A) Data inadequate | B) 8 units |

C) 12 units | D) None of these |

Explanation:

One of AB, AD and CD must have given.

So, the data is inadequate.

A) 1080 cm | B) 180 m |

C) 108 m | D) 118 cm |

Explanation:

We know the rule that,

At particular time for all object , ratio of height and shadow are same.

Let the height of the pole be 'H'

Then

=> H = 108 m.

A) 14 cm | B) 13.5 cm |

C) 12.5 cm | D) 11.4 cm |

Explanation:

Let the required height of the building be x meter

More shadow length, More height(direct proportion)

Hence we can write as

(shadow length) 40.25 : 28.75 :: 17.5 : x

⇒ 40.25 × x = 28.75 × 17.5

⇒ x = 28.75 × 17.5/40.25

= 2875 × 175/40250

= 2875 × 7/1610

= 2875/230

= 575/46

= 12.5 cm

A) 42.2 mts | B) 33.45 mts |

C) 66.6 mts | D) 58.78 mts |

Explanation:

From above diagram

AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )

angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)

∵ BD = CE and Substitute the value of CE from equation 1

100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

A) 21.6 m | B) 23.2 m |

C) 24.72 m | D) None of these |

Explanation:

Draw BE CD

Then, CE = AB = 1.6 m,

BE = AC =

CD = CE + DE = (1.6 + 20) m = 21.6 m.

A) 173 m | B) 200 m |

C) 273 m | D) 300 m |

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.