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TRIANGLES

90709 A vertical stick 20m long casts a shadow 10m long on the ground. At the same time, a tower casts a shadow 50m long on the ground. The height of the tower is:

1 100m.

2 120m.

3 25m.

4 200m.

Explanation:

A100m.
Height of a stick = 20m
and length of its shadow = 10m
At the same time
Let height of tower = x m
and its shadow = 50m
20 : x = 10 : 50
x × 10 = 20 × 50
$\Rightarrow x = \frac{20 \times 50}{10} = 100$
Height of tower = 100m.

TRIANGLES

90710 Two poles of height 6m and 11m stand vertically upright on a plane ground. If the distance between their foot is 12m, the distance between their tops is:

1 12m.

2 14m.

3 13m.

4 11m.

Explanation:

C13m.
Het AB and CD be two poles and distance blw their root is 12m.

ED = CD - CE = 11 - 6 = 5cm
In right $△ ADE,$
AD $^{2}$ = AE $^{2}$ + DE $^{2}$
$\Rightarrow$ AD $^{2}$ = (12) $^{2}$ + (5) $^{2}$
$\Rightarrow$ AD $^{2}$ = 144 + 25
$\Rightarrow$ AD $^{2}$ = 169
$\Rightarrow$ AD = 13m
Thus distance blw their tops is 13m.

TRIANGLES

90713 In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$ The angle B is:

1

120^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

C $90^{\circ}$
Given,
In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$
Here, AC is the longest side.
If the square of the hypotenuse is equal to the square of the other two sides, then it is a right angled triangle.
$So, {AC}^{2} = {AB}^{2} + {BC}^{2}$
$(12)^{2} = (6 \sqrt{3})^{2} + (6)^{2}$
$144 = 108 + 36$
$144 = 144$
$∴ △ ABC$ is a right angled triangle and angle opposite to hypotenuse, i.e. opposite to AC is $∠ b$ and is equal to $90^{\circ}$

TRIANGLES

90714 In the given figure, DE || BC. If DE = 5cm, BC = 8cm and AD = 3.5cm then AB =?

1 5.6cm

2 4.8cm

3 5.2cm

4 6.4cm

Explanation:

A5.6cm
$∴ DE | | BC$
$∴ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ (Thales' theorem)
$\Rightarrow \frac{3.5}{AB} = \frac{5}{8}$
$\Rightarrow AB = \frac{3.5 \times 8}{5} = 5.6 cm$

TRIANGLES

90709 A vertical stick 20m long casts a shadow 10m long on the ground. At the same time, a tower casts a shadow 50m long on the ground. The height of the tower is:

1 100m.

2 120m.

3 25m.

4 200m.

Explanation:

A100m.
Height of a stick = 20m
and length of its shadow = 10m
At the same time
Let height of tower = x m
and its shadow = 50m
20 : x = 10 : 50
x × 10 = 20 × 50
$\Rightarrow x = \frac{20 \times 50}{10} = 100$
Height of tower = 100m.

TRIANGLES

90710 Two poles of height 6m and 11m stand vertically upright on a plane ground. If the distance between their foot is 12m, the distance between their tops is:

1 12m.

2 14m.

3 13m.

4 11m.

Explanation:

C13m.
Het AB and CD be two poles and distance blw their root is 12m.

ED = CD - CE = 11 - 6 = 5cm
In right $△ ADE,$
AD $^{2}$ = AE $^{2}$ + DE $^{2}$
$\Rightarrow$ AD $^{2}$ = (12) $^{2}$ + (5) $^{2}$
$\Rightarrow$ AD $^{2}$ = 144 + 25
$\Rightarrow$ AD $^{2}$ = 169
$\Rightarrow$ AD = 13m
Thus distance blw their tops is 13m.

TRIANGLES

90713 In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$ The angle B is:

1

120^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

C $90^{\circ}$
Given,
In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$
Here, AC is the longest side.
If the square of the hypotenuse is equal to the square of the other two sides, then it is a right angled triangle.
$So, {AC}^{2} = {AB}^{2} + {BC}^{2}$
$(12)^{2} = (6 \sqrt{3})^{2} + (6)^{2}$
$144 = 108 + 36$
$144 = 144$
$∴ △ ABC$ is a right angled triangle and angle opposite to hypotenuse, i.e. opposite to AC is $∠ b$ and is equal to $90^{\circ}$

TRIANGLES

90714 In the given figure, DE || BC. If DE = 5cm, BC = 8cm and AD = 3.5cm then AB =?

1 5.6cm

2 4.8cm

3 5.2cm

4 6.4cm

Explanation:

A5.6cm
$∴ DE | | BC$
$∴ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ (Thales' theorem)
$\Rightarrow \frac{3.5}{AB} = \frac{5}{8}$
$\Rightarrow AB = \frac{3.5 \times 8}{5} = 5.6 cm$

TRIANGLES

90709 A vertical stick 20m long casts a shadow 10m long on the ground. At the same time, a tower casts a shadow 50m long on the ground. The height of the tower is:

1 100m.

