Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1
Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1
Question 1.
Can 30°, 60° and 90° be the angles of a triangle?
Solution:
Given angles 30°, 60° and 90°
Sum of the angles = 30° + 60° + 90° = 180°
∴ The given angles form a triangle.
Question 2.
Can you draw a triangle with 25°, 65° and 80° as angles?
Solution:
Given angle 25°, 65° and 80°.
Sum of the angles = 25° + 65° + 80° = 170° ≠ 180
∴ We cannot draw a triangle with these measures.
Question 5.
Observe the figure and find the value of
∠A + ∠N + ∠G + ∠L + ∠E + ∠S.
Solution:
In the figure we have two triangles namely ∆AGE and ∆NLS.
By angle sum property of triangles,
Sum of angles of ∆AGE = ∠A + ∠G + ∠E = 180° …(1)
Also sum of angles of ∆NLS = ∠N + ∠L + ∠S = 180° … (2)
(1) + (2) ∠A + ∠G + ∠E + ∠N + ∠L + ∠S = 180° + 180°
i.e., ∠A + ∠N + ∠G + ∠L + ∠E + ∠S = 360°
Question 6.
If the three angles of a triangle are in the ratio 3 : 5 : 4, then find them.
Solution:
Given three angles of the triangles are in the ratio 3 : 5 : 4.
Let the three angle be 3x, 5x and 4x.
By angle sum property of a triangle, we have
3x + 5x + 4x = 180°
12x = 180°
x = 180∘/12
x = 15°
∴ The angle are 3x = 3 × 15° = 45°
5x = 5 × 15° = 75°
4x = 4 × 15° = 60°
Three angles of the triangle are 45°, 75°, 60°
Question 7.
In ∆RST, ∠S is 10° greater than ∠R and ∠T is 5° less than ∠S , find the three angles of the triangle.
Solution:
In ∆RST. Let ∠R = x.
Then given S is ∠10° greater than ∠R
∴ ∠S = x + 10°
Also given ∠T is 5° less then ∠S.
So ∠T = ∠S – 5° = (x + 10)° – 5° = x + 10° – 5°
By angle sum property of triangles, sum of three angles = 180°.
∠R + ∠S + ∠T = 180°
x + x + 10° + x + 5° = 180°
3x + 15° = 180°
3x = 180° – 15°
x = 165∘/3 = 55°
∠R = x = 55°
∠S = x + 10° = 55° + 10° = 65°
∠T = x + 5° = 55° + 5° = 60°
∴ ∠R = 55°
∠S = 65°
∠T = 60°
Question 9.
In ∆XYZ, if ∠X : ∠Z is 5 : 4 and ∠Y = 72°. Find ∠X and ∠Z.
Solution:
Given in ∆XYZ, ∠X : ∠Z = 5 : 4
Let ∠X = 5x; and ∠Z = 4x given ∠Y = 72°
By the angle sum property of triangles sum of three angles of a triangles is 180°.
∠X + ∠Y + ∠Z = 180°
5x + 72 + 4x = 180°
5x + 4x = 180° – 72°
9x = 108°
x = 108∘/9 = 12°
∠X = 5x = 5 × 12° = 60°
∠Z = 4x = 4 × 12° = 48°
∴ ∠X = 60°
∠Z = 48°
Question 10.
In a right angled triangle ABC, ∠B is right angle, ∠A is x + 1 and ∠C is 2x + 5. Find ∠A and ∠C.
Solution:
Given in ∆ABC ∠B = 90°
∠A = x + 1
∠B = 2x + 5
By angle sum property of triangles
Sum of three angles of ∆ABC = 180°
∠A + ∠B + ∠C = 180°
(x + 1) + 90° + (2x + 5) = 180°
x + 2x + 1° + 90° + 5° = 180°
3x + 96° = 180°
3x = 180° – 96° = 84°
x = 84∘/3 = 28°
∠A = x + 1 = 28 + 1 = 29
∠C = 2x + 5 = 2 (28) + 5 = 56 + 5 = 61
∴ ∠A = 29°
∠C = 61°
Question 11.
In a right angled triangle MNO, ∠N = 90°, MO is extended to P. If ∠NOP = 128°, find the other two angles of ∆MNO.
Solution:
Given ∠N = 90°
MO is extended to P, the exterior angle ∠NOP = 128°
Exterior angle is equal to the sum of interior opposite angles.
∴ ∠M + ∠N = 128°
∠M + 90° = 128°
∠M = 128° – 90°
∠M = 38°
By angle sum property of triangles,
∴ ∠M + ∠N + ∠O = 180°
38° + 90° + ∠O = 180°
∠O = 180° – 128°
∠O = 52°
∴ ∠M = 38° and ∠O = 52°
Question 12.
Find the value of x in each of the given triangles.
Solution:
(i) In ∆ABC, given B = 65°,
AC is extended to L, the exterior angle at C, ∠BCL = 135°
Exterior angle is equal to the sum of opposite interior angles.
∠A + ∠B = ∠BCL
∠A + 65° = 135°
∠A = 135° – 65°
∴ ∠A = 70°
x + ∠A = 180° [∵ linear pair]
x + 70° = 180° [∵ ∠A = 70°]
x = 180° – 70°
∴ x = 110°
(ii) In ∆ABC, given B = 3x – 8°
∠XAZ = ∠BAC [∵ vertically opposite angles]
8x + 7 + ∠BAC
i.e., In ∆ABC, ∠A = 8x + 7
Exterior angle ∠XCY = 120°
Exterior angle is equal to the sum of the interior opposite angles.
∠A + ∠B = 120°
8x + 7 + 3x – 8 = 120°
8x + 3x = 120° + 8 – 7
11x = 121°
x = 121∘/11 = 11°
Question 13.
In ∆LMN, MN is extended to O. If ∠MLN = 100 – x, ∠LMN = 2x and ∠LNO = 6x – 5, find the value of x.
Solution:
Exterior angle is equal to the sum of the opposite interior angles.
∠LNO = ∠MLN + ∠LMN
6x – 5 = 100° – x + 2x
6x – 5 + x – 2x = 100°
6x + x – 2x = 100° + 5°
5x = 105°
x = 105∘/5 = 21°
x = 21°
Question 14.
Using the given figure find the value of x.
Solution:
In ∆EDC, side DE is extended to B, to form the exterior angle ∠CEB = x.
We know that the exterior angle is equal to the sum of the opposite interior angles
∠CEB = ∠CDE + ∠ECD
x = 50° + 60°
x = 110°
Question 15.
Using the diagram find the value of x.
Solution:
Given triangle is an equilateral triangle as the three sides are equal. For an equilateral triangle all three angles are equal and is equal to 60° Also exterior angle is equal to sum of opposite interior angles.
x = 60° + 60°.
x = 120°
Objective Type Questions
Question 16.
The angles of a triangle are in the ratio 2:3:4. Then the angles are
(i) 20,30,40
(ii) 40, 60, 80
(iii) 80, 20, 80
(iv) 10, 15, 20
Answer:
(ii) 40, 60, 80
Question 17.
One of the angles of a triangle is 65°. If the difference of the other two angles is 45°, then the two angles are
(i) 85°, 40°
(ii) 70°, 25°
(iii) 80°, 35°
(iv) 80° , 135°
Answer:
(iii) 80°,35°
Question 18.
In the given figure, AB is parallel to CD. Then the value of b is
(i) 112°
(ii) 68°
(iii) 102°
(iv) 62° A
Answer:
(ii) 68°
Question 19.
In the given figure, which of the following statement is true?
(i) x + y + z = 180°
(ii) x + y + z = a + b + c
(iii) x + y + z = 2(a + b + c)
(iv) x + y + z = 3(a + b + c)
Ans :
(iii) x + y + z = 2(a + b + c)]
Question 20.
An exterior angle of a triangle is 70° and two interior opposite angles are equal. Then measure of each of these angle will be
(i) 110°
(ii) 120°
(iii) 35°
(iv) 60°
Answer:
(iii) 35°
Question 21.
In a ∆ABC, AB = AC. The value of x is _____.
(i) 80°
(ii) 100°
(iii) 130°
(iv) 120°
Answer:
(iii) 130°
Question 22.
If an exterior angle of a triangle is 115° and one of the interior opposite angles is 35°, then the other two angles of the triangle are
(i) 45°, 60°
(ii) 65°, 80°
(iii) 65°, 70°
(iv) 115°, 60°
Answer:
(ii) 65°, 80°
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