Free Triangles 03 Practice Test - 9th Grade

Question 1

If $△ N V Y ≅ △ A J U$ and both the triangles are scalene, then which of the following is true?

NV = JU

VY = AJ

$∠ V N Y = ∠ A U J$

$∠ V Y N = ∠ J U A$

SOLUTION

Solution : D

Corresponding sides and angles of congruent triangles must be equal.

So, if $△ N V Y ≅ △ A J U$ , then

NV = AJ
VY = JU
NY = AU
$∠ N V Y = ∠ A J U$
$∠ V Y N = ∠ J U A$
$∠ V N Y = ∠ J A U$

Question 2

In a $△$ ABC, $∠$ ABC = $30^{\circ}$ and $∠$ ACB = $45^{\circ}$ . 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 $180^{\circ}$ .
$∠$ BAC + $∠$ CAB + $∠$ CBA = $180^{\circ}$
$∠$ BAC + $30^{\circ}$ + $45^{\circ}$ = $180^{\circ}$
$∠$ BAC = $180^{\circ} - (30^{\circ} + 45^{\circ}) = 105^{\circ}$ .

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

Question 3

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ . It can be concluded that BC||ED.

True

False

SOLUTION

Solution : A

Since $Δ A B C ≅ Δ B E D, ∠ D B E = ∠ C A B = 25^{\circ} (CPCT) And, ∠ D E B = ∠ C B A (CPCT) Now, ∠ B D E + ∠ D B E + ∠ B E D = 180^{\circ} \Rightarrow ∠ B D E + 25^{\circ} + 115^{\circ} = 180^{\circ} \Rightarrow ∠ B D E = 40^{\circ} Also, ∠ C B A + ∠ C B D + ∠ D B E = 180^{\circ} (Linear set of angles) \Rightarrow ∠ C B A + 25^{\circ} + 115^{\circ} = 180^{\circ} \Rightarrow ∠ C B A = 40^{\circ} \Rightarrow ∠ B D E = ∠ C B D and they are alternate angles, so B C | | E D .$

Question 4

In the figure, if AB = CD and $∠$ ABO = $35^{\circ}$ . 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 = $35^{\circ}$ (Corresponding parts of congruent triangles)

Question 5

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

True

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.

Question 6

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.

True

False

SOLUTION

Solution : A

In $△$ ABC,

AB+BC $>$ AC

$\Rightarrow$ AB + BC $>$ AD + CD

$\Rightarrow$ AB + BC $>$ AB + CD (Since AB = AD)

$\Rightarrow$ BC $>$ CD

Question 7

$In the given figure, Δ A B C and Δ O B C are both isosceles.$
$It can be concluded that:$

$Δ A O B ≅ Δ A O C$

AB = AO

$∠ A B O = ∠ A C O$

$Δ A O B is not congruent to Δ A O C$

SOLUTION

Solution : A and C

$In Δ A O B and Δ A O C, A B = A C (Sides of isosceles Δ A B C) . O B = O C (Sides of isosceles Δ O B C) . A O i s c o m m o n ∴ Δ A O B ≅ Δ A O C (by SSS congruence)$
And $∠ A B O = ∠ A C O$ ( By CPCT)

Question 8

$In parallelogram A B C D, A P and C Q are perpendicular to diagonal B D . Statement 1: Δ A P B ≅ Δ C Q D Statement 2: SSS congruence is applied$

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

$In Δ A P B and Δ C Q D, A B = C D (opposite sides of a parallelogram) . ∠ A B P = ∠ C D Q (alternate angles) . ∠ A P B = ∠ C Q D (both are right angles) . ∴ Δ A P B ≅ Δ C Q D$
(using AAS criterion)
Hence, statement 1 is correct but statement 2 is false

Question 9

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

$32 {\frac{1}{2}}^{\circ}$

$16 {\frac{1}{2}}^{\circ}$

$16^{\circ}$

$32^{\circ}$

SOLUTION

Solution : C

In given figure,
$∠ B A C + ∠ A B C + ∠ A C B = 180^{\circ} (Angles of same triangle) \Rightarrow ∠ B A C + 65^{\circ} + 33^{\circ} = 180^{\circ} \Rightarrow ∠ B A C = 82^{\circ}$

$∠ C A N = \frac{∠ B A C}{2} = 41^{\circ} (Given that AN is an angle bisector) . Now, ∠ C A N + ∠ A N C + ∠ A C N = 180^{\circ} (Angles of same triangle) \Rightarrow 41^{\circ} + ∠ A N C + 33^{\circ} = 180^{\circ} \Rightarrow ∠ A N C = 106^{\circ}$

$∠ A N C + ∠ A N M = 180^{\circ} (Linear pair) \Rightarrow 106^{\circ} + ∠ A N M = 180^{\circ} \Rightarrow ∠ A N M = 74^{\circ}$

$∠ A N M + ∠ A M N + ∠ M A N = 180^{\circ} \Rightarrow 74^{\circ} + 90^{\circ} + ∠ M A N = 180^{\circ} \Rightarrow ∠ M A N = 16^{\circ} .$

Question 10

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:

Scalene

Isosceles but not right angled

Equilateral

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 $45^{\circ}$ .
or $∠$ PYO + $∠$ OYZ < $90^{\circ}$

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.

Free Triangles 03 Practice Test - 9th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Related ExamsSurface Areas and Volumes 02Introduction to Euclid's Geometry 02Probability 03

Related Exams
Surface Areas and Volumes 02
Introduction to Euclid's Geometry 02
Probability 03