ML Aggarwal Class 8 Maths Solutions Chapter 14 Practical Geometry

Exercise 14.1

Question 1. Construct a quadrilateral PQRS where PQ = 4.5 cm, QR = 6 cm, RS = 5.5 cm, PS = 5 cm and PR = 6.5 cm.
Answer: Follow these steps to draw the quadrilateral:
(i) Start by drawing a line segment PR = 6.5 cm.
(ii) Using P as the center with a 4.5 cm radius, and R as the center with a 6 cm radius, draw two arcs that cross each other at point Q.
(iii) Draw lines joining P to Q and Q to R.
(iv) Using P as the center with a 5 cm radius, and R as the center with a 5.5 cm radius, draw two arcs that meet at point S.
(v) Draw lines joining P to S and S to R.
This gives you the required quadrilateral PQRS with all the specified measurements.
In simple words: Draw the diagonal PR first. Then find Q using distances from P and R. Next find S the same way. Connect all points to get your quadrilateral.

Exam Tip: Always start with a diagonal or known side when given multiple side lengths. Arc intersections determine vertex positions - mark them clearly.

Question 2. Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 5 cm, CD = 5.6 cm, DA = 4 cm, BD = 5.4 cm.
Answer: Follow these construction steps:
(i) Draw a line segment AB = 3.5 cm.
(ii) With A as center and radius 4 cm, draw an arc. With B as center and radius 5.4 cm, draw another arc. These arcs meet at point D. Join AD and BD.
(iii) Using B as center with radius 5 cm, draw an arc. Using D as center with radius 5.6 cm, draw another arc that meets the first arc at point C.
(iv) Join BC and CD.
This creates the required quadrilateral ABCD with all specified side lengths and diagonal.
In simple words: First create a triangle ABD using the diagonal BD. Then use distances from B and D to find point C. Join all sides to finish the shape.

Exam Tip: When a diagonal is given with sides, construct a triangle first using that diagonal, then complete the quadrilateral by finding the fourth vertex.

Question 3. Construct a quadrilateral PQRS in which PQ = 3 cm, QR = 2.5 cm, PS = 3.5 cm, PR = 4 cm and QS = 5 cm.
Answer: Follow this construction method:
(i) Draw a line segment PQ = 3 cm.
(ii) With P as center and radius 4 cm, draw an arc. With Q as center and radius 2.5 cm, draw another arc that meets the first arc at point R. Join P to R and Q to R.
(iii) Using P as center with radius 3.5 cm, draw an arc. Using Q as center with radius 5 cm, draw another arc that meets at point S.
(iv) Join PS, QS, and SR.
This produces the required quadrilateral PQRS with both diagonals specified.
In simple words: Use the two diagonals PR and QS to help position vertices. Draw arcs from known points at specified distances to find where the diagonals meet.

Exam Tip: When both diagonals are given, use them strategically to control the shape. The arc intersections locate the vertices where distances from two centers match the given lengths.

Question 4. Construct a quadrilateral ABCD such that BC = 5 cm, AD = 5.5 cm, CD = 4.5 cm, AC = 7 cm and BC = 5.5 cm.
Answer: Follow these steps:
(i) Draw a line segment CD = 4.5 cm.
(ii) With C as center and radius 5.5 cm, and D as center and radius 7 cm, draw two arcs that meet at point B. Join BC and BD.
(iii) With C as center and radius 5.5 cm, and D as center and radius 5.5 cm, draw two arcs that meet at point A.
(iv) Join AC and AD, then join AB.
This creates the required quadrilateral ABCD with the specified measurements.
In simple words: Start with side CD. Use radius measurements to find B first, then find A using different radii. Connect all vertices to complete the shape.

Exam Tip: Keep track of which center point and which radius applies to each step. Careful arc drawing prevents errors in vertex location.

Question 5. Construct a quadrilateral ABCD given that BC = 6 cm, CD = 4 cm, ∠B = 45°, ∠C = 90° and ∠D = 120°.
Answer: Follow this angle-based construction:
(i) Draw a line segment BC = 6 cm.
(ii) At point B, draw a ray making an angle of 45° with BC. Label this ray BP.
(iii) At point C, draw a ray making an angle of 90° with BC. Label this ray CQ.
(iv) Along ray CQ, measure and mark off CD = 4 cm to locate point D.
(v) At point D, draw a ray making an angle of 120° with DC. Label this ray DR.
(vi) Let rays BP and DR meet at point A.
This construction yields the required quadrilateral ABCD with all specified angles and side measurements.
In simple words: Draw BC first. At each end point and at D, draw rays at the right angles. The rays meet at corners to complete your shape.

Exam Tip: Use a protractor carefully to set angles precisely. The meeting point of two rays determines the fourth vertex - ensure rays are extended sufficiently to intersect clearly.

Question 6. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, ∠P = 60°, ∠Q = 90° and ∠R = 120°.
Answer: Follow this construction:
(i) Draw a line segment QR = 6 cm.
(ii) At point Q, draw a ray QX making an angle of 90°. Measure off QP = 4 cm along this ray to mark point P.
(iii) At point P, construct a ray making an angle of 60°. At point R, construct a ray making an angle of 120°. These two rays meet at point S.
This produces the required quadrilateral PQRS with all given angles and sides.
In simple words: Start with side QR. Draw perpendicular at Q, mark P at distance 4 cm. From P and R, draw rays at given angles until they cross at S.

Exam Tip: The angles guide ray directions - be precise with the protractor. The intersection of rays from P and R locates the final vertex S.

Question 7. Construct a quadrilateral ABCD such that AB = 5 cm, BC = 4.2 cm, AD = 3.5 cm, ∠A = 90° and ∠B = 60°.
Answer: Follow these construction steps:
(i) Draw a line segment AB = 5 cm.
(ii) At point A, construct an angle of 90°. At point B, construct an angle of 60°.
(iii) Along the 60° ray at B, measure and mark BC = 4.2 cm to locate point C.
(iv) Along the 90° ray at A, measure and mark AD = 3.5 cm to locate point D.
(v) Join CD to complete the quadrilateral.
This gives the required quadrilateral ABCD with all specified dimensions and angles.
In simple words: Draw AB. Make a right angle at A and a 60° angle at B. Measure out the two side lengths on these rays, then connect the endpoints.

Exam Tip: Angles control the ray directions from base vertices. The side lengths determine how far along each ray to place the other vertices.

Question 8. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 5 cm, RS = 4.5 cm, ∠Q = 60° and ∠R = 90°.
Answer: Follow this construction:
(i) Draw a line segment QR = 5 cm.
(ii) At point Q, draw a ray QX making an angle of 60°. Measure off QP = 4 cm along this ray to mark point P.
(iii) At point R, draw a ray RY making an angle of 90°. Measure off RS = 4.5 cm along this ray to mark point S.
(iv) Join PS to complete the quadrilateral.
This creates the required quadrilateral PQRS with all specified measurements.
In simple words: Start with QR. Draw rays at 60° and 90° from Q and R. Mark distances along these rays to find P and S, then connect them.

Exam Tip: Use a protractor to set the angles at Q and R correctly. Each ray direction and side length together determine the vertex positions.

Question 9. Construct a quadrilateral BEST where BE = 3.8 cm, ES = 3.4 cm, ST = 4.5 cm, TB = 5 cm and ∠E = 80°.
Answer: Follow this construction method:
(i) Draw a line segment BE = 3.8 cm.
(ii) At point E, draw a ray EX making an angle of 80°. Measure off ES = 3.4 cm along this ray to locate point S.
(iii) With B as center and radius 5 cm, draw an arc. With S as center and radius 4.5 cm, draw another arc that meets the first arc at point T.
(iv) Join TB and TS to complete the quadrilateral.
This produces the required quadrilateral BEST with all specified sides and angle.
In simple words: Draw BE and mark the 80° angle at E. Measure ES along the angled ray. Use arc distances from B and S to find where T is located.

Exam Tip: Combine angle construction with arc intersections - the angle sets one edge direction, while arc radii locate the remaining vertex.

Question 10. Construct a quadrilateral ABCD where AB = 4.5 cm, BC = 4 cm, CD = 3.9 cm, AD = 3.2 cm and ∠B = 60°.
Answer: Follow this construction:
(i) Draw a line segment AB = 4.5 cm.
(ii) At point B, draw a ray making an angle of 60°. Label this ray BP.
(iii) Along ray BP, measure and mark BC = 4 cm to locate point C.
(iv) With C as center and radius 3.9 cm, draw an arc.
(v) With A as center and radius 3.2 cm, draw another arc that meets the first arc at point D.
(vi) Join AD and CD to complete the quadrilateral.
This yields the required quadrilateral ABCD with all specified sides and angle.
In simple words: Draw AB and place a 60° angle at B. Mark C at distance 4 cm along the angled ray. Use arc distances from A and C to find D.

Exam Tip: The angle at B determines side BC direction. Arc radii from A and C must intersect to locate D - if arcs do not meet, check your measurements.

Exercise 14.2

Question 1. Construct a parallelogram ABCD such that AB = 5 cm, BC = 3.2 cm and ∠B = 120°.
Answer: Follow this construction for a parallelogram:
(i) Draw a line segment AB = 5 cm.
(ii) At point B, construct an angle of 120°.
(iii) Along the 120° ray, measure and mark BC = 3.2 cm to locate point C.
(iv) With C as center and radius equal to AB (5 cm), draw an arc.
(v) With A as center and radius 3.2 cm, draw another arc that meets the first arc at point D.
(vi) Join AD and CD to complete the parallelogram.
This yields the required parallelogram ABCD with the given sides and angle. Opposite sides are equal as required: AB = CD = 5 cm and BC = AD = 3.2 cm.
In simple words: A parallelogram has opposite sides equal. Start with AB, use the angle to place C at the right direction, then find D so that opposite sides match.

Exam Tip: In a parallelogram, opposite sides must be equal - use this property to choose your arc radii. The angle at one vertex and equal opposite sides determine the shape uniquely.

Question 2. Construct a parallelogram ABCD such that AB = 4.8 cm, BC = 4 cm and diagonal BD = 5.4 cm.
Answer: Follow this diagonal-based construction:
(i) Start by drawing a triangle ABD using the known sides AB and the diagonal BD.
(ii) With B as center and radius 4 cm, draw an arc. With D as center and radius 4.8 cm, draw another arc that meets the first arc at point C.
(iii) Join CD, BC, and AC to complete the parallelogram.
This creates the required parallelogram ABCD. Since BC = 4 cm and BC is opposite to AD, we have AD = 4 cm. Similarly, AB = 4.8 cm and CD = 4.8 cm as opposite sides.
In simple words: Use the diagonal to build a triangle first. Then find the fourth point by measuring distances from two vertices. Opposite sides become equal automatically.

Exam Tip: When a diagonal is provided, construct a triangle containing that diagonal first. Then use the remaining side lengths to complete the parallelogram.

Question 3. Construct a parallelogram ABCD such that BC = 4.5 cm, BD = 4 cm and AC = 5.6 cm.
Answer: Follow this construction using both diagonals:
(i) Draw a triangle BOC where BC = 4.5 cm. The diagonals bisect each other in a parallelogram, so:
BO = ½ × 4 = 2 cm
OC = ½ × 5.6 = 2.8 cm
(ii) Extend line OC beyond C to point A such that OA = OC = 2.8 cm.
(iii) Extend line BO beyond O to point D such that OD = OB = 2 cm.
(iv) Join AD to complete the parallelogram.
This produces the required parallelogram ABCD where the diagonals bisect each other at O.
In simple words: Diagonals in a parallelogram cut each other in half. Build a triangle with one half of each diagonal. Extend to the opposite sides to find the remaining vertices.

Exam Tip: Remember that diagonals of a parallelogram bisect each other. This property is key - use half-lengths of both diagonals to construct the initial triangle, then extend to find opposite vertices.

Question 4. Construct a parallelogram ABCD such that AC = 6 cm, BD = 4.6 cm and angle between them is 45°.
Answer: Follow this construction:
(i) Draw a line segment AO = ½ AC = 3 cm. Extend this line to point C such that OC = OA = 3 cm. This creates the full diagonal AC = 6 cm.
(ii) At point O, construct an angle of 45° with respect to diagonal AC. Label the ray OP.
(iii) Along ray OP, measure OD = ½ BD = 2.3 cm to locate point D.
(iv) Extend line DO beyond O to point B such that OB = OD = 2.3 cm. This creates the full diagonal BD = 4.6 cm.
(v) Join AB, BC, CD, and DA to complete the parallelogram.
This yields the required parallelogram ABCD with both diagonals and their angle specified.
In simple words: Draw both diagonals by using half-lengths on either side of their intersection point O. Set the angle between them at O. Connect the four endpoints of the diagonals.

Exam Tip: When both diagonals and their included angle are given, construct the diagonals first at the specified angle, then connect their endpoints to form the parallelogram.

Question 5. Construct a rectangle whose adjacent sides are 5.6 cm and 4 cm.
Answer: Follow this rectangle construction:
(i) Draw a line segment AB = 5.6 cm.
(ii) At point B, construct a right angle (90°). Label the ray BP.
(iii) Along ray BP, measure and mark BC = 4 cm to locate point C.
(iv) With C as center and radius 5.6 cm, draw an arc.
(v) With A as center and radius 4 cm, draw another arc that meets the first arc at point D.
(vi) Join AD and CD to complete the rectangle.
This creates the required rectangle ABCD with adjacent sides 5.6 cm and 4 cm. All angles are 90° and opposite sides are equal.
In simple words: A rectangle has right angles at every corner. Start with one side, make right angles at both ends, measure the other side length, then connect opposite ends equally.

Exam Tip: All angles in a rectangle are 90°. Once you set two perpendicular sides, the opposite sides must be equal in length - use arc radii equal to the adjacent sides to find the fourth vertex.

Question 6. Construct a rectangle such that one side is 5 cm and one diagonal is 6.8 cm.
Answer: Follow this diagonal-based rectangle construction:
(i) Draw a line segment AB = 5 cm.
(ii) At point A, construct a right angle. Label the ray AP.
(iii) With B as center and radius 6.8 cm (the diagonal length), draw an arc that meets ray AP at point D.
(iv) With A as center and radius 6.8 cm, draw another arc.
(v) With D as center and radius 5 cm, draw another arc that meets the previous arc at point C.
(vi) Join BC and CD to complete the rectangle.
This yields the required rectangle ABCD with one side 5 cm and diagonal 6.8 cm. The other side can be found using the Pythagorean theorem: if the other side is x, then x² + 5² = 6.8², giving x ≈ 5.2 cm.
In simple words: Use the given side and diagonal to find the other side. Right angle at A guides one side direction. Arc distances from B and D place the remaining vertices.

Exam Tip: When a diagonal is given with one side, use the Pythagorean relationship to find the other side. In a rectangle, the diagonal equals the hypotenuse of the right triangle formed by two adjacent sides.

Question 7. Construct a rectangle ABCD such that AB = 4 cm and ∠BAC = 60°.
Answer: Follow this angle-based rectangle construction:
(i) Draw a line segment AB = 4 cm.
(ii) At point B, construct a right angle. Label the ray BP.
(iii) At point A, draw a line making an angle of 30° with AB (since ∠BAC = 60°, the angle from AB to AC is 60°, which means the angle from AB to the perpendicular direction is 30°). Let this line meet BP at point D.
(iv) With D as center and radius 4 cm, draw an arc.
(v) With A as center and radius BD, draw another arc that meets the previous arc at point C.
(vi) Join AC and CD to complete the rectangle.
This yields the required rectangle ABCD with the specified side and angle measurement.
In simple words: Draw AB and a right angle at B. Use the angle at A to find where D is positioned on the perpendicular. Then arc distances from A and D locate C.

Exam Tip: The angle ∠BAC in a rectangle relates to the diagonal AC and side AB. Use angle constructions combined with right angles to determine vertex positions.

Question 8. Construct a rectangle such that one diagonal is 6.6 cm and angle between two diagonals is 120°.
Answer: Follow this diagonal-angle construction:
(i) Draw a line segment AO = ½ AC = 3.3 cm. Extend to point C such that OC = OA = 3.3 cm. This creates the full diagonal AC = 6.6 cm.
(ii) At point O, construct an angle of 120°. Label the ray OB.
(iii) Along ray OB, measure OB = ½ AC = 3.3 cm to locate point B.
(iv) Extend line BO beyond O to point D such that OD = OB = 3.3 cm. This creates the full second diagonal BD = 6.6 cm (since both diagonals in a rectangle are equal).
(v) Join AB, BC, CD, and DA to complete the rectangle.
This creates the required rectangle ABCD where one diagonal is 6.6 cm and the diagonals meet at 120°.
In simple words: Build both diagonals using half-lengths from their meeting point O. Set the angle between them at O. Connect the four endpoints of the diagonals to get your rectangle.

Exam Tip: In a rectangle, both diagonals are equal and bisect each other. Use the given diagonal length and angle between diagonals to construct both diagonals, then join their endpoints.

Question 9. Construct a rhombus whose one side is 5 cm and one angle is 45°.
Answer: Follow this rhombus construction:
(i) Draw a line segment AB = 5 cm.
(ii) At point A, construct an angle of 45°. Label the ray AP.
(iii) Along ray AP, measure and mark AD = 5 cm to locate point D.
(iv) With B as center and radius 5 cm, draw an arc.
(v) With D as center and radius 5 cm, draw another arc that meets the first arc at point C.
(vi) Join BC and CD to complete the rhombus.
This yields the required rhombus ABCD where all sides equal 5 cm and one angle is 45°. In a rhombus, opposite angles are equal, so ∠C = 45°, while ∠B = ∠D = 135°.
In simple words: A rhombus has all sides equal. Draw one side, make the given angle, measure out an equal side from that angle. Use equal arc distances from two vertices to find the fourth corner.

