NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Give the justification of the construction.
Solution 1:
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1. Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.
Step 2. Locate 13 (= 5 + 8) points, A1, A2, A3, A4 …….. A13, on AX such that AA1 = A1A2 = A2A3 and so on.
Step 3. Join BA13.
Step 4. Through the point A5, draw a line parallel to BA13 (by making an angle equal to ∠AA13B) at A5 intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.
Justification
The construction can be justified by proving that
AC/CB=5/8
By construction, we have A5C || A13B. By applying Basic proportionality theorem for the triangle AA13B, we obtain
AC/CB=AA5/A5A13 … (1)
From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.
AA5/A5A13 =5/8 … (2)
On comparing equations (1) and (2), we obtain
AC/CB=5/8
This justifies the construction.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Give the justification of the construction.
Solution 2:
Step 1. Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.
Step 2. Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3. Locate 3 points A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that AA1= A1A2 = A2A3.
Step 4. Join BA3 and draw a line through A2 parallel to BA3 to intersect AB at point B'.
Step 5. Draw a line through B' parallel to the line BC to intersect AC at C'.
ΔAB'C' is the required triangle.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Give the justification of the construction.
Solution 3:
Step 1. Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
Step 2. Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3. Locate 7 points, A1, A2, A3, A4 A5, A6, A7 (as 7 is greater between 5 and 7), on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
Step 4. Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B'.
Step 5. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1 1/2 times the corresponding sides of the isosceles triangle. Give the justification of the construction.
Solution 4:
Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.A ΔAB'C' whose sides are 3/2 times of ΔABC can be drawn as follows.
Step 1. Draw a line segment AB of 8 cm.Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'.Join OO'. Let OO' intersect AB at D.
Step 2. Taking D as centre,draw an arc of 4 cm radius which cuts the extended line segment OO' at point C.An isosceles ΔABC is formed, having CD (altitude)as 4 cm and AB (base) as 8 cm.
Step 3. Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4. Locate 3 points (as 3 is greater between 3 and 2) A1, A2, and A3 on AX such that AA1= A1A2 = A2A3.
Step 5. Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B'.
Step 6. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC. Give the justification of the construction.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding side of ΔABC. Give the justification of the construction.
