EXERCISE 11.1
Page No 219:
Question 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.Answer:
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
By construction, we have A5C || A13B. By applying Basic proportionality theorem for the triangle AA13B, we obtain
… (1)From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.
… (2)On comparing equations (1) and (2), we obtain
This justifies the construction.
Page No 220:
Question 2:
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are
of the corresponding sides of the first triangle.Give the justification of the construction.
Answer:
Step 1Draw 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.
Justification
The construction can be justified by proving that
By construction, we have B’C’ || BC
∴ ∠A
= ∠ABC (Corresponding angles)In ΔAB’C’ and ΔABC,
∠
= ∠ABC (Proved above)∠
= ∠BAC (Common)∴ Δ
∼ ΔABC (AA similarity criterion)
… (1)In ΔAA2B’ and ΔAA3B,
∠A2AB’ = ∠A3AB (Common)
∠AA2B’ = ∠AA3B (Corresponding angles)
∴ ΔAA2B’ ∼ ΔAA3B (AA similarity criterion)
From equations (1) and (2), we obtain
This justifies the construction.
Question 3:
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
of the corresponding sides of the first triangle.Give the justification of the construction.
Answer:
Step 1Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 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 5and 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.
Justification
The construction can be justified by proving that
In ΔABC and ΔAB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)
… (1)In ΔAA5B and ΔAA7B’,
∠A5AB = ∠A7AB’ (Common)
∠AA5B = ∠AA7B’ (Corresponding angles)
∴ ΔAA5B ∼ ΔAA7B’ (AA similarity criterion)
On comparing equations (1) and (2), we obtain
⇒
This justifies the construction.
Question 4:
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are
times the corresponding sides of the isosceles triangle.Give the justification of the construction.
Answer:
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
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.
Justification
The construction can be justified by proving that
In ΔABC and ΔAB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)
… (1)In ΔAA2B and ΔAA3B’,
∠A2AB = ∠A3AB’ (Common)
∠AA2B = ∠AA3B’ (Corresponding angles)
∴ ΔAA2B ∼ ΔAA3B’ (AA similarity criterion)
On comparing equations (1) and (2), we obtain
⇒
This justifies the construction.
Question 5:
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are
of the corresponding sides of the triangle ABC.Give the justification of the construction.
Answer:
A ΔA’BC’ whose sides areof the corresponding sides of ΔABC can be drawn as follows.Step 1
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3
Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.
Step 4
Join B4C and draw a line through B3, parallel to B4C intersecting BC at C’.
Step 5
Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.
Justification
The construction can be justified by proving
In ΔA’BC’ and ΔABC,
∠A’C’B = ∠ACB (Corresponding angles)
∠A’BC’ = ∠ABC (Common)
∴ ΔA’BC’ ∼ ΔABC (AA similarity criterion)
… (1)In ΔBB3C’ and ΔBB4C,
∠B3BC’ = ∠B4BC (Common)
∠BB3C’ = ∠BB4C (Corresponding angles)
∴ ΔBB3C’ ∼ ΔBB4C (AA similarity criterion)
From equations (1) and (2), we obtain
⇒
This justifies the construction.
Question 6:
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are
times the corresponding side of ΔABC. Give the justification of the construction.Answer:
∠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.
Step 1
Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3
Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.
Step 4
Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C’.
Step 5
Through C’, draw a line parallel to AC intersecting extended line segment at C’. ΔA’BC’ is the required triangle.
Justification
The construction can be justified by proving that
In ΔABC and ΔA’BC’,
∠ABC = ∠A’BC’ (Common)
∠ACB = ∠A’C’B (Corresponding angles)
∴ ΔABC ∼ ΔA’BC’ (AA similarity criterion)
… (1)In ΔBB3C and ΔBB4C’,
∠B3BC = ∠B4BC’ (Common)
∠BB3C = ∠BB4C’ (Corresponding angles)
∴ ΔBB3C ∼ ΔBB4C’ (AA similarity criterion)
On comparing equations (1) and (2), we obtain
⇒
This justifies the construction.
Question 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
times the corresponding sides of the given triangle. Give the justification of the construction.Answer:
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.
Step 1
Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.
Step 2
Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.
Step 3
Draw a ray AX making an acute angle with AB, opposite to vertex C.
Step 4
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.
Step 5
Join A3B. Draw a line through A5 parallel to A3B intersecting extended line segment AB at B’.
Step 6
Through B’, draw a line parallel to BC intersecting extended line segment AC at C’. ΔAB’C’ is the required triangle.
Justification
The construction can be justified by proving that
In ΔABC and ΔAB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)
… (1)In ΔAA3B and ΔAA5B’,
∠A3AB = ∠A5AB’ (Common)
∠AA3B = ∠AA5B’ (Corresponding angles)
∴ ΔAA3B ∼ ΔAA5B’ (AA similarity criterion)
On comparing equations (1) and (2), we obtain
⇒
This justifies the construction.
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