0
365

# Geometry Quadrilaterals Question for SSC CHSl and MTS

Here you can download SSC CHSL & MTS 2022 – important SSC CHSL & MTS Geometry Quadrilaterals Questions PDF by Cracku. Very Important SSC CHSL & MTS 2022 and These questions will help your SSC CHSL & MTS preparation. So kindly download the PDF for reference and do more practice.

Question 1:Â Vertices A, B, C and D of a quadrilateral ABCD lie on a circle. $\angle$A is three times $\angle$C and $\angle$D is two times $\angle$B. What is the difference between the measures of $\angle$D and $\angle$C?

a)Â $55^\circ$

b)Â $65^\circ$

c)Â $75^\circ$

d)Â $45^\circ$

Solution:

Given,

$\angle$A is three times $\angle$C and $\angle$D is two times $\angle$B.

$\angle$A = 3$\angle$C…….(1)

$\angle$D = 2$\angle$B…….(2)

In a cyclic quadrilateral, opposite angles are supplementary.

$\angle$A +Â $\angle$C = 180$^\circ$ andÂ $\angle$B + $\angle$D = 180$^\circ$

$\angle$A + $\angle$C = 180$^\circ$

3$\angle$C +Â $\angle$C =Â 180$^\circ$Â  [From (1)]

4$\angle$C =Â 180$^\circ$

$\angle$C = 45$^\circ$

$\angle$A =Â 3$\angle$C =Â 135$^\circ$

$\angle$B + $\angle$D = 180$^\circ$

$\angle$B + 2$\angle$B = 180$^\circ$Â  [From (2)]

3$\angle$B = 180$^\circ$

$\angle$B = 60$^\circ$

$\angle$D = 2$\angle$B = 120$^\circ$

Difference between the measures of $\angle$D and $\angle$C =Â 120$^\circ$ –Â 45$^\circ$

=Â 75$^\circ$

Hence, the correct answer is Option C

Question 2:Â ABCD is a cyclic quadrilateral such that when sides AB and DC are produced, they meet at E, and sides AD and BC meet at F, when produced. If $\angle$ADE = 80$^\circ$ and $\angle$AED = 50$^\circ$, then what is the measure of $\angle$AFB?

a)Â $40^\circ$

b)Â $20^\circ$

c)Â $50^\circ$

d)Â $30^\circ$

Solution:

From triangle AED,

$\angle$ADE +Â $\angle$AED +Â $\angle$DAE =Â 180$^\circ$

80$^\circ$ +Â 50$^\circ$ +Â $\angle$DAE =Â 180$^\circ$

$\angle$DAE =Â 50$^\circ$

Opposite angles in a cyclic quadrilateral is supplementary.

$\angle$ADC +Â $\angle$ABC =Â 180$^\circ$

80$^\circ$ +Â $\angle$ABC =Â 180$^\circ$

$\angle$ABC =Â 100$^\circ$

From triangle ABF,

$\angle$ABF + $\angle$AFB + $\angle$BAF = 180$^\circ$

100$^\circ$ + $\angle$AFB + 50$^\circ$Â = 180$^\circ$

$\angle$AFB = 30$^\circ$

Hence, the correct answer is Option D

Question 3:Â ABCD is cyclic quadrilateral in which $\angle$A = x$^\circ$, $\angle$B = 5y$^\circ$, $\angle$C = 2x$^\circ$ and $\angle$D = y$^\circ$. What is the value of (3x – y)?

a)Â 120

b)Â 90

c)Â 150

d)Â 60

Solution:

Opposite angles in a cyclic quadrilateral are supplementary.

$\angle$A +Â $\angle$C =Â 180$^\circ$

x +Â 2x =Â 180$^\circ$

3x =Â 180$^\circ$

x =Â 60$^\circ$

$\angle$B + $\angle$D = 180$^\circ$

5y + y = 180$^\circ$

6y = 180$^\circ$

y = 30$^\circ$

3x – y = 3(60) – 30 = 150

Hence, the correct answer is Option C

Question 4:Â Triangle ABC is an equilateral triangle. D and E are points on AB and AC respectively such that DE is parallel to BC and is equal to half the length of BC. If AD +Â CE + BC = 30 cm, then find the perimeter (in cm) of the quadrilateral BCED.

a)Â 37.5

b)Â 25

c)Â 45

d)Â 35

Solution:

Triangle ABC is an equilateral triangle.

Let the length of BC = 2p

BC = AB = AC = 2p

DE is equal to half the length of BC.

Triangle ABC and triangle ADE are similar triangles.

