Q3: In each of these cases, observe 1. how the length XY compares to that of AB and 2. the shape of the 4-sided figure ABYX.

Observation: Based on the results from the previous question, where the distance of X from A is the same as the distance of Y from B (i.e., AX = BY). 1. Comparison of lengths: In all cases where AX = BY, the measured length of XY is 7 cm. The length of AB is also 7 cm. Therefore, the length of XY…

Playing with Constructions Class 6 Mathematics NCERT Solutions

Q1Section 8.1 Questions

Imagine marking all the points of 4 cm distance from the point P. How would they look?

Solution

Observation: If we mark a point P on a sheet of paper and then mark all the points that are 4 cm away from P in all directions, these points will form a path.

Conclusion: When we join all these points, they form a continuous closed curve. This curve is a circle with its center at point P and a radius of 4 cm.

Final Answer: The points form a circle.

Q2Section 8.1 Questions

For the 'Wavy Wave' figure, what radius should be taken in the compass to get this half circle? What should be the length of AX?

Solution

Given: A wavy wave figure is constructed on a central line AB of length 8 cm. The first wave is a half-circle on a segment AX.

Analysis: The figure shows two identical half-circles making up the wave along the 8 cm line. This means the diameter of each half-circle is half the total length of the line AB. Diameter of the half-circle = Length of AX =

\frac{\text{Length of AB}}{2} = \frac{8 \text{ cm}}{2} = 4 \text{ cm}

. The radius of a circle (or half-circle) is half of its diameter. Radius =

\frac{\text{Diameter}}{2} = \frac{4 \text{ cm}}{2} = 2 \text{ cm}

.

Final Answer: The radius to be taken in the compass is 2 cm. The length of AX should be 4 cm.

Q3Section 8.1 Questions

Take a central line of a different length and try to draw the wave on it.

Solution

To Construct: A wavy wave pattern on a central line of a chosen length.

Let: The length of the central line be 10 cm.

Steps of Construction:

Draw a line segment PQ of length 10 cm.
Find the midpoint of PQ, let it be M. The length of PM will be 5 cm.
To draw the first wave (a semi-circle) above the line, place the compass point at the midpoint of PM (which is 2.5 cm from P) and set the radius to 2.5 cm. Draw a semi-circle from P to M.
To draw the second wave (a semi-circle) below the line, place the compass point at the midpoint of MQ (which is 2.5 cm from M) and set the radius to 2.5 cm. Draw a semi-circle from M to Q on the opposite side of the line.

Result: A wavy wave pattern is constructed on the 10 cm line segment.

Q4Section 8.1 Questions

Try to recreate the figure where the waves are smaller than a half circle (as appearing in the neck of the figure, 'A Person'). The challenge here is to get both the waves to be identical. This may be tricky!

Solution

To Construct: Two identical arcs (waves smaller than a half-circle) on a line segment.

Steps of Construction:

Draw a horizontal line segment AB.
To create the top arc, choose a point P below the line segment AB. Place the compass point at P and choose a radius greater than the distance from P to the line. Draw an arc that intersects the line AB at two points.
To create an identical bottom arc, we need to find a corresponding center point. Draw a perpendicular line from P to the line segment AB. Extend this perpendicular line upwards. Mark a point Q on this extended line such that the distance of Q from AB is the same as the distance of P from AB.
With Q as the center and the same radius used in step 2, draw another arc on the opposite side of the line AB.

Result: Two identical waves, smaller than a half-circle, are constructed symmetrically about the line segment AB.

Q1Section 8.2 Questions

Which of the following is not a name for this square? PQSR SPQR RSPQ QRSP

Solution

Rule for Naming Polygons: The vertices of a polygon must be named in a consecutive order, either clockwise or counter-clockwise, starting from any vertex.

Analysis of Options: Let the vertices of the square be P, Q, R, and S in a clockwise order.

PQSR: This name goes from P to Q, then jumps across the diagonal to S, and then to R. This is not a consecutive order. Therefore, it is not a valid name.
SPQR: This name starts at S and moves consecutively to P, Q, and R. This is a valid name (counter-clockwise).
RSPQ: This name starts at R and moves consecutively to S, P, and Q. This is a valid name (counter-clockwise).
QRSP: This name starts at Q and moves consecutively to R, S, and P. This is a valid name (clockwise).

Final Answer: PQSR is not a name for the given square.

Q2Section 8.2 Questions

Draw the rectangle and four squares configuration (shown in Fig. 8.3) on a dot paper. What did you do to recreate this figure so that the four squares are placed symmetrically around the rectangle? Discuss with your classmates.

Solution

To Construct: A central rectangle with four squares placed symmetrically around it on a dot paper.

Method:

First, draw the central rectangle by connecting dots on the dot paper.
To ensure symmetry, the distance from each corner of the rectangle to the nearest corner of the adjacent square must be the same.
A simple way to achieve this is to leave a uniform gap. For example, from each corner of the rectangle, move one dot unit diagonally outwards. This point will be the innermost corner of each of the four squares.
From these starting corners, draw the four squares, ensuring their sides are parallel to the diagonals of the rectangle or aligned with the grid in a rotated manner.

