MCQ
$\text{ABCD}$ is a rhombus in which $\angle\text{C}=60^\circ.$ Then, $AC : BD =$ ? 
- ✓$\sqrt{3}:1$
- B$\sqrt{3}:\sqrt{2}$
- C$3:1$
- D$3:2$
✓
Answer
Correct option: A.
$\sqrt{3}:1$
Given that $\text{ABCD}$ is a rhombus.
$BC = CD [$Since all the sides of a rhombus are equal]
$\Rightarrow\ \angle\text{BDC}=\angle\text{BCD} [$Angles opposite equal sides are equal$]$
In $\triangle\text{BCD},$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{CBD}=180^\circ [$Angle sum property$]$
$\Rightarrow\ \angle\text{BDC}+60^\circ+\angle\text{BDC}=180^\circ$
$\Rightarrow\ 2\angle\text{BDC}=120^\circ$
$\Rightarrow\ \angle\text{BDC}=60^\circ$
$\Rightarrow\ \angle\text{BDC}=\angle\text{BCD}=60^\circ$
So, $\triangle\text{BCD},$ is an equilateral triangle.
$\therefore BD = BC = CD = x ($say$)$
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
In right $\triangle\text{AOB},$
$AB^2 = AO^2+ OB^2 [$By Pythagoras theoram$]$
$\Rightarrow AO^2 = AB^2 - OB^2$
$\Rightarrow\ \text{AO}^2=\text{x}^2-\Big(\frac{\text{x}}{2}\Big)^2$
$\Rightarrow\ \text{AO}^2=\text{x}^2-\Big(\frac{\text{x}^2}{4}\Big)$
$\Rightarrow\ \text{AO}^2=\frac{3\text{x}^2}{4}$
$\Rightarrow\ \text{AO}=\frac{\sqrt{3}\text{x}}{2}$
So, $\text{AC}=\sqrt{3}\text{x}$
Thus,
$\frac{\text{AC}}{\text{BD}}=\frac{\sqrt{3}\text{x}}{\text{x}}=\frac{\sqrt{3}}{1}$
Hence,
$\text{AC}:\text{BD}=\sqrt{3}:1.$
$BC = CD [$Since all the sides of a rhombus are equal]
$\Rightarrow\ \angle\text{BDC}=\angle\text{BCD} [$Angles opposite equal sides are equal$]$
In $\triangle\text{BCD},$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{CBD}=180^\circ [$Angle sum property$]$
$\Rightarrow\ \angle\text{BDC}+60^\circ+\angle\text{BDC}=180^\circ$
$\Rightarrow\ 2\angle\text{BDC}=120^\circ$
$\Rightarrow\ \angle\text{BDC}=60^\circ$
$\Rightarrow\ \angle\text{BDC}=\angle\text{BCD}=60^\circ$
So, $\triangle\text{BCD},$ is an equilateral triangle.
$\therefore BD = BC = CD = x ($say$)$
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
In right $\triangle\text{AOB},$
$AB^2 = AO^2+ OB^2 [$By Pythagoras theoram$]$
$\Rightarrow AO^2 = AB^2 - OB^2$
$\Rightarrow\ \text{AO}^2=\text{x}^2-\Big(\frac{\text{x}}{2}\Big)^2$
$\Rightarrow\ \text{AO}^2=\text{x}^2-\Big(\frac{\text{x}^2}{4}\Big)$
$\Rightarrow\ \text{AO}^2=\frac{3\text{x}^2}{4}$
$\Rightarrow\ \text{AO}=\frac{\sqrt{3}\text{x}}{2}$
So, $\text{AC}=\sqrt{3}\text{x}$
Thus,
$\frac{\text{AC}}{\text{BD}}=\frac{\sqrt{3}\text{x}}{\text{x}}=\frac{\sqrt{3}}{1}$
Hence,
$\text{AC}:\text{BD}=\sqrt{3}:1.$
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
In the give figure, AB and CD are two intersecting chords of a circle. If $\angle\text{CAB}=40^\circ$ and $\angle\text{BCD}=80^\circ$then $\angle\text{CBD}=?$
→Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11cm, BE = 3cm and DE = 3.5cm, then CD = ?
In figure, if $l_1 \ \| \ l_2,$ what isb the value of $y$ ?
→The radii of two cylinders are in the ratio $2 : 3$ and their heights are in the ratio $5 : 3.$ The ratio of their curved surface areas is:
→If AB is a chord of a circle, P and Q are the two points on the circle different from A and B, then:
→- $\angle\text{APB}=\angle\text{AQB}$
- $\angle\text{APB}+\angle\text{AQB}=180^\circ\ \text{or }\angle\text{APB}=\angle\text{AQB}$
- $\angle\text{APB}+\angle\text{AQB}=90^\circ$
- $\angle\text{APB}+\angle\text{AQB}=180^\circ$
The area of an equilateral triangle with sides $2\sqrt3 \ \text{cm}$ is:
→A grouped frequency table with class intervals of equal size using 3-5 (5 included in this interval) as one of the class intervals is constructed for the following data. The frequency of the class 3-5 is:
|
1
|
4
|
7
|
2
|
0
|
3
|
9
|
2
|
3
|
7
|
6
|
3
|
5
|
|
2
|
5
|
5
|
6
|
2
|
3
|
5
|
1
|
0
|
4
|
6
|
4
|
6
|
In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and $\angle\text{CBD}=35^\circ.$ Then, $\angle\text{BAD}=?$
→- 65°
- 70°
- 110°
- 90°
The shape of base of Pyramid is:
If $x + 1$ is a factor of the polynomial $2x^2 + kx,$ then $k =$
→