OP Malhotra Class 9 Maths Solutions Chapter 11 Rectilinear Figures Exercise 11 (D)

S Chand Class 9 ICSE Maths Solutions Chapter 11 Rectilinear Figures Ex 11(D)

Construct the parallelogram ABCD from the given measurements in problems 1-4, without using set squares and protractor as far as possible.

Question 1. AB = 6.5 cm, BC = 5.2 cm and ∠B = 45°.
Answer:
Steps of constructions:
(i) Draw a line segment \( BC = 5.2 \) cm.
(ii) At point B, draw a ray BX making an angle of \( 45^\circ \) with BC. Then, cut off \( BA = 6.5 \) cm on this ray.
(iii) Using C as the center and a radius of \( 6.5 \) cm, draw an arc. Also, using A as the center and a radius of \( 5.2 \) cm, draw another arc. These two arcs will intersect at point D.
(iv) Connect points A to D and C to D to complete the shape.
ABCD is the required parallelogram. This construction creates a parallelogram by defining two sides and the included angle.

🎯 Exam Tip: When constructing parallelograms, remember that opposite sides are equal and parallel. Start by drawing a base and an angle, then use compasses to mark the lengths of the other sides correctly.

Question 2. Diagonals AC = 4.4 cm, diagonal BD = 6.8 cm and the angle between the diagonals = 60°.
Answer:
Steps of constructions:
(i) Draw diagonal \( BD = 6.8 \) cm. Then, find its midpoint, O, by bisecting it.
(ii) From O, draw a ray OX making an angle of \( 60^\circ \) with BD. Extend this ray downwards to Y.
(iii) Cut off \( OA = OC = \frac{4.4}{2} = 2.2 \) cm along the line XY, so that O is the midpoint of AC.
(iv) Connect points A to B, B to C, C to D, and D to A.
ABCD is the required parallelogram. This method uses the property that diagonals of a parallelogram bisect each other.

🎯 Exam Tip: When constructing with diagonals, remember that they bisect each other. Drawing one diagonal and its midpoint is often the first crucial step to ensure accuracy.

Question 3. AB = 6 cm, BC = 4 cm and altitude = 2.9 cm.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 6 \) cm.
(ii) At point A, draw a perpendicular ray AX. On AX, cut off a segment \( AP = 2.9 \) cm, which represents the altitude.
(iii) From point P, draw a line EPF parallel to AB. This line will contain the other two vertices of the parallelogram.
(iv) With B as the center and a radius of \( 4 \) cm, draw an arc that intersects EPF at C. With A as the center and a radius of \( 4 \) cm (since \( AD = BC \) in a parallelogram), draw an arc that intersects EPF at D.
(v) Join A to D and B to C.
ABCD is the required parallelogram. This construction uses the concept of a constant altitude between parallel sides.

🎯 Exam Tip: The altitude of a parallelogram is the perpendicular distance between its parallel sides. When constructing with altitude, drawing a parallel line at the given height is a critical early step.

Question 4. AB = 4.3 cm, ∠B = 120° and the height = 3 cm.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 4.3 \) cm.
(ii) At point B, draw a ray BY such that \( ∠ABC = 120^\circ \).
(iii) At point A, draw a perpendicular AX. On AX, cut off a segment \( AE = 3 \) cm, which is the given height.
(iv) From point E, draw a line EPF parallel to AB. This line will contain the top side of the parallelogram.
(v) With B as the center and a suitable radius (the length of BC, which is not given but can be determined or assumed from other properties if needed for the diagram; in the source diagram, it looks like a standard parallelogram construction), draw an arc intersecting EPF at C. With A as the center and a radius equal to BC, draw an arc intersecting EPF at D.
(vi) Join A to D and B to C.
ABCD is the required parallelogram. This combines a given angle with a specified height.

🎯 Exam Tip: When given an angle and height, construct the angle first, then draw a perpendicular from the vertex to establish the height, which helps in drawing the parallel top side.

Question 5. Using ruler and compasses only, construct a parallelogram ABCD using the following data: AB = 6 cm, AD = 3 cm and ∠DAB = 45°. If the bisector of ∠DAB meets DC at P, prove that ∠APB is a right angle.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 6 \) cm.
(ii) At point A, draw a ray AX making an angle of \( 45^\circ \) with AB. Cut off \( AD = 3 \) cm on this ray.
(iii) With B as the center and a radius of \( 3 \) cm (since \( BC = AD \)), draw an arc. With D as the center and a radius of \( 6 \) cm (since \( DC = AB \)), draw another arc. These arcs will intersect at point C.
(iv) Join B to C and D to C. ABCD is the required parallelogram.
(v) Draw the angle bisector of \( ∠DAB \). Let this bisector meet DC at point P. Join B to P.

Proof: To prove \( ∠APB = 90^\circ \)
In parallelogram ABCD, consecutive interior angles are supplementary.
\( \implies ∠DAB + ∠ABC = 180^\circ \)
Given \( ∠DAB = 45^\circ \).
\( \implies 45^\circ + ∠ABC = 180^\circ \)
\( \implies ∠ABC = 180^\circ - 45^\circ = 135^\circ \)
AP is the bisector of \( ∠DAB \).
\( \implies ∠PAB = ∠DAP = \frac{45^\circ}{2} = 22.5^\circ \)
Since \( AB \parallel DC \), \( ∠PAB = ∠APD \) (alternate interior angles).
\( \implies ∠APD = 22.5^\circ \)
In \( \triangle ADP \), since \( ∠DAP = ∠APD = 22.5^\circ \), \( \triangle ADP \) is an isosceles triangle with \( AD = DP \).
Given \( AD = 3 \) cm, so \( DP = 3 \) cm.
In a parallelogram, opposite sides are equal. So, \( AB = DC = 6 \) cm and \( BC = AD = 3 \) cm.
Now, calculate PC: \( PC = DC - DP = 6 - 3 = 3 \) cm.
Since \( BC = 3 \) cm and \( PC = 3 \) cm, \( \triangle BCP \) is an isosceles triangle with \( BC = PC \).
In a parallelogram, opposite angles are equal. So, \( ∠BCD = ∠DAB = 45^\circ \).
In \( \triangle BCP \):
\( ∠CBP = ∠CPB = \frac{180^\circ - ∠BCD}{2} = \frac{180^\circ - 45^\circ}{2} = \frac{135^\circ}{2} = 67.5^\circ \)
Now, consider \( ∠PBA \). We know \( ∠ABC = ∠PBA + ∠PBC \).
\( \implies ∠PBA = ∠ABC - ∠PBC = 135^\circ - 67.5^\circ = 67.5^\circ \)
Finally, in \( \triangle APB \):
The sum of angles in a triangle is \( 180^\circ \).
\( \implies ∠APB = 180^\circ - (∠PAB + ∠PBA) \)
\( \implies ∠APB = 180^\circ - (22.5^\circ + 67.5^\circ) \)
\( \implies ∠APB = 180^\circ - 90^\circ \)
\( \implies ∠APB = 90^\circ \)
Hence, \( ∠APB \) is a right angle.
In simple words: First, construct the parallelogram. Then, draw a line that cuts the angle DAB exactly in half and extends to side DC at point P. After drawing BP, we can prove that the angle APB is 90 degrees. This happens because of properties like alternate interior angles and isosceles triangles formed within the parallelogram.

🎯 Exam Tip: For proof questions, break down the problem into smaller geometric properties (alternate angles, isosceles triangles, sum of angles in a triangle). Clearly state each step of your reasoning to earn full marks.

Question 6. Using ruler and compasses construct a parallelogram ABCD given that AB = 4 cm, AC = 10 cm, BD = 6 cm and that AC and BD are the diagonals. Measure BC.
Answer:
Steps of constructions:
In a parallelogram, the diagonals bisect each other. Let O be the point where diagonals AC and BD intersect.
\( \implies AO = OC = \frac{AC}{2} = \frac{10}{2} = 5 \) cm
\( \implies BO = OD = \frac{BD}{2} = \frac{6}{2} = 3 \) cm
(i) Draw a line segment \( AB = 4 \) cm.
(ii) With A as the center and a radius of \( 5 \) cm (for AO), draw an arc. With B as the center and a radius of \( 3 \) cm (for BO), draw another arc. These arcs will intersect at point O.
(iii) Join A to O and B to O.
(iv) Extend AO to C such that \( AO = OC \). Extend BO to D such that \( BO = OD \).
(v) Join A to D, D to C, and C to B.
ABCD is the required parallelogram.
Upon measuring, the side BC should be approximately \( 7.2 \) cm. This construction uses the bisection property of diagonals.
In simple words: Since diagonals of a parallelogram cut each other in half, first find the half-lengths of the diagonals. Then, draw one side of the parallelogram. Use compasses to find the point where the half-diagonals meet. Extend these half-diagonals to their full length to find the other corners, then connect them to complete the parallelogram. Finally, measure the length of BC.

🎯 Exam Tip: Remember that in a parallelogram, the point where the diagonals meet is the midpoint of both diagonals. This is key for construction when given diagonal lengths.

Question 7. Use ruler and compasses to construct a parallelogram with diagonals 6 cm and 8 cm in length having given the acute angle between them is 60°. Measure one of the longer sides.
Answer:
Steps of constructions:
Let the diagonals be \( d_1 = 8 \) cm and \( d_2 = 6 \) cm. The diagonals bisect each other.
So, the half-diagonals are \( \frac{d_1}{2} = 4 \) cm and \( \frac{d_2}{2} = 3 \) cm.
(i) Draw a line segment \( AC = 8 \) cm (the longer diagonal) and bisect it at O.
(ii) At O, draw a ray OX making an angle of \( 60^\circ \) with AC. Produce XO to Y, making a straight line XY.
(iii) From O, cut off \( OD = OB = 3 \) cm (half of the shorter diagonal) along the line XY, such that O is the midpoint of BD.
(iv) Join A to B, B to C, C to D, and D to A.
ABCD is the required parallelogram. This construction relies on the fact that diagonals bisect each other and meet at a specific angle.
Upon measuring, one of the longer sides (BC or AD) will be approximately \( 6 \) cm long.

🎯 Exam Tip: When constructing parallelograms using diagonals and the angle between them, always bisect the diagonals first. The intersection point and the angle are crucial for positioning the vertices.

Question 8. The perpendicular distances between the pair of opposite sides of a parallelogram are 3 cm and 4 cm and one of its angles measures 60°. Using ruler and compasses only construct the parallelogram.
Answer:
Steps of constructions:
(i) Draw a line AX. This will be the base line for one side of the parallelogram.
(ii) At point A, draw a perpendicular to AX. On this perpendicular, cut off \( AL = 3 \) cm. This represents one height.
(iii) Through L, draw a line LY parallel to AX. This line will contain the opposite side of the base.
(iv) From A, draw a ray making an angle of \( 60^\circ \) with AX. Let this ray meet LY at point D. This establishes one angle of the parallelogram.
(v) At AD, draw a perpendicular. On this perpendicular, cut off \( AM = 4 \) cm. This represents the other height between the other pair of parallel sides.
(vi) Through M, draw a line parallel to AD. Let this line meet LY at C. Connect A to B and B to C to complete the parallelogram (B is on AX such that AB is the length of the base).
ABCD is the required parallelogram. This is a complex construction involving two altitudes and an angle.
In simple words: First, draw a base line. Then, draw a perpendicular line from one end of the base and mark the first height (3 cm). Draw a line parallel to the base through this height mark. Next, draw a line from the base at a 60-degree angle to meet the parallel line, forming the first slanted side. After that, draw another perpendicular from the slanted side to mark the second height (4 cm). Draw a line parallel to the slanted side through this second height mark to find the final corner, and then connect all points.

🎯 Exam Tip: Constructing a parallelogram using two altitudes and an angle requires careful management of parallel lines and perpendiculars. Begin by establishing one base and its corresponding height, then use the angle to find the next vertex, and then apply the second altitude to complete the figure.

