MCQ
ABCD is a rhombus. If $\angle\text{ACB} = 30^\circ,$ then $\angle\text{ADB}$ is:
- A45º
- B120º
- C30º
- D60º
✓
Answer
- 60º
Solution:
If $\angle\text{ACB} = 30^\circ$ then $\angle\text{CAD} = 30^\circ$ (The diagonals bisect the angles in a Rhombus) and in $\triangle\text{AOD}, \ \angle\text{AOB} = 90^\circ$ diagonals perpendicular to each other, so by angle sum property of a $\angle\text{ADB} = 60^\circ.$
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 a $\triangle A B C$, if $\angle A=60^{\circ}, \angle B=80^{\circ}$ and the bisectors of $\angle B$ and $\angle C$ meet at $O$, then $\angle B O C=$
→→
lf the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length:
The area of an equilateral triangle having side length equal to $\sqrt\frac{3}{4}\text {cm}$ (using Heron’s formula) is:
→An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is:
→The zeros of the polynomial p(x) = 2x2 + 5x -3 are:
→- $\frac{1}{2},3$
- $\frac{1}{2},-3$
- $\frac{-1}{2},3$
- $1,\frac{-1}{2}$
Which of the following pair is a solution of the equation 3x - 2y = 7?
Which of the following is the ascending order of $(343)^3,(2401)^2$ and $(49)^5 ?$
→In a quadrilateral ABCD, it is given that BD = 16cm. If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD}$ such that AL = 9cm and CM = 7cm, then ar(quadrilateral ABCD) = ?
→- 256cm2
- 128cm2
- 64cm2
- 96cm2
Write the correct answer in the following:
Abscissa of all the points on the x-axis is:
Abscissa of all the points on the x-axis is:
- 0
- 1
- 2
- Any number.