MCQ
In an equilateral triangle $\text{ABC}$ if $\text{AD}\perp\text{BC},$ then $AD^2 =$
- A$CD^2$
- B$2CD^2$
- ✓$3CD^2$
- D$4CD^2$
✓
Answer
Correct option: C.
$3CD^2$
In an equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$
In $\triangle ADC$, applying Pythagoras theorem, we get,
$A C^2=A D^2+D C^2$
$B C^2=A D^2+D C^2(\because A C=B C)$
$(2 D C)^2=A D^2+D C^2(\because B C=2 D C)$
$4 D C^2=A D^2+D C^2$
$3 D C^2=A D^2$
$3 C D^2=A D^2$
Hence, the correct option is $C.$
In $\triangle ADC$, applying Pythagoras theorem, we get,
$A C^2=A D^2+D C^2$
$B C^2=A D^2+D C^2(\because A C=B C)$
$(2 D C)^2=A D^2+D C^2(\because B C=2 D C)$
$4 D C^2=A D^2+D C^2$
$3 D C^2=A D^2$
$3 C D^2=A D^2$
Hence, the correct option is $C.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The area of a sector of angle a (in degrees) of a circle with radius R is:
Choose the correct answer from the given four options : One ticket is drawn at random from a bag containing tickets numbered $1$ to $40$. The probability that the selected ticket has a number which is a multiple of $5$ is :
→The circumference of incircle of a square with side 14 cm is -
If $x = a \cos \theta$ and $y = b \sin \theta$, then $b ^2 x ^2+ a ^2 y ^2=$
→Degree of polynomial $y^3-2 y^2-\sqrt{3} y+\frac{1}{2}$ is
→In $\triangle \text{ABC}$, it is given that $A B=9 \ cm$, $BC=6 \ cm$ and $C A=7.5 \ cm$. Also, $\triangle \text{DEF}$ is given such that $\text{EF}=8 \ cm$ and $\triangle \text{DEF} \sim \triangle \text{ABC}$. Then, perimeter of $\triangle \text{DEF}$ is
→$(\operatorname{cosec} \theta-\cot \theta)^2=?$
→In the given figure, if $AB = 8\ cm$ and $PE = 3\ cm,$ then $AE =$
→A point $(x, 1)$ is equidistant from $(0,0)$ and $(2,0)$. The value of $x$ is
→A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and height of its conical part is