Question 9. Using ruler and compasses only, draw a parallelogram whose diagonals are 5 cm and 7 cm long and which cut each other, so that one pair of the vertically opposite angles are each equal to 75°. Measure the length of the shorter side. State your steps of construction briefly.
Answer:
Steps of constructions:
In a parallelogram, diagonals bisect each other. Let the diagonals be \( d_1 = 7 \) cm and \( d_2 = 5 \) cm.
The half-diagonals are \( \frac{d_1}{2} = 3.5 \) cm and \( \frac{d_2}{2} = 2.5 \) cm.
(i) Draw a line segment \( AC = 7 \) cm (the longer diagonal) and bisect it at O.
(ii) At O, draw a line making an angle of \( 75^\circ \) with AC. Produce this line in both directions to form line XY.
(iii) From O, cut off \( OB = OD = 2.5 \) cm (half of the shorter diagonal) along the line XY, such that O is the midpoint of BD.
(iv) Join A to B, B to C, C to D, and D to A.
ABCD is the required parallelogram.
Upon measuring, the shorter side (AB or CD) will be approximately \( 4 \) cm long. This construction leverages the properties of diagonal bisection and the angle between diagonals.

🎯 Exam Tip: When constructing parallelograms with diagonals and their intersection angle, remember to bisect the diagonals first. The vertically opposite angle between the diagonals is essential for placing the other two vertices accurately.

Question 10. Construct a rhombus ABCD on a base AB = 3.6 cm long and having one of its angles equal to 45°.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 3.6 \) cm.
(ii) At point A, draw a ray AX making an angle of \( 45^\circ \) with AB. Since all sides of a rhombus are equal, cut off \( AD = 3.6 \) cm on this ray.
(iii) With B as the center and a radius of \( 3.6 \) cm, draw an arc. With D as the center and a radius of \( 3.6 \) cm, draw another arc. These arcs will intersect at point C.
(iv) Join B to C and C to D.
ABCD is the required rhombus. This construction uses the property that all sides of a rhombus are equal.
In simple words: Start by drawing one side (the base) of 3.6 cm. From one end of the base, draw a line at a 45-degree angle, and mark another point 3.6 cm along this line. Then, from the other end of the base and from the new point, draw arcs of 3.6 cm to find the last corner. Connect all points to form the rhombus.

🎯 Exam Tip: For rhombus constructions, remember that all four sides are equal in length. This means all arcs drawn from the vertices to find the fourth point will use the same side length as the radius.

Question 11. Using ruler and compasses, draw a square on a line segment 3.2 cm long. Draw the diagonals. Measure their lengths and also the angle between their.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 3.2 \) cm.
(ii) At A and B, draw perpendicular rays AX and BY respectively.
(iii) Cut off \( AD = BC = 3.2 \) cm on AX and BY. (Since all sides of a square are equal).
(iv) Join D to C.
ABCD is the required square.
(v) Draw the diagonals AC and BD.
Upon measuring, each diagonal will be approximately \( 4.5 \) cm long, and the angle between the diagonals will be \( 90^\circ \). This construction emphasizes the equal sides and right angles of a square.

🎯 Exam Tip: When drawing a square, constructing perpendiculars at both ends of the base and then marking the equal side lengths ensures all angles are \( 90^\circ \). Remember that a square's diagonals are equal and bisect each other at right angles.

Question 12. Construct the square PQRS given the PR = 5 cm. Measure PQ.
Answer:
Steps of constructions:
We know that the diagonals of a square are equal and bisect each other at right angles.
(i) Draw a line segment \( PR = 5 \) cm. Bisect PR at point O.
(ii) Draw the perpendicular bisector XY of PR, passing through O. This line will contain the other diagonal QS.
(iii) Cut off \( OQ = OS = \frac{5}{2} = 2.5 \) cm on the line XY, such that O is the midpoint of QS.
(iv) Join P to Q, Q to R, R to S, and S to P.
PQRS is the required square.
Upon measuring, the side PQ will be approximately \( 3.6 \) cm long. This construction leverages the diagonal properties of a square.
In simple words: First, draw the given diagonal and find its exact middle point. Then, draw a straight line that cuts through this middle point at a perfect right angle. Measure half the diagonal length (2.5 cm) on both sides along this new line to find the other two corners. Connect all four points to form the square. Finally, measure one of its sides.

🎯 Exam Tip: For constructing a square given only its diagonal, the most efficient method is to draw the diagonal, find its perpendicular bisector, and mark half the diagonal length along the bisector to locate the other two vertices.

Question 13. Construct the rectangle whose sides are 2.5 cm and 4.3 cm.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 4.3 \) cm.
(ii) At A and B, draw perpendicular rays AX and BY respectively. These lines will form the adjacent sides of the rectangle.
(iii) Cut off \( AD = BC = 2.5 \) cm on AX and BY. (Since opposite sides of a rectangle are equal).
(iv) Join D to C.
ABCD is the required rectangle. This construction uses the right angles and side lengths of a rectangle.
In simple words: Draw a bottom line of 4.3 cm. From both ends of this line, draw lines straight up (at 90 degrees). Measure 2.5 cm up on both vertical lines to mark the top corners. Connect these top corners to finish the rectangle.

🎯 Exam Tip: When constructing a rectangle from given side lengths, start with the longer side as the base, then erect perpendiculars at its endpoints. Measure the shorter side along these perpendiculars to locate the other two vertices.

Question 14. Construct the rectangle ABCD given that diagonals intersect at an angle of 37° and that AC = 6 cm.
Answer:
Steps of constructions:
Diagonals of a rectangle are equal and bisect each other. So, \( AC = BD = 6 \) cm.
Also, \( AO = OC = BO = OD = \frac{6}{2} = 3 \) cm.
(i) Draw a line segment \( AC = 6 \) cm. Bisect AC at point O.
(ii) From O, draw a line XY making an angle of \( 37^\circ \) with AC. Produce it to both sides.
(iii) On line XY, cut off \( OB = OD = 3 \) cm from O.
(iv) Join A to B, B to C, C to D, and D to A.
ABCD is the required rectangle. This construction relies on the properties of equal and bisecting diagonals with a specific angle of intersection.
In simple words: First, draw one diagonal (6 cm) and find its center. Then, draw a line through this center, making a 37-degree angle with the first diagonal. On this new line, measure 3 cm (half the diagonal length) on both sides from the center to find the other two corners. Connect all four corners to make the rectangle.

🎯 Exam Tip: For constructing rectangles with given diagonal length and intersection angle, bisecting the diagonal first and then using the angle to position the other diagonal is key. Remember that all half-diagonals in a rectangle are equal.

Question 15. Using ruler and compasses only, construct a square having a diagonal of length 5 cm. Measure its side correct to the nearest millimetre
Answer:
Steps of constructions:
The diagonals of a square are equal, bisect each other, and intersect at right angles.
(i) Draw a line segment \( AC = 5 \) cm. Bisect AC at point O.
(ii) Draw the perpendicular bisector XY of AC, passing through O. This line will contain the other diagonal BD.
(iii) Cut off \( OB = OD = \frac{5}{2} = 2.5 \) cm on the line XY, such that O is the midpoint of BD.
(iv) Join A to B, B to C, C to D, and D to A.
ABCD is the required square.
Upon measuring its side, it is approximately \( 3.6 \) cm (or \( 36 \) millimetres) long. This construction highlights the unique properties of a square's diagonals.
In simple words: Draw one diagonal (5 cm) and find its middle point. Draw a straight line that cuts through this middle point at a 90-degree angle. Measure 2.5 cm (half of 5 cm) from the center along this new line on both sides. These points, along with the ends of the first diagonal, are the corners of the square. Connect them and measure a side.

🎯 Exam Tip: When constructing a square from its diagonal, remember that the diagonals are equal and bisect each other at a right angle. This provides all the necessary information to position the vertices accurately.

