MCQ 511 Mark
If the diagonals of a quadrilateral PQRS bisect each other such that $\angle P=40^{\circ}$, then $\angle Q=$
- ✓
$140^{\circ}$
- B
$50^{\circ}$
- C
$100^{\circ}$
- D
$120^{\circ}$
AnswerCorrect option: A. $140^{\circ}$
(a) $140^{\circ}$
Since the diagonal of quadrilateral PQRS bisect each other. Therefore, PQRS is a parallelogram.$
\therefore \quad \angle P+\angle Q=180^{\circ} \Rightarrow 40^{\circ}+\angle Q=180^{\circ} \Rightarrow \angle Q=140^{\circ}$
View full question & answer→MCQ 521 Mark
One of the diagonals of a rhombus is equal to a side of the rhombus. The angles of the rhombus are
- A
$70^{\circ}, 110^{\circ}, 110^{\circ}, 70^{\circ}$
- B
$80^{\circ}, 100^{\circ}, 100^{\circ}, 80^{\circ}$
- ✓
$120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$
- D
AnswerCorrect option: C. $120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$
(c) $120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$
Let ABCD be a rhombus such that diagonal AC = AB. Then, $\triangle A B C$ is an equilateral triangle. Therefore,
$\angle B=60^{\circ} \Rightarrow \angle D=60^{\circ} \Rightarrow \angle A=\angle C=180^{\circ}-60^{\circ}=120^{\circ}$
Hence, the angles are $120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$.
View full question & answer→MCQ 531 Mark
In a parallelogram ABCD, if $\angle A=75^{\circ}$, then $\angle D-\angle C=$
- A
$5^{\circ}$
- B
$20^{\circ}$
- ✓
$30^{\circ}$
- D
$60^{\circ}$
AnswerCorrect option: C. $30^{\circ}$
(c) $30^{\circ}$
We have, $\angle A=75^{\circ}$
$\therefore \quad \angle C=75^{\circ} \text { and } \angle D=105^{\circ} \Rightarrow \angle D-\angle C=30^{\circ}$
View full question & answer→MCQ 541 Mark
In a rhombus ABCD, if $A B=A C$, then $\angle A B C$ is
- A
$120^{\circ}$
- B
$90^{\circ}$
- ✓
$60^{\circ}$
- D
AnswerCorrect option: C. $60^{\circ}$
(c) $60^{\circ}$
ABCD is a rhombus
$\therefore \quad A B=B C=C D=D A$
Given that $A B=A C$. Therefore, in $\triangle A B C$, we obtain
$A B=B C=A C \Rightarrow \triangle A B C \text { is equilateral. } \Rightarrow \triangle A B C=60^{\circ} .$
View full question & answer→MCQ 551 Mark
D and E are the mid-points of the sides AB and AC of $\triangle A B C$ and O is any point on side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, the DEQP is
Answer(a) a parallelogram
Given that D and E are the mid-points of AB and AC respectively. Therefore,
$D E \| B C$ or, $D E \| P Q$
In $\triangle A B O, D$ and P are the mid-points of AB and OB respectively. Therefore, $D P \| A O$. Similarly, in $\triangle A O C$, we obtain $E Q \| A O$. Therefore, $D P \| E Q$. Hence, DEQP is a parallelogram.
View full question & answer→MCQ 561 Mark
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
- A
- B
diagonals of ABCD are equal
- ✓
diagonals of ABCD are equal and perpendicular
- D
diagonals of ABCD are perpendicular
AnswerCorrect option: C. diagonals of ABCD are equal and perpendicular
View full question & answer→MCQ 571 Mark
Answer(c) 4
Diagonals of a rhombus bisect each other at right angles. Therefore,
$O A=\frac{1}{2} A C, O B=\frac{1}{2} B D$ and $\angle A O B=90^{\circ}$
In right triangle $A O B$, we obtain
$O A^2+O B^2=A B^2$
$\Rightarrow \quad\left(\frac{1}{2} A C\right)^2+\left(\frac{1}{2} B D\right)^2=A B^2 \Rightarrow A C^2+B D^2=4 A B^2 \Rightarrow k A B^2=4 A B^2 \Rightarrow k=4$
View full question & answer→MCQ 581 Mark
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- ✓
$90^{\circ}$
AnswerCorrect option: D. $90^{\circ}$
(d) $90^{\circ}$
ABCD is a parallelogram.
