Selina Solutions Concise Maths Class 7 Chapter 15 Triangles provides students with a clear idea about the basic concepts covered in this chapter. The solutions are prepared by faculty after conducting research on each topic, keeping in mind the understanding capacity of students. Here, the students can download Selina Solutions Concise Maths Class 7 Chapter 15 Triangles free PDF, from the links which are provided here.
Chapter 15 helps students understand the types of triangles based on the angles and length of sides. The solutions improve problem solving and analytical thinking skills among students, which are important from the exam point of view.
Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Download PDF
Exercises of Selina Solutions Concise Maths Class 7 Chapter 15 – Triangles
Access Selina Solutions Concise Maths Class 7 Chapter 15: Triangles
Exercise 15A page: 176
1. State, if the triangles are possible with the following angles :
(i) 20°, 70° and 90°
(ii) 40°, 130° and 20°
(iii) 60°, 60° and 50°
(iv) 125°, 40° and 15°
Solution:
In a triangle, the sum of three angles is 1800
(i) 20°, 70° and 90°
Sum = 200 + 700 + 900 = 1800
Here the sum is 1800 and therefore it is possible.
(ii) 40°, 130° and 20°
Sum = 40° + 130° + 20° = 1900
Here the sum is not 1800 and therefore it is not possible.
(iii) 60°, 60° and 50°
Sum = 60° + 60° + 50° = 1700
Here the sum is not 1800 and therefore it is not possible.
(iv) 125°, 40° and 15°
Sum = 125° + 40° + 15° = 1800
Here the sum is 1800 and therefore it is possible.
2. If the angles of a triangle are equal, find its angles.
Solution:
In a triangle, the sum of three angles is 1800
So each angle = 1800/3 = 600
3. In a triangle ABC, ∠A = 45° and ∠B = 75°, find ∠C.
Solution:
In a triangle, the sum of three angles is 1800
∠A + ∠B + ∠C = 1800
Substituting the values
450 + 750 + ∠C = 1800
By further calculation
1200 + ∠C = 1800
So we get
∠C = 1800 – 1200 = 600
4. In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
Solution:
Consider ∠Q = ∠R = x
∠P = 60°
We can write it as
∠P + ∠Q + ∠R = 1800
Substituting the values
600 + x + x = 1800
By further calculation
600 + 2x = 1800
2x = 1800 – 600 = 1200
So we get
x = 1200/2 = 600
∠Q = ∠R = 600
Therefore, ∠R = 600.
5. Calculate the unknown marked angles in each figure:
Solution:
In a triangle, the sum of three angles is 1800
(i) From figure (i)
900 + 300 + x = 1800
By further calculation
1200 + x = 1800
So we get
x = 1800 – 1200 = 600
Therefore, x = 600.
(ii) From figure (ii)
y + 800 + 200 = 1800
By further calculation
y + 1000 = 1800
So we get
y = 1800 – 1000 = 800
Therefore, y = 800.
(iii) From figure (iii)
a + 900 + 400 = 1800
By further calculation
a + 1300 = 1800
So we get
a = 1800 – 1300 = 500
Therefore, a = 500.
6. Find the value of each angle in the given figures:
Solution:
(i) From the figure (i)
∠A + ∠B + ∠C = 1800
Substituting the values
5x0 + 4x0 + x0 = 1800
By further calculation
10x0 = 1800
x = 180/10 = 180
So we get
∠A = 5x0 = 5 × 180 = 900
∠B = 4x0 = 4 × 180 = 720
∠C = x = 180
(ii) From the figure (ii)
∠A + ∠B + ∠C = 1800
Substituting the values
x0 + 2x0 + 2x0 = 1800
By further calculation
5x0 = 1800
x0 = 1800/5 = 360
So we get
∠A = x0 = 360
∠B = 2x0 = 2 × 360 = 720
∠C = 2x0 = 2 × 360 = 720
7. Find the unknown marked angles in the given figure:
Solution:
(i) From the figure (i)
∠A + ∠B + ∠C = 1800
Substituting the values
b0 + 500 + b0 = 1800
By further calculation
2b0 = 1800 – 500 = 1300
b0 = 1300/2 = 650
Therefore, ∠A = ∠C = b0 = 650.
(ii) From the figure (ii)
∠A + ∠B + ∠C = 1800
Substituting the values
x0 + 900 + x0 = 1800
By further calculation
2x0 = 1800 – 900 = 900
x0 = 900/2 = 450
Therefore, ∠A = ∠C = x0 = 450.
(iii) From the figure (iii)
∠A + ∠B + ∠C = 1800
Substituting the values
k0 + k0 + k0 = 1800
By further calculation
3k0 = 1800
k0 = 1800/3 = 600
Therefore, ∠A = ∠B = ∠C = 600.
(iv) From the figure (iv)
∠A + ∠B + ∠C = 1800
Substituting the values
(m0 – 50) + 600 + (m0 + 50) = 1800
By further calculation
m0 – 50 + 600 + m0 + 50 = 1800
2m0 = 1800 – 600 = 1200
m0 = 1200/2 = 600
Therefore, ∠A = m0 – 50 = 600 – 50 = 550
∠C = m0 + 50 = 600 + 50 = 650
8. In the given figure, show that: ∠a = ∠b + ∠c
(i) If ∠b = 60° and ∠c = 50°; find ∠a.
(ii) If ∠a = 100° and ∠b = 55°; find ∠c.
(iii) If ∠a = 108° and ∠c = 48°; find ∠b.
Solution:
From the figure
AB || CD
b = ∠C and ∠A = c are alternate angles
In triangle PCD
Exterior ∠APC = ∠C + ∠D
a = b + c
(i) If ∠b = 60° and ∠c = 50°
∠a = ∠b + ∠c
Substituting the values
∠a = 60 + 50 = 1100
(ii) If ∠a = 100° and ∠b = 55°
∠a = ∠b + ∠c
Substituting the values
∠c = 100 – 55 = 450
(iii) If ∠a = 108° and ∠c = 48°
∠a = ∠b + ∠c
Substituting the values
∠b = 108 – 48 = 600
9. Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6.
Solution:
In a triangle, the sum of angles of a triangle is 1800
∠A + ∠B + ∠C = 1800
It is given that
∠A: ∠B: ∠C = 4: 5: 6
Consider ∠A = 4x, ∠B = 5x and ∠C = 6x
Substituting the values
4x + 5x + 6x = 1800
By further calculation
15x = 1800
x = 1800/15 = 120
So we get
∠A = 4x = 4 × 120 = 480
∠B = 5x = 5 × 120 = 600
∠C = 6x = 6 × 120 = 720
10. One angle of a triangle is 60°. The, other two angles are in the ratio of 5 : 7. Find the two angles.
Solution:
