M.C.Q

Question 511 Mark

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if:

Answer

Diagonals of ABCD are equal and perpendicular.
Solution:
A quadrilateral formed by joining the mid points of a square is a square. So, ABCD is a square. In Square, diagonals are equal and perpendicular.

Question 521 Mark

Write the correct answer in the following:
D and E are the mid-points of the sides AB and AC respectively of $\Delta\text{ABC}.$ DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is:

Answer

$\text{DE}=\text{EF}$
Solution:
We need $\text{DE}=\text{EF}.$

Hence, (c) is the correct answer.

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Question 531 Mark

A diagonal of a rectangle is inclined to one side of the rectangle at 35º. The acute angle between the diagonals is:

Answer

70º
Solution:
$\angle\text{OAD} = 90^\circ - (\angle\text{OAB})$
$= 90^\circ - 35^\circ = 55^\circ.$
Now, $\angle\text{ODA} = \angle\text{OAD} = 55^\circ$ [$∵$ OA = OD since diagonals of a rectangles are equal and bisect each other].
$\angle\text{AOD} = 180^\circ - (\angle\text{OAD} + \angle\text{ODA})$
$= 180^\circ - (55^\circ + 55^\circ) = 70^\circ.$

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Question 541 Mark

If the diagonals of a quadrilateral bisect each other at right angles then the figure is a:

Answer

Rhombus
Solution:
Rhombus is the correct answer. As we know that from all the quadrilaterals given in other options the diagonals of rhombus bisect each other at right angles.

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Question 551 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?

Answer

168º
Solution:
Given,
ABCD is a parallelogram,
Angles of quadrilateral 4x, 7x, 9x, 10x
⇒ 4x + 7x + 9x + 10x = 360° [angle sum property of quadrilateral]
⇒ 30x = 360°
$\Rightarrow\text{x}=\frac{360^\circ}{30}=12^\circ$
Hence, sum of smallest and largest angles = 4x + 10x = 4 × 12º + 10 × 12º
= 48º + 120º = 168º

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Question 561 Mark

E Divides AB in the ratio 1 : 3 and also, F divides AC in the ratio 1 : 3. EF = 2.8cm, Find BC = ?

Answer

11.2cm
Solution:
Let AE = x and EB = 3x, AF = y and FC = 3y.
EF = 2.8cm
AE + AF = 2.8 implies x + y = 2.8
BC = CF + FA + AE + EB
= 3y + y + x + 3x
= 4(x + y) = 4(2.8) = 11.2cm

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Question 571 Mark

The Parallel sides of a trapezium are ‘a’ and ‘b’ resp. The line joining the mid-points of its non-parallel sides will be.

Answer

$\frac{1}{2}(\text{a+b})$
Solution:
Join one of the diagonals which intersects the line joining mid points of the non parallel sides. This point of intersection become midpoint of the diagonal by using converse of midpoint theorem and we know that by mid point theorem that the line joining midpoints of any two sides of triangle measures half of the third side.so, by combining the results of two triangles we get above result.

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Question 581 Mark

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The smallest of these angles is:

Answer

60°
Solution:
Let tha common multipal be x.
$\therefore$ The angle measure 3x, 4x, 5x and 6x.
Since the sum of the angles of a quadrilateral being 360°, we have
3x + 4x + 5x + 6x = 360°
⇒18x = 360°
⇒ x = 20°
$\therefore$ The angles of the quadrilateral are
3x = 3(20) = 60°,
4x = 4(20) = 80°,
5x = 5(20)= 100°,
6x = 6(20) = 120°,
$\therefore$ The smallest angle is 60°.

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Question 591 Mark

The bisectors of any two adjacent angles of a parallelogram intersect at:

Answer

90°
Solution:

In a parallelogram, sum of adjacent angles = 180°
$\Rightarrow \angle \text{A}+\angle\text{B}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+\frac{\angle\text{B}}{2}=90^\circ\ ...(1)$
$\Rightarrow\angle\text{OAB}=\frac{\angle\text{A}}{2}$ and $\angle\text{OBA}=\frac{\angle\text{B}}{2}$
Thus, $\angle\text{OAB}+\angle\text{OBA}=90^\circ$ [From eq (1)]
$\Rightarrow\angle\text{AOB}=180^\circ-(\angle\text{OAB}+\angle\text{OBA})=180^\circ-90^\circ$
$\Rightarrow\angle\text{AOB}=90^\circ$

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Question 601 Mark

In a parallelogram ABCD, if $\angle\text{DAB}=75^\circ$ and $\angle\text{DBC}=60^\circ,$ then $\angle\text{BDC}=$

Answer

45°
Solution:

In parallelogram ABCD,
$\angle\text{A}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-75^\circ=105^\circ$
$\angle\text{ADB}=\angle\text{DBC}$ (Alternate angles)
$\Rightarrow\angle\text{ADB}=60^\circ$
$\angle\text{BDC}=\angle\text{ADC}-\angle\text{ADB}=105^\circ-60^\circ=45^\circ$

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Question 611 Mark

If one angle of a parallelogram is $24^\circ$ less than twice the smallest angle, then the measure of the largest angle of the largest angle of the parallelogram is:

Answer

Let the smallest angle $=\angle\text{ADC}=\text{x}^\circ$
Other angle $\angle\text{BCD}$
$\Rightarrow\angle\text{BCD}=2\text{x}^\circ-24^\circ$
Also, $\angle\text{ACD}+\angle\text{BCD}=180^\circ ($Sum of adjacent angles in $||$gram $= 180°)$
$\Rightarrow\text{x}^\circ+2\text{x}^\circ-24^\circ=180^\circ$
$\Rightarrow3\text{x}^\circ=204^\circ$
$\Rightarrow\text{x}=68^\circ$
⇒ Largest angle $=\angle\text{BCD}=2\times68^\circ-24^\circ=112^\circ$

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Question 621 Mark

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if:

Answer

Diagonals of ABCD are perpendicular to each other.
Solution:

In $\triangle\text{ABC},$ P and Q are the mid-point of sides AB and BC respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC }...(\text{i})$
In $\triangle\text{ADC},$ R and S are the mid-point of sides CD and AD respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC }...(\text{ii})$
From (i) and (ii),
$\text{PQ ∥ RS}$ and $\text{PQ = RS}$
Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.
So, PQRS is a parallelogram.
Let the diagonals AC and BD intersect at O.
Now, in $\triangle\text{ABD},$ P and S are the mid-points of sides AB and AD respectively.
$\therefore\text{PS || BD}$
$\Rightarrow\text{PN || MO}$
Also, from (i), $\text{PQ || AC}$
$\Rightarrow\text{PM || NO}$
Thus, in quadrilateral PMON, $\text{PM || NO}$ and $\text{PN || MO},$
⇒ PMON is a parallelogram.
$\Rightarrow\angle\text{MPN}=\angle\text{MON}$ (opposite angles of a parallelogram are equal)
$\Rightarrow\angle\text{MPN}=\angle\text{BOA}$ $(\text{Since }\angle\text{BOA}=\angle\text{MON})$
$\Rightarrow\angle\text{MPN}=90^{\circ}$ $(\text{Since }\text{AC}\perp\text{BD},\angle\text{BOA}=90^{\circ})$
$\Rightarrow\angle\text{QPS}=90^{\circ}$
Thus, PQRS is a parallelogram whose one angle, i.e. $\angle\text{QPS}=90^{\circ}.$
Hence, PQRS is a rectangle if $\text{AC}\perp\text{BD}.$

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Question 631 Mark

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle of the parallelogram is:

Answer

112°
Solution:
Let the smallest angle be x°
⇒ its adjacent angle = (2x - 24)
Since sum of the adjacent angles = 180°
⇒ x + 2x - 24 = 180
⇒ 3x = 204
⇒ x = 68°
So, its adjacent angle = 2(68) - 24 = 136 - 24 = 112°.
Hence, the largest angle of the parallelogram is 112°.

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Question 641 Mark

Which of the following is not true for the Parallelogram?

Answer

Opposite angles are bisected by the diagonals.
Solution:
If opposite angles are bisected by diagonals in parallelogram, all four bisected angles become equal which leads to equal adjacent side. That is not true in case of parallelogram.

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Question 651 Mark

A diagonal of a parallelogram divides it into:

Answer

Two congruent triangles.
Solution:
By SSS congruence condition, opposite sides are equal and common diagonal.

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Question 661 Mark

If ABCD is a Parallelogram with 2 Adjacent angles $\angle\text{A}=\angle\text{B},$ then the parallelogram is a:

Answer

Rectangle
Solution:
The sum of the adjacent angles of a parallelogram is 180º. Opposite sides of a parallelogram are equal. Hence it is a rectangle.

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Question 671 Mark

A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is:

Answer

70°
Solution:

$\angle\text{DAO}+\angle\text{OAB}=\angle\text{DAB}$
$\Rightarrow\angle\text{DAO}+35^{\circ}=90^{\circ}$
$\Rightarrow\angle\text{DAO}=55^{\circ}$
ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.
$\text{OA = OD}$
$\Rightarrow\angle\text{ODA}=\angle\text{DAO}$ (angles opposte to equal sides are equal)
$\Rightarrow\angle\text{ODA}=55^{\circ}$
In DODA, by angle sum property,
$\angle\text{ODA}+\angle\text{DAO}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow55^{\circ}+\angle55^{\circ}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow\angle\text{AOD}=70^{\circ}$

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Question 681 Mark

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a:

Answer

Rhombus.
Solution:

PQ || AC (since in $\triangle\text{ABC}$ mid-points of AB & BC are meeting by PQ)
Similarly, SR || AC
⇒ PQ || SR
Now in $\triangle\text{ABD}$ and $\triangle\text{CBD},$
PS || BD and QR || BD
⇒ PS || QR
Hence, PQRS is a parallelogram.
But $\text{PR }\bot \text{ QS}$
⇒ Diagonals cut at 90°
⇒ PQRS is a Rhomus.

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Question 691 Mark

If the diagonals of a rhombus are $18\ cm$ and $24\ cm$ respectively, then its side is equal to:

Answer

Given,
$\text{ABCD}$ is a rhombus

$AC = 24\ cm, BD = 18\ cm$
$AB = BC = CD = DA \ [$side of rhombus$]$
We know that diagonals of rhombus bisect each other at $90^\circ .$
In right $\triangle\text{AOB}$
$AB^2 = BO^2 + AO^2$
$AB^2 = 122 + 92 = 144 + 81 = 225$
$\text{AB} = \sqrt{225}$
$= 15\ cm$
Side of rhombus $= 15\ cm$

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Question 701 Mark

Digonals necessarily bisect opposite angles in a:

Answer

Square.
Solution:
Diagonals necessarily bisect opposite angles in a square.

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Question 711 Mark

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a:

Answer

Parallelogram.
Solution:
The figure formed by joining the mid points of the adjacent sides of a parallelogram is a parallelogram.

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Question 721 Mark

The figure formed by joining the mid-points of the adjacent sides of a Square is a:

Answer

Square.
Solution:

PS || QR, PQ || SR ...(1)
{Because lines joining the mid-points of any two sides of a triangle are parallel to the third side}
$\text{AC } \bot \text{ BD}$ & $\text{BR } \bot \text{ QS}$ (From Figure)
SR || AC and QR || BD
$\text{AC } \bot \text{ BD}$
$\Rightarrow \text{SR }\bot \text{ QR}$
Hence $\angle\text{SRQ}=90^\circ\ ...(2)$
Also $\triangle\text{APS}\cong\triangle\text{DSR}$
$\Rightarrow\text{PS} = \text{SR}\dots(3)$
From equations (1), (2), (3)
PQRS is a square.