$\therefore \quad \angle B=\angle D \Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle D \Rightarrow \angle A BD=\angle A D B$
$\Rightarrow \quad A B=A D \quad$ [Sides opposite to equal angles in $\triangle A B D$ ]
$\Rightarrow \quad \triangle A B D$ is isosceles
$\Rightarrow \quad A O \perp B D \quad[\because O$ is the mid-point of $B D]$
$\Rightarrow \quad \angle A O B=90^{\circ}$
View full question & answer→MCQ 591 Mark
- A
$\frac{1}{2} A C$
- ✓
$\frac{1}{3} A C$
- C
$\frac{2}{3} A C$
- D
$\frac{3}{4} A C$
AnswerCorrect option: B. $\frac{1}{3} A C$
(b) $\frac{1}{3} A C$
Let G be the mid-point of FC. Join DG.
In $\triangle B C F$, and D and G are mid-points of BC and FC respectively.
$D G\|B F \Rightarrow D G\| E F$
In $\triangle A D G, E$ is the mid-point of $A D$ and $E F \| D G$.
$\therefore \quad F \text { is the mid-point of } A G \text {. }$
But, $G$ is the mid-point of $F C$.
$\therefore \quad A F=F G=G C=A F=\frac{1}{3} A C$
View full question & answer→MCQ 601 Mark
- ✓
- B
- C
$\frac{3}{2} A B$
- D
$\frac{5}{4} A B$
Answer(a) 2 AB
In $\triangle E D C$ and EFB, we have
$\angle D E C=\angle F E B$
$\angle D C E=\angle E B F$$\quad$[Alternate angles]
and $\quad B E=E C$
So, by using SAS congruence crikerion, we obtain
$\begin{aligned}& \triangle E D C \equiv \triangle E F B \Rightarrow B E=D C \Rightarrow B F=A B \\
\therefore \quad & A F=A B+B F=A B+A B \Rightarrow A F=2 A B .\end{aligned}$
View full question & answer→MCQ 611 Mark
If $\angle A, \angle B, \angle C$ and $\angle D$ of a quadrilateral ABCD, taken in order, are in the ratio $3: 7: 6: 4$, then ABCD is a
Answer(c) trapezium
Let $\angle A=3 x^{\circ}, \angle B=7 x^{\circ}, \angle C=6 x^{\circ}$ and $\angle D=4 x^{\circ}$. Then,
$\begin{array}{ll} & \angle A-\angle B+\angle C-\angle D=360^{\circ} \Rightarrow 20 x^{\circ}=360^{\circ} \Rightarrow x=18
\\ \therefore \quad & \angle A=55^{\circ}, \angle B=126^{\circ}, \angle C=100^{\circ} \text { and } \angle D=72^{\circ}
\\ \Rightarrow & \angle A-\angle B=180^{\circ} \text { and } \angle C-\angle D=180^{\circ} \Rightarrow A B C D \text { is a traperium. }\end{array}$
View full question & answer→MCQ 621 Mark
The bisectors of any two adjacent angles of parallelogram intersect at
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- ✓
$90^{\circ}$
AnswerCorrect option: D. $90^{\circ}$
(d) $90^{\circ}$
Let ABCD be a parallelogram such that the bisectors of $\angle A$ and $\angle B$ meet at O. Then,
$\angle A O B=\frac{1}{2}(\angle C+\angle D)=\frac{1}{2} \times 180^{\circ}=90^{\circ}[\because$ Adjacent angles are supplementary $]$
View full question & answer→MCQ 631 Mark
If one angle of a parallelogram is $24^{\circ}$ lesses than twice the smallest angle, then the largest angle of the parallelogram is
- A
$68^{\circ}$
- B
$102^{\circ}$
- ✓
$112^{\circ}$
- D
$136^{\circ}$
AnswerCorrect option: C. $112^{\circ}$
(c) $112^{\circ}$
Let $A B C D$ be a parallelogram with smallest angle $\angle B$. It is given that $\angle A=2 \angle B-24^{\circ}$.$
\begin{array}{ll}
\text { But, } & \angle A+\angle B=180^{\circ} \Rightarrow 2 \angle B-24^{\circ}+\angle B=180^{\circ} \Rightarrow 3 \angle B=204^{\circ} \Rightarrow \angle B=68^{\circ} \\
\therefore & \angle A=180^{\circ}-68^{\circ}=112^{\circ}\end{array}$
View full question & answer→MCQ 641 Mark
If an angle of a parallelogram is two thirds of its adjacent angle, the smallest angle of the parallelogram is
- A
$108^{\circ}$
- ✓
$72^{\circ}$
- C
$54^{\circ}$
- D
$81^{\circ}$
AnswerCorrect option: B. $72^{\circ}$
(b) $72^{\circ}$
Let $A B C D$ be a parallelogram such that $\angle A=\frac{2}{3} \angle B$. But,
$\angle A+\angle B=180^{\circ}$$\quad$[ $\because$ Adjacent angles are supplementary]
$\Rightarrow \quad \frac{2}{3} \angle B+\angle B=180^{\circ} \Rightarrow \frac{5}{3} \angle B=180^{\circ} \Rightarrow\angle B=108^{\circ} \Rightarrow \angle A=72^{\circ}$
Thus, smallest angle is $72^{\circ}$.
