RBSE Solutions Class 8 Maths Chapter 7 Construction of Quadrilaterals Exercise 7.6

Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.6

Question 1. Construct a square in which :
(i) Perimeter is 20 cm.
(ii) Sum of two adjacent sides is 9 cm.
Answer:
(i) First, we find the side length of the square. If the perimeter of a square is 20 cm, then each side is \( \frac{20}{4} = 5 \) cm. A square has all four sides equal and all angles equal to \( 90^\circ \). To construct it, we start by drawing one side, then mark right angles at its ends, and cut off the side length on those perpendicular lines. Finally, we join the top two points to complete the square.
Steps of Construction:
1. Draw a line segment AB \( = 5 \) cm.
2. At point A, construct \( \angle \text{BAX} = 90^\circ \). At point B, construct \( \angle \text{ABY} = 90^\circ \).
3. From ray AX, cut AD \( = 5 \) cm. From ray BY, cut BC \( = 5 \) cm.
4. Join points C and D.
The figure ABCD is the required square.

(ii) We are given that the sum of two adjacent sides of a square is 9 cm. Since all sides of a square are equal, each side will be half of this sum, which is \( \frac{9}{2} = 4.5 \) cm. The construction steps are similar to part (i). We begin by sketching the square to visualize the final shape before drawing it accurately.
Steps of Construction:
1. Draw a line segment AB \( = 4.5 \) cm.
2. At point A, construct \( \angle \text{BAX} = 90^\circ \). At point B, construct \( \angle \text{ABY} = 90^\circ \).
3. From ray AX, cut AD \( = 4.5 \) cm. From ray BY, cut BC \( = 4.5 \) cm.
4. Join points C and D.
The figure ABCD is the required square.

In simple words: For a square, first find the side length using the given information (perimeter or sum of sides). Then, draw one side, make 90-degree angles at both ends, measure the side length along these new lines, and connect the last two points to finish the square.

🎯 Exam Tip: Remember that all sides of a square are equal and all its internal angles are \( 90^\circ \). A good rough sketch helps in planning the construction steps accurately.

Question 2. Construct a square PQRS if the length of half a diagonal is 2.8 cm.
Answer: First, we need to draw a rough sketch to plan the construction. A square's diagonals bisect each other at \( 90^\circ \) and are equal in length. If half a diagonal (PO) is \( 2.8 \) cm, then the full diagonal (PR) is \( 2 \times 2.8 = 5.6 \) cm. This property is key to building the square using its diagonals.
Steps of Construction:
1. Draw a line segment PR \( = 5.6 \) cm (representing a diagonal).
2. Draw the perpendicular bisector XY of PR, which passes through the midpoint O.
3. Taking O as the center, draw arcs with a radius of \( 2.8 \) cm (half diagonal length) on both sides of PR. These arcs will cut the perpendicular bisector XY at points Q and S.
4. Join points PQ, QR, RS, and SP.
The figure PQRS is the required square.

In simple words: To draw a square using its diagonal, first find the full diagonal length. Then draw that diagonal and its perpendicular bisector. Mark points on the bisector using half the diagonal length. Connect these points to form the square.

🎯 Exam Tip: Remember that the diagonals of a square are equal in length and they cut each other in half at a 90-degree angle. This property is crucial for constructing a square when diagonal information is provided.

Question 3. Construct a rectangle ABCD where length AB = 8 cm and breadth AD = 6 cm.
Answer: We need to construct a rectangle with a given length and breadth. First, a rough sketch helps to visualize the final shape. Rectangles have opposite sides equal and parallel, and all internal angles are \( 90^\circ \). So, we will use these properties for construction.
Steps of Construction:
1. Draw a line segment AB \( = 8 \) cm.
2. At point A, construct a ray AX such that \( \angle \text{BAX} = 90^\circ \).
3. From ray AX, cut a line segment AD \( = 6 \) cm.
4. At point B, construct a ray BY such that \( \angle \text{ABY} = 90^\circ \). From ray BY, cut a line segment BC \( = 6 \) cm.
5. Join points C and D.
The figure ABCD is the required rectangle.

In simple words: To make a rectangle with given length and width, first draw the length. Then, draw 90-degree lines upwards from both ends. Measure the width along these upward lines, marking the top two points. Finally, join these top points to complete the rectangle.

🎯 Exam Tip: Always construct angles accurately using a protractor or compass for correct \( 90^\circ \) corners in a rectangle. Ensure opposite sides are measured to be equal.

Question 4. Construct a rectangle PQRS while PQ = 5.5 cm. and diagonal QS = 6.2 cm.
Answer: We need to construct a rectangle given one side and a diagonal. First, draw a rough sketch to understand the shape and dimensions. In a rectangle, each angle is \( 90^\circ \), opposite sides are equal, and diagonals are also equal. This means we can use Pythagoras theorem if needed to find the other side or simply use arcs for precise construction. The diagonal QS will help us fix the position of point S.
Steps of Construction:
1. Draw a line segment PQ \( = 5.5 \) cm.
2. At point P, construct a ray PX such that \( \angle \text{QPX} = 90^\circ \).
3. At point Q, construct a ray QY such that \( \angle \text{PQY} = 90^\circ \).
4. With Q as the center and radius \( 6.2 \) cm (diagonal QS), draw an arc that intersects ray PX at point S.
5. With S as the center and radius \( 5.5 \) cm (opposite side SR = PQ), draw an arc. With P as the center and radius equal to QS, draw an arc to intersect the previous arc at R. (This step in source is `At point P, draw an arc of radius 6.2 cm. intersecting ray QY at point R. Join PR. At point Q, draw an arc of radius 6.2 cm. intersecting ray PX at S. Join QS and SR.`. This is slightly confusing. Let's use standard method.)
Simplified (and corrected) construction for R:
5. From S, draw an arc of radius equal to PQ (5.5 cm). From Q, draw an arc of radius equal to PS (which can be found by Pythagoras from PQS, or by making SR parallel to PQ). For simplicity, since PR = QS = 6.2 cm, use P as center, 6.2 cm radius, intersect QY at R.
6. Join PS, SR, and QR.
The figure PQRS is the required rectangle.

