CBSE Class 7 Mathematics Practice Assignment Set 06

Question 1. Draw line segments of the following lengths. Construct their perpendicular bisectors. Measure the lengths of two parts. (i) 5.6 cm (ii) 6.8 cm (iii) 4 cm
Answer:
Steps of Construction:
1. Draw a line segment using a ruler of the given length: (i) \(AB = 5.6\) cm, (ii) \(AB = 6.8\) cm, and (iii) \(AB = 4\) cm.
2. Taking A as the centre and radius greater than half of \(AB\), draw arcs above and below the line segment.
3. With the same radius and taking B as the centre, draw arcs above and below the line segment to intersect the previous arcs.
4. Join the intersection points of the arcs. This line is the perpendicular bisector.
5. Let the perpendicular bisector intersect \(AB\) at point M.
Measurement:
On measuring, it is found that \(AM = MB\) for each case. For example, in (iii) where \(AB = 4\) cm, \(AM = 2\) cm and \(MB = 2\) cm.
In simple words: A perpendicular bisector divides a line segment into two equal halves at a right angle (90 degrees).

Exam Tip: Ensure the compass radius is strictly greater than half the length of the segment so the arcs intersect clearly.

Question 2. Draw the following angles with your protractor. Draw angles congruent to them using compass. Bisect the angles. Measure the half angles to see if your construction is accurate. (i) 90\(^\circ\) (ii) 60\(^\circ\) (iii) 130\(^\circ\) (iv) 30\(^\circ\)
Answer:
Steps of Construction:
1. Draw the given angles using a protractor: (i) 90\(^\circ\), (ii) 60\(^\circ\), (iii) 130\(^\circ\), and (iv) 30\(^\circ\).
2. To draw a congruent angle using a compass, draw a ray and use standard compass arcs to copy the angle opening.
3. To bisect any angle (say \(\angle ABC\)), place the compass pointer at vertex B and draw an arc cutting both arms at points P and Q.
4. With P and Q as centres, draw intersecting arcs inside the angle. Join the vertex to this intersection point to get the angle bisector.
Measurement:
Measuring the half angles with a protractor yields: (i) 45\(^\circ\), (ii) 30\(^\circ\), (iii) 65\(^\circ\), and (iv) 15\(^\circ\), confirming accurate construction.
In simple words: Congruent angles have the exact same measure, and an angle bisector cuts the angle into two equal parts.

Exam Tip: Keep the compass width uniform when drawing arcs from the arms of the angle during a bisector construction.

Question 3. Draw a line, and mark points X and Y on it such that \(XY = 6\) cm. Mark a point A above XY and another point B below XY. Draw perpendiculars from A and B to XY. Check to see if the perpendiculars are parallel to each other.
Answer:
Steps of Construction:
1. Draw a line \(l\) and mark points X and Y on it such that \(XY = 6\) cm.
2. Mark point A in the upper half-plane and point B in the lower half-plane.
3. Using a set-square or compass, draw a perpendicular line from A to line \(l\) and another perpendicular from B to line \(l\).
Verification:
Measure the distance between the two perpendicular lines at multiple points along their length. The distance remains constant everywhere, proving that the two perpendicular lines are parallel to each other.
In simple words: Two lines perpendicular to the same base line are always parallel to each other.

Exam Tip: Use a set-square aligned properly against the baseline to guarantee precise 90-degree perpendiculars.

Question 4. Draw a line segment PQ equal to \(8.2\) cm. Construct a perpendicular bisector to PQ and let this bisector cut PQ at O. Mark any point R on this bisector. Join PR and QR. What kind of a triangle is RPQ? Explain why.
Answer:
Steps of Construction:
1. Draw line segment \(PQ = 8.2\) cm.
2. Construct its perpendicular bisector intersecting PQ at O.
3. Mark any point R on the perpendicular bisector and join PR and QR.
Conclusion:
\(\triangle RPQ\) is an isosceles triangle.
Explanation:
Since R lies on the perpendicular bisector of PQ, any point on the perpendicular bisector is equidistant from the endpoints of the segment. Therefore, \(PR = QR\), making \(\triangle RPQ\) an isosceles triangle.
In simple words: Any point on the perpendicular bisector of a line segment is at the same distance from both ends of the segment.

Exam Tip: Remember the fundamental theorem: every point on the perpendicular bisector is equidistant from the segment endpoints.

