CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 08

1. Construction should be neat and clean and as per scale given in question.
2. Steps of construction should be provided only to those questions where it is mentioned.

QUESTIONS

Question 1. Draw a line segment AB = 7 cm. Take a point P on AB such that AP : PB = 3 : 4.
Answer: Steps of construction:
1. Draw a line segment \(AB = 7\) cm using a scale.
2. Draw a ray AX making an acute angle with \(AB\) downwards.
3. Along the ray AX, mark \(3 + 4 = 7\) equal points at regular intervals: \(A_1, A_2, A_3, A_4, A_5, A_6, A_7\).
4. Join \(A_7\) to \(B\).
5. Through \(A_3\), draw a line parallel to \(A_7B\) intersecting \(AB\) at point \(P\).
Thus, \(P\) is the point on \(AB\) dividing it in the ratio \(3 : 4\).
In simple words: Divide a line of 7 cm into 7 equal parts using a parallel line method to get a 3:4 ratio split.

Exam Tip: Always make sure the acute angle drawn on ray AX is neat and your compass markings are equidistant.

Question 2. Draw a line segment PQ = 10 cm. Take a point A on PQ such that \(\frac{PA}{PQ} = \frac{2}{5}\). Measure the length of PA and AQ.
Answer: Steps of construction:
1. Draw a line segment \(PQ = 10\) cm.
2. Draw any ray PX making an acute angle with \(PQ\).
3. Locate 5 points \(P_1, P_2, P_3, P_4, P_5\) on PX at equal distances.
4. Join \(P_5\) to \(Q\).
5. Through \(P_2\), draw a line parallel to \(P_5Q\) intersecting \(PQ\) at point \(A\).
6. Measuring the segments with a scale gives \(PA = 4\) cm and \(AQ = 6\) cm.
In simple words: Divide the 10 cm line segment into 5 equal parts and locate point A at the 2nd division mark.

Exam Tip: Verify your measured lengths algebraically: \(PA = \frac{2}{5} \times 10 = 4\) cm and \(AQ = 10 - 4 = 6\) cm.

Question 3. Construct a \(\Delta ABC\) in which BC = 6.5 cm, AB = 4.5 cm and \(\angle ACB = 60^\circ\). Construct another triangle similar to \(\Delta ABC\) such that each side of new triangle is \(\frac{4}{5}\) of the corresponding sides of \(\Delta ABC\).
Answer: Steps of construction:
1. Draw a line segment \(BC = 6.5\) cm.
2. At point C, construct an angle \(\angle BCX = 60^\circ\).
3. With center B and radius 4.5 cm, draw an arc to intersect CX at point A. Join AB. \(\Delta ABC\) is formed.
4. Draw a ray BY making an acute angle with BC on the side opposite to vertex A.
5. Locate 5 points \(B_1, B_2, B_3, B_4, B_5\) on BY at equal distances.
6. Join \(B_5\) to C.
7. Through \(B_4\), draw a line parallel to \(B_5C\) intersecting BC produced at \(C'\).
8. Through \(C'\), draw a line parallel to CA intersecting AB at \(A'\).
\(\Delta A'BC'\) is the required similar triangle.
In simple words: First draw the base triangle with given sides and angle, then use a scale factor of 4/5 to construct the smaller similar triangle inside it.

Exam Tip: Since the scaling factor is less than 1 (\(\frac{4}{5}\)), the new triangle will lie entirely inside the original triangle.

Question 4. Draw a triangle XYZ such that XY = 5 cm, YZ = 7 cm and \(\angle XYZ = 75^\circ\). Now construct a \(\Delta X'YZ' \sim \Delta XYZ\) with its sides \(\frac{3}{2}\) times of the corresponding sides of \(\Delta XYZ\).
Answer: Steps of construction:
1. Draw line segment \(YZ = 7\) cm.
2. At point Y, construct an angle of \(75^\circ\) using a compass or protractor.
3. Cut off \(YX = 5\) cm along this ray and join XZ to form \(\Delta XYZ\).
4. Draw a ray YP making an acute angle with YZ on the side opposite to vertex X.
5. Mark 3 points \(Y_1, Y_2, Y_3\) on YP at equal distances (since the greater number in \(\frac{3}{2}\) is 3).
6. Join \(Y_2\) to Z.
7. Through \(Y_3\), draw a line parallel to \(Y_2Z\) intersecting the extended line YZ at \(Z'\).
8. Through \(Z'\), draw a line parallel to XZ intersecting the extended line YX at \(X'\).
\(\Delta X'YZ'\) is the required triangle.
In simple words: Draw the original triangle, extend its sides, and use division points to construct a larger similar triangle outside it since the scale factor is greater than 1.

