## Maharashtra State Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Practice Set 4.1

Question 1.

∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that \(\frac { BC }{ MN } \) = \(\frac { 5 }{ 4 } \)

Solution:

Analysis:

Question 2.

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that \(\frac { PQ }{ LT } \) = \(\frac { 3 }{ 4 } \)

Solution:

Analysis:

As shown in the figure, Let R – P – L and R – Q – T.

∆PQR ~ ∆LTR … [Given]

∴ ∠PRQ ≅ ∠LRT … [Corresponding angles of similar triangles]

\(\frac { PQ }{ LT } \) = \(\frac { QR }{ TR } \) = \(\frac { PR }{ LR } \) …(i)[Corresponding sides of similar triangles]

But, \(\frac { PQ }{ LT } \) = \(\frac { 3 }{ 4 } \) ….(ii) [Given]

∴ \(\frac { PQ }{ LT } \) = \(\frac { QR }{ TR } \) = \(\frac { PR }{ LR } \) = \(\frac { 3 }{ 4 } \) …[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 R_{1}, R_{2}, R_{3}, and R_{4}, such that RR_{1} = R_{1}R_{2} = R_{2}R_{3} = R_{3}R_{4}.

iii. Join R_{3}Q. Draw line parallel to R_{3}Q through R_{4} 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.

∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm. Construct ∆RST and ∆XYZ, such that \(\frac { RS }{ XY } \) = \(\frac { 3 }{ 5 } \).

Solution:

Analysis:

∆RST ~ ∆XYZ … [Given]

∴ ∠RST ≅ ∠XYZ = 40° … [Corresponding angles of similar triangles]

Question 4.

∆AMT ~ ∆ANE. In ∆AMT, AM = 6.3 cm, ∠TAM = 500, AT = 5.6 cm. \(\frac { AM }{ AH } \) = \(\frac { 7 }{ 5 } \) Construct ∆AHE.

Solution:

Analysis:

As shown in the figure,

Let A – H – M and A – E – T.

∆AMT ~ ∆AHE … [Given]

∴ ∠TAM ≅ ∠EAH … [Corresponding angles of similar triangles]

\(\frac { AM }{ AH } \) = \(\frac { MT }{ HE } \) = \(\frac { AT }{ AE } \) ….. (i)[Corresponding sides of similar triangles]

But, \(\frac { AM }{ AH } \) = \(\frac { 7 }{ 5 } \) …(ii)[Given]

∴ \(\frac { AM }{ AH } \) = \(\frac { MT }{ HE } \) = \(\frac { AT }{ AH } \) = \(\frac { 7 }{ 5 } \) …[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

AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}.

iii. Join A_{7}M. Draw line parallel to A_{7}M through A_{5} 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.

If length of side AB is \(\frac { 11.6 }{ 2 } \) cm, then by dividing the line segment of length 11.6 cm in three equal parts, draw segment AB. (Textbook pg. no. 93)

Solution:

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 A_{1}, A_{2} and A_{3 }on ray AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3}

iv. Join A_{3}D.

v. Through A_{1}, A_{2} draw lines parallel to A_{3}D intersecting AD at B and C, wherein

AB = \(\frac { 11.6 }{ 3 } \) cm

Question 2.

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:

Let ∆ABC be any triangle constructed such that AB = 7 cm, BC = 7 cm and AC = 6 cm.

Question 3.

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:

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 B_{1}, B_{2}, B_{3}, B_{4}, B_{5} starting from B at equal distance.

Join B_{3}C” (ie 3rd part)

Draw a line parallel to AB_{5} through B_{3} to 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.