CBSE Class 8 Mathematics Practical Geometry Worksheet Set 02

Fill in the Blanks

Question. The length of _______________ sides may be enough to construct a triangle.
Answer : Three

Question. _______________ sides and the _______________ angle are enough to construct a triangle.
Answer : Two, included

Question. _______________ angles and the _______________ side are enough to construct a triangle.
Answer : Two, included

Question. _______________ side and length of _______________ is enough to construct a right angled triangle.
Answer : One, Hypotenuse

Question. The lengths of the three line segments are 4 cm, 5 cm, and 6 cm. Can a triangle be constructed with these?
Answer : Yes

Question. The lengths of the three line segments are 3 cm, 4 cm, 8 cm. Is it possible to construct a triangle?
Answer : No

Question. How many lines can be drawn parallel to a given line?
Answer : Many lines

Question. What should be the relation between the three sides of a triangle to construct a triangle?
Answer : Sum of the lengths of any two sides of a triangles is greater than the third side.

Question. If length of a side and length of hypotenuse is given then what type of triangle can be constructed?
Answer : Right Angled Triangle

Question. Is it possible to construct a triangle with the triplet of three line segments 2 cm, 4 cm and 6 cm?
Answer : No

Question. How many minimum number of elements are required to construct a triangle? Why?
Answer : Three, because three vertices have to be located

Question. Give a triplet of lengths using which triangle can be constructed?
Answer : Any triplet which satisfies the condition that sum of the lengths of any two sides of a triangle is greater than the third side.

Question. What should be the sum of three angles so that a triangle can be constructed?
Answer : 180°

Question. If R 50 , S 70 and RS = 5.5 cm then what type of triangle can be constructed?
Answer : Scalene Acute Angled Triangle

Question. Which geometrical instrument can be used to draw an arc?
Answer : Compass

Question. How many lines can be drawn parallel to a given line and through a point outside the given line?
Answer : Only one line

Question. If three sides measure 5.2 cm, 4 cm, 6 cm then what type of triangle can be constructed?
Answer : Scalene

Chap. 4. PRACTICAL GEOMETRY

Question 1. Construct a quadrilateral ABCD in which \(AB = 4.5\text{ cm}\), \(BC = 6\text{ cm}\), \(CD = 5.6\text{ cm}\), \(DA = 7.8\text{ cm}\) and diagonal \(AC = 9\text{ cm}\).
Answer: Steps of construction:
1. Draw a line segment \(AC = 9\text{ cm}\).
2. With \(A\) as center and radius \(4.5\text{ cm}\), draw an arc.
3. With \(C\) as center and radius \(6\text{ cm}\), draw another arc to intersect the previous arc at point \(B\). Join \(AB\) and \(BC\).
4. With \(A\) as center and radius \(7.8\text{ cm}\), draw an arc on the other side of \(AC\).
5. With \(C\) as center and radius \(5.6\text{ cm}\), draw another arc to intersect this arc at point \(D\). Join \(AD\) and \(CD\).
6. Thus, quadrilateral \(ABCD\) is required.
In simple words: Construct two triangles \(ABC\) and \(ADC\) sharing the common base \(AC\).

Exam Tip: Always draw a rough sketch first to visualize the given measurements and relative positions of vertices.

Question 2. Construct a quadrilateral ABCD in which \(AB = 4\text{ cm}\), \(BC = 6\text{ cm}\), \(CD = 5\text{ cm}\), \(DA = 3.5\text{ cm}\) and diagonal \(AC = 7\text{ cm}\).
Answer: Steps of construction:
1. Draw a line segment \(AC = 7\text{ cm}\).
2. With \(A\) as center and radius \(4\text{ cm}\), draw an arc.
3. With \(C\) as center and radius \(6\text{ cm}\), draw an arc cutting the previous arc at \(B\). Join \(AB\) and \(BC\).
4. With \(A\) as center and radius \(3.5\text{ cm}\), draw an arc on the opposite side of \(AC\).
5. With \(C\) as center and radius \(5\text{ cm}\), draw an arc cutting this arc at \(D\). Join \(AD\) and \(CD\).
6. \(ABCD\) is the required quadrilateral.
In simple words: Draw diagonal \(AC\) first, then locate vertex \(B\) and vertex \(D\) using compass arcs of given side lengths.

Exam Tip: Ensure compass measurements are checked accurately against the ruler before drawing arcs.

Question 3. Construct a quadrilateral ABCD in which \(AB = 8\text{ cm}\), \(BC = 9\text{ cm}\), \(CD = 5\text{ cm}\), \(DA = 6.7\text{ cm}\) and diagonal \(BD = 7\text{ cm}\).
Answer: Steps of construction:
1. Draw a line segment \(BD = 7\text{ cm}\).
2. With \(B\) as center and radius \(9\text{ cm}\), draw an arc.
3. With \(D\) as center and radius \(6.7\text{ cm}\), draw an arc cutting the previous arc at \(A\). Join \(AB\) and \(AD\).
4. With \(B\) as center and radius \(8\text{ cm}\), draw an arc on the other side of \(BD\).
5. With \(D\) as center and radius \(5\text{ cm}\), draw an arc cutting this arc at \(C\). Join \(BC\) and \(DC\).
6. \(ABCD\) is the required quadrilateral.
In simple words: When diagonal \(BD\) is given instead of \(AC\), start with \(BD\) as the base line and construct triangles \(ABD\) and \(CBD\).

Exam Tip: Check whether the triangle inequality theorem holds for the given side lengths before construction.

Question 4. Construct a quadrilateral PQRS in which \(PQ = 5\text{ cm}\), \(QR = 5.5\text{ cm}\), \(RS = 6\text{ cm}\), \(SP = 8.9\text{ cm}\) and \(\angle P = 60^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(PQ = 5\text{ cm}\).
2. At point \(P\), construct an angle of \(60^{\circ}\) using a compass.
3. With \(P\) as center and radius \(8.9\text{ cm}\), draw an arc on the ray of the angle to locate point \(S\).
4. With \(S\) as center and radius \(6\text{ cm}\), draw an arc.
5. With \(Q\) as center and radius \(5.5\text{ cm}\), draw an arc to intersect the previous arc at point \(R\).
6. Join \(SR\) and \(QR\). \(PQRS\) is the required quadrilateral.
In simple words: Draw side \(PQ\), make \(60^{\circ}\) at \(P\) to locate \(S\), then use compass arcs from \(S\) and \(Q\) to find vertex \(R\).

Exam Tip: Use a protractor to verify the angle at \(P\) after constructing it with a compass.

Question 5. Construct a quadrilateral ABCD in which \(AB = 7\text{ cm}\), \(BC = 4.1\text{ cm}\), \(AD = 6.3\text{ cm}\), diagonal \(BD = 8.7\text{ cm}\) and diagonal \(AC = 8.9\text{ cm}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 7\text{ cm}\).
2. With \(A\) as center and radius equal to diagonal \(AC = 8.9\text{ cm}\), draw an arc.
3. With \(B\) as center and radius equal to \(BC = 4.1\text{ cm}\), draw an arc intersecting at \(C\). Join \(BC\) and \(AC\).
4. With \(A\) as center and radius equal to \(AD = 6.3\text{ cm}\), draw an arc.
5. With \(B\) as center and radius equal to diagonal \(BD = 8.7\text{ cm}\), draw an arc intersecting at \(D\). Join \(AD\), \(BD\), and \(CD\).
6. \(ABCD\) is the required quadrilateral.
In simple words: Use both diagonals along with adjacent sides to locate all four vertices step by step.

Exam Tip: Label all intersecting arcs clearly to avoid confusion when drawing final lines.

