Class 10: Maths Chapter 4 solutions. Complete Class 10 Maths Chapter 4 Notes.

Contents

## Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 4- Geometric Constructions

Maharashtra Board 10th Maths Chapter 4, Class 10 Maths Chapter 4 solutions

#### 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 BCMN = 54Solution:**

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 PQLT = 34Solution:**

Analysis:

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

∆PQR ~ ∆LTR … [Given]

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

PQLT = QRTR = PRLR …(i)[Corresponding sides of similar triangles]

But, PQLT = 34 ….(ii) [Given]

∴ PQLT = QRTR = PRLR = 34 …[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 RSXY = 35.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. AMAH = 75 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]

AMAH = MTHE = ATAE ….. (i)[Corresponding sides of similar triangles]

But, AMAH = 75 …(ii)[Given]

∴ AMAH = MTHE = ATAH = 75 …[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.

**Question 1.If length of side AB is 11.62 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 = 11.63 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.

#### Practice Set 4.2

**Question 1.Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.Solution:**

Analysis:

seg PM ⊥ line l ….[Tangent is perpendicular to radius]

The perpendicular to seg PM at point M will give the required tangent at M.

**Question 2.Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.Solution:**

Analysis:

seg OM ⊥ line l …[Tangent is perpendicular to radius]

The perpendicular to seg OM at point M will give the required tangent at M.

**Question 3.Draw a circle of radius 3.6 cm. Draw a tangent to the circle at any point on it without using the centre.Solution:**

Analysis:

As shown in the figure, line lis a tangent to the circle at point K.

seg BK is a chord of the circle and LBAK is an inscribed angle.

By tangent secant angle theorem,

∠BAK = ∠BKR

By converse of tangent secant angle theorem,

If we draw ∠BKR such that ∠BKR = ∠BAK, then ray KR

i.e. (line l) is a tangent at point K.

**Question 4.Draw a circle of radius 3.3 cm. Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q. Write your observation about the tangents.Solution:Analysis:**

seg OP ⊥ line l …[Tangent is perpendicular to radius]

seg OQ ⊥ line m

The perpendicular to seg OP and seg OQ at points P and Q

respectively will give the required tangents at P and Q.

Radius = 3.3 cm

∴ Diameter = 2 × 3.3 = 6.6 cm

∴ Chord PQ is the diameter of the circle.

∴ The tangents through points P and Q (endpoints of diameter) are parallel to each other.

**Question 5.Draw a circle with radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. Construct tangents at points M and N to the circle.Solution:**

Analysis:

seg ON ⊥ linel l

seg OM ⊥ Iine m …….[Tangent is perpendicular to radius]

The perpendicular to seg ON and seg 0M at points N and M respectively will give the required tangents at N and M.

**Question 6.Draw a circle with centre P and radius 3.4 cm. Take point Q at a distance 5.5 cm from the centre. Construct tangents to the circle from point Q.Solution:**

Analysis:

As shown in the figure, let Q be a point in the exterior of circle at a distance of 5.5 cm.

Let QR and QS be the tangents to the circle at points R and S respectively.

∴ seg PR ⊥ tangent QR …[Tangent is perpendicular to radius]

∴ ∠PRQ = 90°

∴ point R is on the circle having PQ as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly, point S also lies on the circle having PQ as diameter.

∴ Points R and S lie on the circle with PQ as diameter.

On drawing a circle with PQ as diameter, the points where it intersects the circle with centre P, will be the positions of points R and S respectively.

Ray QR and QS are the required tangents to the circle from point Q.

**Question 7.Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.Solution:**

Analysis:

As shown in the figure, let Q be a point in the exterior of circle at a distance of 5.5 cm.

Let QR and QS be the tangents to the circle at points R and S respectively.

∴ seg PR ⊥ tangent QR …[Tangent is perpendicular to radius]

∴ ∠PRQ = 90°

∴ point R is on the circle having PQ as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly, point S also lies on the circle having PQ as diameter.

∴ Points R and S lie on the circle with PQ as diameter.

On drawing a circle with PQ as diameter, the points where it intersects the circle with centre P, will be the positions of points R and S respectively.

Ray QR and QS are the required tangents to the circle from point Q.

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Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 4- Geometric Constructions

**Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 2) :**

- Chapter 1- Similarity
- Chapter 2- Pythagoras Theorem
- Chapter 3- Circle
- Chapter 4- Geometric Constructions
- Chapter 5- Co-ordinate Geometry
- Chapter 6- Trigonometry
- Chapter 7- Mensuration

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