Final Answer: To place the four squares symmetrically, I drew the central rectangle first and then positioned the squares by leaving a uniform distance (e.g., one dot diagonally) from each corner of the rectangle.

Q3Section 8.2 Questions

Identify if there are any squares in this collection. Use measurements if needed.

Solution

Given: A collection of four-sided figures (quadrilaterals) labeled A, B, C, D.

To Find: Which of the figures is a square.

Properties of a Square:

All four sides are equal in length.
All four angles are right angles ( $90^\circ$ ).

Analysis: By observing or measuring the figures:

Figure A has all sides of equal length and all angles appear to be $90^\circ$ .
Figure B is a rectangle, as its opposite sides are equal but adjacent sides are not.
Figure C is a parallelogram, as its opposite sides are parallel but its angles are not $90^\circ$ .
Figure D is a general quadrilateral.

Final Answer: Figure A is a square.

Q4Section 8.2 Questions

Is it possible to reason out if the sides are equal or not, and if the angles are right or not without using any measuring instruments in the above figure? Can we do this by only looking at the position of corners in the dot grid?

Solution

Answer: Yes, it is possible.

Reasoning:

Side Lengths: On a dot grid, the distance between adjacent dots (horizontally or vertically) can be considered as one unit. For diagonal lines, we can compare their lengths by considering the 'rise' and 'run' between their endpoints. For example, a line segment that goes 3 dots across and 4 dots up has the same length as another segment with the same rise and run. We can use this to check if all sides are equal.
Right Angles: We can check for right angles by observing the slopes of the lines. A horizontal line and a vertical line are perpendicular. For rotated figures, if a line has a 'rise' of 'a' and a 'run' of 'b', a line perpendicular to it will have a 'rise' of 'b' and a 'run' of '-a' (or vice versa). By counting the dots, we can determine if the lines forming the corners are perpendicular, thus forming right angles.

Conclusion: Yes, by analyzing the positions of the corners on the dot grid, we can determine the equality of side lengths and the presence of right angles without direct measurement.

Q5Section 8.2 Questions

Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that their corners are on the dots. Verify if the squares and rectangles that you have drawn satisfy their respective properties.

Solution

To Construct: Three rotated squares and rectangles on a dot grid.

Steps for a Rotated Square:

Pick a starting dot, A.
To draw a side AB, move from A a certain number of units horizontally (run) and a certain number of units vertically (rise). For example, move 2 dots right and 1 dot up to reach point B.
To draw the next side BC, which must be perpendicular and equal in length to AB, start from B and apply a perpendicular movement. The original movement was (run=2, rise=1). The perpendicular movement will be (run=-1, rise=2) or (run=1, rise=-2). Let's move 1 dot left and 2 dots up to reach point C.
From C, move 2 dots left and 1 dot down to reach point D.
Join D back to A. This forms a rotated square.

Steps for a Rotated Rectangle:

Follow steps 1 and 2 from above to create a side AB (e.g., run=2, rise=1).
For the adjacent side BC, choose a different perpendicular movement. For example, from B move 2 dots left and 4 dots up to reach C (run=-2, rise=4).
From C, move 2 dots left and 1 dot down to reach D.
Join D to A. This forms a rotated rectangle.

Verification:

Square: Verify that all sides have the same 'rise' and 'run' components (e.g., all are formed by a (2,1) type movement), confirming equal length. Verify that adjacent sides have perpendicular 'rise' and 'run' components (e.g., (2,1) and (-1,2)), confirming right angles.
Rectangle: Verify that opposite sides have the same 'rise' and 'run', confirming they are equal and parallel. Verify that adjacent sides have perpendicular 'rise' and 'run' components, confirming right angles.

Q1Section 8.3 Construct

Draw a rectangle with sides of length 4 cm and 6 cm. After drawing, check if it satisfies both the rectangle properties.

Solution

To Construct: A rectangle ABCD with sides AB = 4 cm and BC = 6 cm.

Steps of Construction:

Draw a line segment AB of length 4 cm.
At point B, construct a perpendicular line to AB using a protractor or compass. Let this line be BX.
With B as the center and a radius of 6 cm, draw an arc that intersects the line BX at point C.
At point A, construct another perpendicular line to AB. Let this line be AY.
With A as the center and a radius of 6 cm, draw an arc that intersects the line AY at point D.
Join points C and D.

Verification:

Opposite sides equal: Measure the sides. We find AB = CD = 4 cm and BC = AD = 6 cm. The property is satisfied.
All angles are $90^\circ$ : Measure the angles using a protractor. We find $\angle A = \angle B = \angle C = \angle D = 90^\circ$ . The property is satisfied.

Result: The constructed figure ABCD is a rectangle with the given side lengths.

Q2Section 8.3 Construct

Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if it satisfies both the rectangle properties.

Solution

To Construct: A rectangle PQRS with sides PQ = 10 cm and PS = 2 cm.

Steps of Construction:

Draw a line segment PQ of length 10 cm.
At point P, construct a perpendicular line to PQ. Let this line be PX.
With P as the center and a radius of 2 cm, draw an arc that intersects the line PX at point S.
At point Q, construct another perpendicular line to PQ. Let this line be QY.
With Q as the center and a radius of 2 cm, draw an arc that intersects the line QY at point R.
Join points S and R.

Verification:

Opposite sides equal: Measure the sides. We find PQ = SR = 10 cm and PS = QR = 2 cm. The property is satisfied.
All angles are $90^\circ$ : Measure the angles using a protractor. We find $\angle P = \angle Q = \angle R = \angle S = 90^\circ$ . The property is satisfied.

Result: The constructed figure PQRS is a rectangle with the given side lengths.

Q3Section 8.3 Construct

Is it possible to construct a 4-sided figure in which- all the angles are equal to $90^\circ$ but opposite sides are not equal?

Solution

Answer: No, it is not possible.

Reasoning: Let the 4-sided figure be ABCD with

\angle A = \angle B = \angle C = \angle D = 90^\circ

.

Since $\angle A = \angle B = 90^\circ$ , the sum of these adjacent interior angles is $180^\circ$ . This implies that the sides AD and BC are parallel.
Similarly, since $\angle B = \angle C = 90^\circ$ , the sum of these adjacent interior angles is $180^\circ$ . This implies that the sides AB and DC are parallel.
A quadrilateral with both pairs of opposite sides parallel is a parallelogram.
A key property of a parallelogram is that its opposite sides are equal in length.
Therefore, if a 4-sided figure has all angles equal to $90^\circ$ , it must be a parallelogram (specifically, a rectangle), and its opposite sides must be equal.
It is impossible for it to have opposite sides that are not equal.

Final Answer: No.

Q1Section 8.4 An Exploration in Rectangles

Is there a shorthand way of writing it down? In all the sentences, only the position of X, Y and the length XY changes. So we could write this as:
Distance of X from A Distance of Y from B Length of XY

Solution

Given: A rectangle ABCD with AB = 7 cm and BC = 4 cm. Point X is on AD and point Y is on BC.

Measurements: The table is filled by constructing the rectangle, placing points X and Y as described, and measuring the distance XY.

Final Answer:

Distance of X from A	Distance of Y from B	Length of XY
5 mm	3 cm	7.4 cm
1 cm	1 cm	7 cm
2 cm	4 cm	7.3 cm

Q2Section 8.4 An Exploration in Rectangles

Have you checked what happens to the length XY when X and Y are placed at the same distance away from A and B, respectively? For example, as in the cases like these:
Distance of X from A Distance of Y from B Length of XY
5 mm 5 mm
1 cm 1 cm
1 cm 5 mm 1 cm 5 mm

Distance of X from A	Distance of Y from B	Length of XY
5 mm	5 mm
1 cm	1 cm
1 cm 5 mm	1 cm 5 mm

Solution

Given: A rectangle ABCD with AB = 7 cm and BC = 4 cm. Point X is on AD and point Y is on BC. The distance AX is equal to the distance BY.

Measurements: The table is filled by placing X and Y at equal distances from A and B respectively and measuring XY.

Final Answer:

Distance of X from A	Distance of Y from B	Length of XY
5 mm	5 mm	7 cm
1 cm	1 cm	7 cm
1 cm 5 mm	1 cm 5 mm	7 cm

Q3Section 8.4 An Exploration in Rectangles

In each of these cases, observe how the length XY compares to that of AB and the shape of the 4-sided figure ABYX.

Solution

Observation: Based on the results from the previous question, where the distance of X from A is the same as the distance of Y from B (i.e., AX = BY).

Comparison of lengths: In all cases where AX = BY, the measured length of XY is 7 cm. The length of AB is also 7 cm. Therefore, the length of XY is equal to the length of AB.
Shape of ABYX: The figure ABYX has sides AB and XY which are parallel and equal. The sides AX and BY are also parallel (since they are parts of the parallel sides of the rectangle) and are set to be equal in length. The angles at A and B are $90^\circ$ . This means the angles at Y and X will also be $90^\circ$ . A quadrilateral with all angles equal to $90^\circ$ is a rectangle.

Final Answer:

The length of XY is equal to the length of AB (XY = AB).
The 4-sided figure ABYX is a rectangle.

Q4Section 8.4 An Exploration in Rectangles

How does the farthest distance between X and Y compare with the length of AC? BD?

Solution

Analysis:

The points X and Y are on opposite sides AD and BC of the rectangle ABCD.
The distance between X and Y will be maximized when they are at opposite corners of the rectangle.
There are two such cases:
1. When X is at point A and Y is at point C. The distance is the length of the diagonal AC.
2. When X is at point D and Y is at point B. The distance is the length of the diagonal BD.
In a rectangle, the diagonals are equal in length (AC = BD).

Conclusion: The farthest distance between points X and Y is achieved when the line segment XY is a diagonal of the rectangle.

Final Answer: The farthest distance between X and Y is equal to the length of the diagonal AC or BD.