Question 6. Construct a quadrilateral ABCD in which \(AB = 6.5\text{ cm}\), \(BC = 5.5\text{ cm}\), \(DA = 3.8\text{ cm}\), \(\angle A = 90^{\circ}\) and \(\angle B = 70^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 6.5\text{ cm}\).
2. At point \(A\), construct an angle of \(90^{\circ}\).
3. With \(A\) as center and radius \(3.8\text{ cm}\), draw an arc on the \(90^{\circ}\) ray to mark point \(D\).
4. At point \(B\), construct an angle of \(70^{\circ}\) using a protractor.
5. With \(B\) as center and radius \(5.5\text{ cm}\), draw an arc on the \(70^{\circ}\) ray to mark point \(C\).
6. Join \(DC\). \(ABCD\) is the required quadrilateral.
In simple words: Construct the base \(AB\), draw the given angles at both ends \(A\) and \(B\), cut off the given side lengths, and join the remaining vertices.

Exam Tip: For angles not easily constructible with a compass like \(70^{\circ}\), use a protractor accurately.

Question 7. Construct a quadrilateral ABCD with \(BC = 5.5\text{ cm}\), \(CD = 4\text{ cm}\), \(DA = 4\text{ cm}\), \(\angle C = 90^{\circ}\) and \(\angle D = 85^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(CD = 4\text{ cm}\).
2. At point \(C\), construct an angle of \(90^{\circ}\).
3. With \(C\) as center and radius \(5.5\text{ cm}\), draw an arc on the \(90^{\circ}\) ray to locate point \(B\).
4. At point \(D\), construct an angle of \(85^{\circ}\) using a protractor.
5. With \(D\) as center and radius \(4\text{ cm}\), draw an arc on the \(85^{\circ}\) ray to locate point \(A\).
6. Join \(AB\). \(ABCD\) is the required quadrilateral.
In simple words: Start with base \(CD\), draw the given angles at \(C\) and \(D\), measure off the given side lengths, and complete the figure.

Exam Tip: Ensure the base side is positioned horizontally to keep your construction neat.

Question 8. Construct a quadrilateral PQRS in which \(PQ = 4.5\text{ cm}\), \(QR = 7.5\text{ cm}\), \(\angle P = 100^{\circ}\), \(\angle Q = 85^{\circ}\) and \(\angle R = 75^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(PQ = 4.5\text{ cm}\).
2. At point \(Q\), construct an angle of \(85^{\circ}\).
3. With \(Q\) as center and radius \(7.5\text{ cm}\), draw an arc on the \(85^{\circ}\) ray to mark point \(R\).
4. At point \(R\), construct an angle of \(75^{\circ}\).
5. At point \(P\), construct an angle of \(100^{\circ}\) to meet the ray from \(R\) at point \(S\).
6. \(PQRS\) is the required quadrilateral.
In simple words: When three angles and two adjacent sides are given, build outward from the base using the angle rays.

Exam Tip: Extend the intersection rays carefully so that they meet precisely at vertex \(S\).

Question 9. Construct a quadrilateral PQRS in which the sides are \(PQ = 5\text{ cm}\), \(QR = 8.3\text{ cm}\), \(RS = 6.4\text{ cm}\), \(SP = 9\text{ cm}\) and \(\angle S = 90^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(SP = 9\text{ cm}\).
2. At point \(S\), construct an angle of \(90^{\circ}\).
3. With \(S\) as center and radius \(6.4\text{ cm}\), draw an arc on the \(90^{\circ}\) ray to locate point \(R\).
4. With \(P\) as center and radius \(5\text{ cm}\), draw an arc.
5. With \(R\) as center and radius \(8.3\text{ cm}\), draw an arc to intersect the previous arc at point \(Q\).
6. Join \(PQ\) and \(RQ\). \(PQRS\) is the required quadrilateral.
In simple words: Build the right angle at \(S\) to find \(R\), then use compass intersections from \(P\) and \(R\) to find \(Q\).

Exam Tip: Double-check the order of vertices so that the quadrilateral is named correctly in cyclic order.

Question 10. Construct a quadrilateral LMNO in which \(LM = 5\text{ cm}\), \(MN = 6.5\text{ cm}\), \(\angle L = 75^{\circ}\), \(\angle M = 105^{\circ}\) and \(\angle N = 120^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(LM = 5\text{ cm}\).
2. At point \(M\), construct an angle of \(105^{\circ}\).
3. With \(M\) as center and radius \(6.5\text{ cm}\), draw an arc on the \(105^{\circ}\) ray to mark point \(N\).
4. At point \(N\), construct an angle of \(120^{\circ}\).
5. At point \(L\), construct an angle of \(75^{\circ}\) to intersect the ray from \(N\) at point \(O\).
6. \(LMNO\) is the required quadrilateral.
In simple words: Draw base \(LM\), make the angle at \(M\) to locate \(N\), and then draw the given angles at \(N\) and \(L\) to complete vertex \(O\).

Exam Tip: Use a protractor for angles like \(75^{\circ}\), \(105^{\circ}\), and \(120^{\circ}\) if compass construction is complex, or construct using standard standard angles (\(60^{\circ}\), \(90^{\circ}\), \(120^{\circ}\)).

Question 11. Construct a quadrilateral ABCD in which \(AB = 4.1\text{ cm}\), \(BC = 3.8\text{ cm}\), \(\angle A = 60^{\circ}\), \(\angle C = 82^{\circ}\) and \(\angle B = 125^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 4.1\text{ cm}\).
2. At point \(A\), construct an angle of \(60^{\circ}\).
3. At point \(B\), construct an angle of \(125^{\circ}\).
4. With \(B\) as center and radius \(3.8\text{ cm}\), draw an arc on the \(125^{\circ}\) ray to mark point \(C\).
5. At point \(C\), construct an angle of \(82^{\circ}\) to intersect the ray from \(A\) at point \(D\).
6. \(ABCD\) is the required quadrilateral.
In simple words: Start with \(AB\), draw angles at \(A\) and \(B\), mark \(BC\), and complete the final angle at \(C\) to find \(D\).

Exam Tip: Ensure protractor readings are precise to the nearest degree.

Question 12. Construct a parallelogram, one of whose sides is \(5.5\text{ cm}\) and whose diagonals are \(5.8\text{ cm}\) and \(7\text{ cm}\).
Answer: Steps of construction:
1. Since diagonals of a parallelogram bisect each other, draw diagonal \(AC = 7\text{ cm}\).
2. Find the midpoint \(O\) of \(AC\) by drawing its perpendicular bisector.
3. With \(O\) as center and radius equal to half of the second diagonal (\(\frac{5.8}{2} = 2.9\text{ cm}\)), draw arcs above and below \(AC\).
4. With \(A\) as center and radius equal to the side \(5.5\text{ cm}\), draw arcs to intersect the previous arcs at points \(B\) and \(D\).
5. Join \(AB\), \(BC\), \(CD\), and \(DA\).
6. \(ABCD\) is the required parallelogram.
In simple words: Draw one diagonal, bisect it, and use half the second diagonal along with the side length to construct the vertices.

Exam Tip: Remember that diagonals of a parallelogram bisect each other at their intersection point.

Question 13. Construct a parallelogram, one of whose sides is \(6\text{ cm}\) and whose diagonals are \(6\text{ cm}\) and \(8\text{ cm}\).
Answer: Steps of construction:
1. Draw diagonal \(AC = 8\text{ cm}\).
2. Construct the perpendicular bisector of \(AC\) to find its midpoint \(O\).
3. With \(O\) as center and radius equal to half of the second diagonal (\(\frac{6}{2} = 3\text{ cm}\)), draw arcs on both sides of \(AC\).
4. With \(A\) as center and radius \(6\text{ cm}\), draw arcs to intersect the previous arcs at \(B\) and \(D\).
5. Join \(AB\), \(BC\), \(CD\), and \(DA\).
6. The required parallelogram \(ABCD\) is formed.
In simple words: Construct using the diagonal bisecting property and side length as done in the previous problem.

