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THE TRIANGLE AND ITS PROPERTIES

298398 The perimeter of the rectangle whose length is 60cm and a diagonal is 61cm is:

1 120cm

2 122cm

3 71cm

4 142cm

Explanation:

142cm
Given, length of rectangle = 60cm and its diagonal = 61cm.
2
Let the breadth of a rectangle be xcm.
In right angled $△ ABC,$
(AC) $^{1}$ = (AB) $^{1}$ + (BC) $^{1}$
(BC) $^{1}$ = (AC) $^{1}$ + (AB) $^{1}$ [by pythagoras theoram]
x $^{1}$ = (61) $^{1}$ - (60) $^{1}$ = 3721 - 3600 = 121
$\Rightarrow x = \sqrt{121} = 11 cm$
$∴$ Breadth of rectangle = 11cm and length of rectangle = 60cm.
Now, perimeter of rectangle = 2(l + b)
= 2(60 + 11) = 2 × 71
= 142cm.

THE TRIANGLE AND ITS PROPERTIES

298399 A triangle has altitudes:

1 2

2 3

3 1

4 None of these

THE TRIANGLE AND ITS PROPERTIES

298400 In the following figure, find $∠ x$ and $∠ y$ if $∠ x - ∠ y - 10^{\circ}$
3

1 65º, 55º

2 55º, 45º

3 45º, 35º

4 60º, 60º

Explanation:

65º, 55º
$∠ x + ∠ y = 120^{\circ}, ∠ x - ∠ y = 10^{\circ}$
$∠ x = 65^{\circ}, ∠ = 55^{\circ}$

THE TRIANGLE AND ITS PROPERTIES

298401 In Fig. if AB || CD, the values of x and y are:
4

1 x = 21, y = 28

2 x = 21, y = 38

3 x = 38, y = 21

4 x = 22, y = 38

Explanation:

x = 21, y = 38
$∠ AEC + ∠ AEB = 180^{\circ}$ [Linear angles]
$\Rightarrow 79^{\circ} + ∠ AEB = 180^{\circ}$
$\Rightarrow ∠ AEB = 101^{\circ}$
Since, AB || CD
$∠ ABE = ∠ ECD = 58^{\circ}$ [Alternate angles]
Now, In $△ AEB$
$∠ AEB + ∠ EAB + ∠ ABE = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 101^{\circ} + 58^{\circ} + x^{\circ} = 180^{\circ}$
$\Rightarrow x^{\circ} = 21^{\circ}$
$\Rightarrow x = 21$
Now, In $△ AEB$
$∠ AEC + ∠ CAE + ∠ CEA = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 79^{\circ} + y^{\circ} + 3 x^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 3 (21)^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 63^{\circ} = 180^{\circ}$
$\Rightarrow y^{\circ} = 38^{\circ}$
$\Rightarrow y = 38$
Hence, the correct answer is option (b).

THE TRIANGLE AND ITS PROPERTIES

298398 The perimeter of the rectangle whose length is 60cm and a diagonal is 61cm is:

1 120cm

2 122cm

3 71cm

4 142cm

Explanation:

142cm
Given, length of rectangle = 60cm and its diagonal = 61cm.
2
Let the breadth of a rectangle be xcm.
In right angled $△ ABC,$
(AC) $^{1}$ = (AB) $^{1}$ + (BC) $^{1}$
(BC) $^{1}$ = (AC) $^{1}$ + (AB) $^{1}$ [by pythagoras theoram]
x $^{1}$ = (61) $^{1}$ - (60) $^{1}$ = 3721 - 3600 = 121
$\Rightarrow x = \sqrt{121} = 11 cm$
$∴$ Breadth of rectangle = 11cm and length of rectangle = 60cm.
Now, perimeter of rectangle = 2(l + b)
= 2(60 + 11) = 2 × 71
= 142cm.

THE TRIANGLE AND ITS PROPERTIES

298399 A triangle has altitudes:

1 2

2 3

3 1

4 None of these

THE TRIANGLE AND ITS PROPERTIES

298400 In the following figure, find $∠ x$ and $∠ y$ if $∠ x - ∠ y - 10^{\circ}$
3

1 65º, 55º

2 55º, 45º

3 45º, 35º

4 60º, 60º

Explanation:

65º, 55º
$∠ x + ∠ y = 120^{\circ}, ∠ x - ∠ y = 10^{\circ}$
$∠ x = 65^{\circ}, ∠ = 55^{\circ}$

THE TRIANGLE AND ITS PROPERTIES

298401 In Fig. if AB || CD, the values of x and y are:
4

1 x = 21, y = 28

2 x = 21, y = 38

3 x = 38, y = 21

4 x = 22, y = 38

Explanation:

x = 21, y = 38
$∠ AEC + ∠ AEB = 180^{\circ}$ [Linear angles]
$\Rightarrow 79^{\circ} + ∠ AEB = 180^{\circ}$
$\Rightarrow ∠ AEB = 101^{\circ}$
Since, AB || CD
$∠ ABE = ∠ ECD = 58^{\circ}$ [Alternate angles]
Now, In $△ AEB$
$∠ AEB + ∠ EAB + ∠ ABE = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 101^{\circ} + 58^{\circ} + x^{\circ} = 180^{\circ}$
$\Rightarrow x^{\circ} = 21^{\circ}$
$\Rightarrow x = 21$
Now, In $△ AEB$
$∠ AEC + ∠ CAE + ∠ CEA = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 79^{\circ} + y^{\circ} + 3 x^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 3 (21)^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 63^{\circ} = 180^{\circ}$
$\Rightarrow y^{\circ} = 38^{\circ}$
$\Rightarrow y = 38$
Hence, the correct answer is option (b).

THE TRIANGLE AND ITS PROPERTIES

298398 The perimeter of the rectangle whose length is 60cm and a diagonal is 61cm is:

1 120cm

2 122cm

3 71cm

4 142cm

Explanation:

142cm
Given, length of rectangle = 60cm and its diagonal = 61cm.
2
Let the breadth of a rectangle be xcm.
In right angled $△ ABC,$
(AC) $^{1}$ = (AB) $^{1}$ + (BC) $^{1}$
(BC) $^{1}$ = (AC) $^{1}$ + (AB) $^{1}$ [by pythagoras theoram]
x $^{1}$ = (61) $^{1}$ - (60) $^{1}$ = 3721 - 3600 = 121
$\Rightarrow x = \sqrt{121} = 11 cm$
$∴$ Breadth of rectangle = 11cm and length of rectangle = 60cm.
Now, perimeter of rectangle = 2(l + b)
= 2(60 + 11) = 2 × 71
= 142cm.

THE TRIANGLE AND ITS PROPERTIES

298399 A triangle has altitudes:

1 2

2 3

3 1

4 None of these

THE TRIANGLE AND ITS PROPERTIES

298400 In the following figure, find $∠ x$ and $∠ y$ if $∠ x - ∠ y - 10^{\circ}$
3

1 65º, 55º

2 55º, 45º

3 45º, 35º

4 60º, 60º

Explanation:

65º, 55º
$∠ x + ∠ y = 120^{\circ}, ∠ x - ∠ y = 10^{\circ}$
$∠ x = 65^{\circ}, ∠ = 55^{\circ}$

