Mix Examples - Quadrilaterals Questions in English

(A) In a parallelogram $ABCD$,the sum of adjacent angles is $180^o$. Therefore,$\angle A + \angle B = 180^o$.
Given that $\angle A = 5 \angle B$,we substitute this into the equation:
$5 \angle B + \angle B = 180^o$
$6 \angle B = 180^o$
$\angle B = 30^o$.
Since $\angle A = 5 \angle B$,then $\angle A = 5 \times 30^o = 150^o$.
In a parallelogram,opposite angles are equal,so $\angle C = \angle A$.
Therefore,$\angle C = 150^o$.

(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:
$PQ^2 = PO^2 + QO^2$
$PQ^2 = 20^2 + 21^2$
$PQ^2 = 400 + 441$
$PQ^2 = 841$
$PQ = \sqrt{841} = 29 \, cm$.

(C) rhombus is a quadrilateral where all sides are equal and diagonals bisect each other at right angles.
By the properties of a rhombus,the diagonals $XZ$ and $YW$ are perpendicular to each other.
Since the diagonals intersect at point $P$,the angle formed at the intersection,$\angle XPY$,must be $90^o$.
Therefore,$\angle XPY = 90^o$.

(D) In a parallelogram,the diagonals bisect each other.
This means that the point of intersection $P$ is the midpoint of both diagonals $AC$ and $BD$.
Therefore,$BD = 2 \times PB$.
Given that $PB = 5.2 \, cm$,we have:
$BD = 2 \times 5.2 \, cm = 10.4 \, cm$.
Thus,the length of diagonal $BD$ is $10.4 \, cm$.

(A) In a rhombus $ABCD$,adjacent angles are supplementary,meaning $\angle A + \angle B = 180^{\circ}$.
Given that $\angle A = \angle B - 30^{\circ}$,we can substitute this into the equation:
$(\angle B - 30^{\circ}) + \angle B = 180^{\circ}$
$2\angle B = 210^{\circ}$
$\angle B = 105^{\circ}$.
Since $\angle A = \angle B - 30^{\circ}$,then $\angle A = 105^{\circ} - 30^{\circ} = 75^{\circ}$.
In a rhombus,opposite angles are equal,so $\angle C = \angle A$.
Therefore,$\angle C = 75^{\circ}$.

(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,$\angle D = 180^{\circ} - 70^{\circ} = 110^{\circ}$.

(C) In a parallelogram $PQRS$,the opposite sides are equal in length.
Therefore,$PQ = RS = 8 \, \text{cm}$ and $QR = PS = 5 \, \text{cm}$.
The perimeter of a parallelogram is the sum of all its sides.
Perimeter $= PQ + QR + RS + PS$
Perimeter $= 8 \, \text{cm} + 5 \, \text{cm} + 8 \, \text{cm} + 5 \, \text{cm} = 26 \, \text{cm}$.
Alternatively,the perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides}) = 2 \times (PQ + QR) = 2 \times (8 + 5) = 2 \times 13 = 26 \, \text{cm}$.

(D) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the length of the third side.
In $\Delta ABC$,$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $AC$.
Therefore,$PQ = \frac{1}{2} \times BC$.
Given $BC = 8.2 \text{ cm}$.
$PQ = \frac{1}{2} \times 8.2 = 4.1 \text{ cm}$.

(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 $RS = 3x = 3 \times 2.5 = 7.5\, cm$.

(B) In $\Delta ABC$,$P$ is the midpoint of $AB$ and $R$ is the midpoint of $AC$. By the Midpoint Theorem,$PR \parallel BC$ and $PR = \frac{1}{2} BC = BQ = QC$.
Similarly,$Q$ is the midpoint of $BC$ and $R$ is the midpoint of $AC$,so $QR \parallel AB$ and $QR = \frac{1}{2} AB = AP = PB$.
Also,$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$,so $PQ \parallel AC$ and $PQ = \frac{1}{2} AC = AR = RC$.
Thus,in quadrilateral $APQR$,$AP \parallel QR$ and $AR \parallel PQ$,making $APQR$ a parallelogram.
In a parallelogram,opposite angles are equal. Therefore,$\angle BAC = \angle PQR$.

(C) According to the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of its length.
In $\Delta PQR$,since $A$ and $C$ are mid-points of $PQ$ and $RP$ respectively,$AC = \frac{1}{2} QR$.
Similarly,$AB = \frac{1}{2} RP$ and $BC = \frac{1}{2} PQ$.
The perimeter of $\Delta ABC = AB + BC + AC = \frac{1}{2} RP + \frac{1}{2} PQ + \frac{1}{2} QR = \frac{1}{2} (PQ + QR + RP)$.
Given that the perimeter of $\Delta ABC = 18.6 \, cm$.
Therefore,$18.6 = \frac{1}{2} \times (\text{Perimeter of } \Delta PQR)$.
Perimeter of $\Delta PQR = 18.6 \times 2 = 37.2 \, cm$.

(D) In a trapezium,the line segment joining the midpoints of the non-parallel sides is parallel to the parallel sides and its length is equal to half the sum of the lengths of the parallel sides.
Mathematically,for a trapezium $ABCD$ with $AB || CD$,where $P$ and $Q$ are midpoints of $AD$ and $BC$ respectively,the formula is:
$PQ = \frac{1}{2} (AB + CD)$
Given $AB = 18 \, cm$ and $PQ = 15 \, cm$,we substitute these values into the formula:
$15 = \frac{1}{2} (18 + CD)$
Multiply both sides by $2$:
$30 = 18 + CD$
Subtract $18$ from both sides:
$CD = 30 - 18 = 12 \, cm$.
Therefore,the length of $CD$ is $12 \, cm$.

(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.

(N/A) $(1)$ The diagonals of a rhombus are perpendicular to each other.
$(2)$ The diagonals of a parallelogram bisect each other.

(A) According to the Midpoint Theorem and properties of quadrilaterals:
$(1)$ When the midpoints of the sides of a rhombus are joined in order,the resulting quadrilateral is a rectangle because the diagonals of a rhombus bisect each other at right angles.
$(2)$ When the midpoints of the sides of a rectangle are joined in order,the resulting quadrilateral is a rhombus because 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 $120^\circ$.

(B) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
In $\Delta ABC$,$X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.
Therefore,$XY \parallel BC$ and $XY = \frac{1}{2} BC$.
In quadrilateral $XBCY$,we have one pair of opposite sides $XY$ and $BC$ that are parallel to each other $(XY \parallel BC)$.
$A$ quadrilateral with at least one pair of parallel opposite sides is defined as a trapezium.
Thus,$XBCY$ is a trapezium.

(A) In a parallelogram $ABCD$,the adjacent angles are supplementary,meaning their sum is $180^{\circ}$.
Therefore,$\angle A + \angle B = 180^{\circ}$.
Given that $\angle A = \angle B - 30^{\circ}$,we can express $\angle B$ as $\angle B = \angle A + 30^{\circ}$.
Substituting this into the sum equation: $\angle A + (\angle A + 30^{\circ}) = 180^{\circ}$.
$2\angle A + 30^{\circ} = 180^{\circ}$.
$2\angle A = 180^{\circ} - 30^{\circ} = 150^{\circ}$.
$\angle A = 150^{\circ} / 2 = 75^{\circ}$.

(B) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of its length.
Let the sides of $\Delta ABC$ be $AB$,$BC$,and $AC$.
Since $P$,$Q$,and $R$ are midpoints of sides $AB$,$BC$,and $AC$ respectively,the sides of $\Delta PQR$ are $PQ = \frac{1}{2} AC$,$QR = \frac{1}{2} AB$,and $PR = \frac{1}{2} BC$.
The perimeter of $\Delta PQR = PQ + QR + PR = \frac{1}{2} AC + \frac{1}{2} AB + \frac{1}{2} BC = \frac{1}{2} (AB + BC + AC)$.
Given that the perimeter of $\Delta ABC = AB + BC + AC = 23 \, cm$.
Therefore,the perimeter of $\Delta PQR = \frac{1}{2} \times 23 = 11.5 \, cm$.

(C) The perimeter of a parallelogram is given by the formula $P = 2(a + b)$,where $a$ and $b$ are the lengths of adjacent sides.
Given,$AB = 12.5 \, cm$ and $BC = 7 \, cm$.
Here,$a = 12.5 \, cm$ and $b = 7 \, cm$.
Perimeter $P = 2(12.5 + 7) \, cm$.
$P = 2(19.5) \, cm$.
$P = 39 \, cm$.
Therefore,the perimeter of parallelogram $ABCD$ is $39 \, cm$.

(D) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
In $\Delta XYZ$,$A$ and $B$ are the midpoints of $XY$ and $XZ$ respectively.
Therefore,$AB = \frac{1}{2} \times YZ$.
Given $AB = 7.5 \, cm$.
Substituting the value: $7.5 = \frac{1}{2} \times YZ$.
$YZ = 7.5 \times 2 = 15 \, cm$.

(A) In a trapezium $ABCD$,the sides $AB$ and $CD$ are parallel $(AB \parallel CD)$.
According to the properties of parallel lines,the consecutive interior angles (also known as co-interior angles) between parallel lines are supplementary.
Therefore,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.
Given $\angle A = 80^{\circ}$,we can find $\angle D$ as follows:
$\angle A + \angle D = 180^{\circ}$
$80^{\circ} + \angle D = 180^{\circ}$
$\angle D = 180^{\circ} - 80^{\circ}$
$\angle D = 100^{\circ}$.

(B) According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
In $\Delta ABC$,since $P$ and $Q$ are midpoints of $AB$ and $AC$ respectively,we have $PQ = \frac{1}{2} BC$.
Given that $BC + PQ = 21 \text{ cm}$.
Substituting $PQ = \frac{1}{2} BC$ into the equation,we get $BC + \frac{1}{2} BC = 21$.
This simplifies to $\frac{3}{2} BC = 21$.
Multiplying both sides by $\frac{2}{3}$,we get $BC = 21 \times \frac{2}{3} = 14 \text{ cm}$.

(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,$AB = 5x = 5 \times 7 = 35 \, cm$.