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Quadrilaterals

Quadrilaterals

MODULES

Properties of Polygons
Types of Polygons
Area of Regular Polygons
Quadrilateral & Parallelogram
Rhombus & Rectangle
Square
Trapezium & Kite
Symmetrical Shapes
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Quadrilaterals : Level 1
Quadrilaterals : Level 2
Quadrilaterals : Level 3
ALL MODULES

CAT 2025 Lesson : Quadrilaterals - Trapezium & Kite

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3.6 Trapezium

A quadrilateral where one pair of sides are parallel while the other pair of sides are not is a trapezium.

1) The triangles formed by the Diagonals along parallel sides of a trapezium are similar triangles, i.e., △AOB∼△COD \triangle AOB \sim \triangle COD△AOB∼△COD

2) Given the similarity stated above, the diagonals divide each other in the same proportion, i.e., OAOC=OBOD\dfrac{\text {OA}}{\text {OC}} = \dfrac{\text {OB}}{\text {OD}}OCOA​=ODOB​

3) AC2+BD2=AD2+BC2+2AB.CDAC^2 + BD^2 = AD^2 + BC^2 + 2 AB.CDAC2+BD2=AD2+BC2+2AB.CD
4) Mid-point Theorem: The median that passes through the mid-points of the non-parallel sides is parallel to the other 2 sides and half of the sum of the other two sides. AB | | PQ | | CD

Median === PQ =12(AB+CD)= \dfrac{1}{2}(AB + CD)=21​(AB+CD)

5) Where h is the height and b1 and b2 are the parallel sides, Area =12h(b1+b2)=12AE(AB+CD)= \dfrac{1}{2}h(b_1 + b_2)= \dfrac{1}{2}AE(AB + CD)=21​h(b1​+b2​)=21​AE(AB+CD)
Right-angled Trapezium: A trapezium where two adjacent angles are 90o90^{\mathrm{o}}90o. In a right-angled trapezium, one of the non-parallel sides is the height of the trapezium.
Isosceles Trapezium: A trapezium where the two non-parallel sides are equal.

1) Angles formed along a parallel line are equal.

2) A circle circumscribing an isosceles trapezium can be drawn.

3) When mid-points of adjacent sides are joined we get a rhombus.
4) Diagonals are equal and OA === OB, OC === OD.

5) Of the 4 triangles formed by the diagonals,
– triangles along the two equal non-parallel sides are congruent, i.e., △AOD≅△BOC \triangle AOD \cong \triangle BOC△AOD≅△BOC – triangles along the parallel sides are similar, i.e.△AOD∼△COD\triangle AOD \sim \triangle COD△AOD∼△COD

Example 19

In a trapezium ABCD, AB | | CD. If AB === 8, CD === 14 and AD === BC === 5, then what is the area of ABCD?

Solution

In this isosceles trapezium, we drop perpendiculars from A and B to meet CD at E and F respectively.

As AD === BC, AE === BF, ∠\angle∠AED === ∠\angle∠BFC =90o= 90^{\mathrm{o}}=90o, applying RHS,

△\triangle△ AED ≅\cong≅ △\triangle△ BFC

∴ DE === FC === 3

Applying Pythagoras theorem, AE =52−32=4= \sqrt{5^2 - 3^2}= 4=52−32​=4

Area of ABCD =12h(b1+b2)= \dfrac{1}{2}h(b_1 + b_2)=21​h(b1​+b2​)
= 12×4(8+14)\dfrac{1}{2} \times 4(8 + 14)21​×4(8+14) = 44

Answer: 444444


Example 20

In a the figure below, PQ, TU and RS are parallel lines. If PQ : TU : RS = 4 : 7 : 11, then what is the ratio of PT : TR?


Solution

TP and UQ are extended to meet at O.