View full question & answer→MCQ 651 Mark
- A
$45^{\circ}$
- B
$55^{\circ}$
- ✓
$60^{\circ}$
- D
$75^{\circ}$
AnswerCorrect option: C. $60^{\circ}$
(c) $60^{\circ}$
We have $\angle A=75^{\circ}$. Therefore, $\angle C=75^{\circ}$.
Using angle sum property in $\triangle B C D$, we obtain: $\angle C B D=60^{\circ}$.
View full question & answer→MCQ 661 Mark
The diagonals AC and BD of a parallelogram ABCD intersect each other at O such that $\angle D A C=30^{\circ}$ and $\angle A O B=70^{\circ}$. Then, $\angle D B C=$
- ✓
$40^{\circ}$
- B
$35^{\circ}$
- C
$45^{\circ}$
- D
$50^{\circ}$
AnswerCorrect option: A. $40^{\circ}$
(a) $40^{\circ}$
Given $\angle A O B=70^{\circ}$. Therefore, $\angle A O D=110^{\circ}$.
Using angles sum property in $\triangle B O D$, we obtain $\angle A D B=40^{\circ}$. But,
$\angle D B C=\angle A D B$$\quad$[Alternate angles]
$\Rightarrow \quad \angle D B C=40^{\circ}$
View full question & answer→MCQ 671 Mark
Angles A,B,C,D of a quadrilateral ABCD are in the ratio $3: 4: 4: 7$. If the bisectors of angles A and B intersect al O, then $\angle A O B=$
- A
$70^{\circ}$
- B
$80^{\circ}$
- ✓
$110^{\circ}$
- D
$100^{\circ}$
AnswerCorrect option: C. $110^{\circ}$
(c) $110^{\circ}$
Let the measures of angles A,B,C,D be $(3 x)^{\circ},(4 x)^{\circ},(4 x)^{\circ}$ and $(7 x)^{\circ}$ respectively. The bisectors of $\angle A$ and $\angle B$ meet at O.
$\therefore \quad \angle A O B=\frac{1}{2}(\angle C+\angle D)=\frac{1}{2}\left\{(4 x)^{\circ}+(7x)^{\circ}\right\}=\left(\frac{11 x}{2}\right)^{\circ}$
Now, $\angle A+\angle B+\angle C+\angle D=360^{\circ} \Rightarrow(3 x)^{\circ}+(4 x)^{\circ}+(4 x)^{\circ}+(7x)^{\circ}=360^{\circ} \Rightarrow(18 x)^{\circ}=360^{\circ} \Rightarrow x=20$
$\therefore \quad \angle A O B=\left(\frac{11}{2} \times 20\right)^{\circ}=110^{\circ}$
View full question & answer→MCQ 681 Mark
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If $\angle A B D=50^{\circ}$, then $\angle D P C=$
View full question & answer→MCQ 691 Mark
In a quadrilateral ABCD, $\angle A+\angle C$ is 2 times $\angle B+\angle D$. If $\angle A=140^{\circ}$ and $\angle D=60^{\circ}$, then $\angle B=$
View full question & answer→MCQ 701 Mark
ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =
- A
$\frac{3}{2} A B$
- ✓
- C
- D
$\frac{5}{4} A B$
View full question & answer→MCQ 711 Mark
In $\triangle A B C$, E is the mid-point of median AD such that BE produced meets AC at F. If AC = 10.5cm then AF =
View full question & answer→MCQ 721 Mark
P is the mid-point of side BC of a parallelogram ABCD such that $\angle B A P=\angle D A P$. If AD = 10cm then CD =
View full question & answer→MCQ 731 Mark
Diagonals of a quadrilateral ABCD bisect each other. If $\angle A=45^{\circ}$, then $\angle B=$
View full question & answer→MCQ 741 Mark
ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12cm MN = 14cm then CD =
MCQ 751 Mark
The diagonals of a parallelogram ABCD intersect at O. If $\angle B O C=90^{\circ}$ and $\angle B D C=50^{\circ}$, then $\angle O A B=$
View full question & answer→MCQ 761 Mark
In $\triangle A B C, \angle A=30^{\circ}, \angle B=40^{\circ}$ and $\angle C=110^{\circ}$. The angles of the triangle formed by joining the mid-points of the sides of this triangle are
View full question & answer→MCQ 771 Mark
In a rhombus ABCD, if $\angle A C B=40^{\circ}$, then $\angle A D B=$
View full question & answer→MCQ 781 Mark
ABCD is a parallelogram in which diagonal AC bisects $\angle B A D$. If $\angle B A C=35^{\circ}$, then $\angle A B C=$
View full question & answer→MCQ 791 Mark
If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
MCQ 801 Mark
If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10x, what is the sum of the measures of the smallest angle and largest angle?
MCQ 811 Mark
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
MCQ 821 Mark
ABCD is a parallelogram, M is the mid-point of BD and BM bisects $\angle B$. Then, $\angle A M B=$
View full question & answer→MCQ 831 Mark
ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =
MCQ 841 Mark
In a parallelogram ABCD, if $\angle D A B=75^{\circ}$ and $\angle D B C=60^{\circ}$, then $\angle B D C=$
View full question & answer→MCQ 851 Mark
If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is
MCQ 861 Mark
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
MCQ 871 Mark
The figure formed by joining the mid-points of the adjacent sides of a square is a
MCQ 881 Mark
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
MCQ 891 Mark
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
MCQ 901 Mark
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
MCQ 911 Mark
The bisectors of the angle of a parallelogram enclose a
MCQ 921 Mark
The bisectors of any two adjacent angles of a parallelogram intersect at
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- ✓
$90^{\circ}$
AnswerCorrect option: D. $90^{\circ}$
View full question & answer→MCQ 931 Mark
We get a rhombus by joining the mid-points of the sides of a
MCQ 941 Mark
The two diagonals are equal in a
MCQ 951 Mark
Diagonals necessarily bisect opposite angles in a
MCQ 961 Mark
Which of the following quadrilateral is not a rhombus?
- A
- B
Diagonals bisect each other
- C
Diagonals bisect opposite angles
- ✓
One angle between the diagonals is 60°
AnswerCorrect option: D. One angle between the diagonals is 60°
View full question & answer→MCQ 971 Mark
PQRS is a quadrilateral. PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?
- ✓
$\angle P=100^{\circ}, \angle Q =80^{\circ}, \angle R=100^{\circ}$
- B
$\angle P=85^{\circ}, \angle Q =85^{\circ}, \angle R=95^{\circ}$
- C
PQ = 7cm QR = 7cm RS = 8 cm, SP = 8cm
- D
OP = 6.5cm OQ = 6.5cm OR = 5.2cm OS = 5.2cm
AnswerCorrect option: A. $\angle P=100^{\circ}, \angle Q =80^{\circ}, \angle R=100^{\circ}$
View full question & answer→MCQ 981 Mark
The consecutive sides of a quadrilateral have
- A
- ✓
- C
- D
infinitely many common points
View full question & answer→MCQ 991 Mark
The opposite sides of a quadrilateral have
- ✓
- B
- C
- D
infinitely many common points
View full question & answer→
