The following figure is a rectangle in which $x: y = 3: 7$; find the values of $x$ and $y$.
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Special Types of Quadrilaterals question types
38 questions across 3 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
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Sample Questions
Special Types of Quadrilaterals questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
$ABCD$ is a parallelogram. What kind of quadrilateral is it if: $AC = BD$ but $AC$ is not perpendicular to $BD$?
View full solution →$ABCD$ is a parallelogram. What kind of quadrilateral is it if: $AC$ is perpendicular to $BD$ but is not equal to it?
View full solution →$ABCD$ is a parallelogram. What kind of quadrilateral is it if : $AC = BD$ and $AC$ is perpendicular to $BD$?
View full solution →Given: Parallelogram $ABCD$ in which diagonals $AC$ and $BD$ intersect at $M$. Prove: $M$ is the mid$-$point of $LN$.
View full solution →$ABCD$ is a rhombus. If $\angle BCA = 35^\circ$ . find $\angle ADC.$
View full solution →$ABCD$ is a rectangle, if $\angle BPC = 124^\circ$ Calculate : $(i)\ \angle BAP\ (ii) \ \angle ADP$
View full solution →In rhombus $ABCD; (i)$ if $\angle A = 74^\circ$ ; find $\angle B$ and $\angle C.(ii)$ if $AD = 7.5 \ cm $; find $BC$ and $CD.$
View full solution →Prove that the diagonals of a parallelogram bisect each other.
$\text{PQRS}$ is a parallelogram whose diagonals intersect at $M$. If $\angle PMS = 54^\circ , \angle QSR = 25^\circ$ and $\angle SQR = 30^\circ$ ; find : $(i)\ \angle RPS\ (ii)\ \angle PRS\ (iii)\ \angle PSR.$
View full solution →$\text{ABCD}$ is a rhombus. If $\angle BAC = 38^\circ$ , find : $(i)\ \angle ACB\ (ii)\ \angle DAC\ (iii)\ \angle ADC.$
View full solution →In square $PQRS : (i)$ if $PQ = 3x – 7$ and $QR = x + 3$ ; find $PS\ (ii)$ if $PR = 5x$ and $QR = 9x – 8$. Find $QS$
View full solution →In parallelogram $PQRS, \angle Q = (4x – 5)^\circ$ and $\angle S = (3x + 10)^\circ $. Calculate : $\angle Q$ and $\angle R.$
View full solution →In the given figure, $AB || EC, AB = AC$ and $AE$ bisects $\angle DAC$. Prove that:
$(i) \angle EAC = \angle ACB(ii)\text{ ABCE}$ is a parallelogram.
View full solution →$(i) \angle EAC = \angle ACB(ii)\text{ ABCE}$ is a parallelogram.
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