RBSE Solutions Class 9 Maths Chapter 9 Quadrilaterals Exercise 9.4

Question 1. Construct a parallelogram ABCD, given that side AB = 4.7 cm, side AD = 3.5 cm, and diagonal AC = 7 cm.
Answer: Here are the steps to construct the parallelogram ABCD:
1. Draw a line segment \( \text{AB} = 4.7 \) cm.
2. With point \( \text{A} \) as the center, draw an arc with a radius of 7 cm (for diagonal \( \text{AC} \)).
3. With point \( \text{B} \) as the center, draw an arc with a radius of 3.5 cm (for side \( \text{BC} \)). These two arcs should intersect at point \( \text{C} \). Join \( \text{BC} \). We use the property that opposite sides of a parallelogram are equal, so \( \text{BC} = \text{AD} \).
4. With point \( \text{A} \) as the center, draw another arc with a radius of 3.5 cm (for side \( \text{AD} \)).
5. With point \( \text{C} \) as the center, draw an arc with a radius of 4.7 cm (for side \( \text{CD} \)). This arc will intersect the arc drawn in step 4 at point \( \text{D} \). Join \( \text{AD} \) and \( \text{CD} \).
6. The figure \( \text{ABCD} \) is the required parallelogram.
In simple words: First, draw one side, then use the diagonal and the other side to find the third corner. After that, use the lengths of the opposite sides to find the last corner and complete the shape.

🎯 Exam Tip: Always start by drawing the base side. Then, use the given diagonal and adjacent side lengths to form a triangle, which helps in locating the third vertex of the parallelogram.

Question 2. Construct parallelogram PQRS, when PQ = 5 cm, diagonal PR = 7.6 cm and diagonal QS = 5.6 cm.
Answer: Here are the steps to construct the parallelogram PQRS:
1. Draw the line segment \( \text{PQ} = 5 \) cm.
2. In a parallelogram, diagonals bisect each other. So, \( \text{PO} = \frac{\text{PR}}{2} = \frac{7.6}{2} = 3.8 \) cm and \( \text{QO} = \frac{\text{QS}}{2} = \frac{5.6}{2} = 2.8 \) cm. Point \( \text{O} \) is where the diagonals intersect.
3. Construct triangle \( \text{PQO} \): With point \( \text{P} \) as the center, draw an arc with a radius of 3.8 cm. With point \( \text{Q} \) as the center, draw another arc with a radius of 2.8 cm. These two arcs will intersect at point \( \text{O} \). This point is the midpoint of both diagonals.
4. Extend the line segment \( \text{PO} \) to \( \text{R} \) such that \( \text{PO} = \text{OR} \). This makes \( \text{PR} = 2 \times \text{PO} = 2 \times 3.8 = 7.6 \) cm.
5. Extend the line segment \( \text{QO} \) to \( \text{S} \) such that \( \text{QO} = \text{OS} \). This makes \( \text{QS} = 2 \times \text{QO} = 2 \times 2.8 = 5.6 \) cm.
6. Join the points \( \text{QR} \), \( \text{RS} \), and \( \text{PS} \).
7. The figure \( \text{PQRS} \) is the required parallelogram.
In simple words: First draw one side. Then, find the middle point where the diagonals cross by using half of each diagonal length to make a small triangle. Extend lines from this middle point to find the other two corners.

🎯 Exam Tip: Remember that the diagonals of a parallelogram bisect each other. This means you should work with half the diagonal lengths to find the point where they cross, then extend to complete the figure.

Question 3. Construct a parallelogram ABCD whose two sides are 4.6 cm and 3 cm long and included angle is 60°.
Answer: Here are the steps to construct the parallelogram ABCD:
1. Draw a line segment \( \text{AB} = 4.6 \) cm. This will be the base of the parallelogram.
2. At point \( \text{A} \) on the line segment \( \text{AB} \), use a protractor to draw an angle of 60°. This angle will be between sides \( \text{AB} \) and \( \text{AD} \).
3. Along the 60° line from \( \text{A} \), measure and mark a point \( \text{D} \) such that \( \text{AD} = 3 \) cm. This is the second given side.
4. Now, with point \( \text{D} \) as the center, draw an arc with a radius of 4.6 cm (because \( \text{CD} \) must be equal to \( \text{AB} \)).
5. With point \( \text{B} \) as the center, draw another arc with a radius of 3 cm (because \( \text{BC} \) must be equal to \( \text{AD} \)). These two arcs will intersect at a point, which is \( \text{C} \).
6. Join \( \text{BC} \) and \( \text{CD} \).
7. The figure \( \text{ABCD} \) is the required parallelogram.
In simple words: Start by drawing the first side and making the given angle. Then draw the second side. Find the last corner by drawing arcs from the other two known corners, using the lengths of the opposite sides.

🎯 Exam Tip: When given two adjacent sides and an included angle, remember that opposite sides of a parallelogram are equal. Use this property to find the last vertex by drawing intersecting arcs.