Question 5. Construct an isosceles triangle whose base is 8 cm and altitude 5 cm and then construct another triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of the given triangle.
Answer: Steps of construction:
1. Draw base \(BC = 8\) cm.
2. Draw the perpendicular bisector of BC intersecting BC at point D.
3. Along the perpendicular bisector, cut off segment \(DA = 5\) cm.
4. Join AB and AC to form the isosceles triangle \(\Delta ABC\).
5. Draw an acute ray BX downwards from B.
6. Mark 4 equal parts \(B_1, B_2, B_3, B_4\) on BX.
7. Join \(B_4\) to C.
8. Draw a line through \(B_3\) parallel to \(B_4C\) meeting BC at \(C'\).
9. Through \(C'\), draw a line parallel to CA meeting AB at \(A'\).
\(\Delta A'BC'\) is the required triangle.
In simple words: Construct an isosceles triangle using its base and height, then shrink it using the ratio 3/4.

Exam Tip: Always construct the perpendicular bisector accurately to find the exact midpoint for the altitude.

Question 6. Draw an isosceles \(\Delta ABC\) with AB = AC and base BC = 7 cm and vertical angle is \(120^\circ\). Construct \(\Delta AB'C' \sim \Delta ABC\) with its sides \(1\frac{1}{3}\) times of the corresponding sides of \(\Delta ABC\).
Answer: Steps of construction:
1. Draw base \(BC = 7\) cm.
2. At B and C, construct angles of \(30^\circ\) each (since base angles of an isosceles triangle with vertical angle \(120^\circ\) are each \(\frac{180^\circ - 120^\circ}{2} = 30^\circ\)).
3. Let the rays intersect at vertex A. \(\Delta ABC\) is formed.
4. Convert mixed fraction scale factor: \(1\frac{1}{3} = \frac{4}{3}\).
5. Draw an acute ray BX and mark 4 equal points \(B_1, B_2, B_3, B_4\).
6. Join \(B_3\) to C and draw a parallel line from \(B_4\) to meet extended BC at \(C'\).
7. Draw a line through \(C'\) parallel to CA to meet extended AB at \(B'\) (or \(A'\)).
In simple words: Find the base angles as 30 degrees, build the triangle, and expand it by a factor of 4/3.

Exam Tip: Calculate base angles carefully: \((180^\circ - 120^\circ)/2 = 30^\circ\) at each base vertex.

Question 7. Draw \(\Delta PQR\) in which \(\angle Q = 90^\circ\), PQ = 6 cm, QR = 8 cm. Construct \(\Delta P'QR' \sim \Delta PQR\) with its sides equal to 2/3rd of corresponding sides of \(\Delta PQR\).
Answer: Steps of construction:
1. Draw line segment \(QR = 8\) cm.
2. At point Q, construct an angle of \(90^\circ\).
3. Cut off \(QP = 6\) cm along the perpendicular ray and join PR to form right-angled \(\Delta PQR\).
4. Draw an acute ray QX downwards from Q.
5. Mark 3 equal points \(Q_1, Q_2, Q_3\) on QX.
6. Join \(Q_3\) to R.
7. Through \(Q_2\), draw a line parallel to \(Q_3R\) intersecting QR at \(R'\).
8. Through \(R'\), draw a line parallel to PR intersecting PQ at \(P'\).
\(\Delta P'QR'\) is the required triangle.
In simple words: Draw a right triangle with base 8 cm and perpendicular 6 cm, then construct a smaller similar triangle inside with scale factor 2/3.

Exam Tip: Hypotenuse length is \(\sqrt{6^2 + 8^2} = 10\) cm, which can be used to cross-verify your drawing accuracy.

Question 8. Construct a right angled triangle in which base is 2 times of the perpendicular. Now construct a triangle similar to it with base 1.5 times of the original triangle.
Answer: Steps of construction:
1. Assume perpendicular = 4 cm, then base = 8 cm (satisfying base is 2 times perpendicular).
2. Draw base \(BC = 8\) cm.
3. At point B, construct a \(90^\circ\) angle and cut off perpendicular \(BA = 4\) cm.
4. Join AC to complete right-angled \(\Delta ABC\).
5. Scale factor \(1.5 = \frac{3}{2}\).
6. Draw an acute ray BY, mark 3 points \(B_1, B_2, B_3\) at equal intervals.
7. Join \(B_2\) to C, and draw a parallel line from \(B_3\) to intersect extended BC at \(C'\).
8. Through \(C'\), draw a line parallel to CA to intersect extended AB at \(A'\).
In simple words: Choose convenient dimensions where base is double the height, draw the right triangle, and scale it up by 1.5 times.

Exam Tip: Scale factor 1.5 is equivalent to the fraction \(\frac{3}{2}\), so divide the construction ray into 3 parts and base connection on the 2nd part.