△\triangle△ OPQ ∼\sim∼ △\triangle△ OTU (∠\angle∠O is common & ∠\angle∠OPQ = ∠\angle∠OTU. AA rule)

∴ TUPQ=OTOP=74⇒OP + PTOP=74⇒1+PTOP=74⇒PTOP\dfrac{\text{TU}}{\text{PQ}} = \dfrac{\text{OT}}{\text{OP}} = \dfrac{\text{7}}{\text{4}} ⇒ \dfrac{\text{OP + PT}}{\text{OP}} = \dfrac{\text{7}}{\text{4}}⇒ 1 + \dfrac{\text{PT}}{\text{OP}} = \dfrac{\text{7}}{\text{4}}⇒ \dfrac{\text{PT}}{\text{OP}}PQTU​=OPOT​=47​⇒OPOP + PT​=47​⇒1+OPPT​=47​⇒OPPT​

=34 \dfrac{\text{3}}{\text{4}}43​-----(1)

(∠\angle∠ O is common & ∠\angle∠ OPQ = ∠\angle∠ ORS. AA rule)

∴ RSPQ=OROP=114⇒OP + PROP=114⇒PROP=74\dfrac{\text{RS}}{\text{PQ}} = \dfrac{\text{OR}}{\text{OP}} = \dfrac{11}{4} ⇒ \dfrac{\text{OP + PR}}{\text{OP}} = \dfrac{11}{4}⇒ \dfrac{\text{PR}}{\text{OP}} = \dfrac{7}{4}PQRS​=OPOR​=411​⇒OPOP + PR​=411​⇒OPPR​=47​ -----(2)

Dividing (2) and (1) ⇒ PRPT=73⇒TR + PTPT=73⇒TRPT=43⇒PT:TR=3:4\dfrac{\text{PR}}{\text{PT}} = \dfrac{7}{3} ⇒ \dfrac{\text{TR + PT}}{\text{PT}} = \dfrac{7}{3} ⇒ \dfrac{\text{TR}}{\text{PT}} = \dfrac{4}{3}⇒ PT : TR = 3 : 4PTPR​=37​⇒PTTR + PT​=37​⇒PTTR​=34​⇒PT:TR=3:4

Answer: 3:43 : 43:4

3.7 Kite

A quadrilateral where exactly 2 pairs of adjacent sides are equal is a kite.

1) AB = BC and AD = CD.

2) Diagonals intersect at 90o90^{\mathrm{o}}90o.

3) The diagonal that divides the kite into 2 isosceles triangles is bisected by the other diagonal. i.e, AO === OC.

4) BD is the angle bisector of ∠\angle∠B and ∠\angle∠D and divides the kite into two congruent triangles, i.e., △\triangle△ ABD ≅\cong≅ △\triangle△ CBD

5) Area of Kite =2×Area(△ABD) = 2 \times \mathrm {Area (\triangle ABD)} =2×Area(△ABD)

=2×12×BD×AO = 2 \times \dfrac{1}{2} \mathrm {\times BD \times AO}=2×21​×BD×AO =12BD×AC= \dfrac{1}{2} \mathrm {BD \times AC}=21​BD×AC =12d1d2= \dfrac{1}{2} d_1 d_2 =21​d1​d2​

=2×12×AB×AD×sinA = 2 \times \dfrac{1}{2} \times \mathrm{ AB \times AD \times sinA}=2×21​×AB×AD×sinA =absinθ= absin\theta=absinθ

Example 21

In quadrilateral ABCD, AC and BD intersect at O. If AB = BC =2132 \sqrt{13}213​ , OC = 5 and ∠ADB = 30°30\degree30°, then what is the area of ABCD?

Solution

As adjacent sides are equal, ABCD is a kite. In a kite, diagonals intersect at right angles and AO = OC = 5.

Applying Pythagoras in △\triangle△BOC,

BO =(2(13)2−52)=27=33= \sqrt{( 2 \sqrt{(13)^2} - 5^2 )} = \sqrt{27} = 3 \sqrt{3}=(2(13)2​−52)​=27​=33​

△\triangle△ AOB is a 30o−60o−90o30^{\mathrm{o}} - 60^{\mathrm{o}} - 90^{\mathrm{o}}30o−60o−90o triangle.

AOOD=13\mathrm{\dfrac{AO}{OD}}= \dfrac{1}{\sqrt{3}}ODAO​=3​1​⇒ OD =53= 5\sqrt{3}=53​

BD === BO + OD =83= 8\sqrt{3}=83​

Area of Kite =12×(product of diagonals)= \dfrac{1}{2} \times \mathrm{(product \space of \space diagonals)}=21​×(product of diagonals)

=12×AC×BD = \dfrac{1}{2} \times \mathrm{ AC \times BD}=21​×AC×BD

=12×10×83=403= \dfrac{1}{2} \times 10 \times 8 \sqrt{3} = 40 \sqrt{3}=21​×10×83​=403​

Answer: 40340 \sqrt{3}403​


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