THE TRIANGLE AND ITS PROPERTIES

298401 In Fig. if AB || CD, the values of x and y are:
4

1 x = 21, y = 28

2 x = 21, y = 38

3 x = 38, y = 21

4 x = 22, y = 38

Explanation:

x = 21, y = 38
$∠ AEC + ∠ AEB = 180^{\circ}$ [Linear angles]
$\Rightarrow 79^{\circ} + ∠ AEB = 180^{\circ}$
$\Rightarrow ∠ AEB = 101^{\circ}$
Since, AB || CD
$∠ ABE = ∠ ECD = 58^{\circ}$ [Alternate angles]
Now, In $△ AEB$
$∠ AEB + ∠ EAB + ∠ ABE = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 101^{\circ} + 58^{\circ} + x^{\circ} = 180^{\circ}$
$\Rightarrow x^{\circ} = 21^{\circ}$
$\Rightarrow x = 21$
Now, In $△ AEB$
$∠ AEC + ∠ CAE + ∠ CEA = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 79^{\circ} + y^{\circ} + 3 x^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 3 (21)^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 63^{\circ} = 180^{\circ}$
$\Rightarrow y^{\circ} = 38^{\circ}$
$\Rightarrow y = 38$
Hence, the correct answer is option (b).

THE TRIANGLE AND ITS PROPERTIES

298398 The perimeter of the rectangle whose length is 60cm and a diagonal is 61cm is:

1 120cm

2 122cm

3 71cm

4 142cm

Explanation:

142cm
Given, length of rectangle = 60cm and its diagonal = 61cm.
2
Let the breadth of a rectangle be xcm.
In right angled $△ ABC,$
(AC) $^{1}$ = (AB) $^{1}$ + (BC) $^{1}$
(BC) $^{1}$ = (AC) $^{1}$ + (AB) $^{1}$ [by pythagoras theoram]
x $^{1}$ = (61) $^{1}$ - (60) $^{1}$ = 3721 - 3600 = 121
$\Rightarrow x = \sqrt{121} = 11 cm$
$∴$ Breadth of rectangle = 11cm and length of rectangle = 60cm.
Now, perimeter of rectangle = 2(l + b)
= 2(60 + 11) = 2 × 71
= 142cm.

THE TRIANGLE AND ITS PROPERTIES

298399 A triangle has altitudes:

1 2

2 3

3 1

4 None of these

THE TRIANGLE AND ITS PROPERTIES

298400 In the following figure, find $∠ x$ and $∠ y$ if $∠ x - ∠ y - 10^{\circ}$
3

1 65º, 55º

2 55º, 45º

3 45º, 35º

4 60º, 60º

Explanation:

65º, 55º
$∠ x + ∠ y = 120^{\circ}, ∠ x - ∠ y = 10^{\circ}$
$∠ x = 65^{\circ}, ∠ = 55^{\circ}$

THE TRIANGLE AND ITS PROPERTIES

298401 In Fig. if AB || CD, the values of x and y are:
4

1 x = 21, y = 28

2 x = 21, y = 38

3 x = 38, y = 21

4 x = 22, y = 38

Explanation:

x = 21, y = 38
$∠ AEC + ∠ AEB = 180^{\circ}$ [Linear angles]
$\Rightarrow 79^{\circ} + ∠ AEB = 180^{\circ}$
$\Rightarrow ∠ AEB = 101^{\circ}$
Since, AB || CD
$∠ ABE = ∠ ECD = 58^{\circ}$ [Alternate angles]
Now, In $△ AEB$
$∠ AEB + ∠ EAB + ∠ ABE = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 101^{\circ} + 58^{\circ} + x^{\circ} = 180^{\circ}$
$\Rightarrow x^{\circ} = 21^{\circ}$
$\Rightarrow x = 21$
Now, In $△ AEB$
$∠ AEC + ∠ CAE + ∠ CEA = 180^{\circ}$ [Angle sum property of triangle]
$\Rightarrow 79^{\circ} + y^{\circ} + 3 x^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 3 (21)^{\circ} = 180^{\circ}$
$\Rightarrow 79^{\circ} + y^{\circ} + 63^{\circ} = 180^{\circ}$
$\Rightarrow y^{\circ} = 38^{\circ}$
$\Rightarrow y = 38$
Hence, the correct answer is option (b).