Solution
TRIANGLES
Class 09 - Mathematics
1.
(c) 5.3 cm, 2.2 cm, 3.1 cm
Explanation:
Put the sides of triangle a, b, c.
For a possible triangle, the following are should possible.
a+b>=c
b+c>=a
a+c>=b
Here 2.2 + 3.1 = 5.3
So a + b = c
So the triangle becomes a straight line.
So we cannot draw a triangle with sides 5.3 cm, 2.2 cm, 3.1 cm.
2.
(b) ∠P
Explanation:
Since, by corresponding part of congruent ∠E of △EFD is equal to the ∠P of △PQR.
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3.
(b) 40°
Explanation:
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Let the other interior opposite angle be x°.
Then, we have x° + 55° = 95°
⇒ x°= 95°- 55° = 40°
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4.
(d) scalene triangle
Explanation:
scalene triangle
5.
(d) AC = BC
Explanation:
In right triangles BCE and CBF,
BC = CB [Common]
BE = CF [Given]
∠ BEC = ∠ CFB [Each 90°]
∴ △BCE = △CBF [By R.H.S. congruency]
⇒ ∠ CBE = ∠ BCF [By C.P.C.T.]
and ∠ ABC = ∠ ACB
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⇒ AC = AB [Sides opposite to equal angles of a △ are equal]
Similarly, △ABD ≅ △BAE
⇒ ∠ ABC = ∠ BAC [By C.P.C.T.]
⇒ AC = BC [Sides opposite to equal angles of a △ are equal]
6.
(b) 13 cm
Explanation:
As per the rule in a triangle, the sum of any 2 sides should be greater than the third side. So, the length of the third side should
be 13, Since with 7, 10 and 13 we have
7 + 10 > 13, 7 + 13 > 10 and 13 + 10 > 7
7.
(d) A is false but R is true.
Explanation:
A is false but R is true.
8.
(d) A is false but R is true.
Explanation:
A is false but R is true.
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9.
(c) A is true but R is false.
Explanation:
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In ΔABC and ΔPQR
AB = PQ
AC = PR
∠ BAC = ∠ QPR
ΔABC ≅ ΔPQR (By SAS Rule)
10. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Since AB = AC,
then ∠ B = ∠ C
11. In △AC E , we have :
∠A + ∠C + ∠E = 180
∘
....(i)
In △BDF , we have:
∠B + ∠D + ∠F = 180
∘
....(ii) [Sum of the angles of a triangle]
Adding (i) and (ii), we get:
+ ∠B + ∠C + ∠D + ∠E + ∠F = (180 + 180)o
∠A
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.
12. We are Given that ABCD is a quadrilateral in which AB || DC such that
(i) AB = CQ
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(ii) DQ = DC + AB
Proof: in △ABP and ΔPCQ , we have
∠PAB = ∠PQC
∠APB = ∠CPQ
BP = PC
Thus by Angle Angel side criterion of congruence, we have
ΔABP ≅ΔPCQ
The corresponding parts of the congruent triangle are equal, we have
AB = C Q
Now, DQ = DC + CQ
= DC + AB
13. We know that in a triangle, the sum of any two sides is always greater than the third side
In △ABE, we have,
AB + AE > BE ⇒ AB + AE > BD + DE
In ΔC DE , We have,
DE + EC > DC
AB + AE + DE + EC > BD + DE + DC
⇒ AB + (AE + EC ) > BD + DC
⇒ (AB + AC ) > (BD + DC )
Hence, (BD + DC ) < (AB + AC )
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14.
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Since AB || DE and AE is the transversal,
∴ ∠ AED = ∠ BAE (alternate interior angles)
⇒ ∠ AED = 35° [∠ BAE = ∠ BAC = 35°]
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⇒ ∠ CED = 35°
In △DCE, we have
∠ CDE + ∠ DCE + ∠ CED = 180° [sum of interior angles of a triangle is 180]
⇒ 53° + ∠ DCE + 35° = 180°
⇒ ∠ DCE = 180° - 88o = 92o
15.
To prove BD = DC
Proof in right-angled ADB and ADC we have
hyp AB = hyp. AC Given
and AD = AD
∴ ΔADB ≅△ADC
Hence the result, BD = DC
16. In △ABC, we have
AB = AC [Given]
⇒ ∠B = ∠C [∵ Angles opp. to equal sides are equal] ...(i)
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Now, in BCE and △BCD, we have
△
∠B = ∠C [From (i)]
∠C EB = ∠BDC [Each equal to 90°]
and BC = BC [Common side]
So, by ASA(Angle Side Angle) criterion of congruence, we obtain
ΔBC E ≅ΔBC D
⇒ BD = CE [∵ Corresponding parts of congruent triangles are equal]
Hence, BD = CE
17. Given:l is a line and P is point not lying on l
N is any point on other than M.
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To prove: In △PMN, ∠ M is the right angle.
∴ N is an acute angle. (Angle sum property )
∴ ∠ M >∠ N
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∴ PN> PM [Side opposite greater angle]
so PM is shortest.
18. Let ∠A + ∠B = 125 and ∠A + ∠C = 113
∘ ∘
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Then,
∘
∠A + ∠B + ∠A + ∠C = (125 + 113)
∘
⇒ (∠A + ∠B + ∠C ) + ∠A = 238
∘ ∘
⇒ 180 + ∠A = 238
∘
⇒ ∠A = 58
∘
∴ ∠B = 125 − ∠A
(125 - 58)o = 67o
∘
∴ ∠C = 113 − ∠A
(113 - 58)o =
55o
19. We need to find the measure of ∠ A
So here, using the corollary, if the bisectors of ∠ ABC and ∠ ACB of a △ABC meet at a point O, then ∠ BOC = 90
∘
+
1
2
∠A
Thus, in △ABC
∠ BOC = 90 +
∘ 1
∠A
2
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∘ ∘ 1
120 = 90 + ∠A
2
∘ ∘ 1
120 − 90 + ∠A
2
∠ A= 2(30o).