From the triangle ABC
Consider ∠A = 600, ∠B: ∠C = 5:7
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
600 + ∠B + ∠C = 1800
By further calculation
∠B + ∠C = 1800 – 600 = 1200
Take ∠B = 5x and ∠C = 7x
Substituting the values
5x + 7x = 1200
12x = 1200
x = 1200/12 = 100
So we get
∠B = 5x = 5 × 100 = 500
∠C = 7x = 7 × 100 = 700
11. One angle of a triangle is 61° and the other two angles are in the ratio 1 ½ : 1 1/3. Find these angles.
Solution:
From the triangle ABC
Consider ∠A = 610
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
610 + ∠B + ∠C = 1800
By further calculation
∠B + ∠C = 1800 – 610 = 1190
∠B: ∠C = 1 ½: 1 1/3 = 3/2: 4/3
Taking LCM
∠B: ∠C = 9/6: 8/ 6
∠B: ∠C = 9: 8
Consider ∠B = 9x and ∠C = 8x
Substituting the values
9x + 8x = 1190
17x = 1190
x = 1190/ 17 = 70
So we get
∠B = 9x = 9 × 70 = 630
∠C = 8x = 8 × 70 = 560
12. Find the unknown marked angles in the given figures:
Solution:
In a triangle, if one side is produced
Exterior angle is the sum of opposite interior angles
(i) From the figure (i)
1100 = x0 + 300
By further calculation
x0 = 1100 – 300 = 800
(ii) From the figure (ii)
1200 = y0 + 600
By further calculation
y0 = 1200 – 600 = 600
(iii) From the figure (iii)
1220 = k0 + 350
By further calculation
k0 = 1220 – 350 = 870
(iv) From the figure (iv)
1350 = a0 + 730
By further calculation
a0 = 1350 – 730 = 620
(v) From the figure (v)
1250 = a + c …… (1)
1400 = a + b …… (2)
By adding both the equations
a + c + a + b = 1250 + 1400
On further calculation
a + a + b + c = 2650
We know that a + b + c = 1800
Substituting it in the equation
a + 1800 = 2650
So we get
a = 265 – 180 = 850
If a + b = 1400
Substituting it in the equation
850 + b = 1400
So we get
b = 140 – 85 = 550
If a + c = 1250
Substituting it in the equation
850 + c = 1250
So we get
c = 125 – 85 = 400
Therefore, a = 850, b = 550 and c = 400.
Exercise 15B page: 180
1. Find the unknown angles in the given figures:
Solution:
(i) From the figure (i)
x = y as the angles opposite to equal sides
In a triangle
x + y + 800 = 1800
Substituting the values
x + x + 800 = 1800
By further calculation
2x = 1800 – 800 = 1000
x = 1000/2 = 500
Therefore, x = y = 500.
(ii) From the figure (ii)
b = 400 as the angles opposite to equal sides
In a triangle
a + b + 400 = 1800
Substituting the values
a + 400 + 400 = 1800
By further calculation
a = 180 – 80 = 1000
Therefore, a = 1000 and b = 400.
(iii) From the figure (iii)
x = y as the angles opposite to equal sides
In a triangle
x + y + 900 = 1800
Substituting the values
x + x + 900 = 1800
By further calculation
2x = 180 – 90 = 900
x = 90/2 = 450
Therefore, x = y = 450.
(iv) From the figure (iv)
a = b as the angles opposite to equal sides are equal
In a triangle
a + b + 800 = 1800
Substituting the values
a + a + 800 = 1800
By further calculation
2a = 180 – 80 = 1000
a = 1000/2 = 500
Here a = b = 500
We know that in a triangle the exterior angle is equal to sum of its opposite interior angles
x = a + 800
So we get
x = 50 + 80 = 1300
Therefore, a = 500, b = 500 and x = 1300.
(v) From the figure (v)
In an isosceles triangle consider each equal angle = x
x + x = 860
2x = 860
So we get
x = 860/2 = 430
For a linear pair
p + x = 1800
Substituting the values
p + 430 = 1800
By further calculation
p = 180 – 43 = 1370
Therefore, p = 1370.
2. Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures:
Solution:
(i) a = 700 as the angles opposite to equal sides are equal
In a triangle
a + 700 + x = 1800
Substituting the values
700 + 700 + x = 1800
By further calculation
x = 180 – 140 = 400
y = b as the angles opposite to equal sides are equal
Here a = y + b as the exterior angle is equal to sum of interior opposite angles
700 = y + y
So we get
2y = 700
y = 700/2 = 350
Therefore, x = 400 and y = 350.
