
ABCD is a rectangular field. A vertical lamp post of height 12 m stands at the corner A. If the angle of elevation of its top from B is 60° and from C is 45°. then the area of the field is
A. $48\sqrt{2\,}$sq m
B. $48\sqrt{3}$ sq m
C. $12\sqrt{3}$sq m
D. $12\sqrt{2}$sq m
Answer
163.8k+ views
Hint:
We use Pythagoras theorem to solve this question. The well-known Pythagorean Theorem states that the square on the hypotenuse of a right triangle equals the sum of the squares on its legs. First we find the values of AB and AC by using simple trigonometric functions. Then by using Pythagoras theorem we find the value of BC and after that we are able to find the area of the field.
Formula Used:
The formula for the Pythagoras theorem is written as $H^{2}=P^{2}+B^{2}$, where
$H^{2}$ is the hypotenuse of the right triangle
and $P$ and $B$ are its other two legs
Complete Step by step Solution:
Let EA be the lamppost at corner A and ABCD be the rectangular field.

Since AE is a vertical pole, all lines in the planes of the rectangular grid are perpendicular to it, i.e., EA is perp. to AB, BC, CD, and DA.
$\therefore \angle \mathrm{EAD}=90^{\circ}$
Given,
$E A=12 \mathrm{~m}$
Join E B, E C, and A C
Also Given that
$\angle \mathrm{EBA}=60^{\circ}, \angle \mathrm{ACE}=45^{\circ} .$
In . $\triangle A B E$,
$\tan 60^{\circ}=\dfrac{A E}{A B}=\dfrac{12}{A B}$
$\Rightarrow \sqrt{3}=\dfrac{12}{A B}$
$\Rightarrow A B=\dfrac{12}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}$
In $\triangle$ ACE,
$\tan 45^{\circ}=\dfrac{A E}{A C}$
$\Rightarrow \dfrac{12}{A C}=1$
$\Rightarrow A C=12 \mathrm{~m}$
In $\triangle A B C$,
According to the Pythagoras theorem
$\mathrm{BC}=\sqrt{A C^{2}-A B^{2}}$
$=\sqrt{144-48}$
$=\sqrt{96} \mathrm{~m}$
$=4 \sqrt{6} \mathrm{~m}$
Area of the rectangular field $=\mathrm{AB} \times \mathrm{BC}$
$=4 \sqrt{3} \times 4 \sqrt{6}$
$=48 \sqrt{2}$ sq. $\mathrm{m}$
So the correct answer is option A.
Note: The formula for the Pythagoras theorem is written as $H^{2}=P^{2}+B^{2}$, where $H^{2}$ is the hypotenuse of the right triangle and $P$ and $B$ are its other two legs. As a result, the Pythagoras equation can be used to any triangle that has one angle that is exactly 90 degrees to create a Pythagoras triangle.
We use Pythagoras theorem to solve this question. The well-known Pythagorean Theorem states that the square on the hypotenuse of a right triangle equals the sum of the squares on its legs. First we find the values of AB and AC by using simple trigonometric functions. Then by using Pythagoras theorem we find the value of BC and after that we are able to find the area of the field.
Formula Used:
The formula for the Pythagoras theorem is written as $H^{2}=P^{2}+B^{2}$, where
$H^{2}$ is the hypotenuse of the right triangle
and $P$ and $B$ are its other two legs
Complete Step by step Solution:
Let EA be the lamppost at corner A and ABCD be the rectangular field.

Since AE is a vertical pole, all lines in the planes of the rectangular grid are perpendicular to it, i.e., EA is perp. to AB, BC, CD, and DA.
$\therefore \angle \mathrm{EAD}=90^{\circ}$
Given,
$E A=12 \mathrm{~m}$
Join E B, E C, and A C
Also Given that
$\angle \mathrm{EBA}=60^{\circ}, \angle \mathrm{ACE}=45^{\circ} .$
In . $\triangle A B E$,
$\tan 60^{\circ}=\dfrac{A E}{A B}=\dfrac{12}{A B}$
$\Rightarrow \sqrt{3}=\dfrac{12}{A B}$
$\Rightarrow A B=\dfrac{12}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}$
In $\triangle$ ACE,
$\tan 45^{\circ}=\dfrac{A E}{A C}$
$\Rightarrow \dfrac{12}{A C}=1$
$\Rightarrow A C=12 \mathrm{~m}$
In $\triangle A B C$,
According to the Pythagoras theorem
$\mathrm{BC}=\sqrt{A C^{2}-A B^{2}}$
$=\sqrt{144-48}$
$=\sqrt{96} \mathrm{~m}$
$=4 \sqrt{6} \mathrm{~m}$
Area of the rectangular field $=\mathrm{AB} \times \mathrm{BC}$
$=4 \sqrt{3} \times 4 \sqrt{6}$
$=48 \sqrt{2}$ sq. $\mathrm{m}$
So the correct answer is option A.
Note: The formula for the Pythagoras theorem is written as $H^{2}=P^{2}+B^{2}$, where $H^{2}$ is the hypotenuse of the right triangle and $P$ and $B$ are its other two legs. As a result, the Pythagoras equation can be used to any triangle that has one angle that is exactly 90 degrees to create a Pythagoras triangle.
Recently Updated Pages
Difference Between Distance and Displacement: JEE Main 2024

IIT Full Form

Uniform Acceleration - Definition, Equation, Examples, and FAQs

Difference Between Metals and Non-Metals: JEE Main 2024

Newton’s Laws of Motion – Definition, Principles, and Examples

Difference Between Pound and Kilogram with Definitions, Relation

Trending doubts
JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

Lami's Theorem

JEE Main 2025 Cut-off For NIT Andhra Pradesh

Reaction of Metals With Acids for JEE

Electrochemical Series: Definition and Its Applications

Convex and Concave Lenses - JEE

Other Pages
NCERT Solutions for Class 10 Maths Chapter 14 Probability

NCERT Solutions for Class 10 Maths In Hindi Chapter 15 Probability

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
