Quadrilaterals

Naming a quadrilateral, StudySmarter Originals

The quadrilateral above has sides \(PQ\), \(QR\), \(RS\) and \(SP\) and vertices \(P\), \(Q\), \(R\) and \(S\). The diagonals are described by \(PR\) and \(QS\).

This quadrilateral can be denoted by \(PQRS\), \(QRSP\), \(RSPQ\) or \(SPQR\). However, we cannot call it \(SQPR\) or \(RPQS\), for example since the order of the vertices here is incorrect.

Find the perimeter of the parallelogram below.

Example 1, StudySmarter Originals

Solution

To solve this, we shall simply add the dimensions of all four sides of this parallelogram.

\[P=17+17+11+11\]

\[\implies P=2(17)+2(11)\]

\[\implies P=34+22\]

\[\implies P=56\,\mathrm{cm}\]

Thus, its perimeter is \(56 \,\mathrm{cm}\).

Calculate the length of the missing side, \(x\) of the trapezoid below given that the perimeter is \(51\,\mathrm{cm}\).

Example 2, StudySmarter Originals

Solution

Using the formula for the perimeter of a quadrilateral, we find that

\[P=8+12+13+x\]

\[\implies 51=8+12+13+x\]

\[\implies 8+12+13+x=51\]

Now bringing \(8\), \(12\) and \(13\) to the right-hand side of this equation, thus making \(x\) the subject, we obtain

\[x=51-8-12-13\]

\[\implies x=18\,\mathrm{cm}\]

Therefore, the length of the missing side, \(x\) is \(18\, \mathrm{cm}\).

Find the area of the trapezium below.

Example 3, StudySmarter Originals

Solution

From the diagram above we can deduce that

\[a = 21\, \mathrm{cm}\]

\[b = 15\, \mathrm{cm}\]

\[h = 13\, \mathrm{cm}\]

Now using the area formula for a trapezium, we obtain

\[A=\dfrac{1}{2}\times (21+15)\times 13\]

\[\implies A=\dfrac{1}{2}\times (36)13\]

\[\implies A=\dfrac{1}{2}\times 468\]

\[\implies A=234\,\mathrm{cm}^2\]

Important note: Don't be confused by the length of the bases and legs! The parallel sides are the bases which are what we will use in our formula.

Thus the area is \(234\, \mathrm{cm}^2\).

Find the length of the horizontal diagonal, d of the rhombus below given that the area is \(123.5\, \mathrm{cm}\).

Example 4, StudySmarter Originals

Solution

Here, the vertical diagonal is \(13\,\mathrm{cm}\). Using the area formula of a rhombus, we find that

\[A=\dfrac{1}{2}\times 13\times d\]

\[\implies 123.5=\dfrac{1}{2}\times 13\times d\]

\[\implies \dfrac{13\times d}{2}=123.5\]

Now multiplying both sides by \(2\), we obtain

\[\cancel{2}\left(\dfrac{13\times d}{\cancel{2}}\right)=2(123.5)\]

\[\implies 13d=247\]

Then dividing both sides by \(13\), we get

\[\dfrac{\cancel{13}d}{\cancel{13}}=\dfrac{247}{13}\]

\[\implies d=19\,\mathrm{cm}\]

Thus, the length of the horizontal diagonal, \(d\) is \(19\, \mathrm{cm}\).

Given the kite \(ABCD\) below, answer the following questions:

1. Find the missing sides \(AD\) and \(AB\), given that the perimeter is \(88\,\mathrm{cm}\).

2. Calculate the missing angles \(M\) and \(N\).

3. Are any of these sides parallel to each other?

Example 5, StudySmarter Originals

Question 1

Firstly, note that \(AD = AB\) since a kite has two pairs of equal adjacent sides (the other being \(CB = CD\)).

\[AD=AB\]

\[\implies x-2=y-4\]

\[\implies y=x-2+4\]

\[\implies y=x+2\]

Now, given the perimeter and the sum of all its sides, we obtain

\[P=AB+AD+CD+CD\]

\[\implies 88=(y-4)+(x-2)+32+32\]

\[\implies 88=y+x-6+64\]

Now substituting for \(y\) and solving this equation for \(x\), we obtain

\[88=(x+2)+x-6+64\]

\[\implies (x+2)+x-6+64=88\]

\[\implies 2x+60=88\]

\[\implies 2x=88-60\]

\[\implies 2x=28\]

Now dividing both sides by \(2\), have

\[x=\dfrac{28}{2}\]

\[\implies x=14\]

Thus, the lengths of \(AB\) and \(AD\) are

\[AB=y-4\]

\[\implies AB=(x+2)-4\]

\[\implies AB=(14+2)-4\]

\[\implies AB=12\,\mathrm{cm}\]

Since \(AB = AD\), thus \(AD\) is also \(12\, \mathrm{cm}\).

Question 2

Recall that a kite has one pair of equal opposite angles that are obtuse. This means that \(M = N\). Furthermore, that measures of the other two angles are given and the sum of all the interior angles of any quadrilateral is \(360º\). From here, we find that

\[\angle DAB+\angle BCD+\angle M+\angle N=360º\]

\[\implies 102º+51+\angle M+\angle N=360º\]

\implies 153º+2\angle M=360º\]

Now solving for \(M\), we obtain

\[2\angle M=360º-153º\]

\[\implies 2\angle M=207º\]

\[\implies \angle M=\dfrac{207º}{2}\]

\[\implies \angle M=103,5º\]

Thus, both \(M\) and \(N\) are equal to \(103.5º\) (since \(M = N\)).

Question 3

By the properties of a kite, there are no sides that are parallel to each other.

Given the parallelogram \(ABCD\) below, answer the following questions:

1. Find the length of \(AB\) given that the perimeter is \(40\, \mathrm{cm}\).

2. Calculate the area of \(ABCD\).

3. Deduce the angle \(Y\).

Example 6, StudySmarter Originals

Question 1

Firstly, note that \(AB = DC\) and \(AD = BC\) since a parallelogram has two pairs of equal opposite sides. Given the perimeter and the sum of all its sides, we obtain

\[P=AB+AD+CD+CD\]

\[\implies 40=x+9+x+9\]

\[\implies 40=2x+18\]

Now solving this equation for \(x\), we obtain

\[2x=40-18\]

\[\implies 2x=22\]

\[\implies x=\dfrac{22}{2}\]

\[\implies x=11\,\mathrm{cm}\]

Thus, the length of \(AB\) is \(11\, \mathrm{cm}\).

Question 2

Here, the height is \(7\, \mathrm{cm}\) and the width is the length of side \(AB\). We know that \(AB\) is equal to \(11\, \mathrm{cm}\). Thus, by the area formula of a parallelogram, we have

\[A=11\times 7\]

\[\implies 77\,\mathrm{cm}^2\]

Thus, the area of \(ABCD\) is \(77\,\mathrm{cm}^2\).

Question 3

In this case, \(\angle BCD\) and \(\angle Y\) are supplementary since both these angles are on a straight line. Recall that two angles are supplementary when the sum of their measures is equal to \(180º\). Using this idea, we find that

\[\angle BCD+\angle Y=180º\]

\[\implies 114º+\angle Y=180º\]

Now, rearranging this equation and solving for our missing angle, we obtain

\[\angle Y=180º-114º\]

\[\implies \angle Y=66º\]

Thus \(Y\) is equal to \(66º\).

Given the rectangle \(PQRS\) below, answer the following questions:

1. Find the perimeter and area of \(PQRS\).

2. Calculate the length of the diagonal \(QS\).

3. How big is \(\angle QSR\)?

Example 7, StudySmarter Originals

Question 1

Firstly, note that \(PQ = SR\) and \(PS = QR\) since a rectangle has two pairs of equal opposite sides. Now, adding the lengths of all its sides, we obtain

\[P=PQ+QR+SR+PS\]

\[\implies P=15+8+15+8\]

\[\implies P=46\,\mathrm{cm}\]

Here, the length is \(15\,\mathrm{cm}\) and the height is \(8\, \mathrm{cm}\). So by the area formula of a rectangle, we find that

\[A=15\times 8\]

\[\implies A=120\,\mathrm{cm}^2\]

Thus, the perimeter is \(46\, \mathrm{cm}\) and the area is \(120\,\mathrm{cm}^2\).

Question 2

By the properties of a rectangle, recall that it has 4 right angles. Notice that the diagonal \(QS\) creates a right-angle triangle \(QRS\) where \(\angle QRS\) is equal to \(90º\). Thus, we can use Pythagoras' Theorem to find the length of \(QS\). Here, \(QS\) is the hypotenuse and the two sides are \(QR = 8 \,\mathrm{cm}\) and \(SR = 15\, \mathrm{cm}\). By doing so, we get

\[QS^2=QR^2+SR^2\]

\[\implies QS^2=8^2+15^2\]

\[\implies QS^2=64+225\]

\[\implies QS^2=289\]

We will only consider the positive root since we are dealing with measurements. Then

\[QS=\sqrt{289}\]

\[\implies QS=17\,\mathrm{cm}\]

Thus, the diagonal, \(QS\) is equal to \(17\,\mathrm{cm}\).

Question 3

Here, \(\angle PSQ\) and \(\angle QSR\) are complementary since both these angles lie at a right angle. Recall that two angles are complementary when the sum of their measures is equal to \(90º\). With that, we have

\[\angle PSQ+\angle QSR=90º\]

\[\implies 66º+\angle QSR=90º\]

Now, rearranging this equation and solving for our missing angle, we obtain

\[\angle QSR=90º-66º\]

\[\implies \angle QSR=24º\]

Thus \(\angle QSR\) is equal to \(24º\).