Question
Solve the following examples.
Find the height of an equilateral triangle having side 2a.

Answer


Let ABC be an equilateral triangle,
Let AP be a perpendicular on side BC from A.
To find : Height of triangle = AP
As, ABC is an equilateral triangle we have
AB = BC = CA = 2a
Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side
$\Rightarrow BP = CP =\frac{1}{2} = 'a'$
Now, In ΔABP, By Pythagoras theorem
$\text { (Hypotenuse }^2=(\text { base })^2+(\text { Perpendicular })^2$
$\Rightarrow A B^2=B P^2+A P^2$
$\Rightarrow(2 a)^2=a^2+A P^2$
$\Rightarrow A P^2=4 a^2-a^2$
$\Rightarrow A P^2=3 a^2$
$\Rightarrow A P=a \sqrt{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Prove the following trigonometric identities.
$\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}=\sec\theta\text{cosec}\theta-2\sin\theta\cos\theta$
In the following, determine whether the given values are solution of the given equation or not:
$2x^2 - x + 9 = x^2 + 4x + 3, x = 2, x = 3$
A man busy a number of pens for $Rs. 180$.
If he had bought $3$ more pens for the same amount, each pen would have cost him $Rs. 3$ less. How many pens did he buy?
Solve the following quadratic equations by factorization:
$\frac{\text{x}-\text{a}}{\text{x}-\text{b}}+\frac{\text{x}-\text{b}}{\text{x}-\text{a}}=\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}$
If $\text{x}=\frac{2}{3}$ and $x = -3$ are the roots of the equation $ax^2 + 7x + b = 0$, find the values of a and b.
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6cm and 9cm respectively. If the area of $\triangle\text{ABC}=54\text{cm}^2$then find the lengths of sides of AB and AC.
Prove the following trigonometric identities.
If $\text{cosec }\theta-\sin\theta=\text{a}^3,\sec\theta-\cos\theta=\text{b}^3,$ prove that $a^2b^2 (a^2 + b^2) = 1$.
Solve : $\frac{5}{x-1}+\frac{1}{y-2}=2 ; \frac{6}{x-1}-\frac{3}{y-2}=1$
Mr. Deepak Pal invested ₹ 1,00,354 in shares of FV ₹ 100 , when the market value is ₹ 50 . Rate of brokerage is $0.3 \%$ and rate of GST on brokerage is $18 \%$, then how many shares were purchased?
Draw a $\triangle\text{ABC}$ in which base $BC = 6cm, AB = 5cm$ and $\angle\text{ABC}=60^\circ.$ Then construct another triangle whose sides are of the corresponding sides of $\triangle\text{ABC}.$