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Question 731 Mark

In Fig. ABCD is a cyclic quadrilateral. If $\angle\text{BAC}=50^\circ$ and $\angle\text{DBC}=60^\circ$ then find $\angle\text{BCD}=\ ?$ "

Answer

70º
Solution:
Here, $\angle\text{BAC}=50^\circ$ (angles in same segment are equal)

In $\triangle\text{BCD},$ we have
$\angle\text{BCD}=180^\circ−(\angle\text{BDC}+\angle\text{DBC})$
$=180^\circ−(50^\circ+60^\circ)$
$= 70^\circ$

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Question 741 Mark

In Fig. the quadrilateral ABCD circumscribes a circle with centre O. If $\angle\text{AOB}=115^\circ,$ then find $\angle\text{COD}=\ ?$

Answer

115º
Solution:
$∵\angle\text{AOB}=\angle\text{COD}$ (vertically opposite angle)
$∴\angle\text{COD}=115^\circ$

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Question 751 Mark

The area of a quadrilateral whose diagonals measure $48m$ and $32m$ respectively and bisect each other at right angles is:

Answer

According to the question,
$\text{Area of given quadrilateral}=\frac{1}{2}\times\text{Product of diagonals}$
$=\frac{1}{2}\times48\times32$
$=768\text{m}^2$

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Question 761 Mark

In a parallelogram ABCD, if $\text{DAB} = 75^\circ$ and $\angle\text{DBC} = 60^\circ,$ then $\angle\text{BDC} =\ ?$

Answer

45º
Solution:

We know that the opposite angles of a parallelogram are equal.
Therefore, $\angle\text{BCD} = \angle\text{BAD} = 75^\circ...\ \text{(i)}$
(i) Now, in $\triangle\text{BCD},$
We have,
$\angle\text{CDB} + \angle\text{DBC} + \angle\text{BCD} = 180^\circ$ [Since, sum of the angles of a triangle is 180º]
$⇒ \angle\text{CDB} + 60^\circ + 75^\circ = 180^\circ$
$⇒ \angle\text{CDB} + 135^\circ = 180^\circ$
$⇒ \angle\text{CDB} = (180^\circ - 135^\circ) = 45^\circ$

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Question 771 Mark

Consecutive angles of a Parallelogram are:

Answer

Supplementary
Solution:
In parallelogram, both pair of opposite sides are parallel. So, consecutive angles form a pair of co-interior angles. Co-interior angles add up to 180. It means they are supplementary.

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Question 781 Mark

The Diagonals AC and BO of a Parallelogram ABCD intersect each other at the point O such that $\angle\text{DAC} = 30^\circ$ and $\angle\text{AOB} = 70^\circ.$ Then $\angle\text{DBC}?$

Answer

40º
Solution:
$\angle\text{DAC} = \angle\text{ACB} = 30$ (alternate angles)
$\angle\text{BOA} = \angle\text{BOC} = 180$ (linear pair)
$\angle\text{BOC}= 180 - 70 = 110$
In $\triangle\text{BOC}, \ \angle\text{BOC} + \angle\text{OCB} + \angle\text{CBO} = 180$ (angle sum property)
$110 + 30 + \angle\text{CBO} = 180$
$\angle\text{CBO} = 180 - 140 =40$
$\Rightarrow \angle\text{DBC}=40^\circ$

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Question 791 Mark

The figure forms by joining the mid-points of the sides of a Rhombus, taken in order are:

Answer

A Rectangle
Solution:
In rhombus, diagonals bisect each other at right angles. Which proves that all the angles of quadrilateral formed by joining the mid points of rhombus is equal to 90. A quadrilateral with both pair of opposite angles equal is a parallelogram.
A parallelogram having one right angle is a rectangle.

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Question 801 Mark

ABCD is a Parallelogram in which $\angle\text{BAO}=35^\circ, \ \angle\text{DAO}=40^\circ$ and $\angle\text{COD}=105^\circ.$ Find $\angle\text{ABO}=\ ?$

Answer

40º
Solution:

Given, ABCD is a parallelogram having $\angle\text{BAO}=35^\circ, \ \angle\text{DAO}=40^\circ$ and $\angle\text{COD}=105^\circ.$
Now, $\angle\text{COD} = \angle\text{AOB} = 105^\circ$ [vertically opposite angles]
In $\angle\text{ABO},$ by angle sum property of triangle,
$⇒ \angle\text{AOB} + \angle\text{OAB} + \angle\text{ABO} = 180^\circ$
$⇒ 105^\circ + 35^\circ + \angle\text{ABO} = 180^\circ$
$⇒ \angle\text{ABO} = 40^\circ$

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Question 811 Mark

PQRS is a quadrilateral. PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?

