NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 10: Construction

Exercise 10.4

Step-by-Step Explanation (Beginner Level)

1. Draw the triangle: Start by drawing \(\triangle ABC\). - Make \(AB = 5\,\text{cm}\). - Make \(AC = 7\,\text{cm}\). - Keep the angle between them, \(\angle BAC = 60^\circ\).
2. Mark point P on AB: We are told \(AP = \dfrac{3}{4}AB\). - Since \(AB = 5\,\text{cm}\), \[ AP = \tfrac{3}{4} \times 5 = 3.75\,\text{cm}. \] So, put point P at a distance of 3.75 cm from A on line AB.
3. Mark point Q on AC: We are told \(AQ = \dfrac{1}{4}AC\). - Since \(AC = 7\,\text{cm}\), \[ AQ = \tfrac{1}{4} \times 7 = 1.75\,\text{cm}. \] So, put point Q at a distance of 1.75 cm from A on line AC.
4. Join PQ: Now draw a straight line between P and Q. We need to find its length.
5. Use coordinate geometry to calculate: To make calculation easier, we place triangle ABC on the coordinate plane. - Take A at (0, 0). - Take B at (5, 0) because AB = 5 cm along x-axis. - Take C at \((7\cos60^\circ, 7\sin60^\circ)\). Since \(\cos60^\circ = 0.5\) and \(\sin60^\circ = \tfrac{\sqrt3}{2}\): \[ C = (3.5, 3.5\sqrt3). \]
6. Coordinates of P: P divides AB in ratio 3:1 (because AP = 3/4 of AB). So, P = (3.75, 0).
7. Coordinates of Q: Q divides AC in ratio 1:3 (because AQ = 1/4 of AC). So, Q = (0.25 × 3.5, 0.25 × 3.5√3) = (0.875, 0.875√3).
8. Find PQ using distance formula: Distance formula: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \] Here, - P = (3.75, 0), - Q = (0.875, 0.875√3). Substituting: \[ PQ^2 = (3.75 - 0.875)^2 + (0 - 0.875\sqrt3)^2. \] Simplify: \[ PQ^2 = \left(\tfrac{23}{8}\right)^2 + \left(\tfrac{7\sqrt3}{8}\right)^2 = \tfrac{529}{64} + \tfrac{147}{64} = \tfrac{676}{64}. \] \[ PQ = \sqrt{\tfrac{676}{64}} = \tfrac{13}{4} = 3.25\,\text{cm}. \]

Therefore, the length of PQ is 3.25 cm.

Step-by-step Explanation (Beginner level)

1. Start with given measurements: Draw side \(AB = 3\,\text{cm}\). At point \(B\), make an angle of \(60^\circ\). On this new line, mark point \(C\) such that \(BC = 5\,\text{cm}\).
2. Complete parallelogram \(ABCD\): From \(A\), draw a line parallel to \(BC\). From \(C\), draw a line parallel to \(AB\). Let these two lines meet at point \(D\). Now \(ABCD\) is a parallelogram (because opposite sides are parallel).
3. Draw diagonal: Join points \(B\) and \(D\) to draw diagonal \(BD\). This divides the parallelogram into two triangles: \(\triangle ABD\) and \(\triangle BCD\).
4. Enlarge triangle \(BDC\): We need to construct another triangle \(BD'C'\) that is similar to \(BDC\) with scale factor \(\tfrac{4}{3}\). - Extend the ray \(BD\) beyond \(D\). - On this extension, locate point \(D'\) such that \(BD':BD = 4:3\). (You can do this using the basic proportionality method: draw an auxiliary ray from \(B\), mark 3 equal parts, then mark 4 equal parts, and connect back to \(D'\).)
5. Find point \(C': Through \(D'\), draw a line parallel to \(DC\). Through \(B\), draw a line parallel to \(BC\). These two lines meet at \(C'\). Now \(\triangle BD'C'\) is similar to \(\triangle BDC\) with scale factor \(\tfrac{4}{3}\) (because they have the same angles).
6. Locate point \(A': Through \(D'\), draw a line parallel to \(DA\). Extend line \(BA\) forward beyond \(A\). Let this parallel from \(D'\) meet the extended \(BA\) at point \(A'\).

Check if \(A'BC'D'\) is a parallelogram

• In the original parallelogram, we know \(DC \parallel AB\) and \(DA \parallel BC\).
• In the new figure, we constructed: - \(D'C' \parallel DC\), so \(D'C' \parallel AB\). Hence, \(A'B \parallel D'C'\). - \(D'A' \parallel DA\), so \(D'A' \parallel BC\).
• Therefore, in quadrilateral \(A'BC'D'\), both pairs of opposite sides are parallel.