$\Rightarrow$Â Â $\frac{AD}{AB}=\frac{DE}{BC}$

$\Rightarrow$Â Â $\frac{AD}{AB}=\frac{p}{2p}$

$\Rightarrow$Â  $AD=\frac{1}{2}AB$

$\Rightarrow$Â  $AD=\frac{1}{2}\times2p$

$\Rightarrow$Â  AD = p

Similarly, AE = p

and EC = AC – AE = 2p – p = p

AD + CE + BC = 30 cm

p + p + 2p = 30

4p = 30

p =Â $\frac{15}{2}$ cm

Perimeter ofÂ the quadrilateral BCED = BD + DE + CE + BC

= p + p + p + 2p

= 5p

=Â $5\times\frac{15}{2}$

= 37.5 cm

Hence, the correct answer is Option A

Question 5:Â In the figure, a circle touches all the four sides of a quadrilateral ABCD whose sides AB = 6.5 cm, BC = 5.4 cm and CD = 5.3 cm. The length of AD is:

a)Â 4.6 cm

b)Â 5.8 cm

c)Â 6.2 cm

d)Â 6.4 cm

Solution:

Given,Â AB = 6.5 cm, BC = 5.4 cm and CD = 5.3 cm

Let the circle touches AB, BC, CD, DA at T, R, Q, S respectively.

Length of tangents to the circle from an external point are equal.

AT = AS

BT = BR

CQ = CR

DQ = DS

AT + BT + CQ + DQ = AS + BR + CR + DS

$\Rightarrow$Â  (AT + BT) + (CQ + DQ) = (AS + DS) + (BR + CR)

$\Rightarrow$Â  AB + CD = AD + BC

$\Rightarrow$Â  6.5 + 5.3 = AD + 5.4

$\Rightarrow$Â  AD = 6.4 cm

Hence, the correct answer is Option D

Question 6:Â The three medians AX, BY and CZ of $\triangle$ABC intersect at point L. If the area of $\triangle$ABC is 30 $cm^2$, then the area of the quadrilateral BXLZ is:

a)Â 10 $cm^2$

b)Â 12 $cm^2$

c)Â 16 $cm^2$

d)Â 14 $cm^2$

Solution:

Given, Area of $\triangle$ABC = 30 $cm^2$

The three medians AX, BY and CZ of $\triangle$ABC intersect at point L as shown in the above figure

The area of six triangles in the above figure are equal and area is equal to one-sixth of the area of triangle ABC.

Area of $\triangle$BLX = $\frac{1}{6}\times$Area ofÂ $\triangle$ABC

= $\frac{1}{6}\times$30

= 5 $cm^2$

Similarly, Area of $\triangle$BLZ = 5 $cm^2$

$\therefore\$Area of quadrilateral =Â Area of $\triangle$BLX +Â Area of $\triangle$BLZ = 5 + 5 = 10Â $cm^2$

Hence, the correct answer is Option A

Question 7:Â In the given figure, PQRS is a cyclic quadrilateral. What is the measure of the angle PQR if PQ is parallel to SR?

a)Â $110^\circ$

b)Â $80^\circ$

c)Â $100^\circ$

d)Â $70^\circ$

Solution:

Sum of opposite angles = 180$^{\circ\ }$

$=$> Â $\angle$SPQ +Â $\angle$SRQ =Â 180$^{\circ\ }$

$=$> Â 110$^{\circ\ }$Â + $\angle$SRQ = 180$^{\circ\ }$

$=$> Â $\angle$SRQ =Â 70$^{\circ\ }$

Given,Â PQ is parallel to SR

RQ is the transversal intersecting the parallel lines PQ and SR

Sum of the interior angles on the same side of the transversal is 180$^{\circ\ }$

$=$> Â $\angle$SRQ +Â $\angle$PQR =Â 180$^{\circ\ }$

$=$> Â 70$^{\circ\ }$ + $\angle$PQR = 180$^{\circ\ }$

$=$> Â $\angle$PQR =Â 110$^{\circ\ }$

Hence, the correct answer is Option A

Question 8:Â ABCD is a cyclic quadrilateral which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If $\angle A = 60^\circ$ and $\angle ABC = 72^\circ$, then \angle PDC –Â $\angle DPC =$

a)Â $24^\circ$

b)Â $30^\circ$

c)Â $36^\circ$

d)Â $40^\circ$

Solution:

In $\triangle ABP$,

$\angle A + \angle ABC + \angle APB = 180\degree$

$\angle APB = 180 – 60 – 72 = 48\degree$

$\angle ADC = 180 – ABC = 180 – 72 = 108\degree$
$\angle PDC = 180 –Â \angle ADC = 180 – 108 = 72\degree$

$\angle PDC – \angle DPC = 72 – 48 = 24\degree$

Question 9:Â In quadrilateral PQRS, RM $\perp$ QS, PN $\perp$ QS and QS = 6 cm. If RM = 3 cm and PN = 2 cm, then the area of PQRS is

a)Â 13 $cm^2$

b)Â 15 $cm^2$

c)Â 14 $cm^2$

d)Â 11 $cm^2$

Solution:

Area of PQRS = area of $\triangle$PQS +Â area of $\triangle$RQS

(Area of triangle = $\frac{1}{2}\times base \times height$)

=Â $\frac{1}{2}\times 6 \times 2$ +Â $\frac{1}{2}\times 6 \times 3$

= $\frac{1}{2}\times 6(2 + 3)$

= $15 cm^2$

Question 10:Â Sides AB and DC of cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BCare produced to meet at F. If $\angle BAD = 102^\circ$ and $\angle BEC = 38^\circ$ then the difference between $\angle ADC$ and $\angle AFB$ is:

a)Â $21^\circ$

b)Â $31^\circ$

c)Â $22^\circ$

d)Â $23^\circ$

Solution:

= 180 âˆ’ (38 + 102)
= 40$\degree$
â‡’âˆ ADC = 40$\degree$

square ABCD is a cyclic quadrilateral.
âˆ´âˆ DCB + âˆ DAB=180
â‡’âˆ DCB = 180 âˆ’ âˆ DAB
âˆ DCBÂ = 180 âˆ’ 102
âˆ DCBÂ = 78$\degree$
In Î”DFC,
âˆ DFC=180 – (âˆ FDC+âˆ FCD)
âˆ DFCÂ = 180 âˆ’ (40 + 78)
âˆ DFCÂ = 180 âˆ’ 118
âˆ DFCÂ = 62$\degree$
âˆ AFB = âˆ DFC =Â 62$\degree$
.

Difference between $\angle BAD$ and $\angle AFB$ = 62 – 40 = 22$\degree$

Question 11:Â Quadrilateral ABCD circumscribes circle. If AB = 8 cm, BC = 7 cm and CD = 6 cm,then the length of AD is:

a)Â 6 cm

b)Â 7.5 cm

c)Â 7cm

d)Â 6.8 cm

Solution:

AB = 8 cm

BC = 7 cm

CD = 6 cm

By the property,

AB + CD = BC + AD

8 + 6 = 7 + AC

AC = 14 – 7 = 7 cm

Question 12:Â ABCD is a cyclic quadrilateral in which AB = 16.5 cm, BC = x cm, CD = 11 cm, AD = 19.8 cm, and BD is bisected by AC at O. What is the value of x ?

a)Â 12.8 cm

b)Â 12.4 cm

c)Â 13.2 cm

d)Â 13.8 cm

Solution:

ABCD is a cyclic quadrilateral in which AB = 16.5Â cm
BC = x cm
CD = 11 cm

By the property,

16.5 $\times x = 19.8 \times 11$

16.5 $\times x = 217.8$

x = 217.8/16.5 = 13.2 cm

Question 13:Â ABCD is a cyclic quadrilateral. The tangents to the circle at the points A and C on it, intersect at P. If $\angle ABC = 98^\circ$, then what is the measure of $\angle APC$?

a)Â $22^\circ$

b)Â $26^\circ$

c)Â $16^\circ$

d)Â $14^\circ$

Solution:

ACD is aÂ cyclic quadrilateral so,

$\angle ABC +Â \angle ADC = 180\degree$

$\angle ADC = 180 – 98 = 82\degree$

$\angle AOC = 2 \timesÂ \angle ADC = 2 \times 82 = 164\degree$

$\angle OAP +Â \angle APC +Â \angle PCO +Â \angle COA = 360\degree$

$\angle OAP =Â \angle PCO = 90\degree$

($\because$ tangent angle)

$\angle APC = 360 – 90 – 90 – 164 = 16\degree$

Question 14:Â ABCD is a cyclic quadrilateral in which AB = 15 cm, BC = 12 cm and CD = 10 cm. If AC bisects BD, then what is the measure of AD?

a)Â 15 cm

b)Â 13.5 cm

c)Â 18 cm

d)Â 20 cm

Solution:

GivenÂ ABCD is a cyclic quadrilateral where AB=15,BC= 12,CD =10

is given below diagram

from the above diagram AC bisects BD

$\triangle is similar \triangle BCD$ \frac{AB}{AD} = \frac{DC}{BC} \Rightarrow \frac{15}{AD} = \frac{10}{12}\Rightarrow AD = \frac{15\times 12} {10} \Rightarrow AD = 18 cm

therefore Option (C) 18 cm Ans

Question 15:Â From a point P which is at a distance of 10 cm from the centre O of a circle of radius 6 cm, a pair of tangents PQ and PR to the circle at point Q and respectively, are drawn. Then the area of the quadrilateral PQOR is equal to

a)Â 30 sq.cm

b)Â 40 sq.cm

c)Â 24 sq.cm

d)Â 48 sq.cm

Solution:

