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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

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}}

3) AC2+BD2=AD2+BC2+2AB.CDAC^2 + BD^2 = AD^2 + BC^2 + 2 AB.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)

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)
Right-angled Trapezium: A trapezium where two adjacent angles are 90o90^{\mathrm{o}}. 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 – triangles along the parallel sides are similar, i.e.△AOD∼△COD\triangle AOD \sim \triangle 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, ∠\angleAED == ∠\angleBFC =90o= 90^{\mathrm{o}}, applying RHS,

△\triangle AED ≅\cong △\triangle BFC

∴ DE == FC == 3

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

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

Answer: 4444


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 (∠\angleO is common & ∠\angleOPQ = ∠\angleOTU. 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}}

=34 \dfrac{\text{3}}{\text{4}}-----(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} -----(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 : 4

Answer: 3:43 : 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}}.

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 ∠\angleB and ∠\angleD 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)}

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

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

Example 21

In quadrilateral ABCD, AC and BD intersect at O. If AB = BC =2132 \sqrt{13} , OC = 5 and ∠ADB = 30°30\degree, 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 △\triangleBOC,

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

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

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

BD == BO + OD =83= 8\sqrt{3}

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

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

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

Answer: 40340 \sqrt{3}


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