Conclusion: By definition, a quadrilateral with both pairs of opposite sides parallel is a parallelogram. So, \(A'BC'D'\) is indeed a parallelogram.

Step-by-step Construction

1. Take a point \(O\) on your paper. This will be the common centre.
2. Using a compass, draw a circle with radius \(3\,\text{cm}\). This is the inner circle.
3. From the same centre \(O\), draw another circle with radius \(5\,\text{cm}\). This is the outer circle. Now you have two concentric circles.
4. Mark a point \(T\) anywhere on the outer circle (radius \(5\,\text{cm}\)).
5. Join the line \(OT\).
6. Now, to construct the tangents from \(T\) to the inner circle:
   • Draw the circle with diameter \(OT\).
   • This new circle will intersect the inner circle (radius \(3\,\text{cm}\)) at two points. Let those points be \(A\) and \(B\).
   • Join \(TA\) and \(TB\). These are the required tangents.

Step-by-step Verification

We know that a radius drawn to the point of contact is perpendicular to the tangent.

So, in triangle \(OTA\):

• \(OA = 3\,\text{cm}\) (radius of inner circle)
• \(OT = 5\,\text{cm}\) (radius of outer circle)
• \(\angle OAT = 90^\circ\)

By Pythagoras theorem:

\[ TA^2 = OT^2 - OA^2 = (5)^2 - (3)^2 = 25 - 9 = 16 \] \[ TA = \sqrt{16} = 4\,\text{cm} \]

Hence, the length of each tangent is \(4\,\text{cm}\).

Detailed Construction Steps (Beginner Level)

1. Draw the given triangle:
   • First, draw a line segment \(BC = 5\,\text{cm}\).
   • With \(B\) as centre and radius \(6\,\text{cm}\), draw an arc.
   • With \(C\) as centre and radius \(6\,\text{cm}\), draw another arc cutting the first one at \(A\).
   • Join \(AB\) and \(AC\). Now triangle \(ABC\) is an isosceles triangle with \(AB=AC=6\,\text{cm}\) and \(BC=5\,\text{cm}\).
2. Understand the requirement: We need another triangle \(PQR\) similar to \(ABC\), but bigger in size. In the similar triangle, side \(PQ\) must be \(8\,\text{cm}\), while in the original triangle, side \(AB = 6\,\text{cm}\).
3. Find the scale factor:
   • Scale factor \(k = \dfrac{PQ}{AB} = \dfrac{8}{6} = \dfrac{4}{3}\).
   • This means every side of the new triangle will be \(\tfrac{4}{3}\) times the corresponding side of the old triangle.
4. Calculate the new sides:
   • \(PR = AC \times k = 6 \times \dfrac{4}{3} = 8\,\text{cm}\).
   • \(QR = BC \times k = 5 \times \dfrac{4}{3} = \dfrac{20}{3}\,\text{cm} \approx 6.67\,\text{cm}\).
5. Construct the similar triangle:
   • On \(AB\), use the standard similarity method (divide a ray into equal parts, use parallel lines) to enlarge \(AB\) into \(PQ=8\,\text{cm}\).
   • Draw parallels through \(Q\) and use the same scale factor to get \(R\).
   • Join \(P, Q, R\) to form \(\triangle PQR\).

Justification

By construction, the ratio of corresponding sides is the same:

\(\dfrac{PQ}{AB} = \dfrac{PR}{AC} = \dfrac{QR}{BC} = \dfrac{4}{3}\).

Since two triangles have their corresponding sides in the same ratio and their included angles are equal, by the AA similarity rule, \(\triangle PQR \sim \triangle ABC\).

Thus the construction is correct.

Step-by-Step Construction (reduction by \(\dfrac{5}{7}\))

1. Draw the given triangle.
   - First, draw a line segment \(AB = 5\,\text{cm}\).
   - At point \(B\), use a protractor to make an angle of \(60^\circ\).
   - From this angle line, mark point \(C\) such that \(BC = 6\,\text{cm}\).
   - Join \(A\) to \(C\). Now, \(\triangle ABC\) is drawn.
2. Draw a helper ray.
   - From point \(B\), draw a ray \(BX\) making an acute angle (less than \(90^\circ\)) with side \(BC\).
3. Mark equal divisions.
   - On \(BX\), use a compass or scale to cut 7 equal parts. Name them \(B_1, B_2, ..., B_7\) starting from \(B\).
   - So, \(B = B_0\), then \(B_1, B_2,...,B_7\) are equally spaced points.
4. Connect the last point to C.
   - Join \(B_7\) to \(C\).
5. Use parallel lines to reduce.
   - Locate \(B_5\) (because the scale factor is \(\tfrac{5}{7}\)).
   - Draw a line from \(B_5\) parallel to \(B_7C\). Let this line meet the extended side \(BC\) at \(C'\).
6. Construct the reduced A'.
   - Through \(C'\), draw a line parallel to \(AC\).
   - This new line meets the line through \(B\) parallel to \(BA\) at point \(A'\).
7. Complete the reduced triangle.
   - Join \(A'\) and \(C'\). The new triangle \(A'BC'\) is the required reduced triangle.