∠ A = 60o
20. i. △ADE and △CFE
DE = EF (By construction)
∠ AED = ∠ CEF (Vertically opposite angles)
AE = EC(By construction)
By SAS criteria △ADE ≅△CFE
ii. △ADE ≅ △CFE
Corresponding part of congruent triangle are equal
∠ EFC = ∠ EDA
alternate interior angles are equal
⇒ AD ∥ FC
⇒ CF ∥ AB
iii. △ADE ≅ △CFE
Corresponding part of congruent triangle are equal.
CF = AD
We know that D is mid point AB
⇒ AD = BD
⇒ CF = BD
OR
DE = BC
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{line drawn from mid points of 2 sides of △ is parallel and half of third side}
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2
DE ∥ BC and DF ∥ BC
DF = DE + EF
⇒ DF = 2DE(BE = EF)
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⇒ DF = BC
21. i. In △APD and △BQC
AD = BC (given)
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AP = CQ (opposite sides of rectangle)
∠ APD = ∠ BQC = 90o
By RHS criteria △APD ≅ △CQB
ii. △APD ≅ △CQB
Corresponding part of congruent triangle
side PD = side BQ
iii. In △ABC and △CDA
AB = CD (given)
BC = AD (given)
AC = AC (common)
By SSS criteria △ABC ≅ △CDA
OR
In △APD
∠ APD + ∠ PAD + ∠ ADP = 180o
⇒ 90o + (180o - 110o) + ∠ ADP = 180o (angle sum property of △)
⇒ ∠ ADP = m = 180o - 90o - 70o = 20o
∠ ADP = m = 20o
22. Since ABD is a line, we have
∠B + ∠C BD = 180 [linear pair]
∘
1 1 ∘
⇒ ∠B + ∠C BD = 90
2 2
¯
¯¯¯¯¯¯
¯
⇒
1
2
∠B + ∠C BO = 90
∘
[∠ CBO = 1
2
(
∠C BD BO is angle bisector)]
∘ 1
⇒ ∠C BO = (90 − ∠B)
2
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Again, ACE is a straight line.
∴ C + ∠BC E = 180 [linear pair] ∘
1 1 ∘
⇒ ∠C + ∠BC E = 90
2 2
¯
¯¯¯¯¯¯
¯
⇒
1
2
∠C + ∠BC O = 90
∘
[∠ BCO = 1
2
∠ BCE(C O is angle bisector)]
⇒ ∠BC O = (90
∘
−
1
2
∠C ) ....(ii)
We know that the sum of the angles of a triangle is 180°.
So, from △OBC, we get
∠C BO + ∠BC O + ∠BOC = 180
⇒ (90
∘
−
1
2
∠B) + (90
∘
−
1
2
∠C ) + ∠BOC = 180
∘
[using (i) and (ii)]
∘ 1 ∘
⇒ 180 − (∠B + ∠C ) + ∠BOC = 180
2
1
⇒ ∠BOC = (∠B + ∠C )
2
1 1 1
⇒ ∠BOC =
2
(∠A + ∠B + ∠C ) −
2
∠A [adding and subtracting 2
∠A ]
1 ∘ 1 ∘
⇒ ∠BOC = ( × 180 ) − ∠A [∵ ∠A + ∠B + ∠C = 180 ]
2 2
∘ 1
⇒ ∠BOC = (90 − ∠A)
2
Hence, ∠BOC = (90 ∘
−
1
2
∠A) .
23. Join M and D and also M and C.
In DMND and DMNC i
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DN = NC ...[Given]
MN = MN ...[Common]
∠ MND = ∠ MNC ...[Each 90°]
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∴ DMND ≅ DMNC ...[By SAS property]
∴ MD = MC ...[c.p.c.t.] ...(1)
and ∠ DMN = ∠ CMN ...[c.p.c.t.] ...(2)
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But, ∠ AMN = ∠ BMN ...[Each 90° ]
∴ ∠ AMD + ∠ DMN = ∠ BMC + ∠ CMN
∴ ∠ AMD = ∠ BMC ...[From (2)] ... (3)
In DAMD and DBMC,
∠ AMD = ∠ BMC ...[From (3)]
AM = MB . . . [Given]
MD = MC. . . [From (1)]
∴ DAMD ≅ DBMC ...[SAS property]
∴ AD = BC ...[c.p.c.t.]
∴ the other sides of the quadrilateral are equal.
24. △ABC and △ADC,
i. AB = AD, BC = DC . . . [Given]
AC = AC . . . [Common]
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△ABC ≅ △ADC . . .[SSS axiom
∠1 = ∠ 2 . . . .[c.p.c.t.]
AC bisects ∠ A and ∠ 3 = ∠ 4 . . .[c.p.c.t.]
AC bisects ∠ C.
Hence, AC bisects each of the angles A and C.
ii. In △ABE and △ADE,
AB = AD . . . [Given]
∠ 1 = ∠ 2 . . .[As proved above]
AE = AE . . . .[Common]
∴ △ABE ≅ △ADE . . . [SAS axiom]
∴ BE = ED . . . [c.p.c.t.]
iii. △ABC≅ △ADC . . . [As proved above]
∴ ∠ ABC = ∠ ADC . . . [c.p.c.t.]
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