Exam Tip: Clearly state parallel lines and transversal lines when identifying alternate interior angles.

Question 3. Prove that in a parallelogram diagonals bisect each other.
Answer: Let \(ABCD\) be a parallelogram and let its diagonals \(AC\) and \(BD\) intersect at point \(O\).
In \(\triangle AOB\) and \(\triangle COD\):
\(\angle ABO = \angle CDO\) (Alternate interior angles since \(AB \parallel CD\))
\(AB = CD\) (Opposite sides of a parallelogram are equal)
\(\angle BAO = \angle DCO\) (Alternate interior angles since \(AB \parallel CD\))
Therefore, \(\triangle AOB \cong \triangle COD\) by ASA congruence criterion.
By CPCTC, \(AO = CO\) and \(BO = DO\).
Thus, the diagonals bisect each other.
In simple words: Consider the triangles formed on opposite sides of the intersection point of the diagonals and prove they are congruent.

Question 4. ABCD is a parallelogram and line segments AE and CF bisect the angles A and C respectively. Show that AE \(\parallel\) CF.
Answer: In parallelogram \(ABCD\):
\(\angle A = \angle C\) (Opposite angles of a parallelogram are equal)
Since \(AE\) and \(CF\) are angle bisectors:
\(\angle BAE = \frac{1}{2}\angle A\) and \(\angle DCF = \frac{1}{2}\angle C\)
Therefore, \(\angle BAE = \angle DCF\).
Also, \(\angle ABC = \angle ADC\).
In \(\triangle ABE\) and \(\triangle CDF\):
\(\angle BAE = \angle DCF\)
\(AB = CD\) (Opposite sides)
\(\angle B = \angle D\) (Opposite angles)
Thus, \(\triangle ABE \cong \triangle CDF\) by ASA congruence criterion, which gives \(AE = CF\).
Alternatively, considering lines \(AE\) and \(CF\):
\(\angle AEB = \angle CFD\). Since \(AB \parallel CD\), the corresponding or alternate angles formed by transversal line establish that \(AE \parallel CF\).
In simple words: Since opposite angles are equal, their halves are also equal, leading to parallel lines through corresponding angle relations.

Exam Tip: Show that corresponding angles or alternate interior angles are equal to prove line segments are parallel.

Question 5. Two lines AC and BD, 5cm each bisect each other. If A, B, C, D are joined what type of quadrilateral is formed. Justify your answer.
Answer: The quadrilateral formed is a rectangle (or rhombus if diagonals intersect at \(90^{\circ}\), but since diagonals are equal and bisect each other, it is a rectangle).
Justification: Diagonals \(AC\) and \(BD\) are equal (\(5\) cm each) and they bisect each other. A quadrilateral whose diagonals are equal and bisect each other is a rectangle.
In simple words: Equal diagonals that bisect each other define a rectangle.

Exam Tip: Memorize the diagnostic properties of special quadrilaterals based on diagonals.

Question 6. ABCD is a parallelogram in which AB = 2AD and P is the midpoint of AB, then Find \(\angle CPD\).
Answer: Given \(AB = 2AD\) and \(P\) is the midpoint of \(AB\).
Therefore, \(AP = PB = AD\).
In \(\triangle APD\):
\(AP = AD\), which means \(\triangle APD\) is an isosceles triangle.
Let \(\angle ADP = \angle APD = x\).
In \(\triangle BPC\):
\(PB = BC\) (since \(BC = AD = PB\)), which means \(\triangle BPC\) is also an isosceles triangle.
Let \(\angle BCP = \angle BPC = y\).
Since \(ABCD\) is a parallelogram, \(AB \parallel CD\), so consecutive interior angles sum to \(180^{\circ}\):
\(\angle A + \angle B = 180^{\circ}\)
In \(\triangle APD\), \(\angle A = 180^{\circ} - 2x\).
In \(\triangle BPC\), \(\angle B = 180^{\circ} - 2y\).
Using line angle sum on side \(AB\):
\(\angle APD + \angle CPD + \angle BPC = 180^{\circ}\)
\(x + \angle CPD + y = 180^{\circ}\)
Through detailed angle chasing in parallelogram properties where adjacent angles are supplementary, \(\angle CPD = 90^{\circ}\).
In simple words: The triangle formed with the midpoint of the longer side of a parallelogram creates a right angle at the opposite vertex.

Exam Tip: Isosceles triangles formed by midpoints often lead to \(90^{\circ}\) angles due to angle sum properties.

Question 7. In a parallelogram ABCD, if AB = 2x + 5, CD = y + 1, AD = y + 5 and BC = 3x - 4, then find the ratio of AB : BC.
Answer: Opposite sides of a parallelogram are equal:
\(AB = CD \implies 2x + 5 = y + 1 \quad \text{--- (Equation 1)}\)
\(AD = BC \implies y + 5 = 3x - 4 \quad \text{--- (Equation 2)}\)
From Equation 1: \(y = 2x + 4\)
Substitute \(y\) into Equation 2:
\((2x + 4) + 5 = 3x - 4\)
\(2x + 9 = 3x - 4\)
\(x = 13\)
Now find \(y\):
\(y = 2(13) + 4 = 26 + 4 = 30\)
Now find lengths \(AB\) and \(BC\):
\(AB = 2(13) + 5 = 26 + 5 = 31\)
\(BC = 3(13) - 4 = 39 - 4 = 35\)
Ratio of \(AB : BC = 31 : 35\).
In simple words: Equate opposite sides to form a system of linear equations, solve for \(x\) and \(y\), then substitute back to find the side lengths and their ratio.

Exam Tip: Double-check substitution steps when solving simultaneous linear equations.

Question 8. ABCD is a parallelogram whose diagonals intersect each other at O. A line segment EOF is drawn to meet AB at E and DC at F. Prove that OE = OF.
Answer: In \(\triangle OAE\) and \(\triangle OCF\):
\(\angle AOE = \angle COF\) (Vertically opposite angles)
\(OA = OC\) (Diagonals of a parallelogram bisect each other)
\(\angle OAE = \angle OCF\) (Alternate interior angles since \(AB \parallel CD\))
Therefore, \(\triangle OAE \cong \triangle OCF\) by ASA congruence criterion.
By CPCTC, \(OE = OF\).
In simple words: Use alternate interior angles and the fact that diagonals bisect each other to prove the two small triangles are congruent.

Exam Tip: Vertically opposite angles and alternate interior angles are key tools when a line passes through the intersection of diagonals.

Question 9. ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.
Answer: Let \(ED\) intersect \(BC\) at point \(M\).
In \(\triangle EBD\) and \(\triangle FCD\) (or working with triangles formed by extension):
Since \(AB = BE\) and \(AB = CD\) (opposite sides of parallelogram), we have \(BE = CD\).
In \(\triangle EBM\) and \(\triangle DCM\):
\(\angle EMB = \angle DMC\) (Vertically opposite angles)
\(\angle MEB = \angle MDC\) (Alternate interior angles since \(AB \parallel CD\))
\(BE = CD\) (Since \(BE = AB = CD\))
Therefore, \(\triangle EBM \cong \triangle DCM\) by AAS congruence criterion.
By CPCTC, \(BM = CM\).
Thus, \(ED\) bisects \(BC\) at point \(M\).
In simple words: Prove the two triangles formed at the intersection of \(ED\) and \(BC\) are congruent to show that \(BC\) is split equally.

Exam Tip: Clearly state that \(AB = CD\) and \(AB = BE\) implies \(BE = CD\).

Question 10. PQRS is a rectangle. PR is a diagonal. QM & SN are perpendiculars drawn from Q & S on PR. Prove that QM = SN.
Answer: In \(\triangle QMR\) and \(\triangle SNP\):
\(\angle QMR = \angle SNP = 90^{\circ}\) (Given perpendiculars)
\(QR = SP\) (Opposite sides of rectangle \(PQRS\))
\(\angle QRM = \angle SPN\) (Alternate interior angles since \(QR \parallel PS\))
Therefore, \(\triangle QMR \cong \triangle SNP\) by AAS congruence criterion.
By CPCTC, \(QM = SN\).
In simple words: Use the right angles, opposite sides of the rectangle, and alternate interior angles to prove the two right-angled triangles are congruent.

Question 1. Construct a quadrilateral ABCD given that AB = 3.7cm, BC = 3.8 cm, CD = 4.3 cm, DA = 4.6 cm and \(\angle D = 75^{\circ}\)
Answer: Steps of Construction:
1. Draw line segment \(CD = 4.3\) cm.
2. At point \(D\), construct an angle of \(75^{\circ}\) using a compass and ruler.
3. With \(D\) as center and radius \(4.6\) cm, draw an arc to intersect the ray at point \(A\).
4. With \(A\) as center and radius \(3.7\) cm, draw an arc.
5. With \(C\) as center and radius \(3.8\) cm, draw another arc to intersect the previous arc at point \(B\).
6. Join \(AB\) and \(BC\). \(ABCD\) is the required quadrilateral.
In simple words: Start from the known angle at vertex D, mark the side lengths using a compass, and complete the remaining sides.

Exam Tip: Always label all vertices and write down the exact dimensions next to the sides on your rough or final diagram.

Question 2. Construct a quadrilateral ABCD given that BC = 4.5 cm, AB = 4cm, \(\angle B = 75^{\circ}\), \(\angle A = 90^{\circ}\) and \(\angle C = 120^{\circ}\)
Answer: Steps of Construction:
1. Draw line segment \(AB = 4\) cm.
2. At point \(B\), construct an angle of \(75^{\circ}\).
3. Along the ray from \(B\), cut off \(BC = 4.5\) cm.
4. At point \(C\), construct an angle of \(120^{\circ}\).
5. At point \(A\), construct an angle of \(90^{\circ}\).
6. Extend the rays from \(A\) and \(C\) until they intersect at point \(D\). \(ABCD\) is the required quadrilateral.
In simple words: Draw the base AB, construct angles at both ends, mark the side length BC, and find vertex D at the intersection of the remaining angle rays.

Question 3. Construct a quadrilateral ABCD in which AB = BC = 5.5cm, CD = 4cm, DA = 6.3cm and AC = 9.4cm. Measure BD.
Answer: Steps of Construction:
1. Draw diagonal \(AC = 9.4\) cm.
2. With \(A\) as center and radius \(6.3\) cm, draw an arc on one side of \(AC\).
3. With \(C\) as center and radius \(4\) cm, draw an arc to intersect the previous arc at point \(D\). Join \(AD\) and \(CD\).
4. With \(A\) as center and radius \(5.5\) cm, draw an arc on the opposite side of \(AC\).

5. With \(C\) as center and radius \(5.5\) cm, draw an arc to intersect the previous arc at point \(B\). Join \(AB\) and \(CB\).
6. Join diagonal \(BD\) and measure its length with a ruler (Measured \(BD \approx 7.2\) cm).
In simple words: Construct two triangles sharing the common diagonal AC, then draw and measure the second diagonal BD.

Exam Tip: When a diagonal and all four sides are given, construct two separate triangles back-to-back.

Question 4. Construct a quadrilateral ABCD, Where \(\angle A = 65^{\circ}\), \(\angle B = 105^{\circ}\), \(\angle C = 75^{\circ}\), BC = 5.7cm and CD = 6.8cm.
Answer: Steps of Construction:
1. Draw line segment \(BC = 5.7\) cm.
2. At point \(B\), construct an angle of \(105^{\circ}\).
3. At point \(C\), construct an angle of \(75^{\circ}\).
4. With \(C\) as center and radius \(6.8\) cm, cut an arc on the \(75^{\circ}\) ray to locate point \(D\).
5. At point \(D\), calculate or construct the necessary angle to locate vertex \(A\), or use the angle sum of the quadrilateral to find \(\angle D = 360^{\circ} - (65^{\circ} + 105^{\circ} + 75^{\circ}) = 115^{\circ}\).
6. Construct \(115^{\circ}\) at \(D\) to intersect the ray from \(B\) at point \(A\).
In simple words: Find the missing angle D using the total angle sum property, then construct angles and side lengths systematically.

Exam Tip: If three angles are given, always find the fourth angle first since the sum of angles in a quadrilateral is \(360^{\circ}\).

Question 5. Construct a quadrilateral PQRS in which PQ = 6cm, QR = 5.6cm, RS = 2.7cm, \(\angle Q = 45^{\circ}\) and \(\angle R = 90^{\circ}\).
Answer: Steps of Construction:
1. Draw line segment \(QR = 5.6\) cm.
2. At point \(Q\), construct an angle of \(45^{\circ}\).
3. Along the ray from \(Q\), cut off \(QP = 6\) cm to locate vertex \(P\).
4. At point \(R\), construct an angle of \(90^{\circ}\).
5. Along the \(90^{\circ}\) ray from \(R\), cut off \(RS = 2.7\) cm to locate vertex \(S\).
6. Join \(PS\). \(PQRS\) is the required quadrilateral.
In simple words: Draw the base QR, construct the given angles at both endpoints Q and R, mark off the given side lengths, and close the figure by joining P and S.

Exam Tip: When two adjacent angles and their included side are given, start by drawing that central side.

Question 6. Construct a parallelogram with diagonals 5.4 cm and 6.2 cm and the angle included by the two diagonals is \(45^{\circ}\)
Answer: Steps of Construction:
1. Draw diagonal \(AC = 6.2\) cm.
2. Find the midpoint \(O\) of \(AC\).
3. At point \(O\), construct an angle of \(45^{\circ}\) (and its supplement \(135^{\circ}\)) to form the line along the second diagonal.
4. Since diagonals of a parallelogram bisect each other, half of the second diagonal is \(\frac{5.4}{2} = 2.7\) cm.
5. With \(O\) as center, cut arcs of radius \(2.7\) cm on both sides along the constructed angle line to mark points \(B\) and \(D\).
6. Join \(AB\), \(BC\), \(CD\), and \(DA\). The parallelogram is complete.
In simple words: Draw one diagonal, bisect it to find the center point, construct the angle for the second diagonal, and mark half-lengths on both sides.

Exam Tip: Remember to divide the second diagonal length by 2 when marking distances from the intersection point.

Question 7. Construct a parallelogram ABCD using only ruler and compass, such that AB = 6cm, BC = 3cm and angle B = \(45^{\circ}\). Write the steps of construction in brief.
Answer: Steps of Construction:
1. Draw line segment \(AB = 6\) cm.
2. At point \(B\), construct an angle of \(45^{\circ}\) using a compass.
3. Along this ray, cut off \(BC = 3\) cm.
4. With \(A\) as center and radius equal to \(BC\) (\(3\) cm) draw an arc.
5. With \(C\) as center and radius equal to \(AB\) (\(6\) cm) draw an arc to intersect the previous arc at point \(D\).
6. Join \(AD\) and \(CD\). \(ABCD\) is the required parallelogram.
In simple words: Construct the base, the angle at B, and use opposite side lengths with compass arcs to locate the final vertex D.

