9th Grade > Mathematics
TRIANGLES MCQs
Total Questions : 56
| Page 6 of 6 pages
Answer: Option A. ->
ΔDAE ≅ ΔCBE
:
A and C
∠EDC = ∠ECD (Angles of an equilateral triangle)
And ∠ADC = ∠BCD (Angles of a rectangle)
∴∠EDC+∠ADC=∠ECD+∠BCDÂ
⇒∠EDA=∠ECB
In ΔDAE and ΔCBE,
DE=CE (Sides of an equilateral triangle)
DA=BC (Sides of a rectangle)
∠EDA = ∠ECB (proved)
Hence, ΔDAE≅ΔCBE (SAS congruence)
Also, EA = EBÂ (C.P.C.T.C.)
⇒ΔEAB is an isosceles triangle.
It is given that, ABCD is a rectangle.
⇒DA≠DCÂ
Also, DEC is an equilateral triangle.
⇒DC=DE
∴DA≠DE
 Hence, DAE is not an isosceles triangle.
Also, it can not be proved that ∠DAE=15∘.
:
A and C
∠EDC = ∠ECD (Angles of an equilateral triangle)
And ∠ADC = ∠BCD (Angles of a rectangle)
∴∠EDC+∠ADC=∠ECD+∠BCDÂ
⇒∠EDA=∠ECB
In ΔDAE and ΔCBE,
DE=CE (Sides of an equilateral triangle)
DA=BC (Sides of a rectangle)
∠EDA = ∠ECB (proved)
Hence, ΔDAE≅ΔCBE (SAS congruence)
Also, EA = EBÂ (C.P.C.T.C.)
⇒ΔEAB is an isosceles triangle.
It is given that, ABCD is a rectangle.
⇒DA≠DCÂ
Also, DEC is an equilateral triangle.
⇒DC=DE
∴DA≠DE
 Hence, DAE is not an isosceles triangle.
Also, it can not be proved that ∠DAE=15∘.
Answer: Option C. ->
AB > AP
:
C
:
C
AP⊥BC
⟹△APB is a right-angled triangle.
⟹ AB is the hypotenuse and hence the longest side, which makes is greater than AP.
∴AB>AP
Answer: Option C. ->
AB > AP
:
In ΔABC and ΔEDC,AC=CE (given)∠BAC=∠DEC (since AB||DE and AE is a transversal, so they are alternate angles)∠ACB=∠ECD (vertically opposite angles)∴ΔABC≅ΔEDC (A.S.A. congruence criteria)∴AB=DE (sides of congruent triangles)∴x+10=2x−5⇒x=15 units
:
In ΔABC and ΔEDC,AC=CE (given)∠BAC=∠DEC (since AB||DE and AE is a transversal, so they are alternate angles)∠ACB=∠ECD (vertically opposite angles)∴ΔABC≅ΔEDC (A.S.A. congruence criteria)∴AB=DE (sides of congruent triangles)∴x+10=2x−5⇒x=15 units
Answer: Option C. ->
A.S.A.
:
A, C, and D
Angle-Angle-Angle (A.A.A.) is not a criterion for congruence.
Knowing the measures of all the angles of two triangles is not sufficient to determine if the two triangles are congruent. Because two triangles having the same set of angles can be of different sizes, as shown below.
The remaining options - Side-Side-Side (SSS), Angle-Side-Angle (ASA) and Side-Angle-Side (SAS) are criteria to determine if the given triangles are congruent.
:
A, C, and D
Angle-Angle-Angle (A.A.A.) is not a criterion for congruence.
Knowing the measures of all the angles of two triangles is not sufficient to determine if the two triangles are congruent. Because two triangles having the same set of angles can be of different sizes, as shown below.
The remaining options - Side-Side-Side (SSS), Angle-Side-Angle (ASA) and Side-Angle-Side (SAS) are criteria to determine if the given triangles are congruent.
Answer: Option C. ->
A.S.A.
:
If ΔABC≅ΔPQR,∠A=∠P=60∘∠B=∠Q=70∘Now, ∠P+∠Q+∠R=180∘(angles of a triangle)⇒60∘+70∘+∠R=180∘⇒∠R=50∘
:
If ΔABC≅ΔPQR,∠A=∠P=60∘∠B=∠Q=70∘Now, ∠P+∠Q+∠R=180∘(angles of a triangle)⇒60∘+70∘+∠R=180∘⇒∠R=50∘
Answer: Option A. ->
SAS
:
A
:
A
In  the given figure,
AB = CF
AB + BF = CF + BFÂ
[Adding BF on both the sides]
AF = CB
In â–³AFE and â–³CBD
AF = CBÂ Â Â Â Â Â Â Â Â Â [Proved above]
EF = BDÂ Â Â Â Â Â Â Â Â Â [Given]
∠AFE=∠DBC   [Given]
∴△AFE≅△CBD [SAS criterion]
Â