Maharashtra Board Class 10 Math Part – 2 Solution Chapter 4 Problem Set 4 – 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 |
Problem Set |
4 |
Problem set 4
(1) The number of tangents that can be drawn to a circle at a point on the circle is
(a) 3
(b) 2
(c) 1
(d) 0
Solution: (c) 1
(2) The maximum number of tangents that can be drawn to a circle from a point outside it is
(a) 2
(b) 1
(c) One and only one
(d) 0
Solution: (a) 2
(3) If △ABC ~ △PQR and AB/PQ = 7/5, Then
(a) △ABC is bigger
(b) △PQR is bigger
(c) Both triangles will be equal
(d) Cannot be decided
Solution: (a) △ABC is bigger.
(2) Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.
Solution:
Circle of radius 3.5 cm
Point P outside the circle at a length 5.7 cm from the Center
PR & PQ are the desired tangents from point P.
(3) Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.
Solution:
Line l is the desired of tangent on point A.
(4) Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.
Solution:
Circle of radius 6.4 cm
Point R is at a diameter of distance from the centre of the circle.
∴Let, Obc Centre of the circle
∴OR = 6.4+(6.4×2)
= 6.4+12.8
= 19.2 cm.
QR & PR are the desired tangents from points R.
Here, on radius = 6.4 cm, OR = 19.2 cm
(5) Draw a circle with centre P. Draw an arc AB of 100°measure. Draw tangents to the circle at point A and point B.
Solution:
Steps to construct:
(6) Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E-F-A and FA = 4.1 cm. Draw tangents to the circle from point A .
Solution:
Circle of centre E of radius 3.4
Point F is on the circle & Point A outside the circle such that E – F – A & FA = 4.1 cm
Here, EF = radius = 3.4 cm.
FA = 4.1 CM, E-F-A.
AM & AN arc the desired tangents from point A outside the circle.
(7) △ABC ~ △LBN. In △ABC, AB=5.1cm, ∠B = 40°, BC = 4.8 cm, AC/LN = 4/7. Construct △ABC and △LBN.
Solution:
Given, △ABC ~ △LBN
AB= 5.1 CM, LA= 40°, BC = 4.8cm.
AC/LN = 4/7, BC/BN= 4/7, AB/LB =4/7
(corresponding sides of similar triangle arc Proportional)
∴BC/BN= 4/7,
Or, 4.8×7/4=BN
Or, BN = 8.4 cm.
From the △ABC we can see that AC = 3.4cm.
∴ AC/LN = 4/7
3.4× 7/4= LN
LN = 5.95
△ABC & △LBN arc the desired triangle.
(8) Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm,
PQ = 5.8 cm. If YZ/YQ = 6/5 then construct ∆XYZ similar to ∆PYQ.
Solution:
Given, In∆PYQ
PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm
YZ/YQ = 6/5
To construct a similar triangle we need to take proportional sides to equal the value of the sides of triangle since corresponding sides of similar triangle are proportional as ∆XYZ.
∴ YZ/YQ = 6/5,
Let, the base of the triangles be YQ & YZ
∴We have to divide YZ into 5 equal parts.
YQ = 7.2
∴ YZ = 6×7.2/5 = 8.6 cm
Each equal parts = 7.2/5 = 1.4 cm
Steps to draw:
(1) Draw ∆PYQ with given dimensions
(2) Divide segment YQ into 5 equal parts
(3) Mark point Z such that YZ is 6 times each part of YQ
(4) Draw a parallel line to side QP, through Z we get point X on triangle XYZ.
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 4 Geometric Constructions Problem Set 4
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.