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SSC > Triangles

Explore popular questions from Triangles for SSC. This collection covers Triangles previous year SSC questions hand picked by experienced teachers.

Q 1.

Correct2

Incorrect-0.5

In, two similar triangles {tex} \triangle {/tex} ABC and {tex} \triangle {/tex} DEF, DE = 3 cm, EF = 5 cm, DF = 4 cm and BC = 20 cm, then length of AB is equal to

A

15 cm

12 cm

C

10 cm

D

6 cm

Explanation



From similar triangles ABC and DEF

{tex} \frac{AB}{BC} = \frac{DE}{EF} {/tex}

{tex} \therefore \frac{AB}{20} = \frac{3}{5} {/tex}

=> {tex} AB = \frac{3 \times 20}{5} {/tex} = 12 cm.

Q 2.

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If angles of one triangle are respectively equal to the angles of another triangle, then ratio of the corresponding sides is ratio of the corresponding

A

medians

B

bisector angles (angle bisector segments)

C

altitude

all of these

Q 3.

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The locus of the vertex A of an isosceles triangle, ABC which has BC as its fixed base is

A

a line parallel to BC

B

a line perpendicular to BC

C

a circle with BC as a diameter

perpendicular bisector of BC

Q 4.

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Incorrect-0.5

If AB and CD are two chords intersecting at a point P inside the circle such that AP = CP, then consider the following statements :
Assertion (A) : AB = CD,
Reason (R) : APC and DPB are similar triangles
Of these statements

both A and R are true, and R is correct explanation of A.

B

both A and R are true, but R is not a correct explanation of A.

C

A is true, but R is false.

D

A is false, but R is true

Q 5.

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If a triangle and a rectangle have equal areas and equal altitude, then base of the triangle is equal to

A

base of the rectangle.

twice the base of the rectangle.

C

thrice the base of the rectangle.

D

four times the base of the rectangle

Explanation

Let h be the common height of triangle and the rectangle.

If a and b respectively be the bases of triangle and rectangle, then by hypothesis

{tex} \frac{1}{2}a \times h = b \times h {/tex}

=> {tex} \frac{1}{2}a = b {/tex}

=> a = 2b

Q 6.

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Incorrect-0.5

Area of an isosceles right-angled triangle is 800 sq. metres. The greatest possible square has been cut out from it. The length of the diagonal of this square will be

A

{tex} 10\ \sqrt{2} m{/tex}

B

{tex} 10\ \sqrt{3} m{/tex}

C

20 m

{tex} 20\ \sqrt{2} m{/tex}

Explanation

Let x be the base and side of isosceles {tex} \triangle ABC {/tex}.

By hypothesis,



{tex} \frac{1}{2}x \times x = 800 {/tex}

=> {tex} x^{2} = 1600 {/tex}

=> x = 40 m

Therefore, Hypotenuse, BC = 2 x = 40 x/2 m

If s be the side of the maximum square, then

{tex} s^{2}+\frac{1}{2} \left(x-s\right)s+\frac{1}{2} \left(x-s\right)2=800 {/tex}

=> {tex} s^{2}+xs-s^{2} = 800 {/tex}

=> {tex} xs = 800 {/tex}

=> {tex} s = \frac{800}{40} = 20 m {/tex}

Therefore, Length of diagonal of the square = {tex} 20 \sqrt{2} m {/tex}

Q 7.

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If perimeter of a triangle is 100 m and its sides are in the ratio 1 : 2 : 2, then area of the triangle (in {tex} m^{2} {/tex}) is

A

{tex} 100 \sqrt{3} {/tex}

{tex} 100 \sqrt{15} {/tex}

C

{tex} 100 \sqrt{5} {/tex}

D

{tex} 100 \sqrt{7} {/tex}

Explanation

Let the Sides. be x, 2x and 2x.

Then x + 2x + 2x = 100

=> x = 20

Hence sides are 20, 40 and 40.

