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It’s given that AB∥CD. 

Required to find the value of x.

We know that, 

Diagonals of a parallelogram bisect each other. 

So, 

\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\) 

⇒ \(\frac{4}{(4x – 2)}\) = \(\frac{(x +1)}{(2x + 4)}\) 

4(2x + 4) = (4x – 2)(x +1) 

8x + 16 = x(4x – 2) + 1(4x – 2) 

8x + 16 = 4x² – 2x + 4x – 2 

-4x² + 8x + 16 + 2 – 2x = 0 

-4x² + 6x + 8 = 0 

4x² – 6x – 18 = 0 

4x² – 12x + 6x – 18 = 0 

4x(x – 3) + 6(x – 3) = 0 

(4x + 6) (x – 3) = 0 

∴ x = – \(\frac{6}{4}\) or x = 3

It’s given that AB∥CD. 

Required to find the value of x.

We know that, 

Diagonals of a parallelogram bisect each other 

So, 

\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\) 

⇒ \(\frac{(6x – 5)}{(2x + 1)}\) = \(\frac{(5x – 3)}{(3x – 1)}\) 

(6x – 5)(3x – 1) = (2x + 1)(5x – 3) 

3x(6x – 5) – 1(6x – 5) = 2x(5x – 3) + 1(5x – 3) 

18x² – 10x² – 21x + 5 + x +3 = 0 

8x² – 16x – 4x + 8 = 0 

8x(x – 2) – 4(x – 2) = 0 

(8x – 4)(x – 2) = 0 

x = \(\frac{4}{8}\) = \(\frac{1}{2}\) or x = -2 

∴ x = \(\frac{1}{2}\)

Given, 

ΔACB ∼ ΔAPQ 

BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm 

Required to find: CA and AQ

We know that, 

ΔACB ∼ ΔAPQ [given] 

\(\frac{BA}{AQ} = \frac{CA}{AP} = \frac{BC}{PQ}\) [Corresponding Parts of Similar Triangles] 

So, 

\(\frac{6.5}{AQ}\) = \(\frac{8}{4}\) 

AQ = \(\frac{(6.5 \times 4)}{8}\)

AQ = 3.25 cm 

Similarly, as 

\(\frac{CA}{AP}\) = \(\frac{BC}{PQ}\) 

\(\frac{CA}{2.8}\) = \(\frac{8}{4}\) 

CA = 2.8 x 2 

CA = 5.6 cm 

Hence, CA = 5.6 cm and AQ = 3.25 cm.

It’s given that AB∥CD. 

Required to find the value of x.

We know that, 

Diagonals of a parallelogram bisect each other 

So, 

\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\) 

\(\frac{(3x – 19)}{(x – 3)}\)= \(\frac{(x–4)}{ 4}\) 

4(3x – 19) = (x – 3) (x – 4) 

12x – 76 = x(x – 4) -3(x – 4) 

12x – 76 = x² – 4x – 3x + 12 

-x² + 7x – 12 + 12x - 76 = 0 

-x² + 19x – 88 = 0 

x² – 19x + 88 = 0 

x² – 11x – 8x + 88 = 0 

x(x – 11) – 8(x – 11) = 0 

∴ x = 11 or x = 8

AO/OC=BO/OD 

3X-19/X-3=X-4/4 

(x-3)(x-4)=4(3x-19) 

X² -4x-3x+12=12x-76 

X² -7x+12-12x+76=0 

X² -19x+88=0 

X² -11x-8x+88=0 

X(x-11)-8(x-11)=0 

(x-11)(x-8)=0 

x-11=0 

x=11 

or x-8=0 

x=8 

x=11 or 8

(i) ∠ TPQ = ∠PQR + ∠ PRQ

(ii) ∠ UQR = ∠ QRP + ∠QPR

A given triangle to be right-angled, if it satisfies Pythagorean Theorem. That is, the sum of the squares of the two smaller sides must be equal to the square of the largest side.

1.6 cm, 3.8 cm, 4 cm

Longest side = 4 cm

(4)² = 16

and (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00 = 17

16 ≠ 17

It is not a right triangle.

A given triangle to be right-angled, if it satisfies Pythagorean Theorem. That is, the sum of the squares of the two smaller sides must be equal to the square of the largest side.

(a- 1) cm, 2√a cm, (a + 1) cm

Longest side = (a + 1) cm

(a + 1)² = a² + 2a + 1

and (a – 1)² + (2 √a )² = a² – 2a + 1 + 4a = a² + 2a + 1

a² + 2a + 1 = a² + 2a + 1

It is a right triangle.

(i) a = 7, b = 24, c =25 

Here a² = 49, b² = 576, c² = 625 

= a² + b²

= 49 + 576 

= 625= c² 

∴ So given triangle is a right angle. 

(ii) a=9, b=16, c= 18 

Here a² = 81, b² = 256, c² = 324

= a² + b²  

= 81 + 256 

= 337 ≠ c² 

So given Triangle is not a right angle. 

(iii) a =1.6, b =3.8, c = 4 

Here a² = 2.56, b² = 14.44, c² = 16 

= a² + b² 

= 2.56 +14.44 

=17 ≠ c² 

So given Triangle is not a right angle. 

(iv) a = 8, b =10, c = 6

Here a² = 64, b² =100, c² = 36 

= a² + c² 

= 64 +36 

= 100 = b² 

So given Triangle is a right angle.

Correct option is: (C) 4 cm

Given: AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3 

Required to find x. 

By using Thales Theorem, [As DE ∥ BC] 

AD/BD = AE/CE 

So, (4x–3)/ (3x–1) = (8x–7)/ (5x–3) 

(4x – 3)(5x – 3) = (3x – 1)(8x – 7) 

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7) 

20x² – 12x – 15x + 9 = 24x² – 29x + 7 

20x² - 27x + 9 = 24x² - 29x + 7 

⇒ -4x² + 2x + 2 = 0 

4x² – 2x – 2 = 0 

4x² – 4x + 2x – 2 = 0 

4x(x – 1) + 2(x – 1) = 0 

(4x + 2)(x – 1) = 0 

⇒ x = 1 or x = -\(\frac{2}{4}\) 

We know that the side of triangle can never be negative. Therefore, we take the positive value. 

∴ x = 1

(i) we have PM=4cm, QM=4.5 cm,PN=4 cm and NR=4.5 cm 

Hence PM/QM=4/4.5=40/45=8/9 

PN/NR=4/4.5=40/45=8/9 

PM/QM= PN/NR 

by the converse of proportionality theorem 

MN∥QR 

(ii) we have PQ=1.28cm, PR=2.56 cm,PM=0.16 cm and PN=0.32 cm

Hence PQ/PR=1.28/2.56=128/256=1/2

PM/PN=0.16/0.32=16/32=1/2

PQ/PR = PM/PN

by the converse of proportionality theorem

MN∥QR

Correct option is (C) 7.5

Given that \(\triangle ABC \sim \triangle PQR \)

Then their corresponding sides are in the same ratio.

\(\therefore\) \(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\) 

\(\because\) \(\frac{AB}{PQ}=\frac{AC}{PR}\)

\(\Rightarrow\) \(PQ=AB\times\frac{PR}{AC}\)

\(=6\times\frac68\)        \((\because AB=6,PR=6\;\&\;AC=8)\)

\(=\frac92=4.5\)

& \(\frac{BC}{QR}=\frac{AC}{PR}\)

\(\Rightarrow\) \(QR=BC\times\frac{PR}{AC}\)

\(=4\times\frac68\)       \((\because BC=4,PR=6\;\&\;AC=8)\)

\(=\frac62=3\)

\(\therefore PQ+QR=4.5+3=7.5\)

Correct option is: (C) 7.5

Correct option is (B) a right angled triangle

\(\because5^2+12^2\) = 25+144

= 169 \(=13^2\)

\(\therefore PQ^2+QR^2=PR^2\) which is a condition that \(\triangle PQR\) is a right angled triangle.

(By converse of Pythagoras theorem)

B) a right angled triangle

1. D) 10.4

2. C) 80°

3. D) ΔABC ≅ ΔPQR

4. B) X

5. A) A. A. A

(b) 40^o

Consider the ΔPQR.

From the exterior angle property = ∠RPU – ∠PRQ + ∠PQR

∠RPU = ∠PRQ + ∠PQR

140^O = 2 ∠PQR … [given PQ = PR]

∠PQR = 140/2

∠PQR = 70^o

Given, ST || QR and QS is transversal.

From the property of corresponding angles, ∠PST = ∠PQR = 70^o

Now, consider the ΔQSR

RS = RQ … [from the question]

So, ∠SQR – ∠RSQ = 70^O

Then, PQ is a straight line.

∠PST + ∠TSR + ∠RSQ = 180^o

70^o + ∠TSR + 70^o = 180^o

140^o + ∠TSR = 180^o

∠TSR = 180^o – 140^o

∠TSR = 40^o

Given,

AM perpendicular to BC

AN is bisector of ∠A

Therefore, ∠NAC = ∠NAB

In ΔABC

∠A + ∠B + ∠C = 180°

∠A + 65° + 33° = 180°

∠A = 180° – 98°

= 82°

∠NAC = ∠NAB = 41°(Therefore, AN is bisector of ∠A)

In ΔAMB

∠AMB + ∠MAB + ∠ABM = 180°

90° + ∠MAB + 65° = 180°

∠MAB + 155° = 180°

∠MAB = 25°

Therefore,

∠MAB + ∠MAN = ∠BAN

25° + ∠MAN = 41°

∠MAN = 41° – 25°

= 16°

(d) 40°, 132°, 18°

We now that, sum of angles of triangle is equal to 180^o.

But, 40^o + 132^o + 18^o = 190^o

So, these triplets cannot be the angles of a triangle.

Given that, exterior angle at ∠C = 105°. 

One of the interior opposite angle ∠A = 65°. 

The other interior opposite angle is ∠B . ∠A + ∠B =105°(∵ Exterior angle property of a triangle) 

65° + ∠B = 105° 

⇒ ∠B = 105° – 65° = 40°

Correct option is A) greater than

False

Sum of any two angles of a triangle is either greater than the third angle or smaller than the third angle.

Consider ∠A and ∠B in a triangle is x^o

We know that the sum of all the angles in a triangle is 180^o.

So we can write it as

∠A + ∠B + ∠C = 180^o

By substituting the values

x^o + x^o + ∠C = 180^o

By addition

2x^o + ∠C = 180^o ….. (1)

According to the question we get

∠C = x^o + 18^o ……. (2)

By substituting (2) in (1) we get

2x^o + x^o + 18^o = 180^o

On further calculation

3x^o + 18^o = 180^o

By subtraction

3x^o = 180^o – 18^o

3x^o = 162^o

By division

x^o = 162/3

x^o = 54^o

By substituting the values of x

∠A = ∠B = 54^o

∠C = 54^o + 18^o = 72^o

Therefore, ∠A = 54^o, ∠B = 54^o and ∠C = 72^o

Given that,

Two angles are equal and third angle is greater than each of those angles by 30°.

Let, x, x, x + 30° be the angles of the triangle.

We know that,

Sum of all angles of triangle is 180°

x + x + x + 30° = 180°

3x + 30° = 180°

3x = 180° – 30°

3x = 150°

x = 50°

Therefore,

The angles are:

x = 50°

x = 50°

x + 30° = 50° + 30°

= 80°

Therefore, the required angles are 50°, 50°, 80°.

Ratio of the interior opposite angles = 2 : 3 

Sum of the terms of the ratio = 2 + 3 = 5

Sum of the interior angles exterior angle = 125° 

∴ First angle = 2/5 x 125° = 500

Second angle = 3/5 x 125° = 75°

Given that ratio of acute angles = 2 : 3 

Sum of the terms of the ratio = 2 + 3 = 5 

Sum of the acute angles = 90° 

∴ 1st acute angle x 900 = 36° 

2nd acute angle = x 900 = 540 

∴ Angles of the triangle = 36°, 54° and 90° 

Interior angles 

∠ABC, ∠BAC, ∠CAB 

Exterior angles

∠ACZ, ∠BAY and ∠CBX 

Correct option is B) 61°

Given two angles are 35° and 84°.

Let the third angle be x°

\(\because\) sum  of angles in a triangle is 180°

\(\therefore\) 35 + 84 + x = 180

⇒ x = 180 - (35 + 84)

 = 180 - 119

 = 61

\(\therefore\) Third angle is 61°.

Given that the angles of the triangle are 

(x + 10)°, (x – 20)° and (x + 40)° 

(x + 10)° + (x – 20)° + (x + 40)° = 180°

⇒ x + 10° + x – 20° + x + 40° = 180° 

⇒ 3x + 30° = 180° 

⇒ 3x – 180° – 30° 

⇒ 3x = 150°

⇒ x = 150°/3 = 50°

The angles are, 

x + 10° = 50° + 10° = 60° 

x – 20° = 50° – 20° = 30° 

x + 40° = 50° + 40° = 90° 

∴ Measure of the angles are 60°, 30° and 90°.

Correct answer is (c) acute

Correct option is B) Right angled Isosceles

Correct option is C) an acute angled

Correct option is B) 1

A right triangle