Answer

$\angle\text{P} = 100^\circ, \angle\text{Q} = 80^\circ, \angle\text{R} = 100^\circ$
Solution:
$\angle\text{P} = 100^\circ, \angle\text{Q} = 80^\circ, \angle\text{R} = 100^\circ$
Since, sum of all the internal angles of a quadrilateral is 360, we have
$\angle\text{P} + \angle\text{Q} + \angle\text{R} + \angle\text{S} = 360$
$⇒ 100 + 80 + 100 + \angle\text{S} = 360$
$⇒ 280 + \angle\text{S} = 360$
$⇒ \angle\text{S} = 80$
So, it follows that opposite angles of the quadrilateral are equal.
Hence it is a parallelogram.

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Question 821 Mark

ABCD is a Rectangle, diagonals AC and BD intersect each other at P. If $\angle\text{APD} = 52^\circ,$ find $\angle\text{ACB}$ and $\angle\text{DBA}=\ ?$

Answer

64º and 26º
Solution:
In Rectangle, diagonals are equal and bisect each other.
In $\triangle\text{APD, AP = PD}$
$⇒ \angle\text{ADP} = \angle\text{PAD} = \text{x}$ (angle opposite to equal sides are equal)
In $\triangle\text{APD}, \angle\text{APD} + \angle\text{PDA} + \angle\text{DAP} = 180^\circ$ (angle sum property)
$52^\circ + \text{x} + \text{x} = 180^\circ$
$2\text{x} = 180^\circ - 52^\circ = 128^\circ$
$\text{x} = 64^\circ$
$\angle\text{DAC} = \angle\text{BCA} = 64^\circ$ (alternate angles)
In $\triangle\text{ADB}, \angle\text{ADB} + \angle\text{DBA} + \angle\text{BAD} = 180^\circ$ (angle sum property)
$64^\circ + \angle\text{DBA} + 90^\circ = 180^\circ$
$\angle\text{DBA} = 180^\circ - 154^\circ = 26^\circ$

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Question 831 Mark

Write the correct answer in the following:
If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a:

Answer

Trapezium.
Solution:
Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4.
Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°.
3x + 7x + 6x + 4x = 360°
⇒ 20x = 360°
$\Rightarrow\frac{\text{x}=360^\circ}{20^\circ=18^\circ}$
$\therefore$ Angles of the quadrilateral are
$\angle\text{A}=3\times18=54^\circ$
$\angle\text{B}=7\times18=126^{\circ} $
$\angle\text{C}=6\times18=108^\circ$
$\angle\text{D}=4\times18=72^\circ$
From figure, $\angle\text{BCE}=180^\circ-\angle\text{BCD}$ [linear pair axiom]
$\Rightarrow\ \angle\text{BCE}=180^\circ-108^\circ=72^\circ$
Here, $\angle\text{BCE}=\angle\text{ADC}=72^\circ$
Since, the of cointerior angles,
$\therefore\ \text{BC}||AD$
Now, sum of cointerior angles,
$\angle\text{A}+\angle\text{B}=126^\circ+54^\circ=180^\circ$
$\angle\text{C}+\angle\text{D}=108^\circ+72^\circ=180^\circ$
Hence, ABCD is a trapezium.

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Question 841 Mark

In the given figure$, \text{ABCD}$ is a Rhombus. Then,

Answer

$\text{ABCD}$ is a rhombus.
$AB = BC = CD = DA$
In Rhombus, diagonals bisect each other at right angles.
So$, AO = CO$ and $BO = DO$
In triangle $AOB, AO^2 + BO^2 = AB^2 ($Pythagoras theorem$)$
$\Big(\frac{1}{2} \text{AC}\Big)^2 + \Big(\frac{1}{2} \text{BD}\Big)^2 = \text{AB}^2$
$\frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} = \text{AB}^2$
$\text{AC}^2 + \text{BD}^2 = 4\text{AB}^2$

