Chapter 17 Geometrical Constructions Practice Set 40 Solutions
Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 17 Geometrical Constructions Practice Set 40. Step-by-step solved exercises, numerical problems, and digest answers.
6 Solved Questions• 397 words
6th Standard Maths Practice Set 40 Answers Chapter 17 Geometrical Constructions
Question 1Maharashtra Board Solution
Draw line l. Take point P anywhere outside the line. Using a set square draw a line PQ perpendicular to line l.
Solution & Step-by-Step Answer:
Step 1:
Step 2:
line PQ ⊥ line l.
Question 2Maharashtra Board Solution
Draw line AB. Take point M anywhere outside the line. Using a compass and ruler, draw a line MN perpendicular to line AB.
Solution & Step-by-Step Answer:
Step 1:
Step 2:
Step 3:
line MN ⊥ line AB.
Question 3Maharashtra Board Solution
Draw a line segment AB of length 5.5 cm. Bisect it using a compass and ruler.
Solution & Step-by-Step Answer:
Step 1:
Step 2:
line MN is the perpendicular bisector of seg AB.
Question 4Maharashtra Board Solution
Take point R on line XY. Draw a perpendicular to XY at R, using a set square.
Solution & Step-by-Step Answer:
Step 1:
Step 2:
line TR ⊥ line XY.
Maharashtra Board Class 6 Maths Chapter 17 Geometrical Constructions Practice Set 40 Questions and Activities
Question 1Maharashtra Board Solution
In the above construction, why must the distance in the compass be kept constant? (Textbook pg. no. 90)
Solution & Step-by-Step Answer:
The point N is at equal distance from points P and Q. If we change the distance of the compass while drawing arcs from points P and Q, we will not get a point which is at equal distance from P and Q. Hence, the distance in the compass must be kept constant.
Question 2Maharashtra Board Solution
The Perpendicular Bisector. (Textbook pg. no. 90)
Solution & Step-by-Step Answer:
Question 3. Take a rectangular sheet of paper. Fold the paper so that the lower edge of the paper falls on its top edge, and fold it over again from right to left. Observe the two folds that have formed on the. paper. Verify that each fold is a perpendicular bisector of the other. Then measure the following distances. (Textbook pg. no. 91) i. l(XP) ii. l(XA) iii. l(XB) iv. l(YP) v. l(YA) You will observe that l(XP) = l(YP), l(XA) = l(YA) and l(XB) = l(YB) Therefore we can conclude that all points on the vertical fold (perpendicular bisector) are equidistant from the endpoints of the horizontal fold. Solution: [Note: Students should attempt this activity on their own.]