Q1Section 8.4 Construct

Construct a rectangle that can be divided into 3 identical squares.

Solution

To Construct: A rectangle that can be divided into 3 identical squares.

Analysis: If a rectangle is composed of 3 identical squares placed side-by-side, its length will be three times the side length of the square, and its breadth will be equal to the side length of the square. So, the ratio of length to breadth must be 3:1.

Let: The side of each square be 2 cm. Then, the breadth of the rectangle = 2 cm. The length of the rectangle = 3

\times

2 cm = 6 cm.

Steps of Construction:

Draw a line segment AB of length 6 cm.
At A and B, construct perpendiculars AY and BZ to AB.
With A as the center and radius 2 cm, draw an arc to cut AY at D.
With B as the center and radius 2 cm, draw an arc to cut BZ at C.
Join DC. ABCD is the required rectangle.
To show the division, mark points E and F on AB such that AE = EF = FB = 2 cm. Draw perpendiculars from E and F to the side DC.

Result: A rectangle of size 6 cm by 2 cm is constructed, which is divisible into 3 identical squares of side 2 cm.

Q2Section 8.4 Construct

Give the lengths of the sides of a rectangle that cannot be divided into— two identical squares; three identical squares.

Solution

Reasoning:

For a rectangle to be divided into 'n' identical squares, the ratio of its length to its breadth must be n:1 (or 1:n).
If the ratio of the sides is not n:1 for an integer n, it cannot be divided into 'n' identical squares.

Examples:

Cannot be divided into two identical squares: The ratio of length to breadth should not be 2:1. For example, a rectangle with sides of length = 4 cm and breadth = 2.5 cm. The ratio is 4:2.5 or 8:5, which is not 2:1.
Cannot be divided into three identical squares: The ratio of length to breadth should not be 3:1. For example, a rectangle with sides of length = 7 cm and breadth = 2 cm. The ratio is 7:2, which is not 3:1.

Final Answer:

For two squares: Any rectangle where Length $\neq$ 2 $\times$ Breadth, e.g., 4 cm and 2.5 cm.
For three squares: Any rectangle where Length $\neq$ 3 $\times$ Breadth, e.g., 7 cm and 2 cm.

Q3Section 8.4 Construct

A Square within a Rectangle: Construct a rectangle of sides 8 cm and 4 cm. How will you construct a square inside, as shown in the figure, such that the centre of the square is the same as the centre of the rectangle?

Solution

To Construct: A rectangle of 8 cm by 4 cm with a concentric square inside.

Analysis:

The center of the rectangle is the intersection of its diagonals. Let this be O.
The square must also be centered at O.
Let the side of the inner square be 's'. The distance from the center O to each side of the square will be s/2.
The distance from the center O to the longer side of the rectangle is 4/2 = 2 cm. The distance from O to the shorter side is 8/2 = 4 cm.
The square must fit within the rectangle. The maximum side length of the square would be limited by the shorter side of the rectangle, which is 4 cm. Let's choose a side length for the square, for example, 2 cm.
The distance between the parallel sides of the square and rectangle will be uniform. Distance vertically = (Breadth of rectangle - Side of square)/2 = (4-2)/2 = 1 cm. Distance horizontally = (Length of rectangle - Side of square)/2 = (8-2)/2 = 3 cm.

Steps of Construction:

Construct the rectangle ABCD with AB = 8 cm and BC = 4 cm.
Draw the diagonals AC and BD to find the center O.
Draw lines through O parallel to the sides of the rectangle.
To construct a square of side 2 cm, measure 1 cm from O along the vertical line in both directions, and 1 cm from O along the horizontal line in both directions. These four points will be the midpoints of the sides of the square.
Draw lines through these points parallel to the axes to form the square PQRS.

Result: A square PQRS is constructed inside the rectangle ABCD, and both have the same center.

Q4Section 8.4 Construct

Falling Squares: Construct the figure.

Solution

To Construct: A sequence of three identical squares, where each is rotated and touches the corner of the previous one.

Let: The side of the square be 3 cm.

Steps of Construction:

First Square: Draw a square ABCD with side 3 cm, with side AB being horizontal.
Second Square: The second square touches the first one at corner C. One of its sides should pass through C. To align it as shown in the textbook, we can draw a square BEFC such that it shares the side BC with the first square and is below it. Then, rotate this new square by $45^\circ$ around point C. A simpler way is to extend the side DC to the right. Place the corner of the second square at C. Draw the sides of the second square such that they make a $45^\circ$ angle with the sides of the first square.
Third Square: The third square touches the second one at its bottom-right corner. Repeat the process from step 2, placing the third square's top-left corner at the bottom-right corner of the second square, and rotating it similarly.

Alternative Method using a Grid:

Draw the first square aligned with the grid lines.
For the second square, start at the bottom-right corner of the first square. Draw its sides diagonally on the grid (e.g., side length corresponding to a diagonal of a smaller grid square).
For the third square, start at the bottom-right corner of the second (rotated) square. Draw its sides aligned with the grid lines again.

Result: A chain of three falling squares is constructed.

