Geometry Quadrilaterals Question for SSC CHSl and MTS

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

a) ${55}^{\circ }$

b) ${65}^{\circ }$

c) ${75}^{\circ }$

d) ${45}^{\circ }$

1) Answer (C)

Solution:

Given,

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

$\mathrm{\angle }$A = 3$\mathrm{\angle }$C…….(1)

$\mathrm{\angle }$D = 2$\mathrm{\angle }$B…….(2)

In a cyclic quadrilateral, opposite angles are supplementary.

$\mathrm{\angle }$A + $\mathrm{\angle }$C = 180${}^{\circ }$ and $\mathrm{\angle }$B + $\mathrm{\angle }$D = 180${}^{\circ }$

$\mathrm{\angle }$A + $\mathrm{\angle }$C = 180${}^{\circ }$

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

4$\mathrm{\angle }$C = 180${}^{\circ }$

$\mathrm{\angle }$C = 45${}^{\circ }$

$\mathrm{\angle }$A = 3$\mathrm{\angle }$C = 135${}^{\circ }$

$\mathrm{\angle }$B + $\mathrm{\angle }$D = 180${}^{\circ }$

$\mathrm{\angle }$B + 2$\mathrm{\angle }$B = 180${}^{\circ }$  [From (2)]

3$\mathrm{\angle }$B = 180${}^{\circ }$

$\mathrm{\angle }$B = 60${}^{\circ }$

$\mathrm{\angle }$D = 2$\mathrm{\angle }$B = 120${}^{\circ }$

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

= 75${}^{\circ }$

Hence, the correct answer is Option C

Question 2: 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 $\mathrm{\angle }$ADE = 80${}^{\circ }$ and $\mathrm{\angle }$AED = 50${}^{\circ }$, then what is the measure of $\mathrm{\angle }$AFB?

a) ${40}^{\circ }$

b) ${20}^{\circ }$

c) ${50}^{\circ }$

d) ${30}^{\circ }$

2) Answer (D)

Solution:

From triangle AED,

$\mathrm{\angle }$ADE + $\mathrm{\angle }$AED + $\mathrm{\angle }$DAE = 180${}^{\circ }$

80${}^{\circ }$ + 50${}^{\circ }$ + $\mathrm{\angle }$DAE = 180${}^{\circ }$

$\mathrm{\angle }$DAE = 50${}^{\circ }$

ABCD is a cyclic quadrilateral.

Opposite angles in a cyclic quadrilateral is supplementary.

$\mathrm{\angle }$ADC + $\mathrm{\angle }$ABC = 180${}^{\circ }$

80${}^{\circ }$ + $\mathrm{\angle }$ABC = 180${}^{\circ }$

$\mathrm{\angle }$ABC = 100${}^{\circ }$

From triangle ABF,

$\mathrm{\angle }$ABF + $\mathrm{\angle }$AFB + $\mathrm{\angle }$BAF = 180${}^{\circ }$

100${}^{\circ }$ + $\mathrm{\angle }$AFB + 50${}^{\circ }$ = 180${}^{\circ }$

$\mathrm{\angle }$AFB = 30${}^{\circ }$

Hence, the correct answer is Option D

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

a) 120

b) 90

c) 150

d) 60

3) Answer (C)

Solution:

ABCD is cyclic quadrilateral.

Opposite angles in a cyclic quadrilateral are supplementary.

$\mathrm{\angle }$A + $\mathrm{\angle }$C = 180${}^{\circ }$

x + 2x = 180${}^{\circ }$

3x = 180${}^{\circ }$

x = 60${}^{\circ }$

$\mathrm{\angle }$B + $\mathrm{\angle }$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: 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 + CE + BC = 30 cm, then find the perimeter (in cm) of the quadrilateral BCED.

a) 37.5

b) 25

c) 45

d) 35

4) Answer (A)

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.

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

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

$\Rightarrow$  $AD=\frac{1}{2}AB$

$\Rightarrow$  $AD=\frac{1}{2}\times 2p$

$\Rightarrow$  AD = p

Similarly, AE = p

and EC = AC – AE = 2p – p = p

AD + CE + BC = 30 cm

p + p + 2p = 30

4p = 30

p = $\frac{15}{2}$ cm

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

= p + p + p + 2p

= 5p

= $5\times \frac{15}{2}$

= 37.5 cm

Hence, the correct answer is Option A

Question 5: 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) 4.6 cm

b) 5.8 cm

c) 6.2 cm

d) 6.4 cm

5) Answer (D)

Solution:

Given, 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

Adding all of the above

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

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

$\Rightarrow$  AB + CD = AD + BC

$\Rightarrow$  6.5 + 5.3 = AD + 5.4

$\Rightarrow$  AD = 6.4 cm

Hence, the correct answer is Option D

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Question 6: The three medians AX, BY and CZ of $\mathrm{△}$ABC intersect at point L. If the area of $\mathrm{△}$ABC is 30 $c{m}^2$, then the area of the quadrilateral BXLZ is:

a) 10 $c{m}^2$

b) 12 $c{m}^2$

c) 16 $c{m}^2$

d) 14 $c{m}^2$

6) Answer (A)

Solution:

Given, Area of $\mathrm{△}$ABC = 30 $c{m}^2$

The three medians AX, BY and CZ of $\mathrm{△}$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 $\mathrm{△}$BLX = $\frac{1}{6}\times$Area of $\mathrm{△}$ABC

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

= 5 $c{m}^2$

Similarly, Area of $\mathrm{△}$BLZ = 5 $c{m}^2$

$\therefore$Area of quadrilateral = Area of $\mathrm{△}$BLX + Area of $\mathrm{△}$BLZ = 5 + 5 = 10 $c{m}^2$

Hence, the correct answer is Option A

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

a) ${110}^{\circ }$

b) ${80}^{\circ }$

c) ${100}^{\circ }$

d) ${70}^{\circ }$

7) Answer (A)

Solution:

In the cyclic quadrilateral PQRS,

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

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

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

$=$>  $\mathrm{\angle }$SRQ = 70${}^{\circ }$

Given, 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${}^{\circ }$

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

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

$=$>  $\mathrm{\angle }$PQR = 110${}^{\circ }$

Hence, the correct answer is Option A

Question 8: 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 $\mathrm{\angle }A={60}^{\circ }$ and $\mathrm{\angle }ABC={72}^{\circ }$, then \angle PDC – $\mathrm{\angle }DPC=$

a) ${24}^{\circ }$

b) ${30}^{\circ }$

c) ${36}^{\circ }$

d) ${40}^{\circ }$

8) Answer (A)

Solution:

In $\mathrm{△}ABP$,

$\mathrm{\angle }A+\mathrm{\angle }ABC+\mathrm{\angle }APB=180\text{degree}$

$\mathrm{\angle }APB=180–60–72=48\text{degree}$

$\mathrm{\angle }ADC=180–ABC=180–72=108\text{degree}$
$\mathrm{\angle }PDC=180–\mathrm{\angle }ADC=180–108=72\text{degree}$

$\mathrm{\angle }PDC–\mathrm{\angle }DPC=72–48=24\text{degree}$

Question 9: 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) 13 $c{m}^2$

b) 15 $c{m}^2$

c) 14 $c{m}^2$

d) 11 $c{m}^2$

9) Answer (B)

Solution:

Area of PQRS = area of $\mathrm{△}$PQS + area of $\mathrm{△}$RQS

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

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

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

= $15c{m}^2$

Question 10: 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 $\mathrm{\angle }BAD={102}^{\circ }$ and $\mathrm{\angle }BEC={38}^{\circ }$ then the difference between $\mathrm{\angle }ADC$ and $\mathrm{\angle }AFB$ is:

a) ${21}^{\circ }$

b) ${31}^{\circ }$

c) ${22}^{\circ }$

d) ${23}^{\circ }$

10) Answer (C)

Solution:

In ΔADE,
∠ADE=180 − (∠AED + ∠EAD)
= 180 − (38 + 102)
= 40$\text{degree}$
⇒∠ADC = 40$\text{degree}$
square ABCD is a cyclic quadrilateral.
∴∠DCB + ∠DAB=180
⇒∠DCB = 180 − ∠DAB
∠DCB = 180 − 102
∠DCB = 78$\text{degree}$
In ΔDFC,
∠DFC=180 – (∠FDC+∠FCD)
∠DFC = 180 − (40 + 78)
∠DFC = 180 − 118
∠DFC = 62$\text{degree}$
∠AFB = ∠DFC = 62$\text{degree}$.

Difference between $\mathrm{\angle }BAD$ and $\mathrm{\angle }AFB$ = 62 – 40 = 22$\text{degree}$

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

a) 6 cm

b) 7.5 cm

c) 7cm

d) 6.8 cm

11) Answer (C)

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: 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) 12.8 cm

b) 12.4 cm

c) 13.2 cm

d) 13.8 cm

12) Answer (C)

Solution:

ABCD is a cyclic quadrilateral in which AB = 16.5 cm
BC = x cm
CD = 11 cm
AD = 19.8 cm

By the property,

AB⋅BC = AD⋅DC

16.5 $\times x=19.8\times 11$

16.5 $\times x=217.8$

x = 217.8/16.5 = 13.2 cm

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

a) ${22}^{\circ }$

b) ${26}^{\circ }$

c) ${16}^{\circ }$

d) ${14}^{\circ }$

13) Answer (C)

Solution:

ACD is a cyclic quadrilateral so,

$\mathrm{\angle }ABC+\mathrm{\angle }ADC=180\text{degree}$

$\mathrm{\angle }ADC=180–98=82\text{degree}$

$\mathrm{\angle }AOC=2\times \mathrm{\angle }ADC=2\times 82=164\text{degree}$

In quadrilateral AOCP-

$\mathrm{\angle }OAP+\mathrm{\angle }APC+\mathrm{\angle }PCO+\mathrm{\angle }COA=360\text{degree}$

$\mathrm{\angle }OAP=\mathrm{\angle }PCO=90\text{degree}$

($\because$ tangent angle)

$\mathrm{\angle }APC=360–90–90–164=16\text{degree}$

Question 14: 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) 15 cm

b) 13.5 cm

c) 18 cm

d) 20 cm

14) Answer (C)

Solution:

Given 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: 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) 30 sq.cm

b) 40 sq.cm

c) 24 sq.cm

d) 48 sq.cm

15) Answer (D)

Solution:

From the given question we draw the diagram

From the $\mathrm{△}$

$x^2=(10{)}^2–(6{)}^2$

$x^2=100–36$

$x^2=64$

$x=8$

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

= $6\times 8$

=   $48c{m}^2$ Ans

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

a) 12 cm

b) 14 cm

c) 16 cm

d) i5 cm

16) Answer (D)

Solution:

Given that  $10\sqrt{3}$ cm

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

and on the $\mathrm{△}DCBisalsogiven=\frac{1}{2}\times a\times h$   …… Eq (2)

then Eq(1) = Eq (2)

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

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

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

$\Rightarrow h = 15 cm Ans

Question 17: ABCD is cyclic quadrilateral. Sides AB and DC, when produced, meet at E, and sides BC and AD, when produced, meet at F. If $\mathrm{\angle }$BFA = ${60}^{\circ }$ and $\mathrm{\angle }$AED = ${30}^{\circ }$, then the measure of $\mathrm{\angle }$ABC is:

a) ${75}^{\circ }$

b) ${65}^{\circ }$

c) ${80}^{\circ }$

d) ${70}^{\circ }$

17) Answer (A)

Solution:

From the given question we draw the diagram is given below

from the above diagram $\mathrm{\angle }BFA={60}^{\circ },$\angle AFD = 30^\circ $

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

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

from the above Equestion (1) and (2)

$\mathrm{\angle }EBC+\mathrm{\angle }ABC=\mathrm{\angle }ABC+\mathrm{\angle }ADC$

$\mathrm{\angle }EBC=\mathrm{\angle }ADC$ …….(3)

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

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

from the equestion (4) and (5)

$\mathrm{\angle }DCF=\mathrm{\angle }BCF$ (Vertically Opposite angle)

$\mathrm{\angle }DFC+\mathrm{\angle }DCF+\mathrm{\angle }CDF=\mathrm{\angle }BCE+\mathrm{\angle }CBF+\mathrm{\angle }CEB$

$\Rightarrow \mathrm{\angle }DFC+\mathrm{\angle }CDF=\mathrm{\angle }CBF+\mathrm{\angle }CEB$

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

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

$\Rightarrow 2\mathrm{\angle }EBC={210}^{\circ }$

$\Rightarrow \mathrm{\angle }EBC={105}^{\circ }$

then $\mathrm{\angle }ABC+\mathrm{\angle }EBC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ABC+{105}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ABC={180}^{\circ }-{105}^{\circ }$

$\Rightarrow \mathrm{\angle }ABC={75}^{\circ }$ Ans

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

a) ${48}^{\circ }$

b) ${54}^{\circ }$

c) ${50}^{\circ }$

d) ${36}^{\circ }$

18) Answer (C)

Solution:

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

Question 19: 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 $\mathrm{\angle }CAE={34}^{\circ }$ and $\mathrm{\angle }E={30}^{\circ }$, then $\mathrm{\angle }CBD$ is equal to:

a) ${36}^{\circ }$

b) ${26}^{\circ }$

c) ${24}^{\circ }$

d) ${34}^{\circ }$

19) Answer (B)

Solution:

By the exterior angle property,
$\mathrm{\angle }DCA$ = 30 + 34 = 64
$\mathrm{\angle }DAC$ = 180 – 90 – 64 = 26$\text{degree}$
$\mathrm{\angle }DAC=\mathrm{\angle }CBD$
$\mathrm{\angle }CBD=26\text{degree}$

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