Question 4. Construct a rectangle ABCD in which AB = 6 cm and diagonal AC =10 cm.
Answer: Here are the steps to construct the rectangle ABCD:
1. Draw a line segment \( \text{AB} = 6 \) cm.
2. At point \( \text{B} \), construct a line perpendicular to \( \text{AB} \) (making a 90° angle). This will be one of the vertical sides of the rectangle.
3. With point \( \text{A} \) as the center, draw an arc with a radius of 10 cm (the length of the diagonal \( \text{AC} \)). This arc will intersect the perpendicular line from \( \text{B} \) at point \( \text{C} \). Join \( \text{AC} \). In a rectangle, all angles are 90 degrees.
4. Now you have points \( \text{A} \), \( \text{B} \), and \( \text{C} \). With point \( \text{C} \) as the center, draw an arc with a radius of 6 cm (since \( \text{CD} \) must be equal to \( \text{AB} \)).
5. With point \( \text{A} \) as the center, draw another arc with a radius equal to the length of \( \text{BC} \). (In the right-angled triangle \( \text{ABC} \), we can find \( \text{BC} \) using Pythagoras theorem: \( \text{BC} = \sqrt{\text{AC}^2 - \text{AB}^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \) cm. So, draw an arc of 8 cm from \( \text{A} \)).
6. The arcs drawn in step 4 and step 5 will intersect at point \( \text{D} \). Join \( \text{AD} \) and \( \text{CD} \).
7. The figure \( \text{ABCD} \) is the required rectangle.
In simple words: Draw the base, then make a straight up-and-down line at one end. Use the diagonal length from the other end of the base to mark the top corner. Then, use the side lengths to find the last corner.

🎯 Exam Tip: Remember that a rectangle has all 90° angles and opposite sides are equal. Use the Pythagorean theorem for right triangles to find missing side lengths if needed.

Question 5. Construct a rhombus ABCD, such that diagonals AC = 7 cm and diagonal BD = 5 cm.
Answer: Here are the steps to construct the rhombus ABCD:
1. Draw one of the diagonals, say \( \text{AC} = 7 \) cm. This will be the base for our construction.
2. Draw the perpendicular bisector of \( \text{AC} \). This line will cut \( \text{AC} \) exactly in half at a 90° angle. Let the point where it intersects \( \text{AC} \) be \( \text{O} \). The diagonals of a rhombus always bisect each other at right angles.
3. The other diagonal, \( \text{BD} \), is 5 cm long. Since the diagonals bisect each other, \( \text{OB} = \text{OD} = \frac{5}{2} = 2.5 \) cm.
4. With point \( \text{O} \) as the center, draw an arc with a radius of 2.5 cm on both sides of \( \text{O} \) along the perpendicular bisector. Mark the points where these arcs cut the bisector as \( \text{B} \) (above \( \text{AC} \)) and \( \text{D} \) (below \( \text{AC} \)).
5. Join \( \text{AB} \), \( \text{BC} \), \( \text{CD} \), and \( \text{DA} \).
6. The figure \( \text{ABCD} \) is the required rhombus.
In simple words: Draw one diagonal first. Then, draw a straight line that cuts it exactly in half and at a perfect right angle. Use half the length of the other diagonal to mark two points on this new line. Join all four points to make your rhombus.

🎯 Exam Tip: The key properties of a rhombus are that its diagonals bisect each other at right angles. This is the most efficient way to construct a rhombus when diagonal lengths are given.

Question 6. Construct a square ABCD given that its diagonal BD = 6 cm.
Answer: Here are the steps to construct the square ABCD:
1. Draw the diagonal \( \text{BD} = 6 \) cm. This will be the first line segment.
2. Draw the perpendicular bisector of \( \text{BD} \). This line will cut \( \text{BD} \) exactly in half at point \( \text{O} \). In a square, diagonals are equal and bisect each other at 90 degrees.
3. The diagonals of a square are equal and bisect each other. So, half the length of diagonal \( \text{BD} \) is \( \frac{6}{2} = 3 \) cm. This means \( \text{AO} = \text{OC} = 3 \) cm.
4. With point \( \text{O} \) as the center, draw an arc with a radius of 3 cm on both sides of \( \text{O} \) along the perpendicular bisector. Mark the points where these arcs cut the bisector as \( \text{A} \) (above \( \text{BD} \)) and \( \text{C} \) (below \( \text{BD} \)).
5. Join \( \text{AB} \), \( \text{BC} \), \( \text{CD} \), and \( \text{DA} \).
6. The figure \( \text{ABCD} \) is the required square.
In simple words: Draw the diagonal first. Then, draw a line that cuts this diagonal in half at a right angle. Using the middle point, measure half the diagonal length along the new line to find the other two corners. Connect all four points to form the square.

🎯 Exam Tip: Remember that in a square, diagonals are equal in length, bisect each other, and intersect at a 90° angle. This allows you to construct the square by focusing on its diagonals.