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:
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
- Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.
- Locate 13 (= 5 + 8) points, A1, A2, A3, A4 …….. A13, on AX such that AA1 = A1A2 = A2A3 and so on.
- Join BA13.
- 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.
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
2. Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3of the corresponding sides of the first triangle. Give the justification of the construction.
Solution:
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 ∆ABC is the required triangle.
2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
3: Locate 3 points A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that AA1 = A1A2 = A2A3.
4: Join BA3 and draw a line through A2 parallel to BA3 to intersect AB at point B’.
5: Draw a line through B’ parallel to the line BC to intersect AC at C’. ∆AB’C’ is the required triangle.
The construction can be justified by proving that
AB’ = 2/3 AB, B’C’ = 2/3 BC,AC’ = 2/3 AC
By construction, we have B’C’ || BC
∴ ∠A = ∠ABC (Corresponding angles)
In ∆AB’C’ and ∆ABC,
∠AB’C’ = ∠ABC (Proved)
∠B’AC’ = ∠BAC (common)
∆AB’C’ ~ ∆ABC (AA similarity criteirion)
That implies,
AB’/AB = B’C’/BC = AC’/AC ——–(1)
In ∆AA2B’ and ∆AA3B,
∠A2AB’ = ∠A3AB (Common)
∠AA2B’ = ∠AA3B (Corresponding angles)
∴ ∆AA2B’ ∼ ∆AA3B (AA similarity criterion)
⇒ AB’/AB = (AA2/AA3)
⇒ AB’ /AB = 2/3 ——–(2)
From equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 2/3
⇒ AB’ =2/3 AB
B’C’ =2/3 BC
AC’ =2/3 AC
3. 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:
- 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. ∆ABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
- Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
- Locate 7 points, A1, A2, A3, A4 A5, A6, A7 (as 7 is greater between 5and 7), on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
- Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B’.
- Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ∆AB’C’ is the required triangle.
The construction can be justified by proving that
AB’ = 7/5 AB, B’C’ = 7/5 BC,AC’ = 7/5 AC
In ∆ABC and ∆AB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (common)
∆ABC ~ ∆AB’C’ (AA similarity criteirion)
That implies,
AB/AB’ = BC/B’C’ = AC/AC’ ——–(1)
In ∆AA5B and ∆AA7B’,
∠AA5 B = ∠AA7 B’ (Common)
∠AA5 B = ∠AA7 B’ (Corresponding angles)
∴ ∆AA5B ∼ ∆AA7B’ (AA similarity criterion)
⇒ AB/AB’ = (AA5/AA7)
⇒ AB /AB’ = 5/7 ——–(2)
From equations (1) and (2), we obtain
AB/AB’ = BC/B’C’ = AC/AC’ =5/7
⇒ AB’ =7/5 AB
B’C’ =7/5 BC
AC’ =7/5 AC
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1 ½ times the corresponding sides of the isosceles triangle. Give the justification of the construction.
Solution:
Let us assume that ∆ABC 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 ∆AB’C’ whose sides are 3/2 times of ∆ABC can be drawn as follows.
- 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.
- Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ∆ABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
- Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
- Locate 3 points (as 3 is greater between 3 and 2) A1, A2, and A3 on AX such that AA1= A1A2 = A2A3.
- Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B’.
- Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ∆AB’C’ is the required triangle.
The construction can be justified by proving that
AB’ = 3/2 AB, B’C’ = 3/2 BC,AC’ = 3/2AC
In ∆ABC and ∆AB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (common)
∆ABC ~ ∆AB’C’ (AA similarity criteirion)
That implies,
AB/AB’ = BC/B’C’ = AC/AC’ ——–(1)
In ∆AA5B and ∆AA7B’,
∠AA2 B = ∠AA3 B’ (Common)
∠AA B = ∠AA3 B’ (Corresponding angles)
∴ ∆AA5B ∼ ∆AA7B’ (AA similarity criterion)
⇒ AB/AB’ = (AA5/AA7)
⇒ AB /AB’ = 2/3 ——–(2)
From equations (1) and (2), we obtain
AB/AB’ = BC/B’C’ = AC/AC’ =2/3
⇒ AB’ =3/2 AB
B’C’ =3/2 BC
AC’ =3/2 AC
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC. Give the justification of the construction
Solution:
A ∆A’BC’ whose sides are ¾ of the corresponding sides of ∆ABC can be drawn as follows.
