# Free Triangles 01 Practice Test - 9th Grade

### Question 1

Given that ACBD is a kite. By which congruency property are the triangles ACB and ADB congruent?

SAS property

SSS property

RHS property

ASA property

#### SOLUTION

Solution :B

In the △ABC and △ABD,

AC = AD (Given)

CB = DB (Given)

AB is the common side.

⟹△ABC≅△ABD [SSS congruency]

Thus, by SSS congruency rule, the two triangles (ABC and ABD) are congruent.

### Question 2

If three angles of a triangle are equal to three angles of another triangle respectively, then the two triangles are congruent.

True

False

#### SOLUTION

Solution :B

The given statement is false. Even when two triangles have all angles same, they can still have sides of different lengths. However, the ratio of lengths of corresponding sides will be same. But in this case, they will be called 'similar' triangles, not congruent.

Congruent triangles are a special case of similar triangles.

For example, in the image below, both triangles are equilateral and have all angles equal but they are not congruent.

### Question 3

In △ABC, if ∠B=∠C=45∘, then which of the following is/are the longest side(s)?

#### SOLUTION

Solution :C

∠A+∠B+∠C=180∘ (angle sum property of triangle)

⇒∠A+45∘+45∘=180∘⇒∠A=90∘

The side opposite to largest angle will be the longest side. Hence BC is the largest side.

### Question 4

In ΔABD, AB = AD and AC is perpendicular to BD. State the congruence rule by which ΔACB≅ΔACD.

SAS congruence rule

SSS congruence rule

RHS congruence rule

ASA congruence rule

#### SOLUTION

Solution :C

In the given triangle,AB=AD(Given)AC is the common side∠ACB=∠ACD=90∘

∴ΔACB≅ΔACD [ RHS criteria]

Hence, by RHS congruence rule, the two triangles are congruent.

### Question 5

If in the given figure, ABCD is a rectangle and DCE is an equilateral triangle, then which of the following statements is/are correct?

#### SOLUTION

Solution :A and C

∠EDC = ∠ECD (Angles of an equilateral triangle)

And ∠ADC = ∠BCD (Angles of a rectangle)

∴∠EDC+∠ADC=∠ECD+∠BCD

⇒∠EDA=∠ECB

In ΔDAE and ΔCBE,

DE=CE (Sides of an equilateral triangle)

DA=BC (Sides of a rectangle)

∠EDA = ∠ECB (proved)

Hence, ΔDAE≅ΔCBE (SAS congruence)

Also, EA = EB (C.P.C.T.C.)

⇒ΔEAB is an isosceles triangle.

It is given that, ABCD is a rectangle.

⇒DA≠DC

Also, DEC is an equilateral triangle.

⇒DC=DE

∴DA≠DE

Hence, DAE is not an isosceles triangle.

Also, it can not be proved that ∠DAE=15∘.

### Question 6

In the figure, AB = AC and AP ⊥ BC. Then

AB = AP

AB < AP

AB > AP

AB < BP

#### SOLUTION

Solution :C

AP⊥BC

⟹△APB is a right-angled triangle.

⟹ AB is the hypotenuse and hence the longest side, which makes is greater than AP.

∴AB>AP

### Question 7

In the given figure, AC = CE and AB ∥ ED. The value of x is

#### SOLUTION

Solution :In ΔABC and ΔEDC,AC=CE (given)∠BAC=∠DEC (since AB||DE and AE is a transversal, so they are alternate angles)∠ACB=∠ECD (vertically opposite angles)∴ΔABC≅ΔEDC (A.S.A. congruence criteria)∴AB=DE (sides of congruent triangles)∴x+10=2x−5⇒x=15 units

### Question 8

If ΔABC≅ΔPQR and it is given that ∠A=60∘, ∠B=70∘. Then the value of ∠R is

#### SOLUTION

Solution :If ΔABC≅ΔPQR,∠A=∠P=60∘∠B=∠Q=70∘Now, ∠P+∠Q+∠R=180∘(angles of a triangle)⇒60∘+70∘+∠R=180∘⇒∠R=50∘

### Question 9

Which of the following options is/are appropriate criteria for congruence?

#### SOLUTION

Solution :A, C, and D

Angle-Angle-Angle (A.A.A.) is not a criterion for congruence.

Knowing the measures of all the angles of two triangles is not sufficient to determine if the two triangles are congruent. Because two triangles having the same set of angles can be of different sizes, as shown below.

The remaining options - Side-Side-Side (SSS), Angle-Side-Angle (ASA) and Side-Angle-Side (SAS) are criteria to determine if the given triangles are congruent.

### Question 10

In the given figure it is given that AB = CF, EF = BD and ∠AFE = ∠DBC. Then by which criterion △AFE≅△CBD ?

SAS

SSS

ASA

None of these

#### SOLUTION

Solution :A

In the given figure,

AB = CFAB + BF = CF + BF

[Adding BF on both the sides]AF = CB

In △AFE and △CBD

AF = CB [Proved above]

EF = BD [Given]∠AFE=∠DBC [Given]

∴△AFE≅△CBD [SAS criterion]