Practice Set 4.1 Geometry 10th Std Maths Part 2 Answers Chapter 4 Geometric Constructions
Question 1
Maharashtra Board Solution
∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that =
Solution & Step-by-Step Answer:
Analysis:
Question 2
Maharashtra Board Solution
∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that =
Solution & Step-by-Step Answer:
Analysis: As shown in the figure, Let R – P – L and R – Q – T. ∆PQR ~ ∆LTR … [Given] ∴ ∠PRQ ≅ ∠LRT … [Corresponding angles of similar triangles] = = …(i)[Corresponding sides of similar triangles] But, = ….(ii) [Given] ∴ = = = …[From (i) and (ii)] ∴ sides of LTR are longer than corresponding sides of ∆PQR. If seg QR is divided into 3 equal parts, then seg TR will be 4 times each part of seg QR. So, if we construct ∆PQR, point T will be on side RQ, at a distance equal to 4 parts from R. Now, point L is the point of intersection of ray RP and a line through T, parallel to PQ. ∆LTR is the required triangle similar to ∆PQR. Steps of construction: i. Draw ∆PQR of given measure. Draw ray RS making an acute angle with side RQ. ii. Taking convenient distance on the compass, mark 4 points R1, R2, R3, and R4, such that RR1 = R1R2 = R2R3 = R3R4. iii. Join R3Q. Draw line parallel to R3Q through R4 to intersects ray RQ at T. iv. Draw a line parallel to side PQ through T. Name the point of intersection of this line and ray RP as L. ∆LTR is the required triangle similar to ∆PQR.
Question 3
Maharashtra Board Solution
∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm. Construct ∆RST and ∆XYZ, such that = .
Solution & Step-by-Step Answer:
Analysis: ∆RST ~ ∆XYZ … [Given] ∴ ∠RST ≅ ∠XYZ = 40° … [Corresponding angles of similar triangles]
Question 4
Maharashtra Board Solution
∆AMT ~ ∆ANE. In ∆AMT, AM = 6.3 cm, ∠TAM = 500, AT = 5.6 cm. = Construct ∆AHE.
Solution & Step-by-Step Answer:
Analysis: As shown in the figure, Let A – H – M and A – E – T. ∆AMT ~ ∆AHE … [Given] ∴ ∠TAM ≅ ∠EAH … [Corresponding angles of similar triangles] = = ….. (i)[Corresponding sides of similar triangles] But, = …(ii)[Given] ∴ = = = …[From (i) and (ii)] ∴ Sides of AAMT are longer than corresponding sides of ∆AHE. ∴ The length of side AH will be equal to 5 parts out of 7 equal parts of side AM. So, if we construct AAMT, point H will be on side AM, at a distance equal to 5 parts from A. Now, point E is the point of intersection of ray AT and a line through H, parallel to MT. ∆AHE is the required triangle similar to ∆AMT. Steps of construction: i. Draw ∆AMT of given measure. Draw ray AB making an acute angle with side AM. ii. Taking convenient distance on the compass, mark 7 points A1, A2, A3, A4, A5, Ag and A7, such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7. iii. Join A7M. Draw line parallel to A7M through A5 to intersects seg AM at H. iv. Draw a line parallel to side TM through H. Name the point of intersection of this line and seg AT as E. ∆AHE is the required triangle similar to ∆AMT.
Maharashtra Board Class 10 Maths Chapter 4 Geometric Constructions Intext Questions and Activities
Question 1
Maharashtra Board Solution
If length of side AB is cm, then by dividing the line segment of length 11.6 cm in three equal parts, draw segment AB. (Textbook pg. no. 93)
Solution & Step-by-Step Answer:
Steps of construction: i. Draw seg AD of 11.6 cm. ii. Draw ray AX such that ∠DAX is an acute angle. iii. Locate points A1, A2 and A3 on ray AX such that AA1 = A1A2 = A2A3 iv. Join A3D. v. Through A1, A2 draw lines parallel to A3D intersecting AD at B and C, wherein AB = cm
Question 2
Maharashtra Board Solution
Construct any ∆ABC. Construct ∆ A’BC’ such that AB : A’B = 5:3 and ∆ ABC ~ ∆ A’BC’. (Textbook pg. no. 93) Analysis: As shown in the figure, Let B – A’ – A and B – C’ -C ∆ ABC – A’BC’ … [Given] ∴ ∠ABC ≅ ∠A’BC’ …[Corresponding angles of similar trianglesi ∴ Sides of ∆ABC are longer than corresponding sides of ∆A’BC’. Rough figure ∴ the length of side BC’ will be equal to 3 parts out of 5 equal parts of side BC. So if we construct ∆ABC, point C’ will be on side BC, at a distance equal to 3 parts from B. Now A’ is the point of intersection of AB and a line through C’, parallel to CA.
Solution & Step-by-Step Answer:
Let ∆ABC be any triangle constructed such that AB = 7 cm, BC = 7 cm and AC = 6 cm.
Question 3
Maharashtra Board Solution
Construct any ∆ABC. Construct ∆A’BC’ such that AB: A’B = 5:3 and ∆ABC ~ ∆A’BC’. ∆A’BC’ can also be constructed as shown in the adjoining figure. What changes do we have to make in steps of construction in that case? (Textbook pg. no. 94)
Solution & Step-by-Step Answer:
Let ∆ABC be any triangle constructed such that AB = 5cm, BC = 5.5 cm and AC = 6 cm.
i. Steps of construction:
Construct ∆ABC, extend rays AB and CB.
Draw line BM making an acute angle with side AB.
Mark 5 points B1, B2, B3, B4, B5starting from B at equal distance.
Join B3C” (ie 3rd part)
Draw a line parallel to AB5through B3to intersect line AB at C”
Draw a line parallel to AC through C” to intersect line BC at A”
ii. Extra construction:
With radius BC” cut an arc on extended ray CB at C’ [C’ – B – C]
With radius BA” cut an arc on extended ray AB at A’ [A’ – B – A]
∆A’BC’ is the required triangle.