Triangles CBSE Questions & Answers
Triangles
This is Mathematics Class 09 Triangles CBSE Questions & Answers. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
In the adjoining figure, the rule by which \(\triangle ABC \cong \triangle ADC\)

- ARHS
- BAAS
- CSSSCorrect
- DSAS
2
In the adjoining figure, BC = AD, CA\( \bot \)AB and BD\( \bot \)AB. The rule by which \(\triangle ABC \cong \triangle BAD\)is

- ARHSCorrect
- BSSS
- CSAS
- DASA
3
In the adjoining figure, AB = AC and AD\( \bot \)BC. The rule by which \(\triangle ABD \cong \triangle ACD\)is

- AASA
- BRHSCorrect
- CSSS
- DSAS
4
In the adjoining figure, AB = AC and AD is bisector of \(\angle \)A. The rule by which \(\triangle ABD \cong \triangle ACD\)

- AASA
- BAAS
- CSSS
- DSASCorrect
5
In the adjoining figure, \(\angle \)B = \(\angle \)C and AD\( \bot \)BC. The rule by which \(\triangle ABD \cong \triangle ADC\)

- AASA
- BRHS
- CSAS
- DAASCorrect
6
In the adjoining figure, AB = FC, EF = BD and \(\angle \)AFE = \(\angle \)CBD. Then the rule by which \(\triangle AFE \cong \triangle CBD\)

- AASA
- BSSS
- CSASCorrect
- DAAS
7
In the adjoining figure, AB = AC and AD is median of \(\triangle ABC\), then \(\angle \)ADC is equal to

- A\(90^\circ \)Correct
- B\(75^\circ \)
- C\(120^\circ \)
- D\(60^\circ \)
8
In the adjoining figure, \(\triangle ABC \cong \triangle ADC\). If \(\angle \)BAC = \(30^\circ \)and \(\angle \)ABC = \(100^\circ \)then \(\angle \)ACD is equal to

- A\(50^\circ \)Correct
- B\(80^\circ \)
- C\(60^\circ \)
- D\(30^\circ \)
9
In the adjoining figure, O is Mid – point of AB. If \(\angle \)ACO = \(\angle \)BDO, then \(\angle \)OAC is equal to

- A\(\angle \)BOD
- B\(\angle \)OCA
- C\(\angle \)OBDCorrect
- D\(\angle \)ODB
10
In the adjoining figure, AB = BC and \(\angle \)ABD = \(\angle \)CBD, then another angle which measures \(30^\circ \) is

- A\(\angle \)BAD
- B\(\angle \)BDACorrect
- C\(\angle \)BCD
- D\(\angle \)BCA
11
In the adjoining figure, AC = BD. If \(\angle \)CAB = \(\angle \)DBA, then \(\angle \)ACB is equal to

- A\(\angle \)BAD
- B\(\angle \)ABC
- C\(\angle \)BDACorrect
- D\(\angle \)ABD
12
In the adjoining figure, ABCD is a quadrilateral in which AD = CB and AB = CD, then \(\angle \)ACB is equal to

- A\(\angle \)ACD
- B\(\angle \)BAC
- C\(\angle \)CADCorrect
- D\(\angle \)BAD
13
In the adjoining figure, AB = AC and \(\angle \)A = \(70^\circ \), then \(\angle \)C is

- A\(70^\circ \)
- B\(110^\circ \)
- C\(55^\circ \)Correct
- D\(40^\circ \)
14
In the adjoining figure, PQ = PR. If \(\angle \)Q = \(70^\circ \), then measure of \(\angle \)P is

- A\(80^\circ \)
- B\(40^\circ \)Correct
- C\(110^\circ \)
- D\(70^\circ \)
15
In the adjoining figure, AB = AC. If \(\angle \)ACD = \(115^\circ \), then the measure of \(\angle \)A is

- A\(65^\circ \)
- B\(57.5^\circ \)
- C\(50^\circ \)Correct
- D\(70^\circ \)