2 120m.

3 25m.

4 200m.

Explanation:

A100m.
Height of a stick = 20m
and length of its shadow = 10m
At the same time
Let height of tower = x m
and its shadow = 50m
20 : x = 10 : 50
x × 10 = 20 × 50
$\Rightarrow x = \frac{20 \times 50}{10} = 100$
Height of tower = 100m.

TRIANGLES

90710 Two poles of height 6m and 11m stand vertically upright on a plane ground. If the distance between their foot is 12m, the distance between their tops is:

1 12m.

2 14m.

3 13m.

4 11m.

Explanation:

C13m.
Het AB and CD be two poles and distance blw their root is 12m.

ED = CD - CE = 11 - 6 = 5cm
In right $△ ADE,$
AD $^{2}$ = AE $^{2}$ + DE $^{2}$
$\Rightarrow$ AD $^{2}$ = (12) $^{2}$ + (5) $^{2}$
$\Rightarrow$ AD $^{2}$ = 144 + 25
$\Rightarrow$ AD $^{2}$ = 169
$\Rightarrow$ AD = 13m
Thus distance blw their tops is 13m.

TRIANGLES

90713 In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$ The angle B is:

1

120^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

C $90^{\circ}$
Given,
In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$
Here, AC is the longest side.
If the square of the hypotenuse is equal to the square of the other two sides, then it is a right angled triangle.
$So, {AC}^{2} = {AB}^{2} + {BC}^{2}$
$(12)^{2} = (6 \sqrt{3})^{2} + (6)^{2}$
$144 = 108 + 36$
$144 = 144$
$∴ △ ABC$ is a right angled triangle and angle opposite to hypotenuse, i.e. opposite to AC is $∠ b$ and is equal to $90^{\circ}$

TRIANGLES

90714 In the given figure, DE || BC. If DE = 5cm, BC = 8cm and AD = 3.5cm then AB =?

1 5.6cm

2 4.8cm

3 5.2cm

4 6.4cm

Explanation:

A5.6cm
$∴ DE | | BC$
$∴ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ (Thales' theorem)
$\Rightarrow \frac{3.5}{AB} = \frac{5}{8}$
$\Rightarrow AB = \frac{3.5 \times 8}{5} = 5.6 cm$

TRIANGLES

90709 A vertical stick 20m long casts a shadow 10m long on the ground. At the same time, a tower casts a shadow 50m long on the ground. The height of the tower is:

1 100m.

2 120m.

3 25m.

4 200m.

Explanation:

A100m.
Height of a stick = 20m
and length of its shadow = 10m
At the same time
Let height of tower = x m
and its shadow = 50m
20 : x = 10 : 50
x × 10 = 20 × 50
$\Rightarrow x = \frac{20 \times 50}{10} = 100$
Height of tower = 100m.

TRIANGLES

90710 Two poles of height 6m and 11m stand vertically upright on a plane ground. If the distance between their foot is 12m, the distance between their tops is:

1 12m.

2 14m.

3 13m.

4 11m.

Explanation:

C13m.
Het AB and CD be two poles and distance blw their root is 12m.

ED = CD - CE = 11 - 6 = 5cm
In right $△ ADE,$
AD $^{2}$ = AE $^{2}$ + DE $^{2}$
$\Rightarrow$ AD $^{2}$ = (12) $^{2}$ + (5) $^{2}$
$\Rightarrow$ AD $^{2}$ = 144 + 25
$\Rightarrow$ AD $^{2}$ = 169
$\Rightarrow$ AD = 13m
Thus distance blw their tops is 13m.

TRIANGLES

90713 In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$ The angle B is:

1

120^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

C $90^{\circ}$
Given,
In $△ ABC, AB = 6 \sqrt{3cm} = AC = 12 cm$ and $BC = 6 cm$
Here, AC is the longest side.
If the square of the hypotenuse is equal to the square of the other two sides, then it is a right angled triangle.
$So, {AC}^{2} = {AB}^{2} + {BC}^{2}$
$(12)^{2} = (6 \sqrt{3})^{2} + (6)^{2}$
$144 = 108 + 36$
$144 = 144$
$∴ △ ABC$ is a right angled triangle and angle opposite to hypotenuse, i.e. opposite to AC is $∠ b$ and is equal to $90^{\circ}$

TRIANGLES

90714 In the given figure, DE || BC. If DE = 5cm, BC = 8cm and AD = 3.5cm then AB =?

1 5.6cm

2 4.8cm

3 5.2cm

4 6.4cm

Explanation:

A5.6cm
$∴ DE | | BC$
$∴ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ (Thales' theorem)
$\Rightarrow \frac{3.5}{AB} = \frac{5}{8}$
$\Rightarrow AB = \frac{3.5 \times 8}{5} = 5.6 cm$

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