Students often turn to OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Chapter Test to clarify doubts and improve problem-solving skills.

## S Chand Class 9 ICSE Maths Solutions Chapter 8 Triangles Chapter Test

Question 1.

(i) Which of the following is not a criterion for congruency of triangles?

(a) SAS

(b) ASA

(c) SSA

(d) SSS

(ii) Which congruence statement correctly indicates that the two given triangles are congruent?

(a) ∆ABC ≅ ∆EFD

(b) ∆ABC ≅ ∆DEF

(c) ∆ABC ≅ ∆FDE

(d) ∆ABC ≅ ∆FED

Solution:

(i) Among the given criterion

SSA is not a criterion

Others are criterion

(ii) Fom the given figure,

AC = FE

AB = FD

BC = DE

∴ ∆ABC ≅ ∆FDE

Question 2.

(i) Show that ∆RST ≅ ∆TUR when x = 18.

(ii) Show that ∆ABC ≅ ∆DBC when x = 4.

Solution:

(i) When x = 18, then

4x – 11 = 4 x 18 – 11 = 72 – 11

= 61 ⇒ RS = UT = 61

and 2x = 2 x 18 = 36°

⇒ ∠SRT = ∠RTU = 36°

RT = RT (common)

Hence ∆RST ≅ ∆TUR (SAS criterion)

(ii) In the figure,

ABDC is a kite in which AB = DB

⇒ 3x – 9 = 3 ⇒ 3x = 3 + 9 = 12

⇒ x = \(\frac { 12 }{ 3 }\) = 4

AC = DC

⇒ 5 = 2 x 4 – 3 ⇒ 5 = 8 – 3 = 5

BC = BC (common)

∴ ∆ABC ≅ ∆DBC (SSS criterion)

Question 3.

D is a point on the side BC of a ∆ABC such that AD bisects ∠BAC. Then,

(a) BD = CD

(b) BA>BD

(c) BD > BA

(d) CD > CA

Solution:

In ∆ABC,

D is apoint on BC and AD bisects ∠BAC

i.e., ∠BAD = ∠CAD

In ∆ADC, then

Ext. ∠ADB > ∠CAD

⇒ ∠ADB > ∠BAD (∵ ∠CAD = ∠BAD)

∴ In ∆ABD

BA > BD

Question 4.

Two sides of a triangle are of lengths 6 cm and 2.6 cm. The lengths of the third side of the triangle cannot be

(a) 4.7 cm

(b) 5 cm

(c) 4.9 cm

(d) 3.2 cm

Solution:

Two sides of ∆ABC,

Let BC = 6 cm and AB = 2.6 cm

Now BC – AB = 6 – 2.6 = 3.4 cm

∴ Third side must be greater than 3.4 cm

∵ Sum of any two sides is greater than third side

∴ 3.2 cm can be the third side.

Question 5.

In ∆ABC, ∠B = 30°, ∠C = 80° and ∠A = 70°, then

(a) AB > BC < AC

(b) AB < BC > AC

(c) AB > BC > AC

(d) AB < BC < AC

Solution:

In ∆ABC, ∠B = 30°, ∠C = 80°

Then ∠A = 180° – (30° + 80°) = 180° – 110° = 70°

Then AB > BC > AC

Question 6.

Two sides of a triangle are of lengths 4 cm, and 10 cm. If the lengths of the third side is ‘a’ cm, then

(a) a > 5

(b) 6 ≤ a ≤ 12

(c) a < 6

(d) 6 < a < 14

Solution:

In ∆ABC, two sides are 4 cm and 10 cm

Let AB = 4 cm and BC = 10 cm ‘a’ is third side

∵ Sum of any two sides > third side

10 < 4 + a ⇒ 10 – 4 < a ⇒ 6 < a

and a < 4 + 10 ⇒ a < 14

∴ 6 < a < 14