In simple words: Draw the given side of the rectangle. From one end, draw a 90-degree line. From the other end of the side, use the diagonal length to cut an arc on the 90-degree line, which helps find the corner point. Then use side lengths to find the final corner.

🎯 Exam Tip: When given a side and a diagonal of a rectangle, you can use the Pythagorean theorem to find the other side if needed, or rely on arc intersections from the corners and the diagonal length for direct construction.

Question 5. Construct a rectangle EFGH while EF = 4.0 cm. and diagonal EG = 5.0 cm.
Answer: Similar to the previous question, we construct a rectangle using one side and a diagonal. The first step is to draw a rough sketch to guide our actual construction. In a rectangle, all angles are \( 90^\circ \), and the diagonals are equal. Using the diagonal length and the given side, we can find the third point. By the Pythagorean theorem, the other side of the rectangle (EH) would be \( \sqrt{EG^2 - EF^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \) cm. This can also be done using arcs.
Steps of Construction:
1. Draw a line segment EF \( = 4.0 \) cm.
2. At point E, construct a ray EX such that \( \angle \text{FEX} = 90^\circ \).
3. At point F, construct a ray FY such that \( \angle \text{EFY} = 90^\circ \).
4. With E as the center and radius \( 5.0 \) cm (diagonal EG), draw an arc that intersects ray FY at point G. Join EG.
5. From G as the center, draw an arc of radius \( 4.0 \) cm. From E as the center, draw an arc of radius \( 3.0 \) cm (since \( \text{EH} = \sqrt{5^2 - 4^2} = 3 \) cm). These arcs will intersect at point H. Join EH and GH.
The figure EFGH is the required rectangle.

In simple words: For a rectangle with a side and a diagonal, first draw the side. Then, draw a 90-degree line from one end. From the other end of the side, draw an arc with the length of the diagonal to find the opposite corner. Use arc intersections to find the last corner, remembering that opposite sides are equal.

🎯 Exam Tip: When constructing rectangles with a diagonal, consider using the Pythagorean theorem to calculate the missing side if it simplifies the construction steps for you.

Question 6. Construct a rhombus DEFG while FG = 4.8 cm and diagonal EG = 3.4 cm.
Answer: We need to construct a rhombus given its side length and one diagonal. A rhombus has all four sides equal in length. Therefore, if FG = 4.8 cm, then DE, EF, and GD are also 4.8 cm. A rough sketch helps visualize the shape before starting the precise construction.
Steps of Construction:
1. Draw the diagonal EG \( = 3.4 \) cm.
2. With E as the center and radius \( 4.8 \) cm (side length), draw an arc above EG. With G as the center and radius \( 4.8 \) cm, draw another arc that intersects the previous arc at point D.
3. Now, with E as the center and radius \( 4.8 \) cm, draw an arc below EG. With G as the center and radius \( 4.8 \) cm, draw another arc that intersects the previous arc at point F.
4. Join points ED, GD, GF, and EF.
The figure DEFG is the required rhombus.

In simple words: Draw the given diagonal first. Since all sides of a rhombus are equal, use the side length to draw arcs from both ends of the diagonal, both above and below it. These arcs will meet to give you the other two corners. Connect all points to make the rhombus.

🎯 Exam Tip: Remember that a rhombus has four equal sides. This means you only need the length of one side and a diagonal to construct the entire shape using arcs.

Question 7. Construct a rhombus ABCD while BC = 4.0 cm. and \( \angle \text{B} = 75^\circ \).
Answer: We need to construct a rhombus given its side length and one angle. All sides of a rhombus are equal, so AB, CD, and DA will also be 4.0 cm. Since a rhombus is a type of parallelogram, adjacent angles add up to \( 180^\circ \). Therefore, \( \angle \text{C} = 180^\circ - \angle \text{B} = 180^\circ - 75^\circ = 105^\circ \). This helps us determine all angles for the construction.
Steps of Construction:
1. Draw a line segment BC \( = 4.0 \) cm.
2. At point B, construct a ray BX such that \( \angle \text{CBX} = 75^\circ \).
3. At point C, construct a ray CY such that \( \angle \text{BCY} = 105^\circ \).
4. With B as the center, cut an arc of \( 4.0 \) cm on ray BX and mark it as point A.
5. With C as the center, cut an arc of \( 4.0 \) cm on ray CY and mark it as point D.
6. Join points A and D.
The figure ABCD is the required rhombus.

In simple words: For a rhombus with one side and one angle, draw the given side. Then, draw the given angle at one end and the calculated adjacent angle (180 minus the given angle) at the other end. Mark the side length on both these new lines. Finally, connect the two marked points to complete the rhombus.

🎯 Exam Tip: When constructing a rhombus using an angle, remember that adjacent angles in a rhombus (and parallelogram) are supplementary (add up to \( 180^\circ \)), and all four sides are equal.