Question 5. Draw a circle with radius \(4\) cm. Mark the centre O. Draw any chord in the circle and label it as AB. Construct the perpendicular bisector of the chord AB. What do you notice about the bisector? Why? Draw any other chord in the same circle and repeat the above steps. What is your conclusion?
Answer:
Steps of Construction:
1. Draw a circle with centre O and radius \(4\) cm.
2. Draw any chord AB and construct its perpendicular bisector.
Observation:
The perpendicular bisector of chord AB passes directly through the centre O of the circle.
Reason:
The perpendicular from the centre of a circle to a chord bisects the chord, and conversely, the perpendicular bisector of any chord always passes through the centre.
Conclusion:
Repeating this with any other chord confirms that the perpendicular bisector of any chord in a circle always passes through the centre of the circle.
In simple words: If you find the perpendicular bisector of any straight line inside a circle connecting two points on the edge, it will always pass through the exact centre.

Exam Tip: State the circle theorem clearly in your reason: perpendicular bisector of a chord passes through the centre.

Question 6. Draw a line segment AB. At any point C on it, draw a line CD, making a linear pair of angles DCA and DCB. Bisect these two angles. Call these bisecting rays CP and CQ. Measure angle PCQ. What do you notice? Why?
Answer:
Steps of Construction:
1. Draw a line segment AB and choose a point C on it.
2. Draw a line CD perpendicular or at an oblique angle to AB, forming a linear pair \(\angle DCA\) and \(\angle DCB$.
3. Draw ray CP as the bisector of \(\angle DCA\) and ray CQ as the bisector of \(\angle DCB$.
Measurement & Observation:
Measuring \(\angle PCQ\) gives exactly \(90^\circ\).
Why:
The sum of angles in a linear pair is \(180^\circ\). Since CP bisects one angle and CQ bisects the adjacent angle, the sum of their halves is \(\frac{1}{2}(180^\circ) = 90^\circ\).
In simple words: The angle formed between the bisectors of a linear pair is always a right angle (\(90^\circ\)).

Exam Tip: Use algebraic justification: if angles are \(x\) and \(180^\circ - x\), their bisectors form \(\frac{x}{2} + \frac{180^\circ - x}{2} = 90^\circ\).

Question 7. Draw an equilateral triangle ABC of side \(5\) cm. Construct the perpendicular bisector of each side. What do you notice? What is your conclusion?
Answer:
Steps of Construction:
1. Draw an equilateral triangle ABC with side \(5\) cm.
2. Construct the perpendicular bisectors for side AB, side BC, and side CA.
Observation:
All three perpendicular bisectors intersect at one single common point (they are concurrent).
Conclusion:
In an equilateral triangle (and in any triangle), the perpendicular bisectors of the sides are concurrent, and their intersection point is the circumcentre of the triangle.
In simple words: The perpendicular bisectors of all three sides of a triangle always meet at the exact same point.

Exam Tip: Mention the term 'concurrent' when three or more lines pass through the same single point.

Question 8. Draw a line segment AB equal in length to \(5\) cm. At A and B, construct perpendicular segments, each equal to \(5\) cm. Call these AD and BC. Join CD. What is measure of CD? What kind of a figure is ABCD? Why?
Answer:
Steps of Construction:
1. Draw line segment \(AB = 5\) cm.
2. At points A and B, construct perpendicular segments \(AD = 5\) cm and \(BC = 5\) cm.
3. Join points C and D.
Measurements & Conclusion:
- The measure of CD is \(5\) cm.
- The figure ABCD is a square.
Why:
All four sides are equal (\(AB = BC = CD = DA = 5\) cm) and all interior angles are right angles (\(90^\circ\)).
In simple words: A quadrilateral with four equal sides and right angles at the corners is a square.

Exam Tip: Verify both side lengths and interior right angles to conclusively prove the shape is a square rather than a rhombus.

Question 9. Draw a line segment PR equal to \(6\) cm. Construct a perpendicular bisector to PR intersecting PR at O. On this bisector on either side of PR, mark points Q and S such that \(OQ = OS = 3\) cm. Join PQ, QR, RS and SP. What kind of a figure is PQRS?
Answer:
Steps of Construction:
1. Draw line segment \(PR = 6\) cm.
2. Construct its perpendicular bisector intersecting PR at O.
3. Mark points Q and S on the bisector such that \(OQ = 3\) cm and \(OS = 3\) cm.
4. Join PQ, QR, RS, and SP.
Conclusion:
The figure PQRS is a rhombus (specifically a square, since diagonals are equal and bisect at \(90^\circ\); because diagonal lengths are \(PR = 6\) cm and \(QS = 6\) cm, the figure is a square).
In simple words: When diagonals bisect each other at right angles and are equal in length, the resulting shape is a square.

Exam Tip: Check diagonal lengths: if perpendicular bisectors are equal and bisect each other, the rhombus becomes a square.

Question 10. Draw an angle AOB equal to \(60^\circ\). With O as centre and any convenient radius, draw an arc cutting OA at P and OB at Q. At P and Q, draw perpendiculars to intersect at X. Join OX. Measure angles XOA and XOB. Measure the lengths XP and XQ. What are your conclusions?
Answer:
Steps of Construction:
1. Draw \(\angle AOB = 60^\circ\).
2. With centre O, draw an arc intersecting OA at P and OB at Q.
3. Construct perpendiculars at P and Q to intersect at X, then join OX.
Measurements & Conclusions:
- \(\angle XOA = 30^\circ\) and \(\angle XOB = 30^\circ\).
- \(XP = XQ\).
- Conclusion: Ray OX is the angle bisector of \(\angle AOB\), and any point on an angle bisector is equidistant from the arms of the angle.
In simple words: OX splits the \(60^\circ\) angle into two equal \(30^\circ\) angles, and the distances from X to both arms are equal.

Exam Tip: This construction method is the standard geometric proof for constructing an angle bisector using perpendiculars.

Question 11. Draw a line segment PQ = \(5.5\) cm. At Y construct a perpendicular YZ equal to \(4\) cm. Join ZX. Bisect angle ZXY to meet ZY in P. Measure XP.
Answer:
Steps of Construction:
1. Draw line segment PQ (letting one endpoint be X and another Q, or following standard vertex labeling where base is PQ or XY; interpreting standard notation: draw base segment, construct perpendicular YZ, and form the triangle).
2. Construct the angle bisector of \(\angle ZXY\) to meet ZY at point P.
3. Measure the length of XP using a ruler.
In simple words: Follow the step-by-step instructions to draw the triangle and measure the required line segment accurately.

Exam Tip: Double-check vertex names if the question uses X, Y, Z and P interchangeably in triangle constructions.

Question 12. Draw a line segment PQ = \(5.5\) cm. At Q draw another line segment QR equal to \(5.5\) cm at an angle of \(60^\circ\) with QP. Join PR. What kind of a triangle will this be? Why?
Answer:
Steps of Construction:
1. Draw line segment \(PQ = 5.5\) cm.
2. At vertex Q, construct an angle of \(60^\circ\) with QP.
3. Cut off \(QR = 5.5\) cm along this ray and join PR.
Conclusion:
\(\triangle PQR\) is an equilateral triangle.
Why:
Since \(PQ = QR = 5.5\) cm and the included angle \(\angle Q = 60^\circ\), the base angles must also be \(60^\circ\) each, making all three sides and all three angles equal.
In simple words: A triangle with two equal sides and an included angle of \(60^\circ\) is always equilateral.

Question 13. Construct a triangle PQR with PQ = \(7.5\) cm, \(\angle P = \angle Q = 45^\circ\). Measure the third angle. What kind of a triangle is it?
Answer:
Steps of Construction:
1. Draw line segment \(PQ = 7.5\) cm.
2. At point P, construct an angle of \(45^\circ\).
3. At point Q, construct an angle of \(45^\circ\).
4. Extend the arms of these angles to intersect at point R.
Measurements & Conclusion:
- Third angle \(\angle R = 180^\circ - (45^\circ + 45^\circ) = 180^\circ - 90^\circ = 90^\circ\).
- The triangle is a right-angled isosceles triangle.
In simple words: Since two angles are \(45^\circ\), the third angle is \(90^\circ\), making it a right-angled isosceles triangle.
Exam Tip: Always apply the angle sum property of triangles (\(180^\circ\) total) to find the third angle.

Question 14. Draw a triangle ABC with AB = \(4.5\) cm, \(\angle A = 35^\circ\), \(\angle B = 65^\circ\).
Answer:
Steps of Construction:
1. Draw line segment \(AB = 4.5\) cm using a ruler.
2. At point A, use a protractor to draw a ray making an angle of \(35^\circ\) with AB.
3. At point B, use a protractor to draw a ray making an angle of \(65^\circ\) with BA.
4. Mark the intersection of these two rays as point C to complete \(\triangle ABC\).
In simple words: Draw the baseline, construct the two given angles at each end, and where their arms meet is the third vertex.

Exam Tip: Ensure you measure the angles from the correct baseline side (interior of the triangle).

Question 15. Construct \(\triangle PQR\) with sides \(2.5\) cm, \(3.5\) cm and \(4\) cm. Construct another triangle XYZ with sides \(5\) cm, \(7\) cm and \(8\) cm. Measure the angle of both the triangles. What do you find?
Answer:
Steps of Construction:
1. Draw \(\triangle PQR\) using SSS (Side-Side-Side) criterion with sides \(2.5\) cm, \(3.5\) cm, and \(4\) cm.
2. Draw \(\triangle XYZ\) using SSS criterion with sides \(5\) cm, \(7\) cm, and \(8\) cm.
Observation:
Measuring the corresponding angles of both triangles with a protractor reveals that the corresponding angles are equal (\(\angle P = \angle X\), \(\angle Q = \angle Y\), \(\angle R = \angle Z\)).
Conclusion:
The triangles are similar (AA similarity criterion), as their corresponding sides are in the same proportion (\(\frac{5}{2.5} = \frac{7}{3.5} = \frac{8}{4} = 2\)).
In simple words: Triangles with sides in the same ratio have identical angles and are similar in shape.

Exam Tip: Check side proportions: if all three side ratios are equal, the triangle angles are identical.

Question 16. Construct triangle ABC with each of its sides equal to \(6\) cm. In the triangle construct (i) the bisector of \(\angle A\) (ii) the perpendicular bisector of BC. What do you find?
Answer:
Steps of Construction:
1. Draw an equilateral triangle ABC with each side equal to \(6\) cm.
2. Construct the angle bisector of \(\angle A$.
3. Construct the perpendicular bisector of side BC.
Observation:
Both the angle bisector of \(\angle A\) and the perpendicular bisector of BC lie on the exact same line segment (the altitude from A to BC).
Conclusion:
In an equilateral triangle, the angle bisector, median, altitude, and perpendicular bisector from any vertex to the opposite side coincide into a single line.왤
In simple words: In an equilateral triangle, the line dividing the top angle in half is also the exact perpendicular line down the middle.

Exam Tip: Remember this key property of equilateral triangles: symmetry makes angle bisectors and perpendicular bisectors identical.

Question 17. Three points A, B and C are given. C is known to lie outside the straight line AB. Complete the parallelogram ABCD.
Answer:
Steps of Construction:
1. Draw straight line AB and mark point C outside this line.
2. Join AC and BC.
3. Using a compass, take radius equal to length BC and with centre A draw an arc.
4. Take radius equal to length AB and with centre C draw an arc intersecting the previous arc at point D.
5. Join AD and CD to complete the parallelogram ABCD.
In simple words: Opposite sides of a parallelogram are equal and parallel; use a compass to transfer side lengths to find the fourth vertex D.

Exam Tip: Use compass arc intersections to accurately transfer opposite side lengths for quadrilateral constructions.

Question 18. Line segment AB = \(7\) cm. Draw another line segment CD parallel to AB and equal to half the length of AB.
Answer:
Steps of Construction:
1. Draw line segment \(AB = 7\) cm.
2. Calculate the required length: half of \(7\) cm is \(3.5\) cm.
3. At any convenient distance above AB, construct parallel line segments using equal corresponding angles or perpendicular offsets.
4. Draw line segment \(CD = 3.5\) cm parallel to AB.
In simple words: Draw a line segment half the length (\(3.5\) cm) at a fixed perpendicular distance above AB so they remain parallel.

Exam Tip: Construct equal perpendicular offsets of the desired distance at both ends to ensure perfect parallelism.

Question 19. How will you construct a hexagon, given one equilateral triangle of side \(4\) cm?
Answer:
Steps of Construction:
1. Construct an equilateral triangle of side \(4\) cm. Let its vertices be O, A, and B, with O as the common vertex (centre).
2. The interior angle of an equilateral triangle at the centre is \(60^\circ\).
3. Since a regular hexagon consists of \(6\) equilateral triangles meeting at a central point (\(360^\circ / 6 = 60^\circ\)), draw six such triangles around the common centre O with side \(4\) cm.
4. Join the outer vertices to form the complete regular hexagon.
In simple words: A regular hexagon is made of six equilateral triangles sharing a central vertex; repeat the triangle 6 times around the centre.

Exam Tip: Use a protractor to mark successive \(60^\circ\) angles around the centre point.

Question 20. Construct an isosceles triangle PQR with base angles Q and R equal to \(50^\circ\) each and the base QR equal to \(5\) cm. Construct the altitude from P to QR. Also bisect the angle P. What do you notice?
Answer:
Steps of Construction:
1. Draw base segment \(QR = 5\) cm.
2. At point Q, construct an angle of \(50^\circ\). At point R, construct an angle of \(50^\circ\).
3. Extend the arms of these angles to intersect at vertex P, forming isosceles \(\triangle PQR\).
4. Construct the altitude from P perpendicular to QR.
5. Construct the angle bisector of \(\angle P\).
Observation:
The altitude from P to QR and the angle bisector of \(\angle P\) lie on the exact same line segment (line of symmetry).
In simple words: In an isosceles triangle, the line drawn straight down from the top vertex to the base is both the angle bisector and the perpendicular altitude.

Exam Tip: Recall that the line of symmetry in an isosceles triangle passes through the vertex angle and bisects the base at \(90^\circ\).