MCQ
$ABCD$ is a Rhombus such that $\angle\text{ACB}= 40^\circ,$ then $\angle\text{ADB}$ is:
- ✓$50^\circ $
- B$60^\circ$
- C$100^\circ$
- D$40^\circ$
✓
Answer
Correct option: A.
$50^\circ $
In Rhombus, diagonals bisect each other right angle. By using angle sum property in any of the four triangles formed by intersection of diagonals, we get $\angle\text{CBD} = 50$ and $\angle\text{CBD} = \angle\text{ADC}$ (alternate angles).
So, $\angle\text{ADC} = 50$
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, $ABCD$ is a cyclic quadrilateral in which $BC = CD$ and $\angle\text{CBD}=35^\circ,$ Then, $\angle\text{BAD}=?$
→The value of $\{2-3(2-3)^3\}^3,$ is:
→The line represented by the equation $8 x+3 y=24$ cuts the coordinate axes at $A$ and $B$. The area of the triangle $A O B$ is
→Two straight lines $AB$ and $CD$ intersect one another at the point $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ,$ then $\angle\text{AOD}=$
→If $2^{-\text{m}}\times\frac{1}{2^{\text{m}}}=\frac{1}{4},$ then $\frac{1}{14}\bigg\{(4^\text{m})^{\frac{1}{2}}\Big(\frac{1}{5^{\text{m}}}\Big)^{-1}\bigg\}$ is equal to:
→The radius of a cylinder is doubled, and the height remains the same. The ratio between the volumes of the new cylinder and the original cylinder is:
$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$ when simplified is:
→$ABCD$ is a parallelogram in which diagonal $AC$ bisects $\angle\text{BAD}.$ if $\angle\text{BAC}=35^\circ,$ then $\angle\text{ABC}=$
→Write the correct answer of the following: In Fig. if parallelogram $ABCD$ and rectangle $ABEF$ are of equal area, then:
→Write the correct answer in the following: The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is
→