Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1

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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 3.
In each of the following triangles, find the value of x.

Solution:
(i) Let ∠G = x
By angle sum property we know that,
∠E + ∠F + ∠G = 180°
80° + 55° + x = 180°
135° + x = 180°
x = 45°

(ii) Let ∠M = x
By angle sum property of triangles we have
∠M + ∠M + ∠O = 180°
x + 96° + 22° = 180°
x + 118° = 180°
X = 180° – 118° = 620

(iii) Let ∠Z = (2x + 1)° and ∠Y = 90°
By the sum property of triangles we have
∠x + ∠y + ∠z = 180°
29° + 90° + (2x + 1)° = 180°
119° + (2x + 1)° = 180°
(2x + 1)° = 180° – 119°
2x + 1° = 61°
2x = 61° – 1°
2x = 60°
x = \(\frac{60^{\circ}}{2}\)
x = 30°

(iv) Let ∠J = x and ∠L – 3x.
By angle sum property of triangles we have
∠J + ∠K + ∠L = 180°
x + 112° + 3x = 180°
4x = 180° – 112°
x = 68°
x = \(\frac{68^{\circ}}{4}\)
x = 17°

(v) Let ∠S = 3x°
Given \(\overline{\mathrm{RS}}\) = Given \(\overline{\mathrm{RT}}\) = 4.5 cm
Given ∠S = ∠T = 3x° [∵ Angles opposite to equal sides are equal]
By angle sum property of a triangle we have,
∠R + ∠S + ∠T = 180°
72° + 3x + 3x = 180°
72° + 6x = 180°
x = \(\frac{108^{\circ}}{6}\)
x = 18°

(vi) Given ∠X = 3x; ∠Y = 2x; ∠Z = ∠4x
By angle sum property of a triangle we have
∠X + ∠Y + ∠Z = 180°
3x + 2x + 4x = 180°
∴ 9x = 180°
x = \(\frac{180^{\circ}}{9}\) = 20°

(vii) Given ∠T = (x – 4)°
∠U = 90°
∠V = (3x – 2)°
By angle sum property of a triang we have
∠T + ∠U + ∠V = 180°
(x – 4)° + 90° + (3x – 2)° = 180°
x – 4° + 90° + 3x – 2° = 180°
x + 3x + 90° – 4° – 2° = 180°
4x + 84° = 180°
4x = 180° – 84°
4x = 96°
x = \(\frac{96^{\circ}}{4}\) = 24°
x = 24°

(viii) Given ∠N = (x + 31)°
∠O = (3x – 10)°
∠P = (2x – 3)°
By angle sum property of a triangle we have
∠N + ∠O + ∠P = O
(x + 31)° + (3x – 10)° + (2x – 3)° = 180°
x + 31°+ 3x – 10° + 2x – 3° = 180°
x + 3x + 2x + 31° – 10° – 3° = 180°
6x + 18° = 180°
6x = 180° + 18°
6x = 162°
x = \(\frac{162^{\circ}}{6}\) = 27°
x = 27°

Question 4.
Two line segments \(\overline{A D}\) and \(\overline{B C}\) intersect at O. Joining \(\overline{A B}\) and \(\overline{D C}\) we get two triangles, ∆AOB and ∆DOC as shown in the figure. Find the ∠A and ∠B.

Solution:
In ∆AOB and ∆DOC,
∠AOB = ∠DOC [∵ Vertically opposite angles are equal]
Let ∠AOB = ∠DOC = y
By angle sum property of a triangle we have
∠A + ∠B + ∠AOB = ∠D + ∠C + ∠DOC = 180°
3x + 2x + y = 70° + 30° + y = 180°
5x + y = 100° + y = 180°
Here 5x + y = 100° + y
5x = 100° + y – y
5x = 100°
x = \(\frac{100^{\circ}}{5}\) = 20°
∠A = 3x = 3 × 20 = 60°
∠B = 2x = 2 × 20 = 40°
∠A = 60°
∠B = 40°

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 = \(\frac{180^{\circ}}{12}\)
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 = \(\frac{165^{\circ}}{3}\) = 55°
∠R = x = 55°
∠S = x + 10° = 55° + 10° = 65°
∠T = x + 5° = 55° + 5° = 60°
∴ ∠R = 55°
∠S = 65°
∠T = 60°

Question 8.
In ∆ABC , if ∠B is 3 times ∠A and ∠C is 2 times ∠A, then find the angles.
Solution:
In ABC, Let ∠A = x,
then ∠B = 3 times ∠A = 3x
∠C = 2 times ∠A = 2x
By angle sum property of a triangles,
Sum of three angles of ∆ABC =180°.
∠A + ∠B + ∠C = 180
x + 3x + 2x = 180°
x (1 + 3 + 2) = 180°
6x = 180°

x = \(\frac{180^{\circ}}{6}\) = 30°
∠A = x = 30°
∠B = 3x = 3 × 30° = 90°
∠C = 2x = 2 × 30° = 60°
∴ ∠A = 30°
∠B = 90°
∠C = 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 = \(\frac{108^{\circ}}{9}\) = 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 = \(\frac{84^{\circ}}{3}\) = 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 = \(\frac{121^{\circ}}{11}\) = 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 = \(\frac{105^{\circ}}{5}\) = 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°