Therefore, s = 50 [Because, 2s = x+b+c]

Area of the triangle

= {tex} \sqrt{s \left(s-a\right) \left(s-b\right) \left(s-c\right) } {/tex}

= {tex} \sqrt{50 \left(50-20\right) \left(50-40\right) \left(50-40\right) } {/tex}

= {tex} \sqrt{50 \times 30 \times 10 \times 10} {/tex}

= {tex} 100 \sqrt{15} m^{2} {/tex}

Q 8.

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In the given figure, AD is the internal bisector and AE is the external bisector of {tex} \angle {/tex}BAC of any {tex} \triangle {/tex} ABC. Then which one of the following statements is not correct?

{tex} AC^{2} = DC \times CE {/tex}

B

{tex} AC \times CD = AC \times BD {/tex}

C

{tex} AB : AC = BE : CE {/tex}

D

{tex} \angle ADE - 90 ^{\circ} {/tex}

Explanation

(a) {tex} \triangle ABC - \triangle ADB - \triangle CDB {/tex}



From {tex} \triangle ABC\ and\ \triangle ADB {/tex}

{tex} \angle ABC = \angle ADB = 90 ^{\circ} {/tex}

AB = AB, common

From A-S-A, {tex} \triangle ABC - \triangle ADB {/tex}

Again, from {tex} \triangle ABC\ and\ \triangle CDB {/tex}

{tex} \angle C = common {/tex}

{tex} BC = common {/tex}

{tex} \angle ABC = \angle CBD {/tex}

From A-S-A, {tex} \triangle ABC - \triangle ADB {/tex}

From (i) and (ii)

{tex} \triangle ABC = \triangle ADB - \triangle CDB {/tex}

(b) From {tex} \triangle ADB - \triangle BDC {/tex}

{tex} \frac{BD}{AD} = \frac{DC}{BD} {/tex}

{tex} \therefore BD^{2} = AD \times DC {/tex}

And from {tex} \triangle ADB - \triangle ABC {/tex}

{tex} \frac{AB}{AD} = \frac{AC}{AB} {/tex}

(c) {tex} \frac{\triangle ADB}{\triangle CDB} = \frac{\frac{1}{2}AD \times BD}{\frac{1}{2}CD \times BD} {/tex}

= {tex} \frac{AD}{CD} = \frac{AB^{2}}{BC^{2}} \ne \frac{AB}{BC} {/tex}

Q 9.

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In the given figure, {tex} \angle {/tex}ABC = {tex} \angle {/tex}ADB = {tex} 90^{\circ} {/tex}, which one of the following statements does not hold good?

A

{tex} \triangle {/tex} ABC, ADB and CDB are similar.

B

{tex} AD \times DC {/tex}

{tex} \triangle ADB : \triangle CDB = AB : BC {/tex}

D

{tex} AB^{2} = AD \times AC {/tex}

Explanation



{tex} AD = \sqrt{AB^{2}-BD^{2}} {/tex}

= {tex} \sqrt{x^{2}-\frac{x^{2}}{4}} {/tex}

= {tex} \frac{\sqrt{3}}{2}x {/tex}

Therefore, Area of {tex} \triangle ABC = \frac{1}{2} \times BC \times AD {/tex}

= {tex} \frac{1}{2} \times x \times \frac{x\sqrt{3}}{2} {/tex}

= {tex} \frac{\sqrt{3}}{4}x^{2} {/tex}

Q 10.

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Area of an equilateral triangle of side x is

A

{tex} \frac{x^{2}}{\sqrt{2}} {/tex}

{tex} \frac{\sqrt{3}}{4} x^{2} {/tex}

C

{tex} \frac{\sqrt{3}}{2} x^{2} {/tex}

D

{tex} \frac{\sqrt{3}}{2} x^{3} {/tex}

Explanation



Here 'O' is the centroid of triangle and centroid of triangle divides line segment in the ratio 2 : 1 Here BO : OM = 2 : 1

Now, in {tex} \triangle ABD {/tex}

{tex} \left(AD\right)^{2} = \sqrt{AB^{2}-BD^{2}} {/tex}

= 36 - 9

=27

{tex} AD = 3\sqrt{3} cm = BM {/tex}

{tex} \because \triangle ABC\ is\ an\ equilateral\ triangle {/tex}

Now BM = BO + OM

{tex} 3\sqrt{3} = 2x+x {/tex}

{tex} \therefore x = \sqrt{3} {/tex}

{tex} \therefore OB = 2\sqrt{3} cm {/tex}

Q 11.

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In the given figure, {tex} \triangle ABC {/tex} is an equilateral triangle. O is the point of intersection of the medians. If AB = 6 cm, then OB is equal to

A

{tex} 3\sqrt{3}cm {/tex}

{tex} 2\sqrt{3}cm {/tex}

C

{tex} \sqrt{3}cm {/tex}

D

{tex} \frac{\sqrt{3}}{2}cm {/tex}

Q 12.

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If an isosceles right triangle has an area 200 sq. cm, then area of a square drawn on hypotenuse is

A

400 sq. cm

B

{tex} 400 \sqrt{2} cm {/tex}

800 sq. cm.

D

{tex} 800 \sqrt{2} cm {/tex}

Explanation



Area of isosceles right triangle

={tex} \frac{1}{2} \times BC \times AB {/tex}

= {tex} \frac{1}{2} \times x \times x {/tex}

{tex} 200 = \frac{1}{2}x^{2} {/tex}

{tex} \therefore x^{2} = 400 {/tex}

{tex} \therefore x = 2- cm{/tex}

{tex} \therefore AC = \sqrt{x^{2}+x^2} = \sqrt{2x^{2}} {/tex}

= {tex} \sqrt{2}x = 20\sqrt{2} cm {/tex}

Now square is drawn on hypotenuse of isosceles right angle triangle.

Hence side of square = {tex} AC = 20\sqrt{2} cm {/tex}

Area of square = {tex} \left(20\sqrt{2}\right)^{2} {/tex} = 800 cm.

Q 13.

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{tex} \triangle ABC\ and\ PQR {/tex} are congruent if

A

{tex} \angle ABC = \angle QPR {/tex} and {tex} \angle BCA = \angle QRP {/tex}

B

{tex} AB = PQ,\ AC=PR {/tex} and {tex} \angle ABC = \angle PQR {/tex}

C

{tex} AB= PQ,\ \angle ABC = \angle PQR {/tex} and {tex} \angle BAC = \angle PRQ {/tex}

{tex} BC = QR,\ AC = PR {/tex} and {tex} \angle ACB = \angle PRQ{/tex}

Explanation

{tex} \triangle PQR\ and\ \triangle LMN {/tex} are similar triangle (given)

{tex} \frac{PQ}{LM} = \frac{QR}{MN} = \frac{PR}{LN} {/tex}

Now {tex} \frac{PQ}{LM} = \frac{1}{3} {/tex}

{tex} \therefore \frac{1}{3} = \frac{QR}{MN} {/tex}

=> {tex} \frac{1}{3} = \frac{QR}{9} {/tex}

{tex} \left(\because MN = 9 cm\right) {/tex}

Therefore, QR = 3 cm.

Q 14.

Correct2

Incorrect-0.5

{tex} \triangle {/tex}PQR and {tex} \triangle {/tex}LMN are similar. If 3 PQ = LM and MN = 9 cm, then QR is equal to

3 cm

B

6 cm

C

9 cm

D

12 cm

Q 15.

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Incorrect-0.5

If D, E, F are mid-points of the sides BC, CA and AB respectively of a triangle ABC, then which one of the following is not correctly matched?

A

If G is centroid of the {tex} \triangle ABC {/tex}, then AG : GD = 2 : 1

B

If G is centroid of the {tex} \triangle ABC {/tex}, then G is centroid of the {tex} \triangle DEF {/tex}.

If {tex} \ \angle A = 90 ^{\circ} {/tex}, then A is orthocentre of the {tex} \triangle ABC {/tex}.

D

If {tex} \ \angle A > 90 ^{\circ} {/tex}, then orthocentre of the {tex} \triangle ABC {/tex} lies inside the {tex} \triangle ABC {/tex}.

Q 16.

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If lengths of two sides of a triangle are given, then its area is greater when

A

both the sides are greater than the third

B

angle between sides is a right angle.

angle between sides is an obtuse angle.

D

angle between sides is an acute angle.

Q 17.

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Incorrect-0.5

Consider following statements relating to the congruency of two right-angled triangles.
1. Equality of two sides of one triangle with same two sides of the second makes the triangle congruent.
2. Equality of hypotenuse and a side of one triangle with the hypotenuse and a side of the second respectively makes the triangles congruent.
3. Equality of hypotenuse and an acute of triangle with the hypotenuse and an angle of the second respectively makes the triangles congruent.
Of these statements

1, 2 and 3 are correct

B

1 and 2 are correct

C

1 and 3 are correct

D

2 and 3 are correct

Q 18.

Correct2

Incorrect-0.5

If {tex} \triangle ABC {/tex} is a right angled triangle with {tex} \angle A = 90 ^{\circ} {/tex}, AN is perpendicular to BC, BC =12cm and AC = 6cm, then the ratio of {tex} \frac{area\ of \ \triangle ANC}{area\ of\ \triangle ABC} {/tex} is

A

1 : 2

B

1 : 3

1 : 4

D

1 : 8

Explanation

{tex} \triangle AMC\ and\ \triangle BAC {/tex} are similar.

{tex} \frac{Area\ of\ \triangle AMC}{Area\ of\ \triangle BAC} = \left(\frac{AC}{BC}\right)^{2} = \left(\frac{6}{12}\right)^{2} = \frac{1}{4} = 1 : 4 {/tex}

Q 19.

Correct2

Incorrect-0.5

In A PQR, the medians QM and RN intersect at O. PO meets QR in L. If OL is 2.5 cm, then PL is equal to

A

5 cm

B

10 cm

C

2.5 cm

7.5 cm

Explanation



Always median of triangle meets at a point called centroid and centroid divides a line segment in the ratio 2 : 1

i.e., here '0' is the centroid

{tex} \therefore PO = OL = 2 : 1 {/tex}

Now, OL = 2.5 cm

{tex} \therefore OP = 2 \times 2.5 = 5 cm {/tex}

{tex} \therefore PL = OP + OL {/tex}

= {tex} \left(5 + 2.5\right) cm = 7.5 cm {/tex}

Q 20.

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If side of an equilateral triangle is {tex} 20 \sqrt{3} cm {/tex}, then numerical value of the radius of the circle circumscribing the triangle is

20 cm

B

{tex} 20 \sqrt{3} cm {/tex}

C

{tex} 20 \pi cm {/tex}

D

{tex} \frac{20}{ \pi } cm {/tex}

Explanation



Here as the triangle is an equilateral triangle, so here the altitude drawn from each vertex 0 triangle meet at a common point and it can be called centroid.

{tex} AB=BC=AC=20\sqrt{3} cm {/tex}

{tex} \therefore BD=DC=EC=EA=AF=BF {/tex}

= {tex} \frac{20\sqrt{3}}{2} = 10\sqrt{3} {/tex}

{tex} Now\ in\ \triangle ADB {/tex}

{tex} \left(AD\right)^{2}= \left(AB\right)^{2}- \left(BD\right)^{2} {/tex}

={tex} \left(20\sqrt{3}\right)^{2}- \left(10\sqrt{3}\right)^{2} {/tex}

={tex} 1200 - 300 = 900 {/tex}

{tex} \therefore AD = 30 cm {/tex}

Now AO : OD = 2 : 1

{tex} \therefore 2x+1x=30 {/tex}

{tex} \therefore AO = r = 2x = 20 cm {/tex}