Question 16. Using ruler and compasses only construct a rectangle each of whose diagonals measures 6 cm and intersect at an angle of 45°.
Answer:
Steps of constructions:
Diagonals of a rectangle are equal and bisect each other. So, if each diagonal measures \( 6 \) cm, then half of each diagonal is \( \frac{6}{2} = 3 \) cm.
(i) Draw a line segment \( AC = 6 \) cm. Bisect AC at point O.
(ii) At O, draw a line XY making an angle of \( 45^\circ \) with AC. Produce this line in both directions to create line XY.
(iii) Cut off \( OB = OD = 3 \) cm on the line XY, such that O is the midpoint of BD.
(iv) Join A to B, B to C, C to D, and D to A.
ABCD is the required rectangle. This construction method uses the properties of a rectangle's diagonals.
In simple words: Draw one diagonal 6 cm long and find its center. Then, draw another line through this center, making a 45-degree angle with the first diagonal. Mark 3 cm (half the diagonal length) on both sides of the center along this new line. Connect these four points to form the rectangle.

🎯 Exam Tip: When constructing a rectangle based on diagonals and their intersection angle, remember that the diagonals are equal and bisect each other. This means all four segments from the center to each vertex are equal.

Question 17. Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Answer:
Steps of constructions:
Diagonals of a rhombus bisect each other at right angles.
(i) Draw a line segment \( PR = 8 \) cm. Bisect PR at point O.
(ii) Draw the perpendicular bisector XY of PR, passing through O. This line will contain the other diagonal QS.
(iii) Cut off \( OQ = OS = \frac{6}{2} = 3 \) cm on the line XY, such that O is the midpoint of QS.
(iv) Join P to Q, Q to R, R to S, and S to P. PQRS is the required rhombus.
(v) To find point X equidistant from PQ and PS: X lies on the angle bisector of \( ∠QPS \), which is the diagonal PR.
(vi) To find point X equidistant from R and S: X lies on the perpendicular bisector of RS.
(vii) Draw the perpendicular bisector of RS. The point where this bisector intersects PR is point X.
Upon measuring, XR will be approximately \( 3.3 \) cm. This construction combines rhombus properties with locus principles.
In simple words: First, draw the rhombus by drawing one diagonal, finding its center, drawing a perpendicular line through the center, marking half the length of the second diagonal on that line, and connecting the points. Then, to find point X, draw a line that cuts the angle QPS in half (this is diagonal PR). Next, draw a line that cuts the side RS exactly in half and is perpendicular to it. Where these two lines cross is point X. Finally, measure the distance from X to R.

🎯 Exam Tip: When locating a point equidistant from two lines, draw the angle bisector. When locating a point equidistant from two points, draw the perpendicular bisector of the segment connecting them. The intersection of these locus lines gives the required point.

Question 18. Construct the trapezium PQRS in which PQ is parallel to RS, PQ = 7 cm, QR = 3.5 cm, RS = 2.5 cm, PS = 3 cm.
Answer:
Steps of constructions:
Since PQ is parallel to RS, we can use the concept of constructing a parallelogram and a triangle.
(i) Draw a line segment \( PQ = 7 \) cm.
(ii) On PQ, take a point E such that \( QE = PQ - RS = 7 - 2.5 = 4.5 \) cm. So, \( PE = RS = 2.5 \) cm.
(iii) With Q as the center and a radius of \( 3.5 \) cm (for QR), draw an arc. With E as the center and a radius of \( 3 \) cm (for PS, as \( ER \) will be parallel to PS), draw another arc. These arcs will intersect at point R.
(iv) Join E to R and Q to R.
(v) With P as the center and a radius of \( 3 \) cm (for PS), draw an arc. With R as the center and a radius of \( 2.5 \) cm (for RS), draw another arc. These arcs will intersect at point S.
(vi) Join R to S and P to S.
PQRS is the required trapezium. This construction combines a parallelogram (PERF where F is on PQ) and a triangle (QER) or similar concepts.
In simple words: Start by drawing the longest parallel side. Then, subtract the length of the shorter parallel side from it. Use this difference to create a triangle with the two slanted sides of the trapezium. Once this triangle's third vertex is found, complete the parallelogram part using the shorter parallel side to locate the final corner. Connect all points to form the trapezium.

🎯 Exam Tip: When constructing a trapezium where parallel sides are known, a common strategy is to draw an auxiliary line to form a parallelogram and a triangle. This allows for the use of known lengths and angles to locate all vertices.