Justification (Why this works)

From the Basic Proportionality Theorem (Thales’ theorem):

\[ \frac{BB_5}{BB_7} = \frac{5}{7} \]

This ratio is carried forward to the sides of the new triangle:

• \(BA' / BA = 5/7\)
• \(BC' / BC = 5/7\)
• \(A'C' / AC = 5/7\)

So, each side of \(\triangle A'BC'\) is \(\tfrac{5}{7}\) of the corresponding side of \(\triangle ABC\). Hence, the triangles are similar with scale factor \(\tfrac{5}{7}\).

Step-by-Step Construction

1. Draw a circle with centre \(O\) and radius \(4\,\text{cm}\).
2. We want the angle between tangents = \(60^\circ\). Remember: The angle made at the centre by joining the points of contact (\(\angle AOB\)) and the angle between tangents (\(\angle ATB\)) are supplementary. So, \(\angle AOB = 180^\circ - 60^\circ = 120^\circ\).
3. Now, mark two points \(A\) and \(B\) on the circle such that \(\angle AOB = 120^\circ\).
4. At point \(A\), draw a line \(t_A\) perpendicular to radius \(OA\). At point \(B\), draw a line \(t_B\) perpendicular to radius \(OB\). These are tangents to the circle.
5. Extend \(t_A\) and \(t_B\) until they meet at point \(T\). Now, \(AT\) and \(BT\) are the required tangents. The angle between them is \(60^\circ\).

Step-by-Step Justification

1. In triangle \(AOT\): - \(\angle A = 90^\circ\) (because tangent is perpendicular to radius). - \(\angle AOT = \tfrac{1}{2}\angle AOB = \tfrac{1}{2} \times 120^\circ = 60^\circ\).
2. Using cosine rule in right triangle: \[ \cos 60^\circ = \frac{OA}{OT} \] Here, \(OA = 4\,\text{cm}\) and \(\cos 60^\circ = \tfrac{1}{2}\).
3. So, \[ \tfrac{1}{2} = \frac{4}{OT} \quad \Rightarrow \quad OT = \frac{4}{1/2} = 8\,\text{cm}. \]

Therefore, the distance from the centre of the circle to the intersection point of tangents is \(8\,\text{cm}\).

Step-by-Step Construction

1. Draw the given triangle:
   • First, draw a line segment \(AB=4\,\text{cm}\).
   • At point \(A\), use a compass to draw an arc of radius \(9\,\text{cm}\).
   • At point \(B\), draw another arc of radius \(6\,\text{cm}\). The arcs meet at \(C\).
   • Join \(AC\) and \(BC\). Now, \(\triangle ABC\) is ready.
2. Prepare for enlargement: From \(A\), draw a ray \(AX\) making an acute angle with \(AB\).
3. Mark equal divisions: Since the scale factor is \(\tfrac{3}{2}\), divide the ray into 2 equal parts using a compass. Mark them as \(A_1\) and \(A_2\). Then, extend one more equal part to get \(A_3\). (So total = 3 equal divisions on \(AX\)).
4. Connect and draw parallel: Join \(A_2\) with \(B\). Through \(A_3\), draw a line parallel to \(A_2B\). This line meets the extension of \(AB\) at \(B'\).
5. Locate the third vertex: Through \(B'\), draw a line parallel to \(BC\). Through \(A\), draw a line parallel to \(AC\). The two lines meet at \(C'\).
6. Complete the enlarged triangle: Join \(B'C'\). The new triangle \(\triangle AB'C'\) is similar to \(\triangle ABC\).

Justification

By construction, each side is enlarged by the scale factor \(\dfrac{3}{2}\): \(AB' = \dfrac{3}{2} \times AB = 6\,\text{cm}\), \(BC' = \dfrac{3}{2} \times BC = 9\,\text{cm}\), \(AC' = \dfrac{3}{2} \times AC = 13.5\,\text{cm}\).

Thus, \(\dfrac{AB'}{AB} = \dfrac{AC'}{AC} = \dfrac{B'C'}{BC} = \dfrac{3}{2}\). This proves that \(\triangle AB'C' \sim \triangle ABC\) by AA similarity.

Congruence Check

For congruence, scale factor must be exactly 1 (sides must be equal). Here, the scale factor is \(\dfrac{3}{2}\ne1\). So, the triangles are similar but not congruent.