Q5Section 8.4 Construct

Shadings: Construct this. Choose measurements of your choice. Note that the larger 4-sided figure is a square and so are the smaller ones.

Solution

To Construct: A large square divided into smaller squares and triangles for a shaded pattern.

Let: The side of the large square be 8 cm.

Steps of Construction:

Draw a large square ABCD with a side length of 8 cm.
Find the midpoints of the four sides AB, BC, CD, and DA. Let them be P, Q, R, and S respectively.
Join the opposite midpoints, i.e., join PR and SQ. These lines will intersect at the center of the square, O, and divide the large square into four smaller identical squares, each with a side length of 4 cm.
Let's consider one of the smaller squares, say APO S. Draw its diagonal AS.
Shade the triangle APS.
Repeat this for the other three small squares, always drawing the diagonal that does not pass through the center O of the large square, and shading the resulting outer triangle.

Result: The required shaded pattern is constructed within the large square.

Q6Section 8.4 Construct

Square with a Hole: Construct this.

Solution

To Construct: A square with a concentric circular hole.

Let: The side of the square be 6 cm and the radius of the circle be 2 cm.

Steps of Construction:

Draw a square ABCD with a side length of 6 cm.
To find the center of the square, draw its two diagonals, AC and BD. The point where they intersect, let's call it O, is the center.
The center of the circular hole is the same as the center of the square, O.
With O as the center and a radius of 2 cm, use a compass to draw a circle.

Result: A square with a concentric circular hole is constructed.

Q7Section 8.4 Construct

Square with more Holes: Construct this.

Solution

To Construct: A square with one central hole and four smaller holes in the corners.

Let: The side of the large square be 8 cm.

Steps of Construction:

Draw a large square ABCD with a side length of 8 cm.
Divide the square into four smaller identical squares of side 4 cm by joining the midpoints of opposite sides. Let the center of the large square be O.
Central Hole: Draw a circle with center O and a suitable radius, for example, 1 cm.
Corner Holes: Find the center of each of the four smaller squares. These centers will be the centers for the four smaller circular holes.
With each of these four centers, draw identical circles with a suitable radius, for example, 1 cm.

Result: A square with one central hole and four symmetrically placed smaller holes is constructed.

Q8Section 8.4 Construct

Square with Curves: This is a square with 8 cm sidelengths.

Solution

To Construct: A square of side 8 cm with four inward-bulging arcs on each side.

Analysis: For the four arcs to bulge uniformly from each side, the center for drawing these arcs must be the center of the square itself.

Steps of Construction:

Draw a square ABCD with a side length of 8 cm.
Find the center of the square, O, by drawing the diagonals.
To draw the arc on side AB, you need to find the correct radius. The arc should appear to bulge from the side. This suggests the compass tip should be at the center O.
Place the compass tip at the center O. Set the radius of the compass equal to the distance from the center to the midpoint of a side. The distance from O to the midpoint of AB is half the side length, which is 4 cm.
With O as the center and radius 4 cm, draw four arcs. However, this would create a single circle inside the square. The figure shows arcs that seem to connect the midpoints of adjacent sides.

Correct Interpretation of the Figure: The arcs connect the midpoints of the sides.

Draw a square ABCD with a side length of 8 cm.
Find the midpoints of the sides AB, BC, CD, and DA. Let them be P, Q, R, and S respectively.
To draw the arc connecting P and Q, the center of the arc must be the common corner B. With B as the center and radius BP = BQ = 4 cm, draw an arc PQ.
Similarly, with C as the center and radius CQ = CR = 4 cm, draw arc QR.
With D as the center and radius DR = DS = 4 cm, draw arc RS.
With A as the center and radius AS = AP = 4 cm, draw arc SP.

Result: The required figure of a square with four internal curves is constructed.

Q1Section 8.5 Questions

How should the rectangle be constructed so that the diagonal divides the opposite angles into equal parts?

Solution

Analysis:

Let ABCD be a rectangle. A diagonal, say AC, divides the opposite angles $\angle A$ and $\angle C$ .
In a rectangle, all angles are $90^\circ$ . So, $\angle A = \angle C = 90^\circ$ .
If the diagonal AC divides these angles into equal parts, then $\angle DAC = \angle BAC = 45^\circ$ and $\angle DCA = \angle BCA = 45^\circ$ .
Consider the triangle $\triangle ABC$ . We have $\angle B = 90^\circ$ and $\angle BCA = 45^\circ$ . The sum of angles in a triangle is $180^\circ$ , so $\angle BAC = 180^\circ - 90^\circ - 45^\circ = 45^\circ$ .
In $\triangle ABC$ , since $\angle BAC = \angle BCA = 45^\circ$ , the sides opposite to these angles must be equal. Therefore, BC = AB.
A rectangle in which adjacent sides are equal is a square.

Final Answer: The rectangle should be constructed as a square (i.e., with all four sides equal).

Q2Section 8.5 Questions

Construct a rectangle in which one of the diagonals divides the opposite angles into $60^\circ$ and $30^\circ$ .

Solution

To Construct: A rectangle ABCD where diagonal AC divides

\angle A

into

30^\circ

and

60^\circ

.

Steps of Construction:

Draw a line segment AB of any arbitrary length.
At point A, construct a ray AX such that $\angle BAX = 60^\circ$ .
At point A, construct a perpendicular line segment AD to AB. The point D lies on this perpendicular.
We know that the angle at A is $90^\circ$ . Since $\angle BAX = 60^\circ$ , then $\angle DAX$ must be $30^\circ$ . This means the diagonal lies on the ray AX.
At point B, construct a perpendicular line to AB. This line will contain the point C.
The point C is the intersection of the ray AX and the perpendicular line from B.
From point C, draw a line parallel to AB, or from A draw a line parallel to BC to find the point D.
Join CD and AD to complete the rectangle ABCD.

Result: The rectangle ABCD is constructed where the diagonal AC divides

\angle A

into

60^\circ

and

30^\circ

.

Q3Section 8.5 Questions

Construct a rectangle where one of its sides is 5 cm and the length of a diagonal is 7 cm.

Solution

To Construct: A rectangle with a side of 5 cm and a diagonal of 7 cm.

Let: The side CD = 5 cm and the diagonal BD = 7 cm.

Steps of Construction:

Draw a line segment CD of length 5 cm.
At point C, construct a perpendicular line CX. The vertex B will lie on this line.
With D as the center and a radius of 7 cm (the length of the diagonal), draw an arc that intersects the line CX.
The point of intersection is the vertex B.
Now, we have three vertices: C, D, and B. To find the fourth vertex A, we can use the properties of a rectangle.
With B as the center and a radius of 5 cm (equal to CD), draw an arc.
With D as the center and a radius equal to BC (measure it), draw another arc to intersect the previous one at point A.
Join BA and DA.

Result: ABCD is the required rectangle.

Q4Section 8.5 Questions

Construct a rectangle in which one of the diagonals divides the opposite angles into $50^\circ$ and $40^\circ$ .

Solution

To Construct: A rectangle ABCD where diagonal AC divides

\angle A

into

50^\circ

and

40^\circ

.

Steps of Construction:

Draw a line segment AB of any length.
At point A, construct a ray AX such that $\angle BAX = 50^\circ$ . This ray represents the diagonal.
At point B, construct a perpendicular line BY to AB. This line will intersect the ray AX at point C.
Now we have the triangle ABC, where $\angle B = 90^\circ$ and $\angle BAC = 50^\circ$ . It follows that $\angle BCA = 40^\circ$ .
To complete the rectangle, draw a line from C parallel to AB and a line from A parallel to BC. The intersection of these lines is point D.
Alternatively, at A, draw a perpendicular AD to AB. From C, draw a line segment CD of length equal to AB and parallel to AB.

Result: The rectangle ABCD is constructed. The diagonal AC divides

\angle A

into

50^\circ

and

\angle C

into

40^\circ

and

50^\circ

.

Q5Section 8.5 Questions

Construct a rectangle in which one of the diagonals divides the opposite angles into $45^\circ$ and $45^\circ$ . What do you observe about the sides?

Solution

To Construct: A rectangle where a diagonal divides the angles into

45^\circ

and

45^\circ

.

Steps of Construction:

Draw a line segment AB of any length, say 5 cm.
At point A, construct a ray AX such that $\angle BAX = 45^\circ$ .
At point B, construct a perpendicular line BY to AB. This line intersects AX at C.
In $\triangle ABC$ , $\angle B = 90^\circ$ and $\angle BAC = 45^\circ$ . Therefore, $\angle BCA = 180^\circ - 90^\circ - 45^\circ = 45^\circ$ .
Since $\angle BAC = \angle BCA$ , the sides opposite to them are equal, so AB = BC.
Complete the figure by drawing AD perpendicular to AB with AD = BC, and then joining CD.

Observation: Since adjacent sides AB and BC are equal, and it is a rectangle, all four sides must be equal. A rectangle with all sides equal is a square.

Final Answer: The resulting figure is a square. I observe that all its sides are equal in length.

Q6Section 8.5 Questions

Construct a rectangle one of whose sides is 4 cm and the diagonal is of length 8 cm.

Solution

To Construct: A rectangle with a side of 4 cm and a diagonal of 8 cm.

Let: The side AB = 4 cm and the diagonal AC = 8 cm.

Steps of Construction:

Draw a line segment AB of length 4 cm.
At point B, construct a perpendicular line BX.
With A as the center and a radius of 8 cm (the length of the diagonal), draw an arc that intersects the line BX.
The point of intersection is the vertex C.
With C as the center and a radius of 4 cm (equal to AB), draw an arc.
With A as the center and a radius equal to BC (measure it), draw another arc to intersect the previous one at point D.
Join AD and CD.

Result: ABCD is the required rectangle.

Q7Section 8.5 Questions

Construct a rectangle one of whose sides is 3 cm and the diagonal is of length 7 cm.

Solution

To Construct: A rectangle with a side of 3 cm and a diagonal of 7 cm.

Let: The side AB = 3 cm and the diagonal AC = 7 cm.

