Question : A tree of height $h$ metres is broken by storm in such a way that its top touches the ground at a distance of $x$ metres from its root. Find the height at which the tree is broken. (Here $h>x$)
Option 1: $\frac{h^2+x^2}{2h}$ metre
Option 2: $\frac{h^2-x^2}{2h}$ metre
Option 3: $\frac{h^2+x^2}{4h}$ metre
Option 4: $\frac{h^2-x^2}{4h}$ metre
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\frac{h^2-x^2}{2h}$ metre
Solution : Given: AB = Height of tree = $h$ metre Let the height at which the tree is broken, AC = $y$ metre BC = CD = Broken part of tree = $(h – y)$ metre ∴ In ∆ ACD, AC 2 + AD 2 = CD 2 ⇒ $y^2 + x^2 = (h – y)^2$ ⇒ $y^2 + x^2 = h^2 + y^2 – 2hy$ ⇒ $x^2 = h^2 – 2hy$ ⇒ $2hy = h^2 – x^2$ $\therefore y =\frac{h^2-x^2}{2h}$ metre Hence, the correct answer is $\frac{h^2-x^2}{2h}$ metre.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If the sides of an equilateral triangle are increased by 1 metre, then its area is increased by $\sqrt3$ sq. metre. The length of any of its sides is:
Question : The radius of the base and curved surface area of a right cylinder are $r$ units and $4\pi rh$ square units respectively. The height of the cylinder is:
Question : If $x>1$ and $x+\frac{1}{x}=2\frac{1}{12}$, then the value of $x^{4}-\frac{1}{x^{4}}$ is:
Question : If $x\cos \theta -y\sin \theta =\sqrt{x^{2}+y^{2}}$ and $\frac{\cos ^2{\theta }}{a^{2}}+\frac{\sin ^{2}\theta}{b^{2}}=\frac{1}{x^{2}+y^{2}},$ then the correct relation is:
Question : If $\frac{x+1}{x-1}=\frac{a}{b}$ and $\frac{1-y}{1+y}=\frac{b}{a}$, then the value of $\frac{x-y}{1+xy}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile