Mix Examples - Quadrilaterals Questions in English

(B) $(1)$ True: In a rhombus, all sides are equal. Perimeter $= 4 \times \text{side} = 4 \times 12 \, \text{cm} = 48 \, \text{cm}$.
$(2)$ False: In a square $PQRS$, the diagonal $PR = s\sqrt{2}$, where $s$ is the side length. Given $PR = 8 \, \text{cm}$, then $s\sqrt{2} = 8 \implies s = 8 / \sqrt{2} = 4\sqrt{2} \, \text{cm}$. The perimeter is $4s = 4 \times 4\sqrt{2} = 16\sqrt{2} \, \text{cm}$, which is not $32 \, \text{cm}$.

(A) $(1)$ False. In a parallelogram $ABCD$,the sides $AB$ and $BC$ are adjacent sides. The diagonal $AC$ forms a triangle $ABC$. By the triangle inequality theorem,the sum of any two sides must be greater than the third side. Here,$AB + BC = 12 + 5 = 17 \text{ cm}$,which is greater than $AC = 13 \text{ cm}$. However,there is no condition that $ABCD$ must be a rectangle. If it were a rectangle,$AC$ would be $\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}$. Since it is only stated as a parallelogram,$AC$ can take various values depending on the angle $\angle B$. Thus,the statement is not universally true.
$(2)$ False. In a trapezium $ABCD$ where $AB \parallel CD$,the parallel sides (bases) do not necessarily have to be equal. If the parallel sides were equal,the figure would be a parallelogram,not necessarily just a trapezium.

(D) rhombus is a quadrilateral in which all four sides are equal in length.
Given that $ABCD$ is a rhombus,all sides are equal,i.e.,$AB = BC = CD = DA = 13 \text{ cm}$.
The perimeter of a rhombus is calculated by the formula: $\text{Perimeter} = 4 \times \text{side length}$.
Substituting the given value: $\text{Perimeter} = 4 \times 13 \text{ cm} = 52 \text{ cm}$.
Therefore,the perimeter of $ABCD$ is $52 \text{ cm}$.

(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $PR$ and $QS$ intersect at point $O$.
Therefore,$PO = \frac{PR}{2} = \frac{40}{2} = 20 \, cm$.
And $QO = \frac{QS}{2} = \frac{42}{2} = 21 \, cm$.
In the right-angled triangle $\triangle POQ$,by the Pythagorean theorem:
$PQ^2 = PO^2 + QO^2$
$PQ^2 = 20^2 + 21^2$
$PQ^2 = 400 + 441$
$PQ^2 = 841$
$PQ = \sqrt{841} = 29 \, cm$.

(B) In a trapezium $ABCD$,we are given that $AB \parallel CD$.
Since $AB \parallel CD$,the consecutive interior angles between the parallel lines are supplementary.
Therefore,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.
Given $\angle A = 70^{\circ}$,we have $70^{\circ} + \angle D = 180^{\circ}$.
Thus,$\angle D = 180^{\circ} - 70^{\circ} = 110^{\circ}$.

(A) In a rectangle $PQRS$,opposite sides are equal. Therefore,$PQ = RS$ and $QR = PS$.
Given the ratio $PQ : QR = 3 : 5$,let $PQ = 3x$ and $QR = 5x$.
Since $PQ = RS$,we have $RS = 3x$.
The perimeter of a rectangle is given by the formula $2 \times (length + width) = 40\, cm$.
Substituting the values: $2 \times (3x + 5x) = 40$.
$2 \times (8x) = 40$.
$16x = 40$.
$x = 40 / 16 = 2.5$.
Now,calculate $RS = 3x = 3 \times 2.5 = 7.5\, cm$.

(A) $(1)$ $A$ quadrilateral is a closed plane figure formed by joining four points in order,where no three points are collinear. Thus,the missing word is $\text{closed}$.
$(2)$ $A$ quadrilateral has four sides. The sides that do not share a common vertex are called opposite sides. There are two such pairs of opposite sides. Thus,the missing word is $\text{two}$.

(N/A) $(1)$ $A$ quadrilateral has $4$ pairs of consecutive angles. In a quadrilateral $ABCD$,the pairs are $(A, B), (B, C), (C, D),$ and $(D, A)$.
$(2)$ The sum of the interior angles of a quadrilateral is $360^{\circ}$. This is derived from the angle sum property of a polygon,where the sum of interior angles is $(n - 2) \times 180^{\circ}$. For a quadrilateral,$n = 4$,so $(4 - 2) \times 180^{\circ} = 2 \times 180^{\circ} = 360^{\circ}$.

(N/A) $(1)$ $A$ quadrilateral in which only one pair of opposite sides are parallel is called a trapezium.
$(2)$ The diagonals of a rectangle are equal in length.

(B) The sum of the interior angles of a quadrilateral is always $360^{\circ}$.
Given that $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.
Substituting the given values: $100^{\circ} + 80^{\circ} + 120^{\circ} + \angle D = 360^{\circ}$.
Summing the known angles: $300^{\circ} + \angle D = 360^{\circ}$.
Therefore,$\angle D = 360^{\circ} - 300^{\circ} = 60^{\circ}$.

(C) Let the angles of the quadrilateral be $3x$,$4x$,$6x$,and $5x$.
We know that the sum of the angles of a quadrilateral is $360^\circ$.
Therefore,$3x + 4x + 6x + 5x = 360^\circ$.
$18x = 360^\circ$.
$x = \frac{360^\circ}{18} = 20^\circ$.
The angles are:
$3 \times 20^\circ = 60^\circ$
$4 \times 20^\circ = 80^\circ$
$6 \times 20^\circ = 120^\circ$
$5 \times 20^\circ = 100^\circ$.
The greatest angle is $120^\circ$.

(C) The perimeter of a rectangle is given by the formula $P = 2(l + w)$,where $l$ is the length and $w$ is the width.
Here,$l = AB$ and $w = BC$.
Given $AB : BC = 5 : 3$,let $AB = 5x$ and $BC = 3x$.
The perimeter is $112 \, cm$.
So,$2(5x + 3x) = 112$.
$2(8x) = 112$.
$16x = 112$.
$x = 112 / 16 = 7$.
Therefore,$AB = 5x = 5 \times 7 = 35 \, cm$.

(N/A) $1$. In a parallelogram,the diagonals bisect each other. Therefore,$O$ is the midpoint of $AC$.
$2$. Consider $\triangle ADC$. Since $O$ is the midpoint of $AC$,$DO$ is the median of $\triangle ADC$.
$3$. $A$ median of a triangle divides it into two triangles of equal area.
$4$. Therefore,$ar(ADO) = ar(CDO)$.

(N/A) Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at point $O$.
By the properties of a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Thus,$AC \perp BD$ and $AO = OC = \frac{1}{2} AC$,$BO = OD = \frac{1}{2} BD$.
The area of the rhombus $ABCD$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BO$.
Area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times OD$.
Total Area $= \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times OD = \frac{1}{2} \times AC \times (BO + OD)$.
Since $BO + OD = BD$,the Area $= \frac{1}{2} \times AC \times BD$.
Therefore,the area of a rhombus is half the product of its diagonals.

(N/A) The area of quadrilateral $ABCD$ can be divided into the sum of the areas of two triangles,$\triangle ABD$ and $\triangle BCD$,by the diagonal $BD$.
$\operatorname{ar}(ABCD) = \operatorname{ar}(\triangle ABD) + \operatorname{ar}(\triangle BCD)$.
The area of a triangle is given by the formula $\frac{1}{2} \times \text{base} \times \text{height}$.
For $\triangle ABD$,the base is $BD$ and the height is $AM$. Thus,$\operatorname{ar}(\triangle ABD) = \frac{1}{2} \times BD \times AM$.
For $\triangle BCD$,the base is $BD$ and the height is $CN$. Thus,$\operatorname{ar}(\triangle BCD) = \frac{1}{2} \times BD \times CN$.
Substituting these into the first equation:
$\operatorname{ar}(ABCD) = (\frac{1}{2} \times BD \times AM) + (\frac{1}{2} \times BD \times CN)$.
Factoring out the common term $\frac{1}{2} \times BD$:
$\operatorname{ar}(ABCD) = \frac{1}{2} \times BD \times (AM + CN)$.
Hence,the area of the quadrilateral is proved.

(D) The perimeter of a square is given by the formula $P = 4 \times \text{side}$.
Given $P = 16 \, cm$,we have $4 \times \text{side} = 16 \, cm$.
Therefore,the side length is $\text{side} = \frac{16}{4} = 4 \, cm$.
The area of a square is given by the formula $ar(ABCD) = \text{side}^2$.
Thus,$ar(ABCD) = 4^2 = 16 \, cm^2$.

(A) The area of a square can be calculated using the length of its diagonal $d$ with the formula: $\text{Area} = \frac{1}{2} \times d^2$.
Given that the diagonal $AC = 16 \text{ cm}$.
Substituting the value into the formula: $\text{Area} = \frac{1}{2} \times (16)^2$.
$\text{Area} = \frac{1}{2} \times 256$.
$\text{Area} = 128 \text{ cm}^2$.
Therefore,the area of square $ABCD$ is $128 \text{ cm}^2$.