Steps of Construction:

Draw a line segment AB of length 3 cm.
At point B, construct a perpendicular line BX.
With A as the center and a radius of 7 cm, draw an arc that intersects the line BX at point C.
With C as the center and a radius of 3 cm, draw an arc.
With A as the center and a radius equal to the length of BC, draw another arc to intersect the previous one at point D.
Join AD and CD.

Result: ABCD is the required rectangle.

Q1Section 8.6 Construct

House: Recreate this figure. Note that all the lines forming the border of the house are of length 5 cm.

Solution

To Construct: A 'house' figure where the square base and triangular roof sides are all 5 cm.

Analysis: The figure consists of a square (the walls) and an equilateral triangle (the roof) on top of it.

Steps of Construction:

Base: Draw a line segment BC of length 5 cm.
Walls: At B and C, construct perpendiculars BX and CY to BC. Cut off BA = 5 cm on BX and CD = 5 cm on CY.
Top: Join AD. Now we have the square base BCDA.
Roof: The peak of the roof, let's call it E, must be 5 cm from A and 5 cm from D.
With A as the center and a radius of 5 cm, draw an arc above the square.
With D as the center and a radius of 5 cm, draw another arc to intersect the first one at point E.
Join EA and ED.

Wait, the figure in the textbook is different. It is a square base with a curved roof. Let's follow the textbook's construction guide.

Correct Construction (as per textbook guide):

Draw the base of the house, a square BCDE with side length 5 cm.
The top point A needs to be located. It is 5 cm from B and 5 cm from C. With B as center and 5 cm radius, draw an arc. With C as center and 5 cm radius, draw another arc. The intersection is point A.
Join AB and AC.
The roof is a single arc from B to C, centered at A. With A as the center and radius AB = AC = 5 cm, draw the arc connecting B and C.

Result: The house figure with a square base and curved roof is constructed, with all defining lengths being 5 cm.

Q2Section 8.6 Construct

Construct a bigger house in which all the sides are of length 7 cm.

Solution

To Construct: A 'house' figure where all defining lines and arcs have a length or radius of 7 cm.

Steps of Construction:

Draw the base of the house, which is part of a square. Start with a horizontal line segment BC of length 7 cm.
At B and C, construct perpendiculars BX and CY downwards.
Mark points E on BX and D on CY such that BE = CD = 7 cm.
Join ED to complete the rectangular part of the house.
Now, locate the center for the roof arc, point A. Point A must be 7 cm from B and 7 cm from C.
With B as the center and a radius of 7 cm, draw an arc above BC.
With C as the center and a radius of 7 cm, draw another arc to intersect the first one at point A.
With A as the center and a radius of 7 cm, draw the arc that connects points B and C. This forms the roof.

Result: A larger house figure with all defining lengths of 7 cm is constructed.

Q3Section 8.6 Construct

Try to recreate 'A Person', 'Wavy Wave', and 'Eyes' from the section 'Artwork', using ideas involved in the 'House' construction.

Solution

Concept: The key idea from the 'House' construction is finding a point equidistant from two other points by intersecting arcs.

Recreating 'A Person':

The curved smile or frown can be drawn using this method. Mark two points for the ends of the smile. To draw the arc, find a center point by drawing intersecting arcs from the two end points with an equal, chosen radius. Then use this center point to draw the smile arc.

Recreating 'Wavy Wave':

Each semi-circle in the wavy wave is an arc. The center of each semi-circle is the midpoint of its diameter. This is a simpler construction not directly using the intersecting arcs method, but relies on finding a center for an arc.

Recreating 'Eyes':

An eye is typically formed by two arcs. Let the two corners of the eye be P and Q.
Top Eyelid: Find a center point A below the line segment PQ by intersecting two arcs of equal radius from P and Q. Use A as the center to draw the top arc.
Bottom Eyelid: Find a center point B above the line segment PQ by intersecting two arcs of equal radius from P and Q. Use B as the center to draw the bottom arc.
To make the eye symmetrical, the centers A and B should lie on the perpendicular bisector of the segment PQ.

Q4Section 8.6 Construct

Is there a 4-sided figure in which all the sides are equal in length but is not a square? If such a figure exists, can you construct it?

Solution

Answer: Yes, such a figure exists. It is called a rhombus.

Properties of a Rhombus:

All four sides are equal in length.
Opposite angles are equal.
Unlike a square, its angles are not necessarily $90^\circ$ .

To Construct: A rhombus with a side length of 5 cm.

Steps of Construction:

Draw a line segment AB of length 5 cm.
At point B, draw a ray BX making an angle with AB that is not $90^\circ$ . For example, construct an angle of $60^\circ$ .
With B as the center and a radius of 5 cm, draw an arc to cut the ray BX at point C.
Now, we need to find the fourth point D, which is 5 cm away from both A and C.
With A as the center and a radius of 5 cm, draw an arc.
With C as the center and a radius of 5 cm, draw another arc to intersect the previous arc at point D.
Join AD and CD.

Result: ABCD is the required rhombus with all sides equal to 5 cm and angles that are not

90^\circ

.

Playing with ConstructionsClass 6 Mathematics NCERT Solutions

Imagine marking all the points of 4 cm distance from the point P. How would they look?

Solution

For the 'Wavy Wave' figure, what radius should be taken in the compass to get this half circle? What should be the length of AX?

Solution

Take a central line of a different length and try to draw the wave on it.

Solution

Try to recreate the figure where the waves are smaller than a half circle (as appearing in the neck of the figure, 'A Person'). The challenge here is to get both the waves to be identical. This may be tricky!

Solution

Which of the following is not a name for this square? PQSR SPQR RSPQ QRSP

Solution

Draw the rectangle and four squares configuration (shown in Fig. 8.3) on a dot paper. What did you do to recreate this figure so that the four squares are placed symmetrically around the rectangle? Discuss with your classmates.

Solution

Identify if there are any squares in this collection. Use measurements if needed.

Solution

Is it possible to reason out if the sides are equal or not, and if the angles are right or not without using any measuring instruments in the above figure? Can we do this by only looking at the position of corners in the dot grid?

Solution

Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that their corners are on the dots. Verify if the squares and rectangles that you have drawn satisfy their respective properties.

Solution

Draw a rectangle with sides of length 4 cm and 6 cm. After drawing, check if it satisfies both the rectangle properties.

Solution

Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if it satisfies both the rectangle properties.

Solution

Is it possible to construct a 4-sided figure in which- all the angles are equal to 90∘90^\circ90∘ but opposite sides are not equal?

Solution

Is there a shorthand way of writing it down? In all the sentences, only the position of X, Y and the length XY changes. So we could write this as: Distance of X from ADistance of Y from BLength of XY

Solution

Have you checked what happens to the length XY when X and Y are placed at the same distance away from A and B, respectively? For example, as in the cases like these: Distance of X from ADistance of Y from BLength of XY5 mm5 mm1 cm1 cm1 cm 5 mm1 cm 5 mm

Solution

In each of these cases, observe how the length XY compares to that of AB and the shape of the 4-sided figure ABYX.

Solution

How does the farthest distance between X and Y compare with the length of AC? BD?

Solution

Construct a rectangle that can be divided into 3 identical squares.

Solution

Give the lengths of the sides of a rectangle that cannot be divided into— two identical squares; three identical squares.

Solution

A Square within a Rectangle: Construct a rectangle of sides 8 cm and 4 cm. How will you construct a square inside, as shown in the figure, such that the centre of the square is the same as the centre of the rectangle?

Solution

Falling Squares: Construct the figure.

Solution

Shadings: Construct this. Choose measurements of your choice. Note that the larger 4-sided figure is a square and so are the smaller ones.

Solution

Square with a Hole: Construct this.

Solution

Square with more Holes: Construct this.

Solution

Square with Curves: This is a square with 8 cm sidelengths.

Solution

How should the rectangle be constructed so that the diagonal divides the opposite angles into equal parts?

Solution

Construct a rectangle in which one of the diagonals divides the opposite angles into 60∘60^\circ60∘ and 30∘30^\circ30∘.

Solution

Construct a rectangle where one of its sides is 5 cm and the length of a diagonal is 7 cm.

Solution

Construct a rectangle in which one of the diagonals divides the opposite angles into 50∘50^\circ50∘ and 40∘40^\circ40∘.

Solution

Construct a rectangle in which one of the diagonals divides the opposite angles into 45∘45^\circ45∘ and 45∘45^\circ45∘. What do you observe about the sides?

Solution

Construct a rectangle one of whose sides is 4 cm and the diagonal is of length 8 cm.

Solution

Construct a rectangle one of whose sides is 3 cm and the diagonal is of length 7 cm.

Solution

House: Recreate this figure. Note that all the lines forming the border of the house are of length 5 cm.

Solution

Construct a bigger house in which all the sides are of length 7 cm.

Solution

Try to recreate 'A Person', 'Wavy Wave', and 'Eyes' from the section 'Artwork', using ideas involved in the 'House' construction.

Solution

Is there a 4-sided figure in which all the sides are equal in length but is not a square? If such a figure exists, can you construct it?

Solution

Is it possible to construct a 4-sided figure in which- all the angles are equal to $90^\circ$ but opposite sides are not equal?

Is there a shorthand way of writing it down? In all the sentences, only the position of X, Y and the length XY changes. So we could write this as:
Distance of X from A Distance of Y from B Length of XY

Have you checked what happens to the length XY when X and Y are placed at the same distance away from A and B, respectively? For example, as in the cases like these:
Distance of X from A Distance of Y from B Length of XY
5 mm 5 mm
1 cm 1 cm
1 cm 5 mm 1 cm 5 mm

Construct a rectangle in which one of the diagonals divides the opposite angles into $60^\circ$ and $30^\circ$ .

Construct a rectangle in which one of the diagonals divides the opposite angles into $50^\circ$ and $40^\circ$ .

Construct a rectangle in which one of the diagonals divides the opposite angles into $45^\circ$ and $45^\circ$ . What do you observe about the sides?