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The correct answer is (A) 0

The correct answer is (D) mote than 4

The correct answer is (A) X

The correct answer is (C) E

1. A scalene Triangle 

2. The letter F

In a sphere, edges – 0, faces – 0 and vertices – 0.

(i) Let ∠S = 3x° 

Given = \(\overline{RS}\)

Given = \(\overline{RT}\)

= 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 = \(\frac{108°}{6}\)

x = 18°

(ii) 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 = \(\frac{180°}{9}\)

x = 20°

(iii) 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 = \(\frac{96°}{4}\)

= 24° 

x = 24°

(iv) 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 = \(\frac{162°}{6}\)

= 27° 

x = 27°

(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° 

= 62° 

(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 = \(\frac{60°}{2}\)

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 = \(\frac{68° }{4}\)

x = 17°

Given angle 25°, 65° and 80°. 

Sum of the angles = 25° + 65° + 80° 

= 170° ≠ 180 

∴ We cannot draw a triangle with these measures.

Given angles 30°, 60° and 90° 

Sum of the angles = 30° + 60° + 90° 

= 180° 

∴ The given angles form a triangle.

Let the third angle be x. 

Using the angle sum property of a triangle we have, 

30° + 80° + x = 180° x + 110° 

= 180° x 

= 180° – 110° 

= 70° 

Third angle = 70°.

No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be 90°, 45°, 45°. 

Here sum of two angles 45° + 45° = 90°.

(i) Since angles y and 120° form a linear pair. 

y + 120° = 180°

y = 180° – 120°

y = 60° 

Now using the angle sum property of a triangle, we have 

x + y + 50° = 180° x + 60° + 50° 

= 180° x + 110° 

= 180° x 

= 180° – 110° 

= 70°

x = 70° 

y = 60 

(ii) Using the angle sum property of triangle, we have 

50° + 60° + y = 180° 

110° + y = 180° 

y = 180° – 110° 

y = 70° 

Again x and y form a linear pair

∴ x + y = 180° x + 70° 

= 180° x 

= 180° – 70° 

= 110° 

∴ x = 110°; y = 70°

Let the angles of the triangle be x, 2x, x.

Using the angle sum property, we have

x + 2x + x = 180°

4x = 180°

x = \(\frac{180°}{4}\)

x = 45°

2x = 2 x 45° 

= 90°

Thus the three angles of the triangle are 45°, 90°, 45°.

Its two angles are equal. It is an isoscales triangle. 

Its one angle is 90°.

∴ It is a right angled triangle.

Given two angles of the triangle are same and is equal to 46°. If two angles are equal the sides opposite to equal angles are equal. Therefore it will be an isoscales triangle.

The correct answer is AB,BC,CD,DE,EA.

(i) Obtuse Angle.

(ii) Zero Angle.

(iii) Straight Angle.

(iv) Acute Angle.

(v) Right Angle.

The correct answer is X,Y,Z.

We know that the tangents drawn from are external point to a circle are equal.

Therefore AD = AF = x say.

BD = BE = y say and

CE = CF = z say

Now, AB = 12 cm, BC = 8 cm, and CA= 10 cm.

x + y = 12, y + z = 8 and z + x = 10

(x + y) + (y + z) + (z + x) = 12 + 8 + 10

2(x + y + z) = 30

x + y + z = 15

Now, x + y = 12 and x + y + z = 15

12 + z = 15 ⇒ z = 3

y + z = 8 and x + y + z = 15

x + 8 = 15 ⇒ x = 7

and z + x = 10 and x + y + z = 15

10 + y = 15 ⇒ y = 5

Hence, AD = x = 7 cm, BE = y = 5 cm and CF = z = 3 cm.

(a) ∠ADB = ∠CDB (b) ∠ABD = ∠CBD

(c) ∠ADC = ∠BDC, ∠CAD = 90°, ∠CBD = 90°

The correct answer is Two,AC andAD

We have given, ∠BAC = ∠CAD = ∠DAE

∴ There are two trisectors namely, AC and AD.

Since D and E are the points of trisection of BC, therefore BD = DE = CE

Let BD = DE = CE = x

Then BE = 2x and BC = 3x

In right triangles ABD, ABE and ABC, (using Pythagoras theorem)

We have AD² = AB² + BD²

⇒ AD² = AB² + x² … (1)

AE² = AB² + BE²

⇒ AB² + (2x)²

⇒ AE² = AB² + 4x² … (2)

And AC² = AB² + BC² = AB² + (3x)²

AC² = AB² + 9x²

Now 8 AE² – 3 AC² – 5 AD² = 8 (AB² + 4x²) – 3 (AB² + 9x²) – 5 (AB² + x²)

= 8AB² + 32x² – 3AB² – 27x² – 5AB² – 5x²

= 0

∴ 8 AE² – 3 AC² – 5 AD² = 0

8 AE² = 3 AC² + 5 AD².

Hence it is proved.

(i) ∠ADB = ∠CDB, ∠ABD = ∠CBD, BD = BD

Proof:

We are given a right triangle ABC right angled at B.

We need to prove that AC² = AB² + BC²

Let us draw BD ⊥ AC

Now, ΔADB ~ ΔABC

AD/AB = AB/AC (sides are proportional)

AD.AC = AB² … (1)

Also, ΔBDC ~ ΔABC

CD/BC = BC/AC

CD.AC = BC² … (2)

Adding (1) and (2)

AD.AC + CD.AC = AB² + BC²

AC(AD + CD) = AB² + BC²

AC.AC = AB² + BC²

AC² = AB² + BC²

If a line segment of length 5cm is reflected in a line of symmetry (mirror), then its reflection (image) is a Line segment of length b

(A) False

(B) True

(C) False