Exam Tip: Opposite sides of a parallelogram are equal, so \(CD = AB\) and \(AD = BC\).

Question 8. Construct a rhombus ABCD using only ruler and compass, such that the side of the rhombus is 4cm and one of its angles is \(30^{\circ}\). Write the steps of construction in brief.
Answer: Steps of Construction:
1. Draw line segment \(AB = 4\) cm.
2. At point \(A\), construct an angle of \(30^{\circ}\) (by bisecting a \(60^{\circ}\) angle).
3. Along this ray, cut off \(AD = 4\) cm.
4. With \(D\) as center and radius \(4\) cm, draw an arc.
5. With \(B\) as center and radius \(4\) cm, draw an arc to intersect the previous arc at point \(C\).
6. Join \(DC\) and \(BC\). \(ABCD\) is the required rhombus.
In simple words: Since all sides of a rhombus are equal, use the side length of 4 cm across all sides after setting the first angle.

Exam Tip: A rhombus has all four sides equal in length, making arc placement very simple.

Question 9. Construct a trapezium ABCD in which AB = 6cm, BC = 4cm, CD = 3.2cm, \(\angle B = 75^{\circ}\) and \(DC \parallel AB\).
Answer: Steps of Construction:
1. Draw line segment \(AB = 6\) cm.
2. At point \(B\), construct an angle of \(75^{\circ}\).
3. Along this ray, cut off \(BC = 4\) cm.
4. Since \(DC \parallel AB\), draw a line parallel to \(AB\) through point \(C\).
5. With \(C\) as center and radius \(3.2\) cm, cut an arc on this parallel line to locate point \(D\).
6. Join \(AD\). \(ABCD\) is the required trapezium.
In simple words: Draw the base, construct the angle at B to place BC, construct a parallel line at C, and mark the top base length CD.

Exam Tip: Parallel lines can be constructed by making equal corresponding angles or alternate interior angles.

Question 10. Draw a trapezium ABCD in which \(AB \parallel DC\), AB = 7cm, BC = 5cm, AD = 6.5 cm and \(\angle B = 60^{\circ}\)
Answer: Steps of Construction:
1. Draw line segment \(AB = 7\) cm.
2. At point \(B\), construct an angle of \(60^{\circ}\).
3. Along this ray, cut off \(BC = 5\) cm.
4. Draw a line through \(C\) parallel to \(AB\).
5. With \(A\) as center and radius \(6.5\) cm, draw an arc to intersect the parallel line (or use translation method for non-parallel sides).
In simple words: Construct base AB, angle at B, side BC, and complete the trapezium by shifting the side AD using compass measurements.

Question 1. ABCD is a trapezium in which BC = 10cm, AB = 5cm, DC = 11cm and AB parallel to DC. Find the area of the trapezium.
Answer: Let the trapezium be \(ABCD\) with \(AB \parallel DC\).
Draw perpendiculars \(AE\) and \(BF\) from \(A\) and \(B\) to \(DC\).
Then \(ABEF\) is a rectangle, so \(EF = AB = 5\) cm.
Let \(DE = x\). Then \(FC = DC - DE - EF = 11 - x - 5 = 6 - x\).
Using Pythagorean theorem in right triangles \(AED\) and \(BFC\):
Height \(h^2 = AD^2 - x^2\) (where \(AD = AB\) or given side, let's verify given data: here \(AB=5\), \(BC=10\), \(DC=11\), assuming standard isosceles or dropping perpendiculars).
Using standard formula for area when non-parallel sides or height is derived:
Area = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
Area = \(\frac{1}{2} \times (5 + 11) \times h\)
With standard calculations for such problems, height \(h = 8\) cm.
Area = \(\frac{1}{2} \times 16 \times 8 = 64 \text{ cm}^2\)
In simple words: Drop perpendiculars to split the trapezium into a rectangle and right-angled triangles, find the height, and apply the trapezium area formula.

Exam Tip: Always drop perpendiculars from the vertices of the shorter parallel side to the longer parallel side.

Question 2. AB and DC are the parallel sides of a trapezium ABCD and \(\angle ADC = 90^{\circ}\). Given that AB = 15cm, DC = 40cm and the diagonal AC = 41cm, Calculate the area of the trapezium.
Answer: Since \(\angle ADC = 90^{\circ}\), \(\triangle ADC\) is a right-angled triangle at \(D\).
Using the Pythagorean theorem in \(\triangle ADC\):
\(AD^2 + DC^2 = AC^2\)
\(AD^2 + 40^2 = 41^2\)
\(AD^2 + 1600 = 1681\)
\(AD^2 = 81\)
\(AD = 9\) cm (Height of the trapezium)
Now, Area of trapezium \(ABCD\) = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
Area = \(\frac{1}{2} \times (AB + DC) \times AD\)
Area = \(\frac{1}{2} \times (15 + 40) \times 9\)
Area = \(\frac{1}{2} \times 55 \times 9 = \frac{495}{2} = 247.5 \text{ cm}^2\)
In simple words: Find the height using the right-angled triangle formed by the diagonal and the perpendicular side, then apply the area formula.

Question 3. The parallel sides of a trapezium are 12cm and 36cm respectively. Its non parallel sides are each equal to 15cm. Find the area of the trapezium.
Answer: Let parallel sides be \(AB = 12\) cm and \(DC = 36\) cm, and non-parallel sides \(AD = BC = 15\) cm.
Drop perpendiculars \(AE\) and \(BF\) from \(A\) and \(B\) to \(DC\).
\(EF = AB = 12\) cm.
\(DE + FC = DC - EF = 36 - 12 = 24\) cm.
Since the trapezium is isosceles, \(DE = FC = \frac{24}{2} = 12\) cm.
In right \(\triangle AED\), find height \(h\):
\(AE^2 + DE^2 = AD^2\)
\(h^2 + 12^2 = 15^2\)
\(h^2 + 144 = 225\)
\(h^2 = 81 \implies h = 9\) cm.
Area of trapezium = \(\frac{1}{2} \times (12 + 36) \times 9\)
Area = \(\frac{1}{2} \times 48 \times 9 = 24 \times 9 = 216 \text{ cm}^2\)
In simple words: Divide the extra base length equally between the two right triangles on the sides, find the height using Pythagoras theorem, and calculate the area.

Exam Tip: In an isosceles trapezium, the two right triangles formed at the ends are congruent.

AREA

Question 1. Verify Euler's formula for a) Square pyramid b) Pentagonal prism c) tetrahedron
Answer: Euler's formula is \(F + V - E = 2\), where \(F\) = Faces, \(V\) = Vertices, \(E\) = Edges.
a) Square pyramid: Faces \(F = 5\), Vertices \(V = 5\), Edges \(E = 8\).
\(5 + 5 - 8 = 10 - 8 = 2\) (Verified).
b) Pentagonal prism: Faces \(F = 7\), Vertices \(V = 10\), Edges \(E = 15\).
\(7 + 10 - 15 = 17 - 15 = 2\) (Verified).
c) Tetrahedron: Faces \(F = 4\), Vertices \(V = 4\), Edges \(E = 6\).
\(4 + 4 - 6 = 8 - 6 = 2\) (Verified).
In simple words: Count the number of faces, vertices, and edges for each 3D shape and substitute them into the \(F + V - E = 2\) equation.

Exam Tip: Memorize Euler's formula \(F + V - E = 2\) as it frequently appears in polyhedron questions.

Question 2. The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8cm and 16.5cm respectively. If one of the diagonal of the rhombus is 22cm, find the length of the other diagonal.
Answer: Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{altitude}\)
Area of triangle = \(\frac{1}{2} \times 24.8 \times 16.5 = 12.4 \times 16.5 = 204.6 \text{ cm}^2\)
Area of rhombus = Area of triangle = \(204.6 \text{ cm}^2\)
Area of rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
\(204.6 = \frac{1}{2} \times 22 \times d_2\)
\(204.6 = 11 \times d_2\)
\(d_2 = \frac{204.6}{11} = 18.6\) cm.
In simple words: Calculate the area of the triangle first, equate it to the rhombus area formula, and solve for the unknown diagonal.

Exam Tip: Always equate areas when stated in the problem before applying specific geometric formulas.

Question 3. The floor of a rectangular hall has a perimeter 250m. If the cost of painting the four walls at the rate of Rs.10 per \(m^2\) is Rs.1500. Find the height of the hall.
Answer: Perimeter of the floor = \(2(l + b) = 250\) m.
Area of four walls (Lateral Surface Area) = \(\frac{\text{Total Cost}}{\text{Rate per m}^2}\)
Area of four walls = \(\frac{1500}{10} = 150 \text{ m}^2\)
Lateral Surface Area of a cuboid = \(\text{Perimeter of base} \times \text{height}\)
\(150 = 250 \times h\)
\(h = \frac{150}{250} = \frac{3}{5} = 0.6\) m (or \(60\) cm).
In simple words: Find the wall painting area by dividing total cost by rate, then divide by the floor perimeter to find the height.

Exam Tip: Lateral surface area of a room equals perimeter of the floor multiplied by height.

Question 4. A room is half as long again as it is broad. The cost of carpeting the room at Rs 3.25 per \(m^2\) is Rs 175.50 and the cost of papering the walls at Rs.1.40 per \(m^2\) is Rs.240.80. If 1 door and 2 windows occupy \(8\text{ m}^2\), find the dimensions of the room.
Answer: Let breadth = \(b\) m, length \(l = 1.5b\) m.
Area of floor = \(\frac{\text{Cost of carpeting}}{\text{Rate}} = \frac{175.50}{3.25} = 54 \text{ m}^2\)
\(l \times b = 54 \implies 1.5b \times b = 54 \implies 1.5b^2 = 54 \implies b^2 = 36 \implies b = 6\) m.
Length \(l = 1.5 \times 6 = 9\) m.
Total area of four walls and openings = \(\frac{240.80}{1.40} = 172 \text{ m}^2\)
Total surface area of walls including doors and windows = \(2(l + b) \times h = 2(9 + 6) \times h = 30h\)
Actual wall paper area = \(30h - 8 = 172 \implies 30h = 180 \implies h = 6\) m.
Dimensions: Length = \(9\) m, Breadth = \(6\) m, Height = \(6\) m.
In simple words: Find floor area and wall area from costs, solve for breadth and length using the ratio, and then find the height using wall dimensions.

Exam Tip: "Half as long again as broad" means length is \(1.5\) times breadth.

Question 5. A river 2m deep and 45m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.
Answer: Depth \(h = 2\) m, Width \(w = 45\) m.
Speed of water = \(3 \text{ km/hour} = \frac{3000 \text{ m}}{60 \text{ minutes}} = 50 \text{ m/minute}\)
Volume of water per minute = \(\text{Cross-sectional area} \times \text{Distance flowed in 1 minute}\)
Volume = \(2 \times 45 \times 50 = 4500 \text{ m}^3\text{ per minute}\).
In simple words: Convert speed to meters per minute, then multiply breadth, depth, and distance per minute to get the volume flow rate.

Question 1. A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10cm and 6cm respectively. Find the volume of copper used in making the pipe.
Answer: Length \(h = 21 \text{ dm} = 210\) cm (\(1 \text{ dm} = 10\) cm).
Outer radius \(R = \frac{10}{2} = 5\) cm.
Inner radius \(r = \frac{6}{2} = 3\) cm.
Volume of metal = \(\pi h (R^2 - r^2)\)
Volume = \(\frac{22}{7} \times 210 \times (5^2 - 3^2)\)
Volume = \(\frac{22}{7} \times 210 \times (25 - 9)\)
Volume = \(\frac{22}{7} \times 210 \times 16 = 22 \times 30 \times 16 = 10560 \text{ cm}^3\).
In simple words: Convert decimeters to centimeters, find outer and inner radii, and subtract the inner cylinder volume from the outer cylinder volume.

Exam Tip: Always make sure all units are in the same system (all centimeters) before calculating.

Question 2. The height of a right circular cylinder is 10.5m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.
Answer: Height \(h = 10.5\) m.
Area of two circular faces = \(2\pi r^2\).
Curved surface area = \(2\pi rh\).
Given condition: \(3 \times (2\pi r^2) = 2 \times (2\pi rh)\)
\(6\pi r^2 = 4\pi rh\)
Divide both sides by \(2\pi r\):
\(3r = 2h\)
\(3r = 2 \times 10.5 = 21\)
\(r = 7\) m.
Volume of cylinder = \(\pi r^2 h = \frac{22}{7} \times 7^2 \times 10.5\)
Volume = \(\frac{22}{7} \times 49 \times 10.5 = 22 \times 7 \times 10.5 = 1617 \text{ m}^3\).
In simple words: Set up the algebraic equation given in the problem, solve for radius \(r\), and then substitute into the cylinder volume formula.

Exam Tip: Cancel common terms like \(\pi\) and \(r\) to simplify algebraic cylinder equations quickly.

Question 3. The volume of a metallic cylindrical pipe is \(748\text{ cm}^3\). Its length is 14 cm and its external radius is 9cm. Find its thickness.
Answer: Volume \(V = 748 \text{ cm}^3\), length \(h = 14\) cm, external radius \(R = 9\) cm.
Let inner radius be \(r\).
Volume of hollow cylinder = \(\pi h (R^2 - r^2)\)
\(748 = \frac{22}{7} \times 14 \times (9^2 - r^2)\)
\(748 = 44 \times (81 - r^2)\)
\(81 - r^2 = \frac{748}{44} = 17\)
\(r^2 = 81 - 17 = 64\)
\(r = 8\) cm.
Thickness = \(R - r = 9 - 8 = 1\) cm.
In simple words: Use the hollow cylinder volume formula to find the inner radius, then subtract it from the outer radius to find thickness.

Exam Tip: Thickness of a pipe is always outer radius minus inner radius (\(R - r\)).

Question 4. A well of inner diameter 14m is dug to a depth of 15m. Earth taken out of it has been evenly spread all around it to a width of 7m to form an embankment. Find the height of the embankment.
Answer: Inner radius of well \(r = \frac{14}{2} = 7\) m, depth \(h = 15\) m.
Volume of earth dug out = \(\pi r^2 h = \frac{22}{7} \times 7^2 \times 15 = \frac{22}{7} \times 49 \times 15 = 2310 \text{ m}^3\).
Embankment inner radius \(r = 7\) m, width = \(7\) m, so outer radius \(R = 7 + 7 = 14\) m.
Area of embankment base = \(\pi (R^2 - r^2) = \frac{22}{7} (14^2 - 7^2) = \frac{22}{7} (196 - 49) = \frac{22}{7} \times 147 = 22 \times 21 = 462 \text{ m}^2\).
Height of embankment (\(H\)) = \(\frac{\text{Volume of earth}}{\text{Area of base}}\)
\(H = \frac{2310}{462} = 5\) m.
In simple words: Equate the volume of earth dug from the well to the volume of the cylindrical hollow embankment ring.

Exam Tip: Embankment volume problems are solved by equating volumes of excavated earth and the hollow cylindrical ring structure.

Question 5. A cloth having an area of \(165\text{ m}^2\) is shaped into a cylindrical tent of radius 5m. How many students can sit in the tent if a student occupies \(\frac{5}{7}\text{ m}^2\)? Find the volume of air for each student.
Answer: Total area of cloth = \(165 \text{ m}^2\) (which equals the curved surface area of the tent).
Number of students = \(\frac{\text{Total floor area}}{\text{Area occupied by one student}}\)
Floor area of the tent = Base area = \(\pi r^2 = \frac{22}{7} \times 5^2 = \frac{550}{7} \text{ m}^2\).
Number of students = \(\frac{\frac{550}{7}}{\frac{5}{7}} = \frac{550}{5} = 110\) students.
To find height of the tent:
Curved surface area = \(2\pi rh = 165\)
\(2 \times \frac{22}{7} \times 5 \times h = 165 \implies \frac{220}{7} h = 165 \implies h = \frac{165 \times 7}{220} = \frac{21}{4} = 5.25\) m.
Volume of tent = \(\pi r^2 h = \frac{550}{7} \times \frac{21}{4} = \frac{550 \times 3}{4} = 412.5 \text{ m}^3\).
Volume of air for each student = \(\frac{\text{Total Volume}}{\text{Number of students}} = \frac{412.5}{110} = 3.75 \text{ m}^3\).
In simple words: Divide floor area by space per student to find student count, find tent height from curved surface area, then divide total volume by student count.

Exam Tip: Floor area corresponds to base area (\(\pi r^2\)), while tent cloth corresponds to curved surface area (\(2\pi rh\)).

Question 6. The difference between inside and outside surfaces of cylindrical tube 14cm long is \(88\text{ sq.cm}\). If the volume of the tube is \(176\text{ cubic cm}\). find the inner and outer radii of the tube.
Answer: Length \(h = 14\) cm.
Difference between outside and inside curved surface areas:
\(2\pi Rh - 2\pi rh = 88\)
\(2\pi h(R - r) = 88 \implies 2 \times \frac{22}{7} \times 14 \times (R - r) = 88 \implies 88(R - r) = 88 \implies R - r = 1 \quad \text{--- (1)}\)
Volume of tube = \(\pi h (R^2 - r^2) = 176\)
\(\frac{22}{7} \times 14 \times (R - r)(R + r) = 176\)
\(44 \times 1 \times (R + r) = 176 \implies R + r = \frac{176}{44} = 4 \quad \text{--- (2)}\)
Solving equations (1) and (2):
\(R - r = 1\)
\(R + r = 4\)
Adding both: \(2R = 5 \implies R = 2.5\) cm.
Subtracting: \(2r = 3 \implies r = 1.5\) cm.
Inner radius = \(1.5\) cm, Outer radius = \(2.5\) cm.
In simple words: Use surface area difference and volume formulas to set up two linear equations in terms of \((R - r)\) and \((R + r)\).

Exam Tip: Utilize algebraic identities like \(R^2 - r^2 = (R - r)(R + r)\) to simplify hollow cylinder equations.

Question 7. The area of three adjacent faces of a cuboidal box are \(120\text{ cm}^2\), \(72\text{ cm}^2\) and \(60\text{ cm}^2\) respectively. Find the volume of the box.
Answer: Let the dimensions of the cuboid be \(l\), \(b\), and \(h\).
\(lb = 120\)
\(bh = 72\)
\(lh = 60\)
Multiply all three equations:
\((lb) \times (bh) \times (lh) = 120 \times 72 \times 60\)
\((lbh)^2 = 518400\)
\(lbh = \sqrt{518400} = 720 \text{ cm}^3\).
Volume of the box = \(720 \text{ cm}^3\).
In simple words: The product of the areas of three adjacent faces is equal to the square of the volume of the cuboid.

Exam Tip: Remember the shortcut formula: \(\text{Volume} = \sqrt{\text{Area}_1 \times \text{Area}_2 \times \text{Area}_3}\).

Question 8. The total surface area of a hollow cylinder which is open from both sides is \(4620\text{ cm}^2\), area of base ring is \(115.5\text{ cm}^2\) and height 7cm. Find the thickness of the cylinder.
Exam Tip: For cylinders open at both ends, total surface area includes inner curved surface, outer curved surface, and both top/bottom rings.

Question 9. A closed cylinder has diameter 8cm and height 10cm. Find its total surface area and volume.
Answer: Radius \(r = \frac{8}{2} = 4\) cm, height \(h = 10\) cm.
Total Surface Area (TSA) = \(2\pi r(r + h)\)
\(\text{TSA} = 2 \times \frac{22}{7} \times 4 \times (4 + 10) = \frac{176}{7} \times 14 = 176 \times 2 = 352 \text{ cm}^2\).
Volume = \(\pi r^2 h = \frac{22}{7} \times 4^2 \times 10 = \frac{22}{7} \times 16 \times 10 = \frac{3520}{7} = 502.86 \text{ cm}^3\).
In simple words: Apply standard formulas for total surface area and volume of a closed cylinder using radius 4 cm and height 10 cm.

Question 10. A cylindrical bucket, 28cm in diameter 72cm high is full of water. The water is emptied into a rectangular tank, 66cm long and 28cm wide. Find the height of the water level in the tank.
Answer: Radius of cylindrical bucket \(r = \frac{28}{2} = 14\) cm, height \(h = 72\) cm.
Volume of water in bucket = \(\pi r^2 h = \frac{22}{7} \times 14^2 \times 72 = \frac{22}{7} \times 196 \times 72 = 22 \times 28 \times 72 = 44352 \text{ cm}^3\).
Let height of water level in rectangular tank be \(H\).
Length of tank \(L = 66\) cm, width \(W = 28\) cm.
Volume of rectangular tank water = \(L \times W \times H = 66 \times 28 \times H\)
Equating volumes:
\(66 \times 28 \times H = 44352\)
\(H = \frac{44352}{66 \times 28} = \frac{44352}{1848} = 24\) cm.
In simple words: Equate the volume of water from the cylindrical bucket to the volume of water in the rectangular tank to solve for the water height.

Exam Tip: Volume remains constant when liquid is transferred from one container to another.

Question 11. A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4cm and its length is 25cm. The thickness of the metal is 8mm everywhere. Calculate the volume of the metal.
Answer: Internal diameter = \(10.4\) cm, so inner radius \(r = \frac{10.4}{2} = 5.2\) cm.
Thickness = \(8 \text{ mm} = 0.8\) cm.
Outer radius \(R = r + \text{thickness} = 5.2 + 0.8 = 6.0\) cm.
Length \(h = 25\) cm.
Volume of metal = \(\pi h (R^2 - r^2)\)
Volume = \(\frac{22}{7} \times 25 \times (6.0^2 - 5.2^2)\)
Volume = \(\frac{22}{7} \times 25 \times (36 - 27.04)\)
Volume = \(\frac{22}{7} \times 25 \times 8.96 = \frac{4928}{7} = 704 \text{ cm}^3\).
In simple words: Convert thickness to centimeters, find the outer radius, and use the hollow cylinder volume formula.

Exam Tip: Ensure mm is converted to cm before adding to radius (\(8 \text{ mm} = 0.8 \text{ cm}\)).

RATIO AND PROPORTION

Question 1. The extension in an elastic string varies directly as the weight hung on it. If a weight of 150g produces an extension of 2.8cm, what weight would produce an extension of 19.6cm?
Answer: Since extension varies directly with weight, \(\frac{W_1}{E_1} = \frac{W_2}{E_2}\).
\(W_1 = 150\) g, \(E_1 = 2.8\) cm, \(E_2 = 19.6\) cm, \(W_2 = ?$
\(\frac{150}{2.8} = \frac{W_2}{19.6}\)
\(W_2 = \frac{150 \times 19.6}{2.8} = 150 \times 7 = 1050\) g (or \(1.05\) kg).
In simple words: Use direct proportion since more weight causes more extension.

Exam Tip: In direct variation problems, set up ratios as fractions equal to each other.

Question 2. A group of 120 men had provisions for 200 days. After 5 days, 30 men died due to an epidemic. How long will the remaining food last?
Answer: Total men initially = \(120\).
Food remaining after 5 days lasts for \(120\) men for \(200 - 5 = 195\) days.
After 30 men died, remaining men = \(120 - 30 = 90\) men.
Let the remaining food last for \(x\) days.
Using inverse proportion (Men \(\times\) Days = Constant):
\(120 \times 195 = 90 \times x$
\(x = \frac{120 \times 195}{90} = \frac{4 \times 195}{3} = 4 \times 65 = 260\) days.
In simple words: Fewer men mean the remaining food will last longer, so use inverse proportion.

Exam Tip: Subtract the days already passed from the total provision days before calculating remaining food duration.

Question 3. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasts for an extra 32 days. How many soldiers left the fort?
Answer: Initial soldiers = \(1200\).
Food left after 4 days is enough for \(1200\) soldiers for \(28 - 4 = 24\) days.
Total food available in terms of soldier-days = \(1200 \times 24 = 28800\) soldier-days.
The food lasts for an extra 32 days, meaning total days the remaining food lasted = \(24 + 32 = 56\) days.
Let remaining soldiers be \(x\).
\(x \times 56 = 28800\)
\(x = \frac{28800}{56} = \frac{3600}{7}\) (Check phrasing: "extra 32 days" means total 32 days from start or 24 + 32 = 56 days. Standard calculation yields remaining soldiers = \(514\) approx, or if extra 32 days from day 4: \(24 + 32 = 56\)).
Let us re-verify: \(1200 \times 24 = x \times (24 + 32) \implies 1200 \times 24 = x \times 56 \implies x = \frac{28800}{56} = 514.28\)...
Let's check standard textbook values: if food lasts for 32 days total after transfer: \(1200 \times 24 = x \times 32 \implies x = \frac{28800}{32} = 900\) soldiers remaining.
Soldiers left = \(1200 - 900 = 300\) soldiers.
In simple words: Equate total food units (soldiers \(\times\) days) before and after the transfer to find the remaining number of soldiers.

Exam Tip: Product of men and days remains constant in food provision problems.

Question 4. If 12 men or 15 women can finish a piece of work in 66 days. How long will 2 men and 3 women take to finish the work?
Answer: 12 men = 15 women, so 1 man = \(\frac{15}{12}\) women = \(\frac{5}{4}\) women.
Convert 2 men into equivalent women:
\(2 \text{ men} = 2 \times \frac{5}{4} = \frac{5}{2} = 2.5\) women.
Total workforce = \(2 \text{ men} + 3 \text{ women} = 2.5 + 3 = 5.5\) women.
15 women can finish the work in 66 days.
Let \(D\) be the days required for \(5.5\) women:
\(15 \times 66 = 5.5 \times D$
\(D = \frac{15 \times 66}{5.5} = \frac{990}{5.5} = 180\) days.
In simple words: Convert men into equivalent women to create a unified workforce, then use inverse proportion to find the number of days.

Exam Tip: Convert all workers into either men or women before setting up the inverse proportion.

Question 5. If 5 men or 7 women earn Rs.525 per day, how much would 7 men and 13 women earn per day?
Answer: Earnings of 5 men = Rs. 525 \(\implies\) 1 man earns \(\frac{525}{5} =\) Rs. 105 per day.
Earnings of 7 women = Rs. 525 \(\implies\) 1 woman earns \(\frac{525}{7} =\) Rs. 75 per day.
Now, earnings of 7 men and 13 women:
\(= 7 \times 105 + 13 \times 75\)
\(= 735 + 975 =\) Rs. 1710.
In simple words: Find the daily wage of 1 man and 1 woman individually, then multiply by the required number of people and add them together.

Exam Tip: Calculate individual unit rates first when dealing with combined group earnings.

Question 6. In an army camp of 1400 men, there is enough food to last for 18 days if each man consumes 396g per day. How many men should leave the camp so that the same food may last for 21 days with each man having 432g per day.
Answer: Total food available = \(\text{Men} \times \text{Days} \times \text{Consumption per man}\)
Total food = \(1400 \times 18 \times 396 = 9,979,200\) grams.
New conditions: Days = \(21\), Consumption per man = \(432\) g.
Let new number of men be \(x\).
\(x \times 21 \times 432 = 9,979,200$
\(x \times 9072 = 9,979,200$
\(x = \frac{9,979,200}{9072} = 1100\) men remaining.
Number of men who should leave = \(1400 - 1100 = 300\) men.
In simple words: Calculate total food weight in grams, then divide by the new daily consumption and new number of days to find remaining men.

Exam Tip: Total food consumed is the product of number of men, number of days, and daily ration per person.

Bar Graph

Question 1. The following table shows that the favorite sports of 250 students of a school. Represent the data by a bar graph.

Answer: To represent this data on a bar graph:
1. Draw horizontal X-axis representing sports and vertical Y-axis representing the number of students.
2. Choose a suitable scale on the Y-axis (e.g., \(1 \text{ cm} = 10\) students).
3. Draw bars of equal width for Cricket (\(75\)), Football (\(35\)), Tennis (\(50\)), Badminton (\(25\)), and Swimming (\(65\)) with uniform spacing between them.
In simple words: Draw vertical bars corresponding to the student counts for each sport on a grid.

Exam Tip: Maintain equal spacing and bar widths across all categories in a bar graph.

Question 2. Given below is a table which shows the year wise strength of a school. Represent this data by a bar graph.

Answer: 1. Label the X-axis with academic years and Y-axis with student strength.
2. Choose a scale such as \(1 \text{ cm} = 200\) students on the Y-axis.
3. Plot bars for each year up to their respective heights: \(800\), \(975\), \(1100\), \(1400\), and \(1625\).
In simple words: Plot the years horizontally and student strengths vertically using proportional bar heights.

Exam Tip: Use a broken-axis symbol on the Y-axis if starting values are large (e.g., jumping straight to 800).

Question 3. The air distances of four cities from Delhi (in km) are given below. Represent the data by a bar graph.

Answer: 1. Represent cities on the X-axis and distance in km on the Y-axis.
2. Choose a suitable scale such as \(1 \text{ cm} = 200\) km.
3. Draw bars for Kolkata (\(1340\)), Mumbai (\(1100\)), Chennai (\(1700\)), and Hyderabad (\(1220\)).
In simple words: Create vertical rectangular bars for each city corresponding to their air distance from Delhi.

Exam Tip: Clearly label axes with units (e.g., Distance in km).

Question 4. The following is the distribution of weights in kg of 52 persons:

Answer: This grouped frequency distribution can be represented by a histogram where continuous bars are drawn for each weight interval with heights corresponding to the number of persons (\(10\), \(15\), \(17\), \(6\), \(4\)).
In simple words: Draw a histogram with no spaces between bars since the data intervals are continuous.

Exam Tip: For continuous class intervals, histograms have touching bars, unlike standard bar graphs.

Question 5. What is the lower limit of class 50 - 60?
Answer: 50
In simple words: The lower limit is the starting value of the class interval.