Question 19. PQ = 8 cm, QR = 4 cm, RS = 3 cm, SP = 2 cm. Measure ∠QPS.
Answer:
Steps of constructions:
(i) Draw a line segment \( PQ = 8 \) cm.
(ii) On PQ, take a point E such that \( PE = RS = 3 \) cm. This creates a parallelogram for part of the figure. So \( EQ = PQ - PE = 8 - 3 = 5 \) cm.
(iii) With E as the center and a radius of \( 2 \) cm (for SP), draw an arc. With Q as the center and a radius of \( 4 \) cm (for QR), draw another arc. These arcs will intersect at point R.
(iv) Join E to R and Q to R.
(v) With P as the center and a radius of \( 2 \) cm (for SP), draw an arc. With R as the center and a radius of \( 3 \) cm (for RS), draw another arc. These arcs will intersect at point S.
(vi) Join P to S and R to S.
PQRS is the required trapezium. This construction forms a parallelogram (PE'RS, where E' is on PQ) and a triangle (QER).
Upon measuring the angle \( ∠QPS \), it is approximately \( 50^\circ \).
In simple words: Draw the longest base line. Cut a part of this line equal to the shorter parallel side. Use the remaining part of the base and the two slanted sides to form a triangle. Once this is done, use the shorter parallel side to find the last corner, completing the trapezium. Then, measure the angle QPS.

🎯 Exam Tip: For trapezium constructions with given side lengths, if the parallel sides are known, a key strategy is to use one of the non-parallel sides to form a triangle alongside a portion of the longer parallel base. Measure angles carefully when required.

Question 20. Without using set squares or protractor, construct the trapezium ABCD in which AB is parallel to DC, AB = 4.0 cm, ∠ABC = 120°, BC = 2.5 cm, DA = 2.4 cm and ∠DAB is obtuse, Measure AC.
Answer:
Steps of constructions:
(i) Draw a line segment \( AB = 4.0 \) cm.
(ii) At point B, draw a ray BX making an angle of \( 120^\circ \) with AB. Cut off \( BC = 2.5 \) cm on this ray.
(iii) Since AB is parallel to DC, and BC is a transversal, \( ∠ABC + ∠BCD = 180^\circ \). Therefore, \( ∠BCD = 180^\circ - 120^\circ = 60^\circ \). At point C, draw a ray CY making an angle of \( 60^\circ \) with BC.
(iv) With A as the center and a radius of \( 2.4 \) cm (for DA), draw an arc. This arc will intersect ray CY at point D.
(v) Join A to D.
ABCD is the required trapezium. This construction uses parallel line properties to find an angle without a protractor.
Upon measuring AC, it is approximately \( 5.5 \) cm.
In simple words: Draw the base line AB. From point B, draw a line at a 120-degree angle and mark off 2.5 cm for BC. Since the top and bottom lines are parallel, the angle at C will be 60 degrees (180 - 120). Draw a line from C at this 60-degree angle. From point A, use a compass to draw an arc with a radius of 2.4 cm. Where this arc crosses the line from C, that's point D. Connect A to D. Finally, measure the diagonal AC.

🎯 Exam Tip: When constructing a trapezium without a protractor, use the property of parallel lines (consecutive interior angles sum to \( 180^\circ \)) to determine unknown angles and construct them using compass and ruler techniques.

Question 21. Construct the trapezium ABCD in which AB is parallel to DC. AB = 6.5 cm, CD = 3 cm, AC = 7 cm and DB = 5 cm.
Answer: Here are the steps to construct the trapezium ABCD:
(i) First, draw a straight line segment AB that is 6.5 cm long.
(ii) Extend the line AB to a point P such that the length of BP is 3 cm. This extension helps in finding point C.
(iii) Now, with P as the center, draw an arc with a radius of 5 cm. Then, with A as the center, draw another arc with a radius of 7 cm. These two arcs will meet at a point, which will be C.
(iv) Connect points A and C, P and C, and B and C with straight lines.
(v) Next, with B as the center, draw an arc with a radius of 5 cm. Then, with C as the center, draw another arc with a radius of 3 cm. These two arcs will meet at a point, which will be D.
(vi) Finally, join points A and D, B and D, and C and D with straight lines. This completes the trapezium ABCD. A trapezium has at least one pair of parallel sides.
In simple words: Start by drawing the base AB. Then extend it to find a point P that helps locate C using given diagonal lengths. After finding C, use diagonal and side lengths from B and C to locate D. Finally, connect all points to form the trapezium.