(C) The area of a rectangle is given by the formula: $\operatorname{ar}(PQRS) = \text{length} \times \text{breadth}$.
In rectangle $PQRS$,the sides are $PQ$ and $SP$.
Therefore,$\operatorname{ar}(PQRS) = PQ \times SP$.
Given that $\operatorname{ar}(PQRS) = 300 \, cm^2$ and $PQ = 20 \, cm$.
Substituting the values: $300 = 20 \times SP$.
Solving for $SP$: $SP = \frac{300}{20} = 15 \, cm$.
Thus,$SP = 15 \, cm$.

(A) $1$. In $\Delta PQR$,$PM$ is a median,so it divides the triangle into two triangles of equal area: $\text{ar}(PQM) = \text{ar}(PRM) = \frac{1}{2} \text{ar}(PQR)$.
$2$. In $\Delta PQM$,$QN$ is a median because $N$ is the midpoint of $PM$. Therefore,$\text{ar}(PQN) = \text{ar}(QNM) = \frac{1}{2} \text{ar}(PQM)$.
$3$. Given $\text{ar}(PQN) = 36 \text{ cm}^2$,we have $\text{ar}(PQM) = 2 \times \text{ar}(PQN) = 2 \times 36 = 72 \text{ cm}^2$.
$4$. Since $\text{ar}(PQM) = \frac{1}{2} \text{ar}(PQR)$,it follows that $\text{ar}(PQR) = 2 \times \text{ar}(PQM) = 2 \times 72 = 144 \text{ cm}^2$.

(A) The area of a quadrilateral can be calculated by dividing it into two triangles using a diagonal.
$ar(ABCD) = ar(\triangle ABC) + ar(\triangle ADC)$.
The area of a triangle is given by the formula $\frac{1}{2} \times \text{base} \times \text{height}$.
For $\triangle ABC$,the base is $AC = 18 \, cm$ and the height is $BM = 10 \, cm$.
$ar(\triangle ABC) = \frac{1}{2} \times 18 \times 10 = 90 \, cm^2$.
For $\triangle ADC$,the base is $AC = 18 \, cm$ and the height is $DN = 6 \, cm$.
$ar(\triangle ADC) = \frac{1}{2} \times 18 \times 6 = 54 \, cm^2$.
Therefore,$ar(ABCD) = 90 + 54 = 144 \, cm^2$.

(D) The area of a rectangle is given by the formula: $\text{Area} = \text{length} \times \text{breadth}$.
Given that $AB = 12 \, cm$ (length) and $BC = 7 \, cm$ (breadth).
Therefore,$\text{Area}(ABCD) = 12 \, cm \times 7 \, cm = 84 \, cm^2$.
Thus,the correct option is $D$.

(A) square is a quadrilateral where all sides are equal and each interior angle is $90^\circ$.
Given that $XYZW$ is a square and the side length $XY = 17 \text{ cm}$.
The formula for the area of a square is given by $(\text{side})^2$.
Area $= (XY)^2 = 17^2 = 289 \text{ cm}^2$.
Therefore,the area of $XYZW$ is $289 \text{ cm}^2$.

(B) Let the side of the square $ABCD$ be $a \, cm$.
In a square,the diagonal $d$ is related to the side $a$ by the formula $d = a\sqrt{2}$.
Given the diagonal $AC = 16 \, cm$,we have $a\sqrt{2} = 16$.
Squaring both sides,we get $2a^2 = 16^2 = 256$.
Dividing by $2$,we find $a^2 = 128$.
The area of a square is given by $a^2$.
Therefore,the area of square $ABCD = 128 \, cm^2$.

(C) The area of a rhombus is given by the formula: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the diagonals.
Given,$d_1 = AC = 16 \, cm$ and $d_2 = BD = 30 \, cm$.
Substituting the values into the formula:
$\text{Area} = \frac{1}{2} \times 16 \times 30$
$\text{Area} = 8 \times 30 = 240 \, cm^2$.
Therefore,the area of the rhombus $ABCD$ is $240 \, cm^2$.

(D) $1$. The perimeter of the rhombus is $680 \, m$. Since all sides of a rhombus are equal,the side length $a = 680 / 4 = 170 \, m$.
$2$. Let the diagonals be $d_1 = 300 \, m$ and $d_2$. In a rhombus,diagonals bisect each other at right angles.
$3$. Considering a right-angled triangle formed by the sides of the rhombus and half of the diagonals,we have $(d_1/2)^2 + (d_2/2)^2 = a^2$.
$4$. Substituting the values: $(300/2)^2 + (d_2/2)^2 = 170^2$,which gives $150^2 + (d_2/2)^2 = 28900$.
$5$. $22500 + (d_2/2)^2 = 28900$,so $(d_2/2)^2 = 6400$,which means $d_2/2 = 80 \, m$.
$6$. Thus,$d_2 = 160 \, m$.
$7$. The area of the rhombus is $(1/2) \times d_1 \times d_2 = (1/2) \times 300 \times 160 = 24000 \, m^2$.

