# Free Triangles 03 Practice Test - 9th Grade

If  NVYAJU and both the triangles are scalene, then which of the following is true?

A.

NV = JU

B.

VY = AJ

C.

VNY=AUJ

D.

VYN=JUA

#### SOLUTION

Solution : D

Corresponding sides and angles of congruent triangles must be equal. So, if NVYAJU, then

NV = AJ
VY = JU
NY = AU
NVY=AJU
VYN=JUA
VNY=JAU

In a ABC, ABC = 30 and ACB = 45. Which is the longest side in this triangle? A. AB
B. BC
C. CA
D. All are same in length

#### SOLUTION

Solution : B

Sum of all angles in a triangle is 180.
BAC + CAB + CBA = 180
BAC + 30 +45180
BAC = 180(30+45)=105.

The side opposite to the largest angle will be the longest.
Side opposite to BAC = BC.
Hence, BC is the longest side.

In the figure below, ABC  BED, DEB = 115 and CAB = 25. It can be concluded that BC||ED. A.

True

B.

False

#### SOLUTION

Solution : A

Since ΔABCΔBED,DBE=CAB=25 (CPCT)And, DEB=CBA (CPCT)Now, BDE+DBE+BED=180BDE+25+115=180BDE=40Also, CBA+CBD+DBE=180(Linear set of angles)CBA+25+115=180CBA=40BDE=CBD and they are alternate angles, so BC||ED.

In the figure, if AB = CD and ABO = 35. What is the value of DCO in degrees? ___

#### SOLUTION

Solution : In AOB and  DOC,

OB = OC (Radii of the same circle)

OA = OD (Radii of the same circle)

AB = CD (Given)

AOB   DOC (By SSS congruence condition)

So, ABO = DCO = 35(Corresponding parts of congruent triangles)

The diagonal of a rectangle divides it into 2 congruent triangles.

A.

True

B.

False

#### SOLUTION

Solution : A In ABC and  CDA, we have:

AB = CD (Opposite sides of a parallelogram)

BC = AD (Opposite sides of a parallelogram)

AC = AC (Common)

By SSS congruence condition, ABC  CDA. So, the diagonal of the parallelogram divides it into two congruent triangles.

In the figure below, if AC>AB and D is point on AC such that AB = AD, then the relation between BC and CD is BC>CD. A.

True

B.

False

#### SOLUTION

Solution : A In ABC,

AB+BC > AC

AB + BC >  AD + CD

AB + BC >  AB + CD     (Since AB = AD)

BC >  CD

In the given figure, ΔABC and ΔOBC are both isosceles.
It can be concluded that: A.

ΔAOBΔAOC

B.

AB = AO

C.

ABO=ACO

D.

ΔAOB is not congruent toΔAOC

#### SOLUTION

Solution : A and C

In ΔAOB and ΔAOC,AB=AC(Sides of isosceles ΔABC).OB=OC(Sides of isosceles ΔOBC).AO is commonΔAOBΔAOC(by SSS congruence)
And ABO=ACO ( By CPCT)

In parallelogram ABCD,AP and CQ are perpendicular to diagonal BD.Statement 1:ΔAPBΔCQDStatement 2: SSS congruence is applied Pick the correct option.

A. Both statements are correct
B. Both statements are false
C. Statement 1 is correct but 2 is false
D. Statement 1 is false but 2 is correct

#### SOLUTION

Solution : C

In ΔAPB and ΔCQD,AB=CD (opposite sides of a parallelogram).ABP=CDQ (alternate angles).APB=CQD (both are right angles).ΔAPBΔCQD
(using AAS criterion)
Hence, statement 1 is correct but statement 2 is false

In the given figure, AM  BC and AN is the bisector of A. Then MAN is A.

3212

B.

1612

C.

16

D.

32

#### SOLUTION

Solution : C

In given figure,
BAC+ABC+ACB=180(Angles of same triangle)BAC+65+33=180BAC=82

CAN=BAC2=41(Given that AN is an angle bisector).Now, CAN+ANC+ACN=180(Angles of same triangle)41+ANC+33=180ANC=106

ANC+ANM=180(Linear pair)106+ANM=180ANM=74

ANM+AMN+MAN=18074+90+MAN=180MAN=16.

O is any point on the bisector of the acute angle XYZ. From O, a line is extended to join XY such that OP is parallel to ZY. Then, YPO is: A.

Scalene

B.

Isosceles but not right angled

C.

Equilateral

D.

Right & isosceles

#### SOLUTION

Solution : B

POY = OYZ (alternate angles)

Y is bisected, so  POY =  PYO
Hence, PY = PO
So, YPO is isosceles

Also, it is given that XYZ is acute, so any angle which is half of it (bisected by OY) is less than 45.
or  PYO + OYZ < 90

Hence, the third angle of the YPO i.e.  YPO will be obtuse to satisfy angle-sum property of a triangle.

Hence, YPO is isosceles but not right angled.