Exam Tip: All four sides of a rhombus are equal. Once two adjacent sides are drawn at the correct angle, the remaining two sides follow from arc intersections with the same radius length.

Question 10. Construct a rhombus whose one side is 4.5 cm and one diagonal is 5 cm.
Answer: Follow this diagonal-based rhombus construction:
(i) Draw a line segment AB = 4.5 cm.
(ii) With A as center and radius 4.5 cm, draw an arc. With B as center and radius 5 cm, draw another arc that meets the first arc at point D. This arc of radius 5 cm from B represents the diagonal BD.
(iii) With B as center and radius 4.5 cm, draw another arc.
(iv) With D as center and radius 4.5 cm, draw another arc that meets the previous arc at point C.
(v) Join AD, BC, and CD to complete the rhombus.
This creates the required rhombus ABCD with all sides 4.5 cm and one diagonal 5 cm. In a rhombus, diagonals bisect each other at right angles, and the given diagonal measurement helps determine the shape uniquely.
In simple words: Start with one side AB. Use arc distances equal to the side length from two vertices, but let one arc represent the diagonal. This controls the shape and determines the remaining vertices.

Exam Tip: When one diagonal is given for a rhombus, it determines the shape completely along with the side length. Arc intersections using the side length as radius locate the remaining vertices.

Question 11. Construct a rhombus whose diagonals are 6.8 cm and 5.2 cm.
Answer: Follow this diagonal-based construction:
(i) Draw a line segment AC = 6.8 cm.
(ii) Construct the perpendicular bisector of AC. Let it meet AC at point O. This creates two equal segments: AO = OC = 3.4 cm.
(iii) Along the perpendicular bisector, measure and mark OB = ½ × 5.2 = 2.6 cm in one direction and OD = 2.6 cm in the opposite direction.
(iv) Join AB, BC, CD, and DA to complete the rhombus.
This yields the required rhombus ABCD where the diagonals are 6.8 cm and 5.2 cm and they bisect each other at right angles at point O.
In simple words: Draw one diagonal. Find its midpoint and draw the other diagonal perpendicular to it at that midpoint. Use half-lengths of each diagonal on either side to locate the four vertices.

Exam Tip: In a rhombus, diagonals are perpendicular bisectors of each other. Use this property - draw one diagonal, construct its perpendicular bisector, then mark off half of the second diagonal on both sides of the first diagonal.

Question 12. Construct a square whose one side is 4.3 cm.
Answer: Follow this square construction:
(i) Draw a line segment BC = 4.3 cm.
(ii) At point B, construct a right angle (90°). Label the ray BP.
(iii) Along ray BP, measure and mark BA = 4.3 cm to locate point A.
(iv) With C as center and radius 4.3 cm, draw an arc.
(v) With A as center and radius 4.3 cm, draw another arc that meets the first arc at point D.
(vi) Join AD and CD to complete the square.
This creates the required square ABCD with all sides equal to 4.3 cm and all angles 90°. All four sides are equal and all angles are right angles by construction.
In simple words: A square has all sides equal and all angles 90 degrees. Start with one side, make a right angle, measure the same length again, then use equal arc distances from two vertices to find the fourth corner.

Exam Tip: In a square, all sides are equal and all angles are 90°. Once two perpendicular sides of equal length are drawn, arc intersections with that same radius length automatically locate the opposite corners.

Question 13. Construct a square whose one diagonal is 6.2 cm.
Answer: Follow this diagonal-based square construction:
(i) Draw a line segment AC = 6.2 cm.
(ii) Construct the perpendicular bisector of AC. Let it meet AC at point O. This creates: AO = OC = 3.1 cm.
(iii) Along the perpendicular bisector, measure and mark OB = ½ AC = 3.1 cm in one direction and OD = 3.1 cm in the opposite direction.
(iv) Join AB, BC, CD, and DA to complete the square.
This yields the required square ABCD where the diagonal AC = 6.2 cm. In a square, both diagonals are equal and bisect each other at right angles. The side length works out to be 6.2 / √2 ≈ 4.38 cm.
In simple words: Draw the diagonal. Find its center point. Draw the other diagonal perpendicular to it at that center, using the same length. Connect the four endpoints of the diagonals to get your square.

Exam Tip: In a square, the diagonals are equal, perpendicular, and bisect each other. Use half of the given diagonal length on both sides of the center to place all four vertices equidistant from the center.