(ii) From the figure (ii)
Each angle is 600 in an equilateral triangle
In an isosceles triangle
Consider each base angle = a
a + a + 1000 = 1800
By further calculation
2a = 180 – 100 = 800
So we get
a = 800/2 = 400
x = 600 + 400 = 1000
y = 600 + 400 = 1000
(iii) From the figure (iii)
1300 = x + p as the exterior angle is equal to the sum of interior opposite angles
It is given that the lines are parallel
Here p = 600 is the alternate angles and y = a
In a linear pair
a + 1300 = 1800
By further calculation
a = 180 – 130 = 500
Here x + p = 1300
Substituting the values
x + 600 = 1300
By further calculation
x = 130 – 60 = 700
Therefore, x = 700, y = 500 and p = 600.
(iv) From the figure (iv)
x = a + b
Here b = y and a = c as the angles opposite to equal sides are equal
a + c + 300 = 1800
Substituting the values
a + a + 300 = 1800
By further calculation
2a = 180 – 30 = 1500
a = 150/2 = 750
We know that
b + y = 900
Substituting the values
y + y = 900
2y = 900
y = 90/2 = 450
where b = 450
Therefore, x = a + b = 75 + 45 = 1200 and y = 450.
(v) From the figure (v)
a + b + 400 = 1800
So we get
a + b = 180 – 40 = 1400
The angles opposite to equal sides are equal
a = b = 140/2 = 700
x = b + 400 = 700Â + 400= 1100
Here the exterior angle of a triangle is equal to the sum of its interior opposite angles
In the same way
y = a + 400
Substituting the values
y = 700Â + 400 = 1100
Therefore, x = y = 1100.
3. The angle of vertex of an isosceles triangle is 100°. Find its base angles.
Solution:
Consider ∆ ABC
Here AB = AC and ∠B = ∠C
We know that
∠A = 1000
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
1000 + ∠B + ∠B = 1800
By further calculation
2∠B = 1800 – 1000 = 800
∠B = 80/2 = 400
Therefore, ∠B = ∠C = 400.
4. One of the base angles of an isosceles triangle is 52°. Find its angle of vertex.
Solution:
It is given that the base angles of isosceles triangle ABC = 520
Here ∠B = ∠C = 520
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
∠A + 520 + 520 = 1800
By further calculation
∠A = 180 – 104 = 760
Therefore, ∠A = 760.
5. In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.
Solution:
Consider the vertical angle of an isosceles triangle = x
So the base angle = 4x
In a triangle
x + 4x + 4x = 1800
By further calculation
9x = 1800
x = 180/9 = 200
So the vertical angle = 200
Each base angle = 4x = 4 × 200 = 800
6. The vertical angle of an isosceles triangle is 15° more than each of its base angles. Find each angle of the triangle.
Solution:
Consider the angle of the base of isosceles triangle = x0
So the vertical angle = x + 150
In a triangle
x + x + x + 150 = 1800
By further calculation
3x = 180 – 15 = 1650
x = 165/3 = 550
Therefore, the base angle = 550
Vertical angle = 55 + 15 = 700.
7. The base angle of an isosceles triangle is 15° more than its vertical angle. Find its each angle.
Solution:
Consider the vertical angle of the isosceles triangle = x0
Here each base angle = x + 150
In a triangle
x + 150 + x + 150Â + x = 1800
By further calculation
3x + 300 = 1800
3x = 180 – 30 = 1500
x = 150/3 = 500
Therefore, vertical angle = 500 and each base angle = 50 + 15 = 650.
8. The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.
Solution:
Consider each base angle of an isosceles triangle = x
Vertical angle = 3 (x + x) = 3 (2x) = 6x
In a triangle
6x + x + x = 1800
By further calculation
8x = 1800
x = 180/8 = 22.50
Therefore, each base angle = 22.50 and vertical angle = 3 (22.5 + 22.5) = 3 × 45 = 1350.
9. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.
Solution:
It is given that the ratio between a base angle and the vertical angle of an isosceles triangle = 1: 4
Consider base angle = x
Vertical angle = 4x
In a triangle
x + x + 4x = 1800
By further calculation
6x = 1800
x = 180/6 = 300
Therefore, each base angle = x = 300 and vertical angle = 4x = 4 × 300 = 1200.
10. In the given figure, BI is the bisector of ∠ABC and CI is the bisector of ∠ACB. Find ∠BIC.
Solution:
In ∆ ABC
BI is the bisector of ∠ABC and CI is the bisector of ∠ACB
Here AB = AC
∠B = ∠C as the angles opposite to equal sides are equal
We know that ∠A = 400
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
400 + ∠B + ∠B = 1800
By further calculation
400 + 2∠B = 1800
2∠B = 180 – 40 = 1400
∠B = 140/2 = 700
Here BI and CI are the bisectors of ∠ABC and ∠ACB
∠IBC = ½ ∠ABC = ½ × 700 = 350
∠ICB = ½ ∠ACB = ½ × 700 = 350
In ∆ IBC
∠BIC + ∠IBC + ∠ICB = 1800
Substituting the values
∠BIC + 350 + 350 = 1800
By further calculation
∠BIC = 180 – 70 = 1100
Therefore, ∠BIC = 1100.
11. In the given figure, express a in terms of b.
Solution:
From the ∆ ABC
BC = BA
∠BCA = ∠BAC
Here the exterior ∠CBE = ∠BCA + ∠BAC
a = ∠BCA + ∠BCA
a = 2∠BCA …… (1)
Here ∠ACB = 1800 – b
Where ∠ACD and ∠ACB are linear pair
∠BCA = 1800 – b ……. (2)
We get
a = 2 ∠BCA
Substituting the values
a = 2 (1800 – b)
a = 3600 – 2b
12. (a) In Figure (i) BP bisects ∠ABC and AB = AC. Find x.
(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects ∠ABC and ∠ADB = 70°.
Solution:
(a) From the figure (i)
AB = AC and BP bisects ∠ABC
AP is drawn parallel to BC
Here PB is the bisector of ∠ABC
∠PBC = ∠PBA
∠APB = ∠PBC are alternate angles
x = ∠PBC ….. (1)
In ∆ ABC
∠A = 600
Since AB = AC we get ∠B = ∠C
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
600 + ∠B + ∠C = 1800
We get
600 + ∠B + ∠B = 1800
By further calculation
2∠B = 180 – 60 = 1200
∠B = 120/2 = 600
½ ∠B = 60/2 = 300
∠PBC = 300
So from figure (i) x = 300
(b) From the figure (ii)
DA = DB = DC
Here BD bisects ∠ABC and ∠ADB = 700
In a triangle
∠ADB + ∠DAB + ∠DBA = 1800
Substituting the values
700 + ∠DBA + ∠DBA = 1800
By further calculation
700 + 2∠DBA = 1800
2∠DBA = 180 – 70 = 1100
∠DBA = 110/2 = 550
Here BD is the bisector of ∠ABC
So ∠DBA = ∠DBC = 550
In ∆ DBC
DB = DC
∠DCB = ∠DBC
Hence, x = 550.
13. In each figure, given below, ABCD is a square and ∆ BEC is an equilateral triangle.
Find, in each case: (i) ∠ABE (ii) ∠BAE
Solution:
The sides of a square are equal and each angle is 900
In an equilateral triangle all three sides are equal and all angles are 600
In figure (i) ABCD is a square and ∆ BEC is an equilateral triangle
(i) ∠ABE = ∠ABC + ∠CBE
Substituting the values
∠ABE = 900 + 600= 1500
(ii) In ∆ ABE
∠ABE + ∠BEA + ∠BAE = 1800
Substituting the values
1500 + ∠BAE + ∠BAE = 1800
By further calculation
2∠BAE = 180 – 150 = 300
∠BAE = 30/2 = 150
In figure (ii) ABCD is a square and ∆ BEC is an equilateral triangle
(i) ∠ABE = ∠ABC – ∠CBE
Substituting the values
∠ABE = 900 – 600 = 300
(ii) In ∆ ABE
∠ABE + ∠BEA + ∠BAE = 1800
Substituting the values
300 + ∠BAE + ∠BAE = 1800
By further calculation
2∠BAE = 180 – 30 = 1500
∠BAE = 150/2 = 750
14. In ∆ ABC, BA and BC are produced. Find the angles a and h. if AB = BC.
Solution:
In ∆ ABC, BA and BC are produced
∠ABC = 540 and AB = BC
In ∆ ABC
∠BAC + ∠BCA + ∠ABC = 1800
Substituting the values
∠BAC + ∠BAC + 540 = 1800
2∠BAC = 180 – 54 = 1260
∠BAC = 126/2 = 630
∠BCA = 630
In a linear pair
∠BAC + b = 1800
Substituting the value
630 + b = 1800
So we get
b = 180 – 63 = 1170
In a linear pair
∠BCA + a = 1800
Substituting the value
630Â + a = 1800
So we get
a = 180 – 63 = 1170
Therefore, a = b = 1170.
Exercise 15C page: 185
1. Construct a ∆ABC such that:
(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm
(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm
(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm
Solution:
(i) Steps of Construction
1. Construct a line segment BC = 4 cm.
2. Taking B as centre and 6 cm as radius construct an arc.
3. Taking C as centre and 5.5 cm as radius construct another arc which intersects the first arc at the point A.
4. Now join AB and AC.
Therefore, ∆ABC is the required triangle.
(ii) Steps of Construction
1. Construct a line segment CB = 6.5 cm
2. Taking C as centre and 4.2 cm as radius construct an arc.
3. Taking B as centre and 5.1 cm as radius construct another arc which intersects the first arc at the point A.
4. Now join AC and AB.
Therefore, ∆ABC is the required triangle.
(iii) Steps of Construction
1. Construct a line segment BC = 4 cm
2. Taking B as centre and 3.5 cm as radius construct an arc.
3. Taking C as centre and 5 cm as radius construct another arc which intersects the first arc at the point A.
4. Now join AB and AC.
Therefore, ∆ABC is the required triangle.
2. Construct a ∆ ABC such that:
(i) AB = 7 cm, BC = 5 cm and ∠ABC = 60°
(ii) BC = 6 cm, AC = 5.7 cm and ∠ACB = 75°
(iii) AB = 6.5 cm, AC = 5.8 cm and ∠A = 45°
Solution:
(i) Steps of Construction
1. Construct a line segment AB = 7 cm.
2. At the point B construct a ray which makes an angle 600 and cut off BC = 5cm.
3. Now join AC.
Therefore, ∆ABC is the required triangle.
(ii) Steps of Construction
1. Construct a line segment BC = 6 cm.
2. At the point C construct a ray which makes an angle 750 and cut off CA = 5.7 cm.
3. Now join AB.
Therefore, ∆ABC is the required triangle.
(iii) Steps of Construction
1. Construct a line segment AB = 6.5 cm.
2. At the point A construct a ray which makes an angle 450 and cut off AC = 5.8 cm.
3. Now join CB.
Therefore, ∆ABC is the required triangle.
3. Construct a ∆ PQR such that :
(i) PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R.
(ii) QR = 4.4 cm, ∠R = 30° and ∠Q = 75°. Measure PQ and PR.
(iii) PR = 5.8 cm, ∠P = 60° and ∠R = 45°.
Measure ∠Q and verify it by calculations
Solution:
(i) Steps of Construction
1. Construct a line segment PQ = 6 cm.
2. At point P construct a ray which makes an angle 450.
3. At point Q construct another ray which makes an angle 600 which intersect the first ray at point R.
Therefore, ∆ PQR is the required triangle.
By measuring ∠R = 750.
(ii) Steps of Construction
1. Construct a line segment QR = 4.4 cm.
2. At point Q construct a ray which makes an angle 750.
3. At point R construct another ray which makes an angle 300 which intersect the first ray at point R.
Therefore, ∆ PQR is the required triangle.
By measuring the length, PQ = 2.1 cm and PR = 4.4 cm.
(iii) Steps of Construction
1. Construct a line segment PR = 5.8 cm.
2. At point P construct a ray which makes an angle 600.
3. At point R construct another ray which makes an angle 450 which intersect the first ray at point Q.
Therefore, ∆ PQR is the required triangle.
By measuring ∠Q = 750.
Verification –
∠P + ∠Q + ∠R = 1800
Substituting the values
600 + ∠Q + 450 = 1800
By further calculation
∠Q = 180 – 105 = 750
4. Construct an isosceles ∆ ABC such that:
(i) base BC = 4 cm and base angle = 30°
(ii) base AB = 6.2 cm and base angle = 45°
(iii) base AC = 5 cm and base angle = 75°.
Measure the other two sides of the triangle.
Solution:
(i) Steps of Construction
In an isosceles triangle the base angles are equal
1. Construct a line segment BC = 4 cm.
2. At the points B and C construct rays which makes an angle 300 intersecting each other at the point A.
Therefore, ∆ ABC is the required triangle.
By measuring the equal sides, each is 2.5 cm in length approximately.
(ii) Steps of Construction
In an isosceles triangle the base angles are equal
1. Construct a line segment AB = 6.2 cm.
2. At the points A and B construct rays which makes an angle 450 intersecting each other at the point C.
Therefore, ∆ ABC is the required triangle.
By measuring the equal sides, each is 4.3 cm in length approximately.
(iii) Steps of Construction
In an isosceles triangle the base angles are equal
1. Construct a line segment AC = 5 cm.
2. At the points A and C construct rays which makes an angle 750 intersecting each other at the point B.
Therefore, ∆ ABC is the required triangle.
By measuring the equal sides, each is 9.3 cm in length approximately.
5. Construct an isosceles ∆ABC such that:
(i) AB = AC = 6.5 cm and ∠A = 60°
(ii) One of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles.
(iii) BC = AB = 5-8 cm and ZB = 30°. Measure ∠A and ∠C.
Solution:
(i) Steps of Construction
1. Construct a line segment AB = 6.5 cm.
2. At point A construct a ray which makes an angle 600.
3. Now cut off AC = 6.5cm
4. Join BC.
Therefore, ∆ ABC is the required triangle.
(ii) Steps of Construction
1. Construct a line segment AB = 6 cm.
2. At point A construct a ray which makes an angle 450.
3. Now cut off AC = 6cm
4. Join BC.
Therefore, ∆ ABC is the required triangle.
By measuring ∠B and ∠C, both are equal to 67 ½ 0.
(iii) Steps of Construction
1. Construct a line segment BC = 5.8 cm.
2. At point B construct a ray which makes an angle 300.
3. Now cut off BA = 5.8cm
4. Join AC.
Therefore, ∆ ABC is the required triangle.
By measuring ∠C and ∠A is equal to 750.
6. Construct an equilateral triangle ABC such that:
(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.
(ii) Each side is 6 cm.
Solution:
(i) Steps of Construction
1. Construct a line segment AB = 5cm.
2. Taking A and B as centres and 5 cm radius, construct two arcs which intersect each other at the point C.
3. Now join AC and BC where ∆ ABC is the required triangle.
4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point p.
5. Join PA, PB and PC.
By measuring each is 2.8 cm.
(ii) Steps of Construction
1. Construct a line segment AB = 6cm.
2. Taking A and B as centres and 6 cm radius, construct two arcs which intersect each other at the point C.
3. Now join AC and BC
Therefore, ∆ ABC is the required triangle.
7. (i) Construct a ∆ ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.
(ii) Construct an isosceles ∆PQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using ruler and compasses only construct a circumcircle to this triangle.