🎯 Exam Tip: When constructing geometric figures, always use a sharp pencil and precise measurements with your ruler and compass. Accuracy in drawing arcs and lines is crucial for correct construction.

Question 22. Hexagon on a side of 4 cm. Find out its area.
Answer: First, we construct the hexagon using these steps:
(i) Draw a circle with the center at O and a radius of 4 cm.
(ii) Pick any point, let's call it A, on the edge of this circle. Then, using a compass set to 4 cm (the same as the radius), draw arcs starting from A. These arcs will cut the circle at points B, C, D, E, and F.
(iii) Join all these points (AB, BC, CD, DE, EF, and FA) with straight lines. This creates the required regular hexagon ABCDEF.
For a regular hexagon, the radius of its circumcircle is equal to the length of its side. So, for a hexagon with a side of 4 cm, the area is calculated using the formula for a regular hexagon:
Area of a regular hexagon \( = 6 \times \frac{\sqrt{3}}{4} (\text{side})^2 \)
Now, substitute the side length, which is 4 cm:
\( = 6 \times \frac{\sqrt{3}}{4} (4)^2 \)
\( = 6 \times \frac{\sqrt{3}}{4} \times 16 \)
\( = 6 \times \sqrt{3} \times 4 \)
\( = 24\sqrt{3} \)
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\( = 24 \times 1.732 \)
\( = 41.568 \text{ cm}^2 \)
In simple words: To draw this hexagon, draw a circle with a radius equal to the side length (4 cm). Mark points around the circle by stepping off the radius. Connect these points to make the hexagon. The area is found by a special formula that multiplies 6 (for 6 triangles) by the area of one equilateral triangle formed by the side length.

🎯 Exam Tip: Remember that a regular hexagon can be divided into six equilateral triangles, which simplifies area calculations. Knowing this property is very helpful for problems involving hexagons.

Question 23. Hexagon on a side of 3.2 cm. Find out its area.
Answer: Here are the steps to construct the hexagon:
(i) First, draw a circle with its center at O and a radius of 3.2 cm.
(ii) Pick a starting point, A, anywhere on the circle's edge. This will be the first vertex of your hexagon.
(iii) Keeping your compass open to 3.2 cm (the side length), place the compass point on A and draw an arc that cuts the circle. This new intersection point is B. Repeat this step, placing the compass point on B to find C, then on C to find D, and so on, until you have marked points E and F.
(iv) Connect all the points you marked (AB, BC, CD, DE, EF, and FA) with straight lines. This forms the required regular hexagon ABCDEF. The side length of a regular hexagon is equal to the radius of its circumcircle.
Now, let's find the area. The formula for the area of a regular hexagon is:
Area of a regular hexagon \( = 6 \times \frac{\sqrt{3}}{4} (\text{side})^2 \)
Given the side length is 3.2 cm, substitute this value into the formula:
\( = 6 \times \frac{\sqrt{3}}{4} (3.2)^2 \)
\( = 6 \times \frac{\sqrt{3}}{4} \times 10.24 \)
\( = 6 \times \sqrt{3} \times 2.56 \)
\( = 15.36\sqrt{3} \)
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\( = 15.36 \times 1.732 \)
\( \approx 26.60 \text{ cm}^2 \)
In simple words: Draw a circle with a radius of 3.2 cm. Use the same 3.2 cm length to step around the circle with a compass, marking six points. Connect these points to form the hexagon. The total area is found by multiplying 6 (for the number of triangles) by the area of one small triangle inside.

🎯 Exam Tip: Always state the formula you are using for area calculations. For construction questions, precise measurements and clear arcs are essential to get full marks.