Maharashtra Board Class 10 Math Part 2 Solution Chapter 4 Practice Set 4.1 Geometric Constructions

Maharashtra Board Class 10 Math Part – 2 Solution Chapter 4 Practice Set 4.1 – Geometric Constructions

Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 4: Geometric Constructions. Marathi or English Medium Students of Class 10 get here Geometric Constructions full Exercise Solution.

Std

Maharashtra Class 10
Subject

Math Part 2 Solution

Chapter

Geometric Constructions
Practice Set

4.1

Practice set 4.1

 

(1) ∆ABC ~∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6cm, CA = 4.5 cm.

Construct ∆ABC and ∆LMN such that BC/MN = 5/4.

Solution:

Given, ∆ABC, AB = 5.5cm, BC = 6cm, CA = 4.5 cm.

BC/MN = 5/4; ∴ AB/LM = 5/4 ; CA/NL = 5/4 [Corresponding sides of similar ∆ are proportional]

Or, MN = 6×6/5 = 4.8 cm ∵ LM = 4×5.5/5 = 4.4 cm;

NL = 4×4.5/5 = 3.6 cm

Steps to draw a triangle:

(1) Draw a base with given length here BC & MN

(2) Draw two lines of length of other two sides.

(3) With your compass take the length of each side and draw on arc from each vertex of the base corresponding the length of the other two sides above the base.

(4) The intersection of the two arc is the remaining vertex here a & c.

 

(2) PQR ~ LTR. In PQR, PQ= 4.2 cm, QR=5.4 cm, pr= 4.8 cm. Construct PQR and LTR, such that PQ/LT= ¾.

Solution:

Given, △PQR ~ △LTR,

In △ PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm.

PQ/LT = ¾ , QR/TR = 3/4,

∴ PR/LR = ¾

(∵Corresponding sides of similar single are congruent)

Now, LT = 4× PQ/3

4×4.2/3= 5.6 cm

TR= 4×QR/3 = 4×5.4/3=7.2 cm

LR = 4×PR/3 = 4×4.8/3=6.4 cm

(1) Draw a base with given length here BC & MN

(2) Draw two lines of length of other two sides.

(3) With your compass take the length of each side and draw on arc from each vertex of the base corresponding the length of the other two sides above the base.

(4) The intersection of the two arc is the remaining vertex here a & c.

 

(3) RST ~ XYZ. In RST, RS= 4.5 CM, RST=40°, ST=5.7 cm Construct RST and XYZ, such that RS/XY = 3/5.

Solution:

Given, RST = 40°

△RST~△XYZ, RS=4.5cm, ST= 5.7 cm

RS/XY=3/5, ST/YZ=3/5, RT/XZ=3/5

(Corresponding sides of similar △ are Proportional)

∴ XY = 5×RS/3 = 4.5×5/3= 7.5 cm

YZ = 5×ST/3 = 5×5.7/3=4.5 cm

Steps to draw 40°on base:

(1) Draw a base of given length here ST & YZ

(2) Take the compass and draw an are from S & Y of any radius.

(3) With some radius cut the arc from the point where it intersects the base.

(4) Now bisect the smaller arc with the base point and the former are cut with same radius.

(5) Now bisect the ever smaller are using same method.

(6) Now, again bisect the new smaller arc using same method and draw a ray and cut it using the compass with the given length. Join three vertex.

 

(4) AMT ~ AHE. In AMT, AM = 6.3 cm, TAM = 50°, AT = 5.6 cm. AM/AH=7/5. Construct AHE.

Solution:

Given, △ AMT ~ △ AHE

AM = 6.3 CM, ∠TAM = 50 = 5.6 cm.

AM/AM = 7/5, ∴ AT/AE = 7/5

(Corresponding sides of similar △ arc proportional)

AH = 5×AM/7 = 5×6.3/ 7= 4.5 cm

AE = 5×AT/7 = 5×5.6/7 = 4 cm

∠TAM = ∠AM = 50

(Corresponding angles of similar △ are congruent)

Steps to draw 50° angle:

(1) Draw a base of given length.

(2) With any radius draw an arc from first point of the base.

(3) With same radius make two consecutive areas along the base.

(4) Bisect the two areas along the base with same radius you get a 20° angle from the base draw a line over.

(5) Now, from the original arc cut a point on the arc with the compass from the point where the arc cuts the base.

(6) Now from the new point on the arc cut another arc on the original arc using same radius.

(7) Now bisect the arc between the two cut points on the original are you get a 90° angle draw a line over it.

(8) Take the length of the there between 90° line cutting the arc and the point where the first arc was cut on the original arc.

(9) Now cut the original using the length from the point where the 20° line meets the original arc.

(10) The desired cut is a 50° angles now draw a ray and cut it with the given length which is the remaining vertex.

(11) Join all vertices to get the desired triangle.

 

Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 4 Geometric Constructions Practice Set 4.1

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.

Updated: April 7, 2022 — 3:44 pm

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