(A) $(1)$ True: The diagonal $d$ of a square with side $a$ is given by $d = a \sqrt{2}$. For $a = 20 \, cm$,$d = 20 \sqrt{2} \, cm$. Thus,the statement is true.
$(2)$ False: The area of an equilateral triangle with side $a$ is $A = \frac{\sqrt{3}}{4} a^2$. If the side is doubled $(a' = 2a)$,the new area $A'$ becomes $A' = \frac{\sqrt{3}}{4} (2a)^2 = \frac{\sqrt{3}}{4} (4a^2) = 4 \times A$. The area becomes $4$ times the original area,not twice.

(B) Let the side of the square $PQRS$ be $a \, cm$.
In a square,the diagonal $d$ is related to the side $a$ by the formula $d = a\sqrt{2}$.
Given that the diagonal $PR = 10 \, cm$,we have $a\sqrt{2} = 10$.
Solving for $a$,we get $a = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, cm$.
The area of a square is given by the formula $\text{Area} = a^2$.
Substituting the value of $a$,we get $\text{Area} = (5\sqrt{2})^2 = 25 \times 2 = 50 \, cm^2$.
Alternatively,the area of a square can be calculated using the diagonal $d$ as $\text{Area} = \frac{1}{2} \times d^2$.
Substituting $d = 10 \, cm$,we get $\text{Area} = \frac{1}{2} \times (10)^2 = \frac{1}{2} \times 100 = 50 \, cm^2$.

(C) The area of a quadrilateral can be calculated by dividing it into two triangles using a diagonal.
Area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC)$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8 = 80 \, cm^2$.
Area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 12 = 120 \, cm^2$.
Total Area $= 80 + 120 = 200 \, cm^2$.

(D) Let the side of the rhombus be $a$. Since the perimeter is $100\,cm$,we have $4a = 100\,cm$,which gives $a = 25\,cm$.
Let the diagonals of the rhombus be $d_1 = 48\,cm$ and $d_2$. The diagonals of a rhombus bisect each other at right angles.
This forms a right-angled triangle with the side of the rhombus as the hypotenuse and half of the diagonals as the other two sides.
Using the Pythagorean theorem: $(a)^2 = (d_1/2)^2 + (d_2/2)^2$.
Substituting the values: $(25)^2 = (48/2)^2 + (d_2/2)^2$.
$625 = (24)^2 + (d_2/2)^2$.
$625 = 576 + (d_2/2)^2$.
$(d_2/2)^2 = 625 - 576 = 49$.
$d_2/2 = \sqrt{49} = 7$.
Therefore,$d_2 = 7 \times 2 = 14\,cm$.

(C) The area of a square is calculated by multiplying the length of its side by itself. Therefore, the formula for the area of a square is $\text{Area} = (\text{side})^{2}$.

(A) The area of a rectangle is defined as the product of its length and breadth.
Therefore,Area of a rectangle $=$ $\text{length} \times \text{breadth}$.

(A) The area of a rhombus is calculated by the formula: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the two diagonals of the rhombus.
Therefore,the area of a rhombus is $\frac{1}{2} \times$ product of its diagonals.

(B) The area of a square can be calculated using its side length as $Area = side^2$. However,if the length of the diagonal $(d)$ is given,the area is calculated as $Area = \frac{d^2}{2}$. Therefore,the missing term is the diagonal.

(D) The area of a rhombus can be calculated using the formula: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the diagonals.
Given: $d_1 = 40 \, cm$ and $d_2 = 42 \, cm$.
Substituting the values into the formula:
$\text{Area} = \frac{1}{2} \times 40 \times 42$
$\text{Area} = 20 \times 42 = 840 \, cm^2$.
Thus,the correct option is $D$.

(A) Let the diagonals of the rhombus be $d_1 = 16\,cm$ and $d_2 = 30\,cm$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the side of the rhombus be $a$. The diagonals divide the rhombus into four congruent right-angled triangles.
The legs of these triangles are half the lengths of the diagonals: $d_1/2 = 8\,cm$ and $d_2/2 = 15\,cm$.
Using the Pythagorean theorem: $a^2 = (8)^2 + (15)^2$.
$a^2 = 64 + 225 = 289$.
$a = \sqrt{289} = 17\,cm$.
The perimeter of a rhombus is $4 \times \text{side} = 4 \times 17 = 68\,cm$.