Question 14. Construct a parallelogram ABCD in which \(AB = 4.5\text{ cm}\), \(BC = 4\text{ cm}\) and diagonal \(AC = 6.2\text{ cm}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 4.5\text{ cm}\).
2. With \(A\) as center and radius \(6.2\text{ cm}\), draw an arc.
3. With \(B\) as center and radius \(4\text{ cm}\), draw an arc to intersect the first arc at point \(C\). Join \(BC\) and \(AC\).
4. With \(A\) as center and radius \(4\text{ cm}\) (since opposite side \(AD = BC\)), draw an arc.
5. With \(C\) as center and radius \(4.5\text{ cm}\) (since opposite side \(CD = AB\)), draw an arc to intersect at point \(D\). Join \(AD\) and \(CD\).
6. \(ABCD\) is the required parallelogram.
In simple words: Construct triangle \(ABC\) first, then use opposite side lengths (\(AD = BC\) and \(CD = AB\)) to locate vertex \(D\).

Exam Tip: Utilize the property that opposite sides of a parallelogram are equal in length.

Question 15. Construct a parallelogram ABCD in which \(AB = 6.5\text{ cm}\), \(AD = 3\text{ cm}\) and diagonal \(BD = 4.9\text{ cm}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 6.5\text{ cm}\).
2. With \(A\) as center and radius \(3\text{ cm}\) (equal to side \(AD\)), draw an arc.
3. With \(B\) as center and radius \(4.9\text{ cm}\) (equal to diagonal \(BD\)), draw an arc to intersect the previous arc at \(D\). Join \(AD\) and \(BD\).
4. With \(D\) as center and radius \(6.5\text{ cm}\) (equal to \(AB\)), draw an arc.
5. With \(B\) as center and radius \(3\text{ cm}\) (equal to \(AD\)), draw an arc intersecting at \(C\). Join \(DC\) and \(BC\).
6. \(ABCD\) is the required parallelogram.
In simple words: Form triangle \(ABD\) using side \(AB\), side \(AD\), and diagonal \(BD\), then complete the parallelogram using opposite sides.

Exam Tip: Double check that opposite sides \(AB = CD\) and \(AD = BC\).

Question 16. Construct a parallelogram OKAY in which \(OK = 5\text{ cm}\), \(KA = 6\text{ cm}\) and \(\angle Y = 50^{\circ}\).
Answer: Steps of construction:
1. In parallelogram \(OKAY\), opposite sides are equal, so \(OY = KA = 6\text{ cm}\) and \(AY = OK = 5\text{ cm}\). Also, opposite angles are equal, so \(\angle O = \angle Y = 50^{\circ}\).
2. Draw line segment \(OK = 5\text{ cm}\).
3. At point \(O\), construct an angle of \(50^{\circ}\).
4. With \(O\) as center and radius \(6\text{ cm}\) (side \(OY\)), draw an arc on the \(50^{\circ}\) ray to locate point \(Y\).
5. With \(Y\) as center and radius \(5\text{ cm}\) (side \(YA\)), draw an arc.
6. With \(K\) as center and radius \(6\text{ cm}\) (side \(KA\)), draw an arc to intersect at \(A\). Join \(YA\) and \(KA\). \(OKAY\) is the required parallelogram.
In simple words: Use the parallelogram property that opposite angles are equal to find angle \(O\), then construct using side lengths.

Exam Tip: State the property of parallelograms clearly before starting the construction steps.

Question 17. Construct a rectangle ABCD whose sides are \(6\text{ cm}\) and \(3\text{ cm}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 6\text{ cm}\).
2. At points \(A\) and \(B\), construct right angles (\(90^{\circ}\)).
3. With \(A\) as center and radius \(3\text{ cm}\), draw an arc on the perpendicular at \(A\) to mark point \(D\).
4. With \(B\) as center and radius \(3\text{ cm}\), draw an arc on the perpendicular at \(B\) to mark point \(C\).
5. Join \(DC\).
6. \(ABCD\) is the required rectangle.
In simple words: Draw base \(AB\), erect \(90^{\circ}\) angles at both ends, cut off the height of \(3\text{ cm}\), and join the top ends.

Exam Tip: A rectangle has all interior angles equal to \(90^{\circ}\) and opposite sides equal.

Question 18. Construct a parallelogram PQRS in which \(QR = 4\text{ cm}\), \(PQ = 5.5\text{ cm}\) and \(\angle Q = 78^{\circ}\).
Answer: Steps of construction:
1. Draw line segment \(PQ = 5.5\text{ cm}\).
2. At point \(Q\), construct an angle of \(78^{\circ}\) using a protractor.
3. With \(Q\) as center and radius \(QR = 4\text{ cm}\), draw an arc on the \(78^{\circ}\) ray to mark point \(R\).
4. With \(R\) as center and radius equal to \(PQ = 5.5\text{ cm}\), draw an arc.
5. With \(P\) as center and radius equal to \(QR = 4\text{ cm}\), draw an arc to intersect at point \(S\).
6. Join \(RS\) and \(PS\). \(PQRS\) is the required parallelogram.
In simple words: Draw base \(PQ\), make \(78^{\circ}\) at \(Q\) to mark \(R\), and then use opposite sides to complete vertex \(S\).

Exam Tip: Use a protractor carefully for non-standard angles like \(78^{\circ}\).

Question 19. Construct a parallelogram whose diagonals are \(4.4\text{ cm}\) and \(7.4\text{ cm}\), and an angle between them is \(75^{\circ}\).
Answer: Steps of construction:
1. Draw diagonal \(AC = 7.4\text{ cm}\).
2. Find the midpoint \(O\) of \(AC\) by drawing its perpendicular bisector.
3. At point \(O\), construct an angle of \(75^{\circ}\) with \(AC\). Extend the line on both sides of \(O\).
4. With \(O\) as center and radius equal to half of the second diagonal (\(\frac{4.4}{2} = 2.2\text{ cm}\)), draw arcs on both branches of the angle line to mark points \(B\) and \(D\).
5. Join \(AB\), \(BC\), \(CD\), and \(DA\).
6. The required parallelogram is formed.
In simple words: Draw one diagonal, bisect it, construct the given angle at the midpoint, and mark half lengths of the second diagonal on both sides.

Exam Tip: Remember that the diagonals of a parallelogram bisect each other, so you only need half their lengths from the intersection point.

Question 20. Construct a rectangle ABCD whose one side is \(4\text{ cm}\) and diagonal is \(5\text{ cm}\).
Answer: Steps of construction:
1. Draw line segment \(AB = 4\text{ cm}\).
2. At point \(A\), construct a right angle (\(90^{\circ}\)).
3. With \(B\) as center and radius equal to the diagonal \(5\text{ cm}\), draw an arc to intersect the perpendicular line at point \(D\). Join \(AD\) and \(BD\).
4. With \(D\) as center and radius \(4\text{ cm}\), draw an arc.
5. With \(B\) as center and radius equal to \(AD\), draw an arc to intersect at point \(C\). Join \(DC\) and \(BC\).
6. \(ABCD\) is the required rectangle.
In simple words: Construct right-angled triangle \(ABD\) using side \(AB\), right angle at \(A\), and diagonal \(BD\), then complete the rectangle.

Exam Tip: Use Pythagoras theorem mentally (\(3\text{--}4\text{--}5\) triangle) to verify that the missing side will be \(3\text{ cm}\).

Question 21. Construct a parallelogram ABCD in which \(BC = 3.5\text{ cm}\), \(\angle C = 80^{\circ}\) and \(CD = 4\text{ cm}\).
Answer: Steps of construction:
1. Draw line segment \(CD = 4\text{ cm}\).
2. At point \(C\), construct an angle of \(80^{\circ}\) using a protractor.
3. With \(C\) as center and radius \(3.5\text{ cm}\) (side \(CB\)), draw an arc on the \(80^{\circ}\) ray to locate point \(B\).
4. With \(B\) as center and radius \(4\text{ cm}\) (side \(BA = CD\)), draw an arc.
5. With \(D\) as center and radius \(3.5\text{ cm}\) (side \(DA = CB\)), draw an arc to intersect at point \(A\). Join \(BA\) and \(DA\).
6. \(ABCD\) is the required parallelogram.
In simple words: Draw side \(CD\), make \(80^{\circ}\) at \(C\) to mark \(B\), and then use opposite side lengths to complete vertex \(A\).

Exam Tip: Ensure that opposite sides are equal in length before drawing the final intersection arcs.