From the given question we draw the diagram

From the $\triangle$

$x^2 = (10)^2 – (6)^2$

$x^2 = 100 – 36$

$x^2 = 64$

$x = 8$

then areaÂ quadrilateral PQOR =$2 \times \frac {1}{2} \times 6 \times 8$

= $6\times 8$

=Â  Â $48 cm^2$ Ans

Question 16:Â Two equilateral triangles of side $10\sqrt{3}$ cm are joined to form a quadrilateral. What is the altitude of the quadrilateral?

a)Â 12 cm

b)Â 14 cm

c)Â 16 cm

d)Â i5 cm

Solution:

Given thatÂ Â $10\sqrt{3}$ cm

We know the area of equilateral triangle = $\frac{\sqrt {3}} {4} a^2$Â  Â …… Eq (1)

and on the $\triangle DCBÂ Â isÂ Â alsoÂ Â given = \frac{1}{2} \times a \times h$Â  Â …… Eq (2)

then Eq(1) = Eq (2)

$\frac{\sqrt {3}} {4} a^2Â =Â Â \frac{1}{2} \times a \times h$

$\Rightarrow h =Â \frac{\sqrt{3}} {2} a$

$\Rightarrow h =Â \frac{\sqrt{3}} {2} \timesÂ 10\sqrt{3}$

$\Rightarrow h = 15 cm Ans Question 17:Â ABCD is cyclic quadrilateral. Sides AB and DC, when produced, meet at E, and sides BC and AD, when produced, meet at F. If$\angle$BFA =$60^\circ$and$\angle$AED =$30^\circ$, then the measure of$\angle$ABC is: a)Â$75^\circ$b)Â$65^\circ$c)Â$80^\circ$d)Â$70^\circ$17)Â AnswerÂ (A) Solution: From the given question we draw the diagram is given below from the above diagram$\angle BFA = 60^\circ , $\angle AFD = 30^\circ$

then $\angle EBC + \angle ABC = 180^\circ$ (straight line) ………….(1)

$\angle ABC + \angle ADC = 180^\circ$ (Opposite angle of cyclic quadrilateral)….. (2)

from the above Equestion (1) and (2)

$\angle EBC + \angle ABC = \angle ABC + \angle ADC$

$\angle EBC = \angle ADC$ …….(3)

$\angle DFC + \angle DCF + \angle CDF = 180^\circ$ (angle sum property of a triangle) ……. (4)

$\angle BCE + \angle CBE + \angle CEB = 180^\circ$ (angle sum property of a triangle) ………(5)

from the equestion (4) and (5)

$\angle DCF = \angle BCF$ (Vertically Opposite angle)

$\angle DFC + \angle DCF + \angle CDF = \angle BCE + \angle CBF + \angle CEB$

$\Rightarrow \angle DFC + \angle CDF = \angle CBF + \angle CEB$

$\Rightarrow 60^\circ + 180^\circ – \angle EBC = \angle EBC + \angle CEB$

$\Rightarrow 60^\circ + 180^\circ = 2 \angle EBC + 30^\circ$

$\Rightarrow 2 \angle EBC = 210^\circ$

$\Rightarrow \angle EBC= 105^\circ$

then $\angle ABC + \angle EBC = 180^\circ$

$\Rightarrow \angle ABC + 105^\circ = 180^\circ$

$\Rightarrow \angle ABC = 180^\circ -105^\circ$

$\Rightarrow \angle ABC = 75^\circ$ Ans

Question 18:Â In quadrilateral $ABCD, \angle C = 72^\circ$ and $\angle D = 28^\circ$. The bisectors of $\angle A$ and $\angle B$ meet in O. What is the measure of $\angle AOB$?

a)Â $48^\circ$

b)Â $54^\circ$

c)Â $50^\circ$

d)Â $36^\circ$

Solution:

In quadrilateral $ABCD$,
$\angle A +Â \angle B +Â \angle C +Â \angle D$ = 360
$\angle A + \angle B = 360 – 72 – 28 = 260\degree$
$\frac{1}{2}(\angle A + \angle B) =Â 130\degree$
In $\triangle$ AOB,
$\frac{1}{2}(\angle A + \angle B) + \angle AOBÂ = 180$
$\angle AOB = 180 – 130 = 50\degree$

Question 19:Â In a circle with centre O, ABCD isa cyclic quadrilateral and AC is the diameter. Chords AB and CD are produced to meet at E. If $\angle CAE = 34^\circ$ and $\angle E = 30^\circ$, then $\angle CBD$ is equal to:

a)Â $36^\circ$

b)Â $26^\circ$

c)Â $24^\circ$

d)Â $34^\circ$

$\angle DCA$ = 30 + 34 = 64
$\angleÂ DAC$Â = 180 – 90 – 64 = 26$\degree$
$\angleÂ DAC = \angleÂ CBD$
$\angleÂ CBD = 26\degree$