CBSE Class 8 Mathematics Practical Geometry Worksheet Set 01

Question. How many measurement can determine a quadrilateral uniquely?
Answer : 5

Question. In the given figure which angles are equal.

Answer : ∠E and ∠O

Question. Which property is used to construct a parallelogram. If its one side and two diagonals are given.
Answer : Diagonals are bisects to each other.

Question. In the given figure ZEAL is a parallelogram. If ∠Z = 90°. What the new shape obtained.

Answer : Rectangle

Question. In the given figure TRUE is a parallelogram find TE + EU.

Answer : 8 cm

Question. How many measurement can determine a square?
Answer : 1

Question. How many measurements can determine a parallelogram uniquely?
Answer : 3

Question. How many measurement can determine a rhombus.
Answer : 2

Question. What property is used to construct a rhombus. If its two diagonals are given.
Answer : Diagonals of a rhombus bisect each other at right angle.

Question. HOME is a rhombus. If DM = 4 cm and ED = 3 cm, then what is HM + OE

Answer : 14 cm

Practical Geometry

Question 1. Name the quadrilateral whose opposite sides are parallel.
Answer: Parallelogram
In simple words: A parallelogram has both pairs of opposite sides parallel.

Exam Tip: Remember that rectangles, squares, and rhombuses are also special types of parallelograms.

Question 2. Name the quadrilateral whose diagonals are equal and bisect each other at right angle.
Answer: Square
In simple words: A square has equal diagonals that intersect at 90 degrees.

Exam Tip: A rhombus has perpendicular diagonals, but they are not necessarily equal in length.

Question 3. Why do we call square as a regular quadrilateral?
Answer: A square is called a regular quadrilateral because all its four sides are of equal length and all its four interior angles are equal (each measuring \(90^\circ\)).
In simple words: A polygon is regular if all its sides and angles are equal.

Exam Tip: Regular means equilateral and equiangular combined.

Question 4. Construct the quadrilateral ABCD in which AB = 5 cm, BC = 4 cm, \(\angle A = 60^\circ\), \(\angle B = 105^\circ\) and \(\angle C = 105^\circ\).
Answer:
Steps of construction:
1. Draw a line segment AB = 5 cm.
2. At point A, draw an angle of \(60^\circ\) using a compass.
3. At point B, draw an angle of \(105^\circ\).
4. With B as center and radius 4 cm, cut an arc on the ray of angle \(105^\circ\) to mark point C.
5. At point C, draw an angle of \(105^\circ\) such that its ray meets the ray from A at point D.
6. Join C and D to complete the required quadrilateral ABCD.
In simple words: Draw the base, construct the given angles at the vertices, cut off side lengths, and close the shape.

Exam Tip: Always construct angles carefully using a compass where possible for accuracy.

Question 5. The adjacent sides SP and PQ of a parallelogram PQRS are 4 cm each. State the measure of all the sides. What is another name of this figure?
Answer:
In a parallelogram, opposite sides are equal.
Given adjacent sides SP = 4 cm and PQ = 4 cm.
Therefore, all four sides are equal: PQ = QR = RS = SP = 4 cm.
Another name for this figure is a rhombus.
In simple words: A parallelogram with all equal sides is a rhombus.

Exam Tip: A parallelogram with adjacent sides equal is always a rhombus.

Question 6. The ratio of two adjacent sides of a parallelogram is 4:5. If its perimeter is 72 cm, find its adjacent sides.
Answer:
Let the adjacent sides be \(4x\) and \(5x\).
Perimeter of a parallelogram = \(2 \times (\text{sum of adjacent sides})\)
\(2(4x + 5x) = 72\)
\(2(9x) = 72\)
\(18x = 72\)
\(x = \frac{72}{18} = 4 \text{ cm}\)
First side = \(4 \times 4 = 16 \text{ cm}\)
Second side = \(5 \times 4 = 20 \text{ cm}\)
In simple words: Use the perimeter formula with the given ratio to find the multiplier x, then find each side.

Exam Tip: Always verify that \(2 \times (16 + 20) = 72\) cm to check your answer.

Question 7. The park in a town is made in the form of a kite. Its perimeter is 90 metres and one side is 10m more than other side. What are the lengths of other sides?
Answer:
In a kite, two pairs of adjacent sides are equal.
Let one pair of unequal sides be \(x\) and \(x + 10\).
Perimeter = \(2(x + (x + 10)) = 90\)
\(2(2x + 10) = 90\)
\(2x + 10 = 45\)
\(2x = 35\)
\(x = 17.5 \text{ m}\)
Other side = \(17.5 + 10 = 27.5 \text{ m}\)
Thus, the lengths of the four sides of the kite are 17.5 m, 17.5 m, 27.5 m, and 27.5 m.
In simple words: A kite has two pairs of equal adjacent sides, and the perimeter is twice the sum of the two distinct adjacent sides.

Exam Tip: Remember a kite has two equal sides of length 17.5 m and two equal sides of length 27.5 m.

Question 8. Construct the trapezium ABCD in which \(AB \parallel CD\), AB = 8 cm, BC = 6 cm, CD = 4 cm and \(\angle B = 60^\circ\).
Answer:
Steps of construction:
1. Draw line segment AB = 8 cm.
2. At point B, construct an angle of \(60^\circ\).
3. With center B and radius 6 cm, draw an arc intersecting the ray at point C.
4. Draw a line parallel to AB through C.
5. With center C and radius 4 cm, cut an arc on the parallel line to get point D.
6. Join A to D to complete the trapezium ABCD.
In simple words: Construct the base, make the given angle and side at B, draw a parallel line for the top base, and cut off the given length for CD.

Exam Tip: Ensure the top and bottom bases are strictly parallel.

Question 9. Construct the quadrilateral ABCD in which AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and AC = 7 cm.
Answer:
Steps of construction:
1. Draw a diagonal AC = 7 cm.
2. With center A and radius 4.5 cm, draw an arc above AC.
3. With center C and radius 5.5 cm, draw an arc to intersect the previous arc at point B. Join AB and BC.
4. With center A and radius 6 cm, draw an arc below AC.
5. With center C and radius 4 cm, draw an arc to intersect the previous arc at point D. Join AD and CD.
6. ABCD is the required quadrilateral.
In simple words: Construct two triangles (ABC and ADC) sharing the common diagonal AC.

Exam Tip: When four sides and one diagonal are given, always start by constructing the triangle containing the diagonal.

Question 10. Construct the quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, RS = 5 cm, PS = 5.5 cm and PR = 7 cm.
Answer:
Steps of construction:
1. Draw diagonal PR = 7 cm.
2. With P as center and radius 4 cm, draw an arc. With R as center and radius 6 cm, draw an arc intersecting the first arc at Q. Join PQ and QR.
3. With P as center and radius 5.5 cm, draw an arc on the opposite side of PR. With R as center and radius 5 cm, draw an arc intersecting the previous arc at S. Join PS and RS.
4. PQRS is the required quadrilateral.
In simple words: Build two triangles, PQR and PRS, using the diagonal PR as the common base.

Question 11. Construct Quadrilateral ABCD in which AB = 3.5 cm, BC = 4 cm, CD = 5 cm, \(\angle B = 45^\circ\) and \(\angle C = 150^\circ\).
Answer:
Steps of construction:
1. Draw line segment BC = 4 cm.
2. At B, construct an angle of \(45^\circ\).
3. With B as center and radius 3.5 cm, cut an arc on the ray to mark point A. Join AB.
4. At C, construct an angle of \(150^\circ\).
5. With C as center and radius 5 cm, cut an arc on the ray to mark point D. Join CD.
6. Join A and D to complete quadrilateral ABCD.
In simple words: Start with side BC, build the angles and adjacent sides at both B and C, then join the remaining endpoints.

Exam Tip: Constructing \(150^\circ\) requires making a \(180^\circ\) angle and subtracting \(30^\circ\), or using a protractor if allowed.

Question 12. Construct the quadrilateral TRUE in which TR = 3.5 cm, RU = 3 cm, UE = 4 cm, \(\angle R = 75^\circ\) and \(\angle U = 120^\circ\).
Answer:
Steps of construction:
1. Draw line segment RU = 3 cm.
2. At R, construct an angle of \(75^\circ\).
3. With R as center and radius 3.5 cm, cut an arc on the ray to mark point T. Join TR.
4. At U, construct an angle of \(120^\circ\).
5. With U as center and radius 4 cm, cut an arc on the ray to mark point E. Join UE.
6. Join T and E to complete quadrilateral TRUE.
In simple words: Draw the base RU, construct the given angles at R and U, mark the side lengths, and connect T to E.

Exam Tip: Ensure angles are measured accurately from the baseline RU.

Question 13. Construct the parallelogram ABCD with AB = 3.5 cm, BC = 4 cm and AC = 6.5 cm.
Answer:
Steps of construction:
1. Draw line segment AB = 3.5 cm.
2. With A as center and radius 6.5 cm, draw an arc. With B as center and radius 4 cm, draw an arc intersecting the first arc at C. Join AC and BC.
3. With A as center and radius 4 cm, draw an arc opposite to C. With C as center and radius 3.5 cm, draw an arc intersecting the previous arc at D. Join AD and CD.
4. ABCD is the required parallelogram.
In simple words: Construct triangle ABC first, then use opposite side lengths to locate point D.

Exam Tip: Opposite sides of a parallelogram are equal, so AD = BC = 4 cm and CD = AB = 3.5 cm.

Question 14. Construct a rhombus with side 4.5 cm and one diagonal 6 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 6 cm.
2. With A as center and radius 4.5 cm, draw an arc above AC and another below AC.
3. With C as center and radius 4.5 cm, draw arcs above and below AC intersecting the previous arcs at B and D respectively.
4. Join AB, BC, CD, and DA.
5. ABCD is the required rhombus.
In simple words: Since all sides of a rhombus are equal, construct two congruent triangles (ABC and ADC) on the diagonal.

Exam Tip: All four sides of a rhombus are equal in length.

Question 15. Construct the quadrilateral ABCD with AB = 4 cm, BC = 6 cm, CD = 5.5 cm, AD = 5 cm and AC = 8 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 8 cm.
2. With A as center and radius 4 cm, and C as center and radius 6 cm, draw arcs intersecting at B. Join AB and BC.
3. With A as center and radius 5 cm, and C as center and radius 5.5 cm, draw arcs intersecting at D. Join AD and CD.
4. ABCD is the required quadrilateral.
In simple words: Draw diagonal AC, then construct triangles ABC and ADC using SSS criterion.

Exam Tip: Verify that the sum of two sides in any triangle is greater than the third side.

Question 16. Construct the quadrilateral HOPE with HO = 4.5 cm, OP = 4 cm, PE = 6.5 cm, EH = 3 cm and OE = 6.5 cm.
Answer:
Steps of construction:
1. Draw diagonal OE = 6.5 cm.
2. With O as center and radius 4.5 cm, and E as center and radius 3 cm, draw arcs intersecting at H. Join OH and EH.
3. With O as center and radius 4 cm, and E as center and radius 6.5 cm, draw arcs intersecting at P. Join OP and EP.
4. HOPE is the required quadrilateral.
In simple words: Use diagonal OE to split the quadrilateral into two triangles, OHE and OPE.

Question 17. Construct the quadrilateral PQRS in which PQ = 4 cm, QR = 3 cm, PS = 2.5 cm, PR = 4.5 cm and QS = 4 cm.
Answer:
Steps of construction:
1. Draw diagonal PR = 4.5 cm.
2. With P as center and radius 4 cm, and R as center and radius 3 cm, draw arcs intersecting at Q. Join PQ and QR.
3. Now use diagonal QS = 4 cm. With R as center and radius equal to RS, or construct triangle PQS using PS = 2.5 cm, PQ = 4 cm, and QS = 4 cm to locate S.
4. Join PS, RS, and QS to complete PQRS.
In simple words: Construct triangle PQR first, then use the second diagonal QS and side PS to locate vertex S.

Exam Tip: When two diagonals and three sides are given, select a base triangle that uses one diagonal and two known sides.

Question 18. ABCD is a trapezium with \(AB \parallel CD\), and \(\angle A = 50^\circ\) and \(\angle B = 50^\circ\). Prove that (i) BC = DA (ii) \(\angle C = \angle D\) and find the measurement of \(\angle C\).
Answer:
(i) Since \(AB \parallel CD\) and AD, BC are transversal lines, consecutive interior angles add up to \(180^\circ\).
\(\angle A + \angle D = 180^\circ \implies 50^\circ + \angle D = 180^\circ \implies \angle D = 130^\circ\)
Similarly, \(\angle B + \angle C = 180^\circ \implies 50^\circ + \angle C = 180^\circ \implies \angle C = 130^\circ\)
Since base angles \(\angle A = \angle B = 50^\circ\), the trapezium is an isosceles trapezium.
Therefore, non-parallel sides are equal: BC = DA.
(ii) From the calculation above, \(\angle C = 130^\circ\) and \(\angle D = 130^\circ\), so \(\angle C = \angle D\).
In simple words: An isosceles trapezium has equal base angles and equal non-parallel legs.

Exam Tip: Consecutive interior angles between parallel lines are supplementary.

Question 19. The perimeter of a parallelogram is 140 cm. If one of the sides is greater than the other by 20 cm, find the lengths of all the sides of the parallelogram.
Answer:
Let the smaller side be \(x\).
Then the adjacent side = \(x + 20\).
Perimeter = \(2(\text{side}_1 + \text{side}_2) = 140\)
\(2(x + x + 20) = 140\)
\(2(2x + 20) = 140\)
\(2x + 20 = 70\)
\(2x = 50\)
\(x = 25 \text{ cm}\)
Smaller sides = 25 cm each.
Larger sides = \(25 + 20 = 45 \text{ cm}\) each.
Thus, the lengths of all four sides are 25 cm, 45 cm, 25 cm, and 45 cm.
In simple words: Set up an equation using the perimeter formula and solve for the unknown side length x.

Exam Tip: Opposite sides of a parallelogram are equal in length.

Question 20. ABCD is a trapezium in which \(AB \parallel DC\) and AD = BC. If CE is drawn parallel to AD, meeting AB at E, prove the following: (i) AECD is a parallelogram. (ii) AD = EC
Answer:
(i) In quadrilateral AECD:
\(AE \parallel DC\) (since \(AB \parallel DC\))
\(AD \parallel CE\) (by construction)
Since both pairs of opposite sides are parallel, AECD is a parallelogram.
(ii) Since AECD is a parallelogram, opposite sides are equal.
Therefore, AD = EC.
In simple words: Constructing a parallel line creates a smaller parallelogram whose opposite sides are equal.
Exam Tip: Drawing auxiliary parallel lines is a powerful technique for solving trapezium problems.

Question 21. Construct the square READ with RE = 5.1 cm.
Answer:
Steps of construction:
1. Draw line segment RE = 5.1 cm.
2. At R and E, construct angles of \(90^\circ\).
3. With centers R and E, draw arcs of radius 5.1 cm on the perpendicular rays to mark points A and D.
4. Join A and D.
5. READ is the required square.
In simple words: All sides of a square are equal and all interior angles are \(90^\circ\).

Exam Tip: Use a set square or compass to ensure accurate \(90^\circ\) corners.

Question 22. Construct a rhombus ABCD where AC = 5.2 cm and BD = 6.4 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 5.2 cm.
2. Draw the perpendicular bisector of AC, intersecting AC at point O.
3. From O, cut off arcs of length \(\frac{6.4}{2} = 3.2 \text{ cm}\) on the perpendicular bisector above and below AC to mark points B and D.
4. Join AB, BC, CD, and DA.
5. ABCD is the required rhombus.
In simple words: Diagonals of a rhombus bisect each other at right angles. Draw one diagonal, find its midpoint, and set off half the length of the second diagonal on the perpendicular bisector.

Exam Tip: Remember that diagonals of a rhombus bisect each other at \(90^\circ\).

Question 23. Construct a rectangle ABCD with AB = 5 cm and BC = 4 cm.
Answer:
Steps of construction:
1. Draw line segment AB = 5 cm.
2. At points A and B, construct angles of \(90^\circ\).
3. With centers A and B, draw arcs of radius 4 cm along the perpendicular rays to mark points D and C.
4. Join D and C.
5. ABCD is the required rectangle.
In simple words: A rectangle has adjacent sides perpendicular to each other and opposite sides equal.

Exam Tip: Opposite sides of a rectangle are equal: AB = CD = 5 cm and BC = AD = 4 cm.

Question 24. Construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, \(\angle A = 55^\circ\), \(\angle B = 110^\circ\) and \(\angle D = 90^\circ\).
Answer:
First, find the missing angle \(\angle C\).
Sum of interior angles of a quadrilateral = \(360^\circ\)
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
\(55^\circ + 110^\circ + \angle C + 90^\circ = 360^\circ\)
\(255^\circ + \angle C = 360^\circ\)
\(\angle C = 360^\circ - 255^\circ = 105^\circ\)
Steps of construction:
1. Draw line segment AB = 6 cm.
2. At B, construct \(\angle B = 110^\circ\) and cut off BC = 5 cm.
3. At C, construct \(\angle C = 105^\circ\).
4. At A, construct \(\angle A = 55^\circ\) such that its ray intersects the ray from C at point D.
5. ABCD is the required quadrilateral.
In simple words: Find the fourth angle using the angle sum property of quadrilaterals, then construct using base side and angles.

Exam Tip: Always find all possible missing angles before starting construction.

Most Important Questions

Question 1. Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm? If not, give reason.
Answer:
No, it is not possible to construct this quadrilateral.
Reason: In triangle ABC, the lengths are AB = 3 cm, AC = 8 cm, and BC is not directly given, but considering triangle ADC with sides AD = 7.5 cm, CD = 3 cm, and AC = 8 cm, the sum of two sides is \(3 + 7.5 = 10.5 \text{ cm}\), which is greater than AC = 8 cm. However, checking triangle inequality for triangle ABD where AB = 3 cm, BD = 4 cm, and DA = 7.5 cm: \(AB + BD = 3 + 4 = 7 \text{ cm}\), which is less than DA = 7.5 cm. Since the sum of two sides (\(AB + BD\)) is less than the third side (DA), triangle ABD cannot be formed.
In simple words: The sum of the lengths of any two sides of a triangle must be greater than the third side. Here, \(3 + 4 < 7.5\), so triangle ABD is impossible.

Exam Tip: Always verify the triangle inequality theorem (\(a + b > c\)) for component triangles before constructing quadrilaterals.

Question 2. Is it possible to construct a quadrilateral ABCD in which AB = 5 cm, BC = 7.5 cm, \(\angle A = 80^\circ\), \(\angle B = 140^\circ\), \(\angle C = 145^\circ\)? If not, give reason.
Answer:
No, it is not possible.
Reason: The sum of the given three angles is \(80^\circ + 140^\circ + 145^\circ = 365^\circ\). Since the sum of all four interior angles of a quadrilateral must be exactly \(360^\circ\), and three angles alone exceed \(360^\circ\), this quadrilateral cannot exist.
In simple words: The sum of interior angles of any quadrilateral is always \(360^\circ\). Here, three angles already sum to \(365^\circ\), which is impossible.

Exam Tip: Check the angle sum property before attempting angle-based constructions.

Question 3. Construct a quadrilateral ABCD in which AB = 4.4 cm, BC = 4 cm, CD = 6.4 cm, DA = 2.8 cm and BD = 6.6 cm.
Answer:
Steps of construction:
1. Draw diagonal BD = 6.6 cm.
2. With B as center and radius 4 cm, and D as center and radius 2.8 cm, draw arcs intersecting at A. Join AB and AD.
3. With B as center and radius 4.4 cm (Wait, using BC = 4 cm, CD = 6.4 cm from D and C), with B as center and radius 4 cm (BC), and D as center and radius 6.4 cm (CD), draw arcs intersecting at C. Join BC and DC.
4. ABCD is the required quadrilateral.
In simple words: Construct triangles ABD and CBD on common diagonal BD using SSS condition.

Exam Tip: Double check which side belongs to which triangle when a diagonal is given.

Question 4. Construct a parallelogram ABCD where AB = 3.6 cm, BC = 4.2 cm and AC = 6.5 cm.
Answer:
Steps of construction:
1. Draw line segment AB = 3.6 cm.
2. With A as center and radius 6.5 cm, draw an arc. With B as center and radius 4.2 cm, draw an arc intersecting the first arc at C. Join AC and BC.
3. With A as center and radius 4.2 cm, and C as center and radius 3.6 cm, draw arcs intersecting at point D. Join AD and CD.
4. ABCD is the required parallelogram.
In simple words: Build triangle ABC first, then use opposite side lengths to complete the parallelogram.

Exam Tip: Opposite sides of a parallelogram are equal in length.

Question 5. Construct a rhombus with side 4.5 cm and one diagonal 6 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 6 cm.
2. With A and C as centers and radius 4.5 cm, draw arcs above and below AC intersecting at B and D.
3. Join AB, BC, CD, and DA.
4. ABCD is the required rhombus.
In simple words: Construct two isosceles triangles with side lengths 4.5 cm sharing the diagonal AC.

Exam Tip: All four sides of a rhombus are equal.

Question 6. Construct a quadrilateral ABCD in which AB = 5 cm, BC = 4 cm, AD = 3 cm, CD = 6 cm and BD = 5 cm.
Answer:
Steps of construction:
1. Draw diagonal BD = 5 cm.
2. With B as center and radius 4 cm, and D as center and radius 6 cm, draw arcs intersecting at C. Join BC and DC.
3. With B as center and radius 5 cm, and D as center and radius 3 cm, draw arcs intersecting at A. Join AB and AD.
4. ABCD is the required quadrilateral.
In simple words: Construct triangles BCD and BAD sharing the diagonal BD.

Exam Tip: Verify triangle side lengths before drawing intersecting arcs.

Question 7. Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, \(\angle A = 90^\circ\) and \(\angle B = 105^\circ\).
Answer:
Steps of construction:
1. Draw line segment AB = 3 cm.
2. At B, construct \(\angle B = 105^\circ\) and cut off BC = 3 cm.
3. At A, construct \(\angle A = 90^\circ\) and cut off AD = 5 cm.
4. Join C and D to complete quadrilateral ABCD.
In simple words: Start with AB, construct the given angles and side lengths at A and B, then connect the endpoints C and D.

Exam Tip: Ensure angles are measured precisely from the base line AB.

Question 8. Construct a rectangle with sides 4.5 cm and 6 cm.
Answer:
Steps of construction:
1. Draw base line AB = 6 cm.
2. At points A and B, construct \(90^\circ\) angles.
3. With centers A and B, cut arcs of radius 4.5 cm along the perpendicular rays to mark points D and C.
4. Join D and C.
5. ABCD is the required rectangle.
In simple운동: A rectangle has four right angles and opposite sides equal in length.

Exam Tip: Opposite sides are equal: AB = CD = 6 cm and AD = BC = 4.5 cm.

Question 9. Construct a quadrilateral ABCD in which AB = 7 cm, AD = 6 cm, AC = 7 cm, BD = 7.5 cm and BC = 5 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 7 cm.
2. With A as center and radius 7 cm, and C as center and radius 5 cm, draw arcs intersecting at B. Join AB and BC.
3. With A as center and radius 6 cm, and B as center and radius 7.5 cm (using diagonal BD), draw arcs intersecting at D. Join AD and BD.
4. Join CD to complete quadrilateral ABCD.
In simple words: Construct triangle ABC, then locate point D using sides AD and BD.

Exam Tip: Choose the best triangle to start with based on the available three side lengths.

Question 10. Construct a quadrilateral ABCD in which AB = 5.5 cm, AD = 4.4 cm, CD = 6.5 cm, AC = 6.5 cm and BD = 7.1 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 6.5 cm.
2. With A as center and radius 5.5 cm, and C as center and radius 6.5 cm, draw arcs intersecting at B. Join AB and BC.
3. With A as center and radius 4.4 cm, and using diagonal BD = 7.1 cm with point C/B, locate point D.
4. Join AD, CD, and BD.
In simple words: Use diagonal AC to construct triangle ABC, then locate vertex D using AD and DC.

Exam Tip: Verify all given dimensions correspond correctly to the constructed figure.

Question 11. Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 2.5 cm, CD = 4 cm, AC = 6.5 cm and BD = 5 cm.
Answer:
Steps of construction:
1. Draw diagonal AC = 6.5 cm.
2. With A as center and radius 5.4 cm, and C as center and radius 2.5 cm, draw arcs intersecting at B. Join AB and BC.
3. With C as center and radius 4 cm (CD), and using diagonal BD = 5 cm from B, locate point D.
4. Join AD, CD, and BD.
In simple words: Build triangle ABC on diagonal AC, then locate point D using side CD and diagonal BD.

Exam Tip: Keep track of which diagonal connects which vertices.

Question 12. Construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, \(\angle A = 55^\circ\), \(\angle B = 110^\circ\) and \(\angle D = 90^\circ\).
Answer:
First, find \(\angle C\):
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
\(55^\circ + 110^\circ + \angle C + 90^\circ = 360^\circ\)
\(255^\circ + \angle C = 360^\circ \implies \angle C = 105^\circ\)
Steps of construction:
1. Draw AB = 6 cm.
2. At B, make \(\angle B = 110^\circ\) and cut BC = 5 cm.
3. At C, make \(\angle C = 105^\circ\).
4. At A, make \(\angle A = 55^\circ\) to meet the ray from C at D.
5. ABCD is the required quadrilateral.
In simple words: Calculate the missing angle first, then construct using base and angles.

Exam Tip: The sum of interior angles in a quadrilateral is always \(360^\circ\).

Question 13. Construct a quadrilateral PQRS in history where \(\angle Q = 45^\circ\), \(\angle R = 90^\circ\), QR = 5 cm, PQ = 4 cm and RS = 3 cm.
Answer:
Steps of construction:
1. Draw line segment QR = 5 cm.
2. At Q, construct an angle of \(45^\circ\) and cut off PQ = 4 cm.
3. At R, construct an angle of \(90^\circ\) and cut off RS = 3 cm.
4. Join P and S to complete quadrilateral PQRS.
In simple words: Draw the base QR, construct the given angles and side lengths at Q and R, then close the shape by joining P and S.

Exam Tip: Ensure angle measurements are exact.

Question 14. Construct a quadrilateral ABCD in which AB = 3.6 cm, BC = 5.5 cm, CD = 5 cm, angle B = 125 and angle C = 80.
Answer:
Steps of construction:
1. Draw line segment BC = 5.5 cm.
2. At B, construct an angle of \(125^\circ\) and cut off AB = 3.6 cm.
3. At C, construct an angle of \(80^\circ\) and cut off CD = 5 cm.
4. Join A and D to complete quadrilateral ABCD.
In simple words: Start with side BC, construct the given angles and side lengths at B and C, then join AD.

Question 15. Construct a quadrilateral ABCD in which AB = 5.1 cm, AD = 4 cm, BC = 2.5 cm, angle A = 60 and angle B = 85.
Answer:
Steps of construction:
1. Draw line segment AB = 5.1 cm.
2. At A, construct an angle of \(60^\circ\) and cut off AD = 4 cm.
3. At B, construct an angle of \(85^\circ\) and cut off BC = 2.5 cm.
4. Join C and D to complete quadrilateral ABCD.
In simple words: Draw base AB, construct angles and side lengths at A and B, then connect C and D.

Exam Tip: Check all side and angle specifications before finishing the drawing.

Question 16. Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 6.5 cm, \(\angle A = 75^\circ\), \(\angle B = 105^\circ\) and \(\angle C = 120^\circ\).
Answer:
First, find the fourth angle \(\angle D\):
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
\(75^\circ + 105^\circ + 120^\circ + \angle D = 360^\circ\)
\(300^\circ + \angle D = 360^\circ \implies \angle D = 60^\circ\)
Steps of construction:
1. Draw AB = 3.5 cm.
2. At B, construct \(\angle B = 105^\circ\) and cut BC = 6.5 cm.
3. At C, construct \(\angle C = 120^\circ\).
4. At A, construct \(\angle A = 75^\circ\) to intersect the ray from C at point D.
5. ABCD is the required quadrilateral.
In simple words: Calculate angle D first using the angle sum property, then construct the shape.

Exam Tip: The sum of all four interior angles in any quadrilateral is \(360^\circ\).

Question 17. Construct a quadrilateral ABCD in which AB = 5.3 cm, AD = 2.9 cm, \(\angle A = 70^\circ\), \(\angle B = 95^\circ\) and \(\angle C = 85^\circ\).
Answer:
First, find \(\angle D\):
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
\(70^\circ + 95^\circ + 85^\circ + \angle D = 360^\circ\)
\(250^\circ + \angle D = 360^\circ \implies \angle D = 110^\circ\)
Steps of construction:
1. Draw AB = 5.3 cm.
2. At A, construct \(\angle A = 70^\circ\) and cut AD = 2.9 cm.
3. At B, construct \(\angle B = 95^\circ\).
4. At D, construct \(\angle D = 110^\circ\) (or use angle C at the other end) to complete ABCD.
In simple words: Determine the missing angle D first, then draw using side AB and the known angles.

Exam Tip: Verify angle sums carefully before proceeding with construction.

Question 18. Construct a quadrilateral in which QR = 7.5 cm, RP = PS = 6 cm, RS = 5 cm and QS = 10 cm.
Answer:
Steps of construction:
1. Draw diagonal QS = 10 cm.
2. With Q as center and radius 7.5 cm (QR), and S as center and radius 5 cm (RS), draw arcs intersecting at R. Join QR and RS.
3. With S as center and radius 6 cm (PS), and R as center and radius 6 cm (RP), draw arcs intersecting at P. Join SP and RP.
4. PQRS is the required quadrilateral.
In simple words: Construct triangles QSR and PSR sharing the diagonal QS.

Exam Tip: Make sure diagonal QS is drawn accurately as the baseline.

Question 19. Construct a trapezium ABCD in which \(AB \parallel CD\), AB = 8 cm, BC = 6.0 cm, CD = 4 cm and \(\angle B = 60^\circ\).
Answer:
Steps of construction:
1. Draw line segment AB = 8 cm.
2. At B, construct an angle of \(60^\circ\) and cut off BC = 6 cm.
3. Draw a line parallel to AB through C.
4. With C as center and radius 4 cm (CD), cut an arc on the parallel line to mark point D.
5. Join A and D to complete trapezium ABCD.
In simple words: Construct the base and side BC with its angle, draw a parallel line for the top base, and mark length CD.

Exam Tip: Ensure the top base CD is parallel to AB.

Question 20. Construct a parallelogram whose two sides and one angle are 4 cm, 5.5 cm and \(70^\circ\) respectively.
Answer:
Steps of construction:
1. Draw line segment AB = 5.5 cm.
2. At A, construct an angle of \(70^\circ\).
3. With A as center and radius 4 cm, cut an arc on the ray to mark point D. Join AD.
4. With D as center and radius 5.5 cm, and B as center and radius 4 cm, draw arcs intersecting at point C.
5. Join DC and BC to complete parallelogram ABCD.
In simple words: Draw one side and the given angle, mark the adjacent side length, then use opposite side lengths to complete the parallelogram.

Exam Tip: Opposite sides of a parallelogram are equal in length.