In $\triangle\text{ABC, BC = AB}$ and $\angle\text{B}=80^{\circ}.$ Then, $\angle\text{A = ?}$
- ✓$50^\circ$
- B$40^\circ$
- C$100^\circ$
- D$80^\circ$
Answer: A.
View full solution →Question types
Congruence of Triangles and Inequalities in a Triangle question types
76 questions across 7 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
76
Questions
7
Question groups
5
Question types
01
MCQ(1M)
18 Q→02Fill in the Blanks.
9 Q→031 Mark Question
5 Q→042 Mark Question
5 Q→053 Mark Question
11 Q→064 Mark Question
6 Q→075 Mark Question
22 Q→Sample Questions
Congruence of Triangles and Inequalities in a Triangle questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
In $\triangle\text{ABC}$ and $\triangle\text{DEF,}$ it is given that $\angle\text{B}=\angle\text{E}$ and $\angle\text{C}=\angle\text{F}.$ In order that $\triangle\text{ABC}\cong\triangle\text{DEF},$ we must have:
- A$AB = DF$
- B$AC = DE$
- ✓$BC = EF$
- D$\angle\text{A}=\angle\text{D}$
Answer: C.
View full solution →If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is:
- AEquilateral
- ✓Isosceles
- CScalene
- DRight $-$ angled
Answer: B.
View full solution →Which of the following is not a criterion for congruence of triangles?
- ✓$\text{SSA}$
- B$\text{SAS}$
- C$\text{ASA}$
- D$\text{SSS}$
Answer: A.
View full solution →In $\triangle\text{ABC,}$ if $\angle\text{C}>\angle\text{B},$ then:
- A$BC > AC$
- ✓$AB > AC$
- C$AB < AC$
- D$BC < AC$
Answer: B.
View full solution →Fill in the blanks.
Medians of an equilateral triangle are .
Medians of an equilateral triangle are .
Fill in the blanks.
Drawing a $\triangle\text{ABC}$ with AB = 3cm, BC = 4cm and CA = 7cm is .
View full solution →Drawing a $\triangle\text{ABC}$ with AB = 3cm, BC = 4cm and CA = 7cm is .
Fill in the blanks with < or >.
(Sum of any two sides of a triangle) (twice the median to the 3rd side).
(Sum of any two sides of a triangle) (twice the median to the 3rd side).
Fill in the blanks with < or >.
(Difference of any two sides of a triangle) (the third side).
(Difference of any two sides of a triangle) (the third side).
Fill in the blanks.
Each angle of an equilateral triangle measures .
Each angle of an equilateral triangle measures .
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
3cm, 4cm, 8cm
3cm, 4cm, 8cm
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
10cm, 5cm, 6cm
10cm, 5cm, 6cm
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
5cm, 4cm, 9cm
5cm, 4cm, 9cm
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
2.5cm, 5cm, 7cm
2.5cm, 5cm, 7cm
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
8cm, 7cm, 4cm
8cm, 7cm, 4cm
In $\triangle\text{ABC, }\angle\text {A}= \angle\text{B}=45^{\circ}.$ Which is its longest side?
View full solution →“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.” Is the statement true? Why?
In $\triangle\text{ABC, }\angle\text {A}=100^{\circ}$and $\angle\text{C}=50^{\circ}.$ Which is its shortest side?
View full solution →In $\triangle\text{ABC, }\angle\text {A}=90^{\circ}.$ Which is its longest side?
View full solution →“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.” Is the statement true? Why?
In the given figure, line l is the bisector of an angle $\angle\text{A}$ and B is any point on l. If BP and BQ are perpendiculars from B to the arms of $\angle\text{A},$ Show that:
View full solution →- $\triangle\text{APB}\cong\triangle\text{AQB}$
- BP = BQ, i.e., B is equidistant from the arms of $\angle\text{A}.$
In the given figure, AD and BC are equal perpendiculars to a line segment AB.
Show that CD bisects AB.
Show that CD bisects AB.
AD is an altitude of an isosceles $\triangle\text{ABC}$ in which AB = AC. Show that:
View full solution →- AD bisects BC,
- AD bisects $\angle\text{A}.$
The bisectors of $\angle\text{B}$ and $\angle\text{C}$ of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that $\angle\text{MOC}=\angle\text{ABC}.$
View full solution →In the given figure, PQ > PR and QS and RS are the bisectors of $\angle\text{Q}$ and $\angle\text{R}$ respectively. Show that SQ > SR.
View full solution →P is a point on the bisector of $\angle\text{ABC}.$ If the line through P, parallel to BA meets BC at Q, prove that $\triangle\text{BPQ}$ is an isosceles triangle.
View full solution →The bisectors of $\angle\text{B}$ and $\angle\text{C}$ of an isosceles $\triangle\text{ABC}$ with AB = AC intersect each other at a point O.Show that the exterior angle adjacent to $\angle\text{ABC}$ is equal to $\angle\text{BOC}.$
View full solution →D is any point on the side AC of $\triangle\text{ABC}$ with AB = AC. Show that CD < BD.
View full solution →In the adjoining figure, explain how one can find the breadth of the river without crossing it.
In the given figure, $\angle\text{B}<\angle\text{A}$ and $\angle\text{C}<\angle\text{D}.$ Show that $\text{AD < BC}.$
View full solution →In the given figure, O is a point in the interior of square ABCD such that $\triangle\text{OAB}$ is an equilateral triangle. Show that $\triangle\text{OCD}$ is an isosceles triangle.
View full solution →In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).
In the adjoining figure, X and Y are respectively two point on equal sides AB and AC of $\triangle\text{ABC}$ such thet AX = AY. Prove that CX = BY.
View full solution →In the given figure, D is a point on side BC of a $\triangle\text{ABC}$ and E is a point such that CD = DE. Prove that AB + AC > BE.
View full solution →In a quadrilateral ABCD, show that
(AB + BC + CD + DA) <2 (BD + AC).
(AB + BC + CD + DA) <2 (BD + AC).
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