# Free Triangles 02 Practice Test - 9th Grade

### Question 1

In △NVY, if NV=VY and VY≠ YN, then which of the following is true?

∠NVY = ∠VYN

∠NVY = ∠YNV

∠NVY+ 2∠VYN = 180∘

∠VYN + 2∠NVY = 180∘

#### SOLUTION

Solution :C

Given, in △NVY, NV = VY.

Then, ∠VYN=∠YNV.

[angles opposite to equal sides of a triangle are equal]

Using angle sum property of a triangle, we get:

∠NVY+∠VYN+∠YNV=180∘

So, ∠NVY+2∠VYN=180∘.

It is also given that VY≠YN. Hence the options ∠NVY=∠VYN and ∠NVY=∠YNV are not correct.

### Question 2

Which of the following is sufficient for two triangles denoted by Δ1 and Δ2 to be congruent ?

Any two sides of Δ1 should be equal to any two sides of Δ2

All angles of Δ1 should be equal to all angles of Δ2

Any two sides of Δ1 and one angle should be equal to any two sides and one angle of Δ2

Any two sides of Δ1 and the included angle should be equal to any two sides and the included angle of Δ2

#### SOLUTION

Solution :D

Two triangles can't be congruent if any two sides and one angle of one are equal to any two sides and one angle of the other.

They will be congruent when the angle is included between the equal pair of sides. This is the SAS condition of congruency of triangles.

The SAS congruence rule states:

Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

### Question 3

In the given figure, O is equidistant from the sides AC and AB. Then the value of x - 3 is

#### SOLUTION

Solution :In ΔAOC and ΔAOB,OC=OB(Given)∠OBC=∠OCA(Both angles are right angles)And AO is common side.Hence, ΔAOC≅ΔAOB(RHS congruence)∴∠CAO=∠BAO(CPCT)⇒2x+24=30⇒x=3⇒x−3=0

### Question 4

In the given figure, if ∠ACB = 30∘, the value of ∠ACD is

#### SOLUTION

Solution :In the given figure, AD=AB(Given)DC=BC(Given)AC is common∴ΔADC≅ΔABC(SSS congruence)Hence, ∠ACD=∠ACB=30∘

### Question 5

If all the sides of △CAT are equal to corresponding sides of △RAT, then △RAT ≅ △CAT.

True

False

#### SOLUTION

Solution :A

If all the sides of △CAT are equal to corresponding sides of △RAT, then △RAT will be congruent to △CAT and the congruence criterion will be SSS.

### Question 6

In the given figure, ΔABD is an isosceles triangle with AB=AD. A median is drawn from A to side BD at C. Which of the following option(s) is/are correct?

AC is perpendicular to BD

∠BAC=∠DAC

ΔABC≅ΔADC

Area of ΔABD=12×Area of ΔABC

#### SOLUTION

Solution :A, B, C, and D

Consider ΔABC and ΔADC,AC=AC (Common side of both triangles)AB=AD (Sides of isosceles triangle)BC=CD (Median divides a side in two equal parts)∴ΔABC≅ΔADC (by SSS congruence criterion) So the option "ΔABC≅ΔADC" is correct.∴∠ACB=∠ACD (CPCT)But they are supplementary angles. So, ∠ACB+∠ACD=180∘⇒∠ACB=∠ACD=90∘∴AC is perpendicular to BD and the hence option "AC is perpendicular to BD" is correct. Also, ∠BAC=∠DAC (corresponding angles of congruent triangles)Hence, the option"∠BAC=∠DAC" is also correct.Also, area of ΔABC=area of ΔADC=12(area of ΔABD).ΔABC≅ΔADC⟹ ar(ΔABC)=ar(ΔADC)Hence, the option "Area of ΔABD=12×Area of ΔABC"is also correct.

### Question 7

In the given figure, ΔABC is equilateral with point O as its circumcentre. Here, area of ΔBOC=13 area of ΔABC.

#### SOLUTION

Solution :A

Concider ΔAOB,ΔAOC and ΔBOC.AB=BC=AC (Sides of equilateral triangle)AO=BO=CO (Radii of circumcircle)∴ΔAOB≅ΔBOC≅ΔCOB (SSS congruence).∴Area of ΔAOB=Area of ΔBOC=Area of ΔAOC.But Area of ΔAOB+Area of ΔBOC+Area of ΔAOC=Area ofΔABC.∴Area of ΔBOC=13Area of ΔABC

Hence, the given statement is true.

### Question 8

In the given parallelogram ABCD, ΔABD≅ΔBCD.

#### SOLUTION

Solution :B

Consider ΔABD and ΔCDB,AB=CD (Opposite sides of parallelogram)AD=CB (Opposite sides of parallelogram)∠ABD=∠CDB (Alternate angles).∴ΔABD≅ΔCDB

Hence, the given statement is false as the vertices are not given in corresponding sequence in question.

### Question 9

In the given figure, if AD = BC and AD || BC, then

#### SOLUTION

Solution :B

In △ACB and △CAD,

AD=BC [Given]

∠CAD=∠ACB [Alternate angles]

CA=AC [common side]

△ACB≅△CAD [SAS congruency]

∴AB=DC [CPCT]

### Question 10

In the given figure AD is the bisector of ∠A and AB = AC. Then, by which criterion △ACD and △ABD are congruent?

SSS

SAS

ASA

None of these

#### SOLUTION

Solution :B

In △ACD and △ABD

∠BAD=∠CAD

[∵ AD is the bisector of ∠A]

AD = AD [common side]

AB = AC [Given]

△ACD≅△ABD [SAS congruency]

The triangles are congruent by SAS congruence rule.