**Question 1.**

Solution:

Solution:

Analysis:

**Question 2.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]

∴ 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.Solution:**

Analysis:

∆RST ~ ∆XYZ … [Given]

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

**Question 4.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]

∴ 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.

**Intext Questions and Activities**

**Question 1.If length of side AB is 11.6/3 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 = 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.