Free Triangles 03 Practice Test - 9th Grade
Question 1
If △NVY≅△AJU and both the triangles are scalene, then which of the following is true?
NV = JU
VY = AJ
∠VNY=∠AUJ
∠VYN=∠JUA
SOLUTION
Solution : D
Corresponding sides and angles of congruent triangles must be equal.
So, if △NVY≅△AJU, then
NV = AJ
VY = JU
NY = AU
∠NVY=∠AJU
∠VYN=∠JUA
∠VNY=∠JAU
Question 2
In a △ABC, ∠ABC = 30∘ and ∠ACB = 45∘. Which is the longest side in this triangle?
SOLUTION
Solution : B
Sum of all angles in a triangle is 180∘.
∠BAC + ∠CAB + ∠CBA = 180∘
∠BAC + 30∘ +45∘ = 180∘
∠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.
Question 3
In the figure below, △ABC ≅ △BED, ∠DEB = 115∘ and ∠CAB = 25∘. It can be concluded that BC||ED.
True
False
SOLUTION
Solution : A
Since ΔABC≅ΔBED,∠DBE=∠CAB=25∘ (CPCT)And, ∠DEB=∠CBA (CPCT)Now, ∠BDE+∠DBE+∠BED=180∘⇒∠BDE+25∘+115∘=180∘⇒∠BDE=40∘Also, ∠CBA+∠CBD+∠DBE=180∘(Linear set of angles)⇒∠CBA+25∘+115∘=180∘⇒∠CBA=40∘⇒∠BDE=∠CBD and they are alternate angles, so BC||ED.
Question 4
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)
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
⇒ AB + BC > AD + CD
⇒ AB + BC > AB + CD (Since AB = AD)
⇒ BC > CD
Question 7
In the given figure, ΔABC and ΔOBC are both isosceles.
It can be concluded that:
ΔAOB≅ΔAOC
AB = AO
∠ABO=∠ACO
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 common∴ΔAOB≅ΔAOC(by SSS congruence)
And ∠ABO=∠ACO ( By CPCT)
Question 8
In parallelogram ABCD,AP and CQ are perpendicular to diagonal BD.Statement 1:ΔAPB≅ΔCQDStatement 2: SSS congruence is applied
Pick the correct option.
SOLUTION
Solution : C
In ΔAPB and ΔCQD,AB=CD (opposite sides of a parallelogram).∠ABP=∠CDQ (alternate angles).∠APB=∠CQD (both are right angles).∴ΔAPB≅ΔCQD
(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
3212∘
1612∘
16∘
32∘
SOLUTION
Solution : C
In given figure,
∠BAC+∠ABC+∠ACB=180∘(Angles of same triangle)⇒∠BAC+65∘+33∘=180∘⇒∠BAC=82∘
∠CAN=∠BAC2=41∘(Given that AN is an angle bisector).Now, ∠CAN+∠ANC+∠ACN=180∘(Angles of same triangle)⇒41∘+∠ANC+33∘=180∘⇒∠ANC=106∘
∠ANC+∠ANM=180∘(Linear pair)⇒106∘+∠ANM=180∘⇒∠ANM=74∘
∠ANM+∠AMN+∠MAN=180∘⇒74∘+90∘+∠MAN=180∘⇒∠MAN=16∘.
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∘.
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.