Question 9. Draw an equilateral triangle PQR with side 5 cm. Now construct \(\Delta PQ'R'\) such that \(\frac{PQ}{PQ'} = \frac{1}{2}\). Measure \(PQ'\).
Answer: Steps of construction:
1. Draw line segment \(QR = 5\) cm.
2. With centers Q and R and radius 5 cm, draw arcs intersecting each other at point P.
3. Join PQ and PR to form equilateral \(\Delta PQR\).
4. Draw an acute ray QX and mark 2 equal points \(Q_1, Q_2\) on it.
5. Join \(Q_1\) to R, and draw a line through \(Q_2\) parallel to \(Q_1R\) meeting extended QR at \(R'\).
6. Through \(R'\), draw a line parallel to PR meeting extended QP at \(P'\).
7. Measuring \(PQ'\) gives \(5 \times 2 = 10\) cm.
In simple words: Construct an equilateral triangle of side 5 cm, then double its size outward as given by the ratio 1:2.

Exam Tip: Since \(\frac{PQ}{PQ'} = \frac{1}{2}\), the new side \(PQ' = 2 \times PQ = 10\) cm.

Question 10. Draw a circle of radius 4 cm with centre O. Take a point P outside the circle such that OP = 6 cm. Draw tangents PA and PB to the circle. Measure the lengths of PA and PB.
Answer: Steps of construction:
1. Draw a circle with centre O and radius 4 cm.
2. Mark a point P outside the circle such that \(OP = 6\) cm.
3. Draw the perpendicular bisector of OP, letting M be the midpoint of OP.
4. Taking M as centre and MP (or MO) as radius, draw a circle that intersects the given circle at points A and B.
5. Join PA and PB. These are the required tangents.
6. Measuring PA and PB gives \(PA = PB \approx 4.47\) cm (calculated as \(\sqrt{6^2 - 4^2} = \sqrt{20} \approx 4.47\) cm).
In simple words: Draw the circle and external point, find the midpoint of the distance between them, draw a temporary circle to locate contact points, and join them.

Exam Tip: Tangents drawn from an external point to a circle are always equal in length (\(PA = PB\)).

Question 11. Draw a line segment AB = 8 cm. Taking AB as diameter a circle is drawn with centre O. Now draw \(OP \perp AB\). Through P draw a tangent to the circle.
Answer: Steps of construction:
1. Draw a line segment \(AB = 8\) cm and find its midpoint O.
2. Draw a circle with centre O and radius 4 cm.
3. Draw radius OP perpendicular to AB (\(OP \perp AB\)) such that P lies on the circle.
4. At point P, construct a line perpendicular to radius OP. This perpendicular line is the required tangent to the circle at P.
In simple words: Draw a circle on an 8 cm diameter, mark a radius straight up, and draw a perpendicular line at that point to form the tangent.

Exam Tip: The tangent at any point on a circle is always perpendicular to the radius through the point of contact.

Question 12. Draw a circle of radius OP = 3 cm. Draw \(\angle POQ = 45^\circ\) such that OQ = 5 cm. Now draw two tangents from Q to given circle.
Answer: Steps of construction:
1. Draw a circle with centre O and radius 3 cm (let P be a point on it).
2. Draw radius OP, and construct an angle \(\angle POQ = 45^\circ\) such that \(OQ = 5\) cm.
3. Draw the perpendicular bisector of OQ to find its midpoint M.
4. With centre M and radius MQ, draw a circle intersecting the given circle at two points (say T and S).
5. Join QT and QS. These are the required tangents from Q.
In simple words: Draw the circle, make a 45 degree angle to point Q at 5 cm distance, and use the standard circle-bisector method to draw tangents from Q.

Question 13. Draw a circle with centre O and radius 3.5 cm. Draw two tangents PA and PB from an external point P such that \(\angle APB = 45^\circ\). What is the value of \(\angle AOB + \angle APB\).
Answer: Answer:
1. In quadrilateral OAPB, the angles at the points of contact A and B are \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\).
2. The sum of angles in quadrilateral OAPB is \(360^\circ\).
\(\angle AOB + \angle APB + \angle OAP + \angle OBP = 360^\circ\)
\(\angle AOB + \angle APB + 90^\circ + 90^\circ = 360^\circ\)
\(\angle AOB + \angle APB + 180^\circ = 360^\circ\)
\(\angle AOB + \angle APB = 180^\circ\)
Thus, the sum \(\angle AOB + \angle APB = 180^\circ\) always (they are supplementary).
In simple words: The opposite angles of a tangent quadrilateral always add up to 180 degrees.

Exam Tip: Remember the standard geometric theorem: angle between tangents and angle subtended at the centre by the chord of contact are supplementary.

Question 14. Draw a circle of radius 4 cm. Now draw a set of tangents from an external point P such that the angle between the two tangents is half of the central angle made by joining the points of contact to the centre.
Answer: Answer:
Let angle between tangents be \(\angle APB = \theta\) and central angle be \(\angle AOB = 2\theta\).
We know that \(\angle AOB + \angle APB = 180^\circ\)
Substitute \(\angle AOB = 2\angle APB\):
\(2\angle APB + \angle APB = 180^\circ\)
\(3\angle APB = 180^\circ \implies \angle APB = 60^\circ\)
Thus, the central angle \(\angle AOB = 120^\circ\).
Steps of construction:
1. Draw a circle of radius 4 cm with centre O.
2. Draw two radii OA and OB such that \(\angle AOB = 120^\circ\).
3. Draw perpendiculars to OA at A and OB at B. These intersect at external point P, forming tangents at an angle of \(60^\circ\).
In simple words: Set the central angle to 120 degrees so that the outer angle between tangents naturally becomes 60 degrees (half of 120).

Exam Tip: Use the supplementary property \(\angle AOB + \angle APB = 180^\circ\) to solve for the exact angles before constructing.

Question 15. Draw a line segment AB = 9 cm. Taking A and B as centres draw two circles of radius 5 cm and 3 cm respectively. Now draw tangents to each circle from the centre of the other.
Answer: Steps of construction:
1. Draw line segment \(AB = 9\) cm.
2. With centre A, draw a circle of radius 5 cm.
3. With centre B, draw a circle of radius 3 cm.
4. Find the midpoint M of AB.
5. With centre M and radius MA (or MB), draw a circle that intersects circle A at points P and Q, and circle B at points R and S.
6. Join BP and BQ (tangents from B to circle A), and join AR and AS (tangents from A to circle B).
In simple words: Draw the two circles sharing the same center-line, find the midpoint, draw a shared reference circle, and connect the center points to intersection points.

Exam Tip: Since distance AB = 9 cm, circle A (radius 5) and circle B (radius 3) are well-separated, making intersection points clear.

Question 16. Draw a circle of radius 3.5 cm with centre O. Take point P such that OP = 6 cm. OP cuts the circle at T. Draw two tangents PQ and PR. Join Q to R. Through T draw AB parallel to QR such that A and B are point on PQ and PR.
Answer: Steps of construction:
1. Draw a circle of radius 3.5 cm with centre O.
2. Take a point P at distance \(OP = 6\) cm, intersecting the circle at point T.
3. Draw tangents PQ and PR from P to the circle.
4. Join Q and R.
5. Through point T, construct a line AB parallel to chord QR such that A lies on PQ and B lies on PR.
In simple words: Draw the circle, external point, and tangents. Find point T where OP crosses the circle and draw a line parallel to the chord of contact through T.

Exam Tip: Ensure the parallel line at T is constructed accurately using alternate interior angles or corresponding angle methods.

Question 17. Draw a circle of diameter 7 cm. Draw a pair of tangents to the circle, which are inclined to each other at an angle of \(60^\circ\).
Answer: Steps of construction:
1. Draw a circle of diameter 7 cm (radius \(r = 3.5\) cm) with centre O.
2. Draw any radius OA.
3. Draw radius OB such that \(\angle AOB = 180^\circ - 60^\circ = 120^\circ\).
4. At points A and B, draw perpendiculars to radii OA and OB respectively.
5. These perpendicular lines will intersect at an external point P at an inclination angle of \(60^\circ\).
In simple words: Draw radii with a 120 degree angle between them, then draw perpendiculars at their outer tips to meet at the required 60 degree tangent angle.

Exam Tip: The angle between tangents and the angle between radii always sum to \(180^\circ\): \(180^\circ - 60^\circ = 120^\circ\).

Question 18. Draw a circle with centre O and radius 3.5 cm. Take a horizontal diameter. Extend it to both sides to point P and Q such that OP = OQ = 7 cm. Draw tangents PA and QB one above the diameter and the other below the diameter. Is PA \(||\) BQ.
Answer: Answer: Yes, \(PA \parallel BQ\).
Explanation:
1. Radius OA is perpendicular to tangent PA (\(OA \perp PA\)), so \(\angle OAP = 90^\circ\).
2. Radius OB is perpendicular to tangent QB (\(OB \perp QB\)), so \(\angle OBQ = 90^\circ\).
3. Since AB is a diameter, angle \(\angle AOB = 180^\circ\), making alternate interior angles or interior angles on the same side of transversal PQ equal (\(\angle OAP + \angle OBQ = 90^\circ + 90^\circ = 180^\circ\)).
4. Thus, line segment PA is parallel to line segment BQ (\(PA \parallel BQ\)).
In simple words: Because the two radii form a straight line and both tangents are perpendicular to it, the two tangents are parallel to each other.

Exam Tip: Whenever tangents are drawn at the endpoints of a diameter, they are always parallel to each other.