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Question 851 Mark

The length of each side of a rhombus is 10cm and one if its diagonals is of length 16cm. The length of the other diagonal is:

Answer

12cm.
Solution:

Let ABCD be the rhombus with diagonals AB = 10cm and Ac = 16cm.
Since the diagonals of a rhombus are perpendicular bisectors of each other,
$\Rightarrow\text{OA}=8\text{cm},\text{BD = 2OB}$ and $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
By Pythagoras theorem,
$\text{AB}^2=\text{OA}^2+\text{OB}^2$
$\Rightarrow(10)^{2}=(8)^2+\text{OB}^2$
$\Rightarrow100=64+\text{OB}^2$
$\Rightarrow\text{OB}^2=36$
$\Rightarrow\text{OB}=\sqrt{36}$
$\Rightarrow\text{OB}=6\text{cm}$
$\Rightarrow\text{BD}=2\times\text{OB}$
$\Rightarrow\text{BD}=2\times6$
$\Rightarrow\text{BD}=12\text{cm}$
Hence, the lenght of the other diagonal is 12cm.

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Question 861 Mark

In fig D is mid-point of AB and $\text{DE || BC}$ then AE is equal to:

Answer

EC
Solution:
By mid-point theorem of a triangle E is the midpoint of AC,
Hence AE = EC

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Question 871 Mark

If area of a || gm with sides a and b is A and that of a rectangle with sides a and b is B, then:

Answer

A < B
Solution:
Let h be the heigth of the parallelogram.
Then, h < b.
We konw that, area of a parallelogram = base × height
If a is the base of the parallelogram, then area of a parallelogram = a × h
⇒ A = a × h
We know that, area of a rectangle = length × breadth
⇒ A = a × b
So, a × h < a × b
Hence, A < B.

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Question 881 Mark

Given 4 points A, B, C, D such that 3 Points A, B, C are collinear. By joining these points in order, we get

Answer

A Triangle
Solution:
If three or more points lie on same line then they are called collinear points. A, B and C are collinear means they form a line. Point D is outside the line. Point D can be join with the end points of the line formed by points A, B, and C. For this we need total three lines to form a closed figure. A closed figure with three sides is called triangle.

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Question 891 Mark

The consecutive sides of a quadrilateral have.

Answer

One common points.
Solution:
One common point.

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Question 901 Mark

Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is:

Answer

120º
Solution:
We know that The sum of angles of quad = 360º
4th angle = 360º - (75º + 90º + 75º)
= 360º - 240º
= 120º

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Question 911 Mark

Two parallelograms stand on same base and between the same parallels. The ratio of their areas is:

Answer

1 : 1
Solution:
Parallelograms on the same base and between the same parallels are equal in area. Hence the ration of two parallelograms will be 1 : 1.

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Question 921 Mark

If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10, what is the sum of the measures of the smallest angle and largest angle?

Answer

168°
Solution:
Sum of all angles of a Quadrilateral = 360°
4x + 7x + 9x + 10x = 360°
30x = 360°
x = 12°
So, sum of smallest and largest angle,
i.e. 4x + 10x = 14x = 14 × 12 = 168°

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Question 931 Mark

Write the correct answer in the following:
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is:

Answer

A rectangle.
Solution:
The figure will be a rectangle.
Hence, (b) is the correct answer.

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Question 941 Mark

The length of each side of a rhombus is 10cm and one of its diagonal is of length 16cm. The Length of the other Diagonal is:

Answer

12cm
Solution:
Use pythagoras theorem in right triangle,
$102 -\Big[\frac{16}{2}\Big]^2 = 100 - 64 = 36 = [6]^2$
Hence, the other diagonal = 6 × 2 = 12cm

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Question 951 Mark

In fig D is mid-point of AB and $\text{DE || BC}$ then AE is equal to:

Answer

EC
Solution:
By mid-point theorem of a triangle E is the midpoint of AC, hence AE = EC.

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Question 961 Mark

One Angle of a quadrilateral is of 108º and the remaining three angles are equal. Find each of the three equal angles.

