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Mathematics Question Bank SA - 1 - CBSE (CCE)
7
TRIANGLES
EXERCISE 7.1
Topics :
Congruence of Triangles.
Criteria for Congruence of Triangles.
Some More Criteria for Congruence of Triangles.
2 MARKS
1.
In two triangles ABC and PQR, it is given that:
AB = 4 cm, BC = 7 cm, B = 30o, R = 30o, QR = 4 cm and RP = 7cm. Are these triangles
congruent? State the used congruence rule and write the triangles in the symbolic
form. (2014)
Ans.Yes, ∆ABC ≅ ∆QRP (SAS)
C
P
7 cm
A
30o
4 cm
7 cm
B
Q
30o
4 cm
R
Two triangles are congruent if two sides and the included angle of one triangle is
equal to the two sides and the included angle of the other.
2.
In the figure, l||m. If ∠ABC = ∠ABD = 40o and ∠A = 90o, then prove that ∆BCD is
isosceles. (2014)
m
l
C
A
40o
40o
B
D
Ans. In ∆ABC and ∆ABD, ∠ABC= ∠ABD = 40o(given)
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AB=AB
(common)
∠BAC=∠BAD = 90o
∴ ∆ABC ≅ ∆ABD
(ASA)
TEACHERS FORUM
Triangles
Mathematics Question
Bank SA - 1 - CBSE (CCE)
⇒ BC = BD
(CPCT)
⇒ ∆BCD is isosceles.
3.
ABC is a right angled triangle in which ∠B = 90o and AB = BC. Find ∠A and ∠C.(2014)
Ans.
since AB
= BC
∠A=∠C = xo (Angles opposite to equal sides are equal)
x + x + 90o =180o A
2x =90o
x =45o
⇒ ∠A=∠C = 45
4.
x
x
B
C
o
Find the area of the right angled triangle in which sides other than hypotenuse are 18
cm and 80 cm. Also, find the perimeter of the triangle.
(2013, 2014)
1
1
bh =
x 80 x 18 = 720 cm2
2
2
Now AC = √182 + 802 = √324 + 6400
Ans.
A
Ar(∆ABC)=
18 cm
B
=√6724 = 82 cm
80 cm
C
∴ Perimeter (∆ABC) = 18 + 80 + 82 = 180 cm
5.
For two triangles to be congruent, write any two congruence rules.
(2014)
Ans.1.
If three sides of one triangle are equal to three sides of the other triangle, then
the two triangles are congruent (SSS congruence rule).
2.
If two sides and the inculded angle of one triangle are equal to two sides and the
inculded angle of the other triangle, then the two triangles are congruent (SAS
congruence rule).
6.
In the figure, OA = OB and OD = OC. Show that
(i) ∆AOD ≅ ∆BOC (ii) AD||BC C
(2011, 2013)
B
O
A
D
Ans. (i) In ∆AOD and ∆BOC,AO = BO
(given)
∠AOD = ∠BOC(vertically opposite angle)
OD = OC
(given)
∴∆AOD ≅ ∆BOC (SAS)
(ii) Now
But these angles form a pair of alternate angles and hence AD||BC.
-219-
∠OAD = ∠OBC(CPCT)
TEACHERS FORUM
Mathematics Question Bank SA - 1 - CBSE (CCE)
7.
In the given figure ∆ABC and ∆DBC are two isosceles triangles on the same base BC.
Show that ∠ABD = ∠ACD. (2012)
Ans.In ∆ABC,
AB=AC
(Given)
∴∠ABC=∠ACB →(i)
In ∆DBC, BD = CD
(Given)
∴∠DBC=∠DCB →(ii)
(i) + (ii)⇒∠ABC + ∠DBC= ∠ACB + ∠DCB
⇒∠ABD = ∠ACD
3 MARKS
8.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule
prove that the triangle ABC is isosceles.
(2014)
Ans. In ∆AFC and ∆AEB, ∠A is common.
∠AFC=∠AEB
FC=EB
∴ ∆AFC ≅ ∆AEB
⇒ AC = AB
(given)
A
(given)
F
(AAS)
(CPCT)
B
E
C
∴ ∆ABC is isosceles.
9.
In the given figure, it is given that AB = CF, EF = BD and ∠AFE = ∠CBD. Prove that
∆AFE ≅ ∆CBD. (2014)
D
A
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F
C
E
Ans. B
Given, AB = CF
i.e., AB + BF = CF + BF
i.e., AF = CB
In ∆AFE and ∆CBD,
AF = CB
→(1)
[From (1)]
TEACHERS FORUM