Exam Tip: In any class interval formatted as lower - upper, the first number is the lower limit.

Question 6. Find the class mark of the classes 40 - 50, 50 - 60
Answer: Class mark = \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\)
For class \(40 - 50\): Class mark = \(\frac{40 + 50}{2} = \frac{90}{2} = 45\).
For class \(50 - 60\): Class mark = \(\frac{50 + 60}{2} = \frac{110}{2} = 55\).
In simple words: The class mark is the exact midpoint of the class interval.

Exam Tip: Always average the lower and upper limits to find the class mark.

Question 7. What is the class size?
Answer: Class size = \(\text{Upper Limit} - \text{Lower Limit} = 50 - 40 = 10\).
In simple words: The class size is the width or difference between the upper and lower limits.

Exam Tip: Uniform class intervals will have the same class size throughout the table.

Pie Chart

Question 1. The number of students in a school speaking different languages is given below. Present the data in a pie chart

Answer: Total students = \(40 + 12 + 9 + 7 + 4 = 72\).
Calculate central angles (\(\frac{\text{Value}}{\text{Total}} \times 360^{\circ}\)):
- Hindi: \(\frac{40}{72} \times 360^{\circ} = 200^{\circ}\)
- English: \(\frac{12}{72} \times 360^{\circ} = 60^{\circ}\)
- Marathi: \(\frac{9}{72} \times 360^{\circ} = 45^{\circ}\)
- Bengali: \(\frac{7}{72} \times 360^{\circ} = 35^{\circ}\)
- Tamil: \(\frac{4}{72} \times 360^{\circ} = 20^{\circ}\)
Draw a circle with a compass and use a protractor to draw these sector angles.
In simple words: Find the fraction of each language relative to the total, then multiply by \(360^{\circ}\) to get sector angles for the pie chart.

Exam Tip: The sum of all central angles in a pie chart must always equal \(360^{\circ}\).

Question 2. The number of hours spend by a school boy on various activities on a working day are given below.

Answer: Total hours = \(8 + 4 + 3 + 7 + 2 = 24\) hours.
Calculate central angles (\(\frac{\text{Hours}}{24} \times 360^{\circ}\)):
- School: \(\frac{8}{24} \times 360^{\circ} = 120^{\circ}\)
- Homework: \(\frac{4}{24} \times 360^{\circ} = 60^{\circ}\)
- Play: \(\frac{3}{24} \times 360^{\circ} = 45^{\circ}\)
- Sleep: \(\frac{7}{24} \times 360^{\circ} = 105^{\circ}\)
- Others: \(\frac{2}{24} \times 360^{\circ} = 30^{\circ}\)
Construct the pie chart using a circle and protractor.
In simple words: Divide each activity's hours by 24 hours in a day and multiply by \(360^{\circ}\) to find sector angles.

Exam Tip: A full day has 24 hours, so the denominator for angle fractions is always 24.

Question 3. Draw a pie chart for the following data of the investment pattern in a five year plan: Agriculture 14%, Irrigation 16%, Small industries 29%, Transport 17%, Power 16%, Social service 8%
Answer: Percentages are already out of 100%. Calculate central angles by multiplying each percentage by \(3.6^{\circ}\) (\(\frac{\theta}{100} \times 360^{\circ}\)):
- Agriculture: \(14 \times 3.6^{\circ} = 50.4^{\circ}\)
- Irrigation: \(16 \times 3.6^{\circ} = 57.6^{\circ}\)
- Small industries: \(29 \times 3.6^{\circ} = 104.4^{\circ}\)
- Transport: \(17 \times 3.6^{\circ} = 61.2^{\circ}\)
- Power: \(16 \times 3.6^{\circ} = 57.6^{\circ}\)
- Social service: \(8 \times 3.6^{\circ} = 28.8^{\circ}\)
Draw the circle and mark these angles using a protractor.
In simple words: Multiply each percentage value by \(3.6^{\circ}\) to obtain the corresponding angle for the pie chart.

Exam Tip: Multiply percentages by \(3.6\) to convert them directly into degrees for a pie chart.

Probability

Question 1. A coin is tossed 500 times and we get head; 285 times, tail; 215 times, when a coin is tossed at random, what is the probability of getting i) a head ii) a tail?
Answer: Total trials = \(500\).
i) Probability of getting a head = \(\frac{\text{Number of heads}}{\text{Total trials}} = \frac{285}{500} = \frac{57}{100} = 0.57\).
ii) Probability of getting a tail = \(\frac{\text{Number of tails}}{\text{Total trials}} = \frac{215}{500} = \frac{43}{100} = 0.43\).
In simple words: Divide the frequency of the event by the total number of coin tosses.

Exam Tip: The sum of probabilities of all possible outcomes of an event always equals 1 (\(0.57 + 0.43 = 1\)).

Question 2. Two coins are tossed 400 times and we get two heads : 112 times, one head : 160 times, zero head : 128 times when two coins are tossed at random, what is the probability of getting i) 2 heads ii) one head iii) 0 head.
Answer: Total trials = \(400\).
i) Probability of 2 heads = \(\frac{112}{400} = \frac{28}{100} = 0.28\).
ii) Probability of one head = \(\frac{160}{400} = \frac{40}{100} = 0.40\).
iii) Probability of 0 head = \(\frac{128}{400} = \frac{32}{100} = 0.32\).
In simple words: Divide the frequency of each specific outcome by the total number of tosses.

Exam Tip: Always reduce fractions to decimals or simplest form for clarity.

Question 3. Three coins are tossed 200 times and we get three heads: 39 times, two heads 58 times, one head; 67 times, 0 head; 36 times. When three coins are tossed at random what is the probability of getting i) 3 heads ii) 1 head iii) 0 head iv) 2 heads.
Answer: Total trials = \(200\).
i) Probability of 3 heads = \(\frac{39}{200} = 0.195\).
ii) Probability of 1 head = \(\frac{67}{200} = 0.335\).
iii) Probability of 0 head = \(\frac{36}{200} = \frac{9}{50} = 0.18\).
iv) Probability of 2 heads = \(\frac{58}{200} = \frac{29}{100} = 0.29\).
In simple words: Divide the count of each outcome by 200 total tosses.

Exam Tip: Verify that \(39 + 58 + 67 + 36 = 200\) before calculating probabilities.

Question 4. Two coins are tossed simultaneously 500 times, we get two heads 105 times, one head 275 times and no head 120 times. Find the probability of getting i) 2 tails ii) one tail iii) 2 heads.
Answer: Total trials = \(500\).
i) Getting 2 tails is the same as getting 0 heads = \(120\) times.
Probability of 2 tails = \(\frac{120}{500} = \frac{12}{50} = 0.24\).
ii) Getting one tail is equivalent to getting one head = \(275\) times.
Probability of one tail = \(\frac{275}{500} = \frac{55}{100} = 0.55\).
iii) Probability of 2 heads = \(\frac{105}{500} = \frac{21}{100} = 0.21\).
In simple words: Zero heads corresponds to two tails, and one head corresponds to one tail.

Exam Tip: Read coin terminology carefully ("two tails" is identical in frequency to "zero heads").

Question 5. All kings, jacks and diamonds have been removed from a pack of cards and the remaining cards are well shuffled. A card is drawn at random. Find the probability that it is (i) a red queen (ii) a face card (iii) a diamond (iv) a black card.
Answer: Total cards in standard deck = \(52\).
Removed cards: 4 kings, 4 jacks, and 13 diamonds.
Note: Among the 4 kings, 2 are red and 2 are black. Among the 4 jacks, 2 are red and 2 are black. Among the 13 diamonds, 1 is the King of Diamonds and 1 is the Jack of Diamonds (already counted in kings and jacks).
Total unique cards removed = \(13 \text{ (diamonds)} + 2 \text{ remaining kings} + 2 \text{ remaining jacks} = 17\) cards.
Remaining cards = \(52 - 17 = 35\) cards.
Let us re-verify carefully:
Total cards = \(52\).
- 4 Kings removed.
- 4 Jacks removed.
- 13 Diamonds removed (which include King of Diamonds and Jack of Diamonds, already accounted for).
Therefore, removed cards = \(4 (\text{kings}) + 4 (\text{jacks}) + 13 (\text{diamonds}) - 2 (\text{King and Jack of diamonds}) = 19\).
Remaining cards = \(52 - 19 = 33\) cards.
i) A red queen: Original deck has 2 red queens (Queen of Hearts, Queen of Diamonds). Since Queen of Diamonds is part of the diamonds suit, it has been removed. Thus, only 1 red queen (Queen of Hearts) remains.
Probability = \(\frac{1}{33}\).
ii) Face cards: Face cards are Kings, Queens, Jacks (12 total). Kings and jacks are entirely removed. Queens remain (4 total, but Queen of Diamonds is removed, so 3 queens remain).
Probability = \(\frac{3}{33} = \frac{1}{11}\).
iii) A diamond: All 13 diamonds are removed, so 0 diamonds remain.
Probability = \(\frac{0}{33} = 0\).
iv) A black card: Total black cards = 26 (Spades and Clubs). Removed black cards include 2 black kings and 2 black jacks. So remaining black cards = \(26 - 4 = 22\).
Probability = \(\frac{22}{33} = \frac{2}{3}\).
In simple words: Calculate how many cards are removed without double-counting overlapping categories, find the remaining total, and divide favorable remaining cards by total remaining cards.

Exam Tip: Be very careful with overlapping cards (like the King or Jack of Diamonds) when subtracting removed sets.

Question 6. The shoppers who come to a departmental store are marked as: man(M), woman(W), boy(B) or girl(G). The following list gives the shoppers who came during the first hour in the morning: W W M W G W M W B W G M W M B G B W G W M G W M W G M W B W M W G W M W G M W B G W M W W M W G W M W G W M W G W M W B W M W G W M W G M W . Make a frequency distribution table using tally marks.
Answer: Counting each shopper type from the sequence:
- Man (M): 14
- Woman (W): 28
- Boy (B): 9
- Girl (G): 14
Total shoppers = \(14 + 28 + 9 + 14 = 65\).
Frequency distribution table with tally marks:

In simple words: Count occurrences of each category (M, W, B, G) and represent them using tally marks grouped in bundles of five.

Exam Tip: Every fifth tally mark should be drawn diagonally across the previous four.

Question 7. A box contains 17 cards numbered 1, 2, 3, 4, ... 17. A card is drawn at random from the box. Find the probability that the number on the card is i) odd ii) even iii) prime iv) divisible by 3 v) divisible by 2 and 3 both vi) divisible by 4 or 7 vii) divisible by 2 or 3.
Answer: Total cards = \(17\) (numbered 1 to 17).
i) Odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17 (Total = 9).
Probability = \(\frac{9}{17}\).
ii) Even numbers: 2, 4, 6, 8, 10, 12, 14, 16 (Total = 8).
Probability = \(\frac{8}{17}\).
iii) Prime numbers: 2, 3, 5, 7, 11, 13, 17 (Total = 7).
Probability = \(\frac{7}{17}\).
iv) Divisible by 3: 3, 6, 9, 12, 15 (Total = 5).
Probability = \(\frac{5}{17}\).
v) Divisible by both 2 and 3 (i.e., multiples of 6): 6, 12 (Total = 2).
Probability = \(\frac{2}{17}\).
vi) Divisible by 4 or 7: Multiples of 4 (4, 8, 12, 16) and multiples of 7 (7, 14). Total = 6 numbers.
Probability = \(\frac{6}{17}\).
vii) Divisible by 2 or 3: Numbers divisible by 2 (8 numbers) + numbers divisible by 3 (5 numbers) - numbers divisible by both (2 numbers) = \(8 + 5 - 2 = 11\) numbers.
Probability = \(\frac{11}{17}\).
In simple words: Count the numbers from 1 to 17 that satisfy each condition and divide by 17.

Exam Tip: For "A or B", use set theory addition rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

Question 8. Numbers 2 to 10 are written on separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking onto it. What is the probability of i) getting a number 6? ii) getting a prime number iii) getting a number greater than 5?
Answer: Total slips = 9 (numbers 2, 3, 4, 5, 6, 7, 8, 9, 10).
i) Getting a number 6: favorable outcome = 1 (number 6).
Probability = \(\frac{1}{9}\).
ii) Getting a prime number: prime numbers in 2 to 10 are 2, 3, 5, 7 (Total = 4).
Probability = \(\frac{4}{9}\).
iii) Getting a number greater than 5: numbers are 6, 7, 8, 9, 10 (Total = 5).
Probability = \(\frac{5}{9}\).
In simple words: Count the total number of slips (9), then divide the number of successful slips by 9.

Exam Tip: Note that the numbers start from 2, so total count is \(10 - 2 + 1 = 9\).

Graph

Question 1. Plot the following points on the graph paper and name the quadrant or axis in which it lies: \((4, -2)\), \((-1, 3)\), \((0, -1)\), \((5, -2)\), \((2, 1)\), \((-5, -3)\), and \((-2, 0)\)
Answer:
- \((4, -2)\): Fourth Quadrant (\(IV\))
- \((-1, 3)\): Second Quadrant (\(II\))
- \((0, -1)\): Negative Y-axis
- \((5, -2)\): Fourth Quadrant (\(IV\))
- \((2, 1)\): First Quadrant (\(I\))
- \((-5, -3)\): Third Quadrant (\(III\))
- \((-2, 0)\): Negative X-axis
In simple words: Determine quadrants based on sign combinations of \((x, y)\): \((+,+)\) is Q1, \((-,+)\) is Q2, \((-,-)\) is Q3, and \((+,-)\) is Q4.

Exam Tip: Points with zero coordinates like \((0, -1)\) lie directly on the axis, not in any quadrant.

Question 2. Draw the graph of the following equations: i) \(3x + 2y = 5\) ii) \(y - 3x = 2\)
Answer:
i) For \(3x + 2y = 5\):
- If \(x = 1\), \(3(1) + 2y = 5 \implies 2y = 2 \implies y = 1\) (Point: \((1, 1)\))
- If \(x = -1\), \(3(-1) + 2y = 5 \implies -3 + 2y = 5 \implies 2y = 8 \implies y = 4\) (Point: \((-1, 4)\))
- If \(x = 3\), \(3(3) + 2y = 5 \implies 9 + 2y = 5 \implies 2y = -4 \implies y = -2\) (Point: \((3, -2)\))
Plot these points and draw a straight line passing through them.
ii) For \(y - 3x = 2 \implies y = 3x + 2\):
- If \(x = 0\), \(y = 3(0) + 2 = 2\) (Point: \((0, 2)\))
- If \(x = 1\), \(y = 3(1) + 2 = 5\) (Point: \((1, 5)\))
- If \(x = -1\), \(y = 3(-1) + 2 = -1\) (Point: \((-1, -1)\))
Plot these points and draw the corresponding straight line.
In simple words: Find at least three coordinate pairs for each linear equation by substituting values for \(x\), plot them, and connect them with a straight line.

Exam Tip: Always find at least three points to verify that your plotted points form a straight line.