Answer

84º, 84º, 84º
Solution:
Let ABCD be a quadrilateral with $\angle\text{A} = 108^\circ$ and $\angle\text{B} = \angle\text{C} = \angle\text{D} = \text{x}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
108º + x + x + x = 360º
3x = 360º - 108º
3x = 252º
x = 84º

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Question 971 Mark

Three angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The smallest of these angles is:

Answer

60º
Solution:
Let the angles be 3x, 4x, 5x and 6x
3x + 4x + 5x + 6x = 360º (Sum of angles of a quadrilateral)
18x = 360º
$\text{x}=\frac{360}{18}$
x = 20º
$∴$ Angles of the quadrilateral are: 3x = 3 × 20º = 60º
4x = 4 × 20º = 80º
5x = 5 × 20º = 100º
6x = 6 × 20º = 120º
Hence, the smallest angle is 60º.

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Question 981 Mark

If the diagonals of a quadrilateral bisect each other, and opposite sides are parallel and equal, then the quadrilateral must be.

Answer

Parallelogram
Solution:
By theorm diagonals of quadrilateral bisect each other if and only if it is a parallelogram.For a quadrilateral to be parallelogram some other properties are required: Opposite sides are equal and parallel. Opposite angles are equal. Sum of any adjacent angles is 180.

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Question 991 Mark

The figure formed by joining the mid-points of the adjacent sides of a square is a:

Answer

Square
Solution:

In a square ABCD, P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
⇒ AB = BC = CD = AD ... [Sides of square are equal]
In $\triangle\text{ADC},$
$\text{SR || AC}$ and $\text{SR}=\frac{1}{2}\text{AC}$ ... [By mid-point theorem] ...(1)
In $\triangle\text{ABC},$
$\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}$ ... [By mid-point theorem] ...(2)
From equation (1) and (2)
$\text{SR || PQ}$ and $\text{SR}=\text{PQ}=\frac{1}{2}\text{AC}\ ...\ (3)$
Similarly, $\text{SP || BD}$ and $\text{BD || RQ}$
$\therefore\ \text{SP || RQ}$ and $\text{SP}=\frac{1}{2}\text{BD}$
And $\text{RQ}=\frac{1}{2}\text{BD}$
$\therefore\ \text{SP}=\text{RQ}=\frac{1}{2}\text{BD}$
Since, diagonals of a square bisect each other at right angle.
$\text{AC = BD}$
$\text{SR}=\text{RQ}=\frac{1}{2}\text{AC}\ ...\ (4)$
From (3) and (4)
SR = PQ = SP = RQ
We know that the diagonals of a square bisect each other at right angles.
$\angle\text{EOF} = 90^\circ.$
Now, $\text{RQ || DB}$
$\text{RE || FO}$
Also, $\text{SR || AC}$
$\Rightarrow\text{FR || OE}$
$\therefore$ OERF is a parallelogram.
So, $\angle\text{FRE} = \angle\text{EOF} = 90^\circ$ (Opposite angles are equal)
Thus, PQRS is a parallelogram with $\angle\text{R} = 90^\circ$ and SR = PQ = SP = RQ.
$\therefore$ PQRS is a square.

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Question 1001 Mark

ABCD is a trpezium in which AB || DC. M and N are then mid-points of AD and BC respectively. If AB = 12cm, MN = 14cm, then CD =

Answer

16cm.
Solution:

Let a line BP is drawn || to AD to meet DC at P.
ABPD is a parallelogram.
AB || PD, AD || BP
So AB = DP
Let BP cuts MN at Q.
MQ is also || to AB || PD
So AB = MQ = PD = 12cm ...(1)
QN = MN - MQ = 14 - 12 = 2cm
Consider $\triangle\text{BPC}.$
Q and N are the mid-points of BP & BC, and the line joining them QN || PC.
Then by property, $\frac{\text{QN}}{\text{PC}}=\frac{1}{2}$
⇒ PC = 2QN = 2 × 2 = 4cm
Now, DC = DP + PC
DP = 12cm [From (1)]
⇒ DC = 12 + 4 = 16cm

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MATHS M.C.Q