NCERT Solutions Class 10 Maths Chapter 11: ConstructionsExercise 11.1In each of the following, give the justification of the construction also: 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. Solution Steps for the construction are as follows:
Justification: A_{5}C is parallel to A_{13}B by construction. Therefore, we can use basic proportionality theorem in triangle AA_{13}B. AC/BC = AA_{5}/A_{5}A_{13} By construction, AA_{5} has 5 equal divisions of segment AA_{13} and A_{5}A_{13} has 8 equal divisions. Thus, AA_{5}/A_{5}A_{13}= 5/8 Therefore, AC/BC = 5/8. Hence, the construction is justified. 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Solution Steps for the construction are as follows:
Justification: B'C' is parallel to BC by construction. Therefore ∠ABC = ∠AB'C' In ΔAB'C' and ΔABC, ∠ABC = ∠AB'C' (Already Proven) ∠BAC = ∠B'AC' (Common angle) Therefore, by the AA similarity criterion ΔAB'C' ∼ ΔABC. This implies that AB'/AB = B'C'/BC = AC'/AC (Corresponding sides of similar triangles) A_{2}B' is parallel to A_{3}B by construction. Therefore ∠AA_{2}B' = ∠AA_{3}B In ΔAA_{2}B' and ΔAA_{3}B, ∠A_{2}AB' = ∠A_{3}AB (Common angle) ∠AA_{2}B' = ∠AA_{3}B (Already Proven) Therefore, by the AA similarity criterion ΔAA_{2}B' ~ ΔAA_{3}B. This implies that AB'/AB = AA_{2}/AA_{3} (Corresponding sides of similar triangles) By construction, AA_{2} has 2 equal divisions of segment AA_{3} and AA_{3} has 3 equal divisions. Thus, AA_{2}/AA_{3} = 2/3 Therefore, AB'/AB = B'C'/BC = AC'/AC = 2/3 AB' = (2/3)AB B'C' = (2/3)BC AC'= (2/3)AC Hence, the construction is justified. 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Solution Steps for the construction are as follows:
Justification: B'C' is parallel to BC by construction. Therefore ∠ABC = ∠AB'C' In ΔAB'C' and ΔABC, ∠ABC = ∠AB'C' (Already Proven) ∠BAC = ∠B'AC' (Common angle) Therefore, by the AA similarity criterion ΔAB'C' ∼ ΔABC. This implies that AB'/AB = B'C'/BC = AC'/AC (Corresponding sides of similar triangles) A_{7}B' is parallel to A_{5}B by construction. Therefore ∠AA_{7}B' = ∠AA_{5}B In ΔAA_{7}B' and ΔAA_{5}B, ∠A_{7}AB' = ∠A_{5}AB (Common angle) ∠AA_{7}B' = ∠AA_{5}B (Already Proven) Therefore, by the AA similarity criterion ΔAA_{7}B' ~ ΔAA_{5}B. This implies that AB'/AB = AA_{7}/AA_{5} (Corresponding sides of similar triangles) By construction, AA_{7} has 7 equal divisions of segment AA_{7} and AA_{5} has 5 equal divisions. Thus, AA_{7}/AA_{5} = 7/5 Therefore, AB'/AB = B'C'/BC = AC'/AC = 7/5 AB' = (7/5)AB B'C' = (7/5)BC AC'= (7/5)AC Hence, the construction is justified. 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1½ times the corresponding sides of the isosceles triangle. Solution Steps for the construction are as follows:
Justification: A'C' is parallel to AC by construction. Therefore ∠A'C'B = ∠ACB In ΔA'BC' and ΔABC, ∠A'BC' = ∠ABC (Common angle) ∠A'C'B = ∠ACB (Already proven) Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC. This implies that A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles) B_{2}C is parallel to B_{3}C' by construction. Therefore ∠BB_{2}C = ∠BB_{3}C' In ΔBB_{2}C and ΔBB_{3}C', ∠CBB_{2} = ∠C'BB_{3} (Common angle) ∠BB_{2}C = ∠BB_{3}C' (Already Proven) Therefore, by the AA similarity criterion ΔBB_{2}C and ΔBB_{3}C'. This implies that BB_{2}/BB_{3} = BC/BC' (Corresponding sides of similar triangles) By construction, BB_{3} has 3 equal divisions of segment BB_{3} and BB_{2} has 2 equal divisions. Thus, BB_{2}/BB_{3} = 2/3 Therefore, A'B/AB = BC'/BC = A'C'/AC = 3/2 A'B= (3/2)AB BC' = (3/2)BC A'C'= (3/2)AC Hence, the construction is justified. 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC. Solution Steps for the construction are as follows:
Justification: A'C' is parallel to AC by construction. Therefore ∠A'C'B = ∠ACB In ΔA'BC' and ΔABC, ∠A'BC' = ∠ABC (Common angle) ∠A'C'B = ∠ACB (Already proven) Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC. This implies that A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles) B_{4}C is parallel to B_{3}C' by construction. Therefore ∠BB_{4}C = ∠BB_{3}C' In ΔBB_{4}C and ΔBB_{3}C', ∠CBB_{4} = ∠C'BB_{3} (Common angle) ∠BB_{4}C = ∠BB_{3}C' (Already Proven) Therefore, by the AA similarity criterion ΔBB_{4}C and ΔBB_{3}C'. This implies that BB_{4}/BB_{3} = BC/BC' (Corresponding sides of similar triangles) By construction, BB_{3} has 3 equal divisions of segment BB_{4} and BB_{4} has 4 equal divisions. Thus, BB_{4}/BB_{3} = 4/3 Therefore, A'B/AB = BC'/BC = A'C'/AC = 3/4 A'B= (3/4)AB BC' = (3/4)BC A'C'= (3/4)AC Hence, the construction is justified. 6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC. Solution Steps for the construction are as follows:
Justification: A'C' is parallel to AC by construction. Therefore ∠A'C'B = ∠ACB In ΔA'BC' and ΔABC, ∠A'BC' = ∠ABC (Common angle) ∠A'C'B = ∠ACB (Already proven) Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC. This implies that A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles) B_{3}C is parallel to B_{4}C' by construction. Therefore ∠BB_{3}C = ∠BB_{4}C' In ΔBB_{3}C and ΔBB_{4}C', ∠CBB_{3} = ∠C'BB_{4} (Common angle) ∠BB_{3}C = ∠BB_{4}C' (Already Proven) Therefore, by the AA similarity criterion ΔBB_{3}C and ΔBB_{4}C'. This implies that BB_{3}/BB_{4} = BC/BC' (Corresponding sides of similar triangles) By construction, BB_{3} has 3 equal divisions of segment BB_{4} and BB_{4} has 4 equal divisions. Thus, BB_{3}/BB_{4} = 3/4 Therefore, A'B/AB = BC'/BC = A'C'/AC = 4/3 A'B= (4/3)AB BC' = (4/3)BC A'C'= (4/3)AC Hence, the construction is justified. 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Solution Steps for the construction are as follows:
Justification: A'C' is parallel to AC by construction. Therefore ∠A'C'B = ∠ACB In ΔA'BC' and ΔABC, ∠A'BC' = ∠ABC (Common angle) ∠A'C'B = ∠ACB (Already proven) Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC. This implies that A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles) B_{3}C is parallel to B_{5}C' by construction. Therefore ∠BB_{3}C = ∠BB_{5}C' In ΔBB_{3}C and ΔBB_{5}C', ∠CBB_{3} = ∠C'BB_{5} (Common angle) ∠BB_{3}C = ∠BB_{5}C' (Already Proven) Therefore, by the AA similarity criterion ΔBB_{3}C and ΔBB_{5}C'. This implies that BB_{3}/BB_{5} = BC/BC' (Corresponding sides of similar triangles) By construction, BB_{3} has 3 equal divisions of segment BB_{5} and BB_{5} has 5 equal divisions. Thus, BB_{3}/BB_{5} = 3/5 Therefore, A'B/AB = BC'/BC = A'C'/AC = 5/3 A'B= (5/3)AB BC' = (5/3)BC A'C'= (5/3)AC Hence, the construction is justified. Exercise 11.2In each of the following, give also the justification of the construction: 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Solution Steps for the construction are as follows:
Justification: We need to connect O with B and C. ∠B is an angle in a semicircle subtended by a diameter. Therefore, it will be a right angle. This means that OB will be perpendicular to AB. Since, the radius OB of the circle is perpendicular to the line AB that touches it at one point. Therefore, AB is a tangent to this circle. Similarly, AC is a tangent to this circle. Hence, the construction is justified. 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. Solution Steps for the construction are as follows:
Verification: ABO forms a right triangle. Therefore, by using Pythagoras Theorem in ABO: AO^{2} = AB^{2} + OB^{2} 6^{2} = AB^{2} + 4^{2} 36  16 = AB^{2} 20 = AB^{2} AB = √20 cm AB = 4.47 cm Tangents to a circle from an external point are equal, so AB = AC = 4.47 cm The result has been verified. Justification: We need to connect O with B and C. ∠B is an angle in a semicircle subtended by a diameter. Therefore, it will be a right angle. This means that OB will be perpendicular to AB. Since, the radius OB of the circle is perpendicular to the line AB that touches it at one point. Therefore, AB is a tangent to this circle. Similarly, AC is a tangent to this circle. Hence, the construction is justified. 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. Solution Steps for the construction are as follow:
Justification: We need to connect O with A, B, C and D. ∠A is an angle in a semicircle subtended by a diameter. Therefore, it will be a right angle. This means that OA will be perpendicular to PA. Since, the radius OA of the circle is perpendicular to the line PA that touches it at one point. Therefore, PA is a tangent to this circle. Similarly, PB, QC, and QD are tangents to this circle. Hence, the construction is justified. 4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. Solution Steps for the construction are as follows:
Justification: By construction we have, ∠B = 90° ∠C = 90° And ∠O = 120° OABC forms a quadrilateral, therefore ∠O + ∠A + ∠B + ∠C = 360° 120° + ∠A + 90° + 90° = 360° ∠A + 300° = 360° ∠A = 60° Hence, the construction is justified. 5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Solution Steps for the construction are as follows:
Justification: We need to connect A with P and Q and connect B with R and S. ∠P is an angle in a semicircle subtended by a diameter. Therefore, it will be a right angle. This means that AP will be perpendicular to PB. Since, the radius AP of the circle is perpendicular to the line PB that touches it at one point. Therefore, PB is a tangent to this circle. Similarly, QB is also a tangent to the circle with centre A. Similarly, we can prove that AR and AS are tangents to the circle with the centre B. Hence, the construction is justified. 6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. Solution Steps for the construction are as follows:
Justification: We need to join O with E. ∠E is an angle in a semicircle subtended by a diameter. Therefore, it will be a right angle. This means that OE will be perpendicular to AE. Since, the radius OE of the circle is perpendicular to the line AE that touches it at one point. Therefore, AE is a tangent to this circle. Similarly, AB is a tangent to this circle. Hence, the construction is justified. 7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. Solution Steps for the construction are as follows:
Justification: We need to join the points Q and R with centre O. ∠Q is an angle in a semicircle subtended by a diameter. Therefore, it will be a right angle. This means that OQ will be perpendicular to PQ. Since, the radius OQ of the circle is perpendicular to the line PQ that touches it at one point. Therefore, PQ is a tangent to this circle. Similarly, PR is a tangent to this circle. Hence, the construction is justified.
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