- Draw a ∆ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
- Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
- Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.
- Join B4C and draw a line through B3, parallel to B4C intersecting BC at C’.
- Draw a line through C’ parallel to AC intersecting AB at A’. ∆A’BC’ is the required triangle.
The construction can be justified by proving that
A’B = 3/4 AB, BC’ = 3/4 BC, A’C’ = 3/4 AC
In ∆A’BC’ and ∆ABC,
∠A’C’B = ∠ACB (Corresponding angles)
∠A’BC’ = ∠BAC (common)
∆A’BC’ ~ ∆ABC (AA similarity criteirion)
That implies,
AB/AB’ = BC/B’C’ = AC/AC’ ——–(1)
In ∆BB5C’ and ∆BB4 C,
∠B3BC’ = ∠B4 BC (Common)
∠BB3C’ = ∠BB4 C (Corresponding angles)
∴ ∆BB3C’ ∼ ∆BB4C (AA similarity criterion)
⇒ BC’/BC = (BB3/BB4)
⇒ BC’ /BC = 3/4 ——–(2)
From equations (1) and (2), we obtain
A’B/AB = BC’/BC = A’C’/AC =3/4
⇒ A’B =3/4 AB
BC’ =3/2 BC
A’C’ =3/4 AC
6. 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 ∆ABC. Give the justification of the construction
Solution:
∠B = 45°, ∠A = 105°
Sum of all interior angles in a triangle is 180°.
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 180° − 150°
∠C = 30°
The required triangle can be drawn as follows.
- Draw a ∆ABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
- Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
- Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.
- Join B3 Draw a line through B4 parallel to B3C intersecting extended BC at C’.
- Through C’, draw a line parallel to AC intersecting extended line segment at C’. ∆A’BC’ is the required triangle.
The construction can be justified by proving that
A’B = 4/3 AB, BC’ = 4/3BC, A’C’ = 4/3AC
In ∆ABC and ∆A’BC’,
∠ABC = ∠A’BC’ (Common angles)
∠ACB = ∠A’C’B (corresponding angles)
∆ABC ~ ∆A’BC’ (AA similarity criteirion)
That implies,
AB/A’B = BC/BC’ = AC/A’C’ ——–(1)
In ∆B3BC and ∆B4B C’,
∠B3BC = ∠B4 BC’ (Common)
∠BB3C = ∠BB4 C’ (Corresponding angles)
∴ ∆BB3C ∼ ∆BB4C’ (AA similarity criterion)
⇒ BC/BC’ = (BB3/BB4)
⇒ BC/BC’ = 3/4 ——–(2)
From equations (1) and (2), we obtain
AB/A’B = BC/BC’ = AC/A’C’ =3/4
⇒ A’B =4/3 AB
BC’ =4/3 BC
A’C’ =4/3 AC
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Give the justification of the construction.
Solution:
It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows.
- Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.
- Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ∆ABC is the required triangle.
- Draw a ray AX making an acute angle with AB, opposite to vertex C.
- Locate 5 points (as 5 is greater in 5 and 3), A1, A2, A3, A4, A5, on line segment AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
- Join A3 Draw a line through A5 parallel to A3B intersecting extended line segment AB at B’.
- Through B’, draw a line parallel to BC intersecting extended line segment AC at C’. ∆AB’C’ is the required triangle
The construction can be justified by proving that
AB’ = 5/3 AB, B’C’ = 5/3BC, AC’ = 5/3 AC
In ∆ABC and ∆AB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (common angles)
∆ABC ~ ∆AB’C’ (AA similarity criteirion)
That implies,
AB/AB’ = BC/B’C’ = AC/AC’ ——–(1)
In ∆AA3B and ∆AA5B’,
∠A3AB = ∠A5 AB’ (Common)
∠AA3B = ∠AA5 B’ (Corresponding angles)
∴ ∆AA3B ∼ ∆AA5B’ (AA similarity criterion)
⇒ AB/AB’ = (AA3/AA5)
⇒ AB/AB’ = 3/5 ——–(2)
From equations (1) and (2), we obtain
AB/AB’ = BC/B’C’ = AC/AC’ =3/5
⇒ AB’ =5/3 AB
B’C’ =5/3 BC
AC’ =5/3 AC
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