(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.
Construct a circumcircle to this triangle.
Solution:
(i) Steps of Construction
1. Construct a line segment BC = 4.5 cm.
2. Taking B as centre and 6 cm radius construct an arc.
3. Taking C as centre and 5.5 cm radius construct another arc which intersects the first arc at point A.
4. Now join AB and AC
Therefore, ∆ ABC is the required triangle.
5. Construct a perpendicular bisector of AB and AC which intersect each other at the point O.
6. Now join OB, OC and OA.
7. Taking O as centre and radius OA construct a cirlce which passes through the points A, B and C.
This is the required circumcircle of ∆ ABC.
(ii) Steps of Construction
1. Construct a line segment PQ = 6.5 cm.
2. At point Q, construct an arc which makes an angle 750.
3. Taking P as centre and radius 6.5 cm construct an arc which intersects the angle at point R.
4. Join PR.
∆ PQR is the required triangle.
4. Construct the perpendicular bisector of sides PQ and PR which intersects each other at the point O.
5. Join OP, OQ and OR.
6. Taking O as centre and radius equal to OP or OQ or OR construct a circle which passes through P, Q and R.
This is the required circumcircle of ∆ PQR.
(iii) Steps of Construction
1. Construct a line segment AB = 5.5 cm.
2. Taking A and B as centres and radius 5.5 cm construct two arcs which intersect each other at point C.
3. Now join AC and BC.
∆ ABC is the required triangle.
4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point O.
5. Now join OA, OB and OC.
6. Taking O as centre and OA or OB or OC as radius construct a circle which passes through A, B and C.
This is the required circumcircle.
8. (i) Construct a ∆ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
(ii) Construct an isosceles ∆ MNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct an incircle to this triangle and measure its radius.
(iii) Construct an equilateral ∆DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
(iv) Construct a ∆ PQR such that PQ = 6 cm, ∠QPR = 45° and angle PQR = 60°. Locate its incentre and then draw its incircle.
Solution:
(i) Steps of Construction
1. Construct a line segment AB = 6 cm.
2. Taking A as centre and 6.5 cm as radius and B as centre and 5.6 cm as radius construct arcs which intersect each other at point C.
3. Now join AC and BC.
4. Construct the angle bisector of ∠A and ∠B which intersect each other at point I.
5. From the point I construct IL which is perpendicular to AB.
6. Taking I as centre and IL as radius construct a circle which touches the sides of ∆ABC internally.
By measuring the required incircle the radius is 1.6 cm.
(ii) Steps of Construction
1. Construct a line segment MN = 5.8 cm.
2. At points M and N construct two rays which make an angle 300 each intersecting each other at point P.
3. Construct the angle bisectors of ∠M and ∠N which intersect each other at point I.
4. From the point I draw perpendicular IL on MN.
5. Taking I as centre and IL as radius construct a circle which touches the sides of ∆ PMB internally.
By measuring the required incircle the radius is 0.6 cm.
(iii) Steps of Construction
1. Construct a line segment BC = 5.5 cm.
2. Taking B and C as centres and 5.5 cm radius construct two arcs which intersect each other at point A.
3. Now join AB and AC.
4. Construct the perpendicular bisectors of ∠B and ∠C which intersect each other at the point I.
5. From the point I construct IL which is perpendicular to BC.
6. Taking I as centre and IL as radius construct a circle which touches the sides of ∆ABC internally.
This is the required incircle.
(iv) Steps of Construction
1. Construct a line segment PQ = 6 cm.
2. At the point P construct rays which make an angle of 450 and at point Q which makes an angle 600Â thats intersects each other at point R.
3. Construct the bisectors of ∠P and ∠Q which intersect each other at point I.
4. From the point I construct IL which is perpendicular to PQ.
5. Taking I as centre and IL as radius construct a circle which touches the sides of ∆PQR internally.
This is the required incircle where the point I is incentre.
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