Access Answers for SAT Maths Chapter 11 – Constructions
Exercise 11.1
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Construction Procedure:
A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows.
1. Draw line segment AB with the length measure of 7.6 cm
2. Draw a ray AX that makes an acute angle with line segment AB.
3. Locate the points i.e.,13 (= 5+8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX such that it
becomes AA1 = A1A2 = A2A3 and so on.
4. Join the line segment and the ray, BA13.
5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B
6. The point A5 which intersects the line AB at point C.
7. C is the point divides line segment AB of 7.6 cm in the required ratio of 5:8.
8. Now, measure the lengths of the line AC and CB. It comes out to the measure of 2.9 cm and 4.7
cm respectively.
Justification:
The construction of the given problem can be justified by proving that
AC/CB = 5/ 8
By construction, we have A5C || A13B. From Basic proportionality theorem for the triangle AA13B, we
get
,AC/CB =AA5/A5A13….. (1)
From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of
line segments respectively.
Therefore, it becomes
AA5/A5A13=5/8… (2)
Compare the equations (1) and (2), we obtain
AC/CB = 5/ 8
Hence, Justified.
2. 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.
Construction Procedure:
1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.
2. Take the point A as centre, and draw an arc of radius 5 cm.
3. Similarly, take the point B as its centre, and draw an arc of radius 6 cm.
4. The arcs drawn will intersect each other at point C.
5. Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.
6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of
vertex C.
7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it
becomes AA1= A1A2 = A2A3.
8. Join the point BA3 and draw a line through A2which is parallel to the line BA3 that intersect AB at
point B’.
9. Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.
10. Therefore, ΔAB’C’ is the required triangle.
, Justification:
The construction of the given problem can be justified by proving that
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
From the construction, we get B’C’ || BC
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAAB’ and ΔAAB,
∠A2AB’ =∠A3AB (Common)
From the corresponding angles, we get,
∠AA2B’ =∠AA3B
Therefore, from the AA similarity criterion, we obtain
ΔAA2B’ and AA3B
So, AB’/AB = AA2/AA3
Exercise 11.1
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Construction Procedure:
A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows.
1. Draw line segment AB with the length measure of 7.6 cm
2. Draw a ray AX that makes an acute angle with line segment AB.
3. Locate the points i.e.,13 (= 5+8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX such that it
becomes AA1 = A1A2 = A2A3 and so on.
4. Join the line segment and the ray, BA13.
5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B
6. The point A5 which intersects the line AB at point C.
7. C is the point divides line segment AB of 7.6 cm in the required ratio of 5:8.
8. Now, measure the lengths of the line AC and CB. It comes out to the measure of 2.9 cm and 4.7
cm respectively.
Justification:
The construction of the given problem can be justified by proving that
AC/CB = 5/ 8
By construction, we have A5C || A13B. From Basic proportionality theorem for the triangle AA13B, we
get
,AC/CB =AA5/A5A13….. (1)
From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of
line segments respectively.
Therefore, it becomes
AA5/A5A13=5/8… (2)
Compare the equations (1) and (2), we obtain
AC/CB = 5/ 8
Hence, Justified.
2. 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.
Construction Procedure:
1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.
2. Take the point A as centre, and draw an arc of radius 5 cm.
3. Similarly, take the point B as its centre, and draw an arc of radius 6 cm.
4. The arcs drawn will intersect each other at point C.
5. Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.
6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of
vertex C.
7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it
becomes AA1= A1A2 = A2A3.
8. Join the point BA3 and draw a line through A2which is parallel to the line BA3 that intersect AB at
point B’.
9. Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.
10. Therefore, ΔAB’C’ is the required triangle.
, Justification:
The construction of the given problem can be justified by proving that
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
From the construction, we get B’C’ || BC
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAAB’ and ΔAAB,
∠A2AB’ =∠A3AB (Common)
From the corresponding angles, we get,
∠AA2B’ =∠AA3B
Therefore, from the AA similarity criterion, we obtain
ΔAA2B’ and AA3B
So, AB’/AB = AA2/AA3
