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Question (1):
ABC is a triangle in which AB=AC. D is a point on BC produced. Prove that AD>AB.
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Answer:
AB=AC
D is a point on BC produced.
AD>AB
AB=AC (given)
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Question (2):
In the adjoining diagram, name the shortest and the longest side. B is the greatest angle.
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Answer:
Angle A is the least angle.
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Question (3):
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Answer:
( exterior angle = sum of int. opposite angle)
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Question (4):
 
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Answer:
 
 
Combining relation (i) and (ii), we can write
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Question (5):
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Answer:
 
 
 
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Question (6):
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Answer:
In quadrilateral ABCD, AD is the largest side and BC is the shortest side.
Join BD and AC.
In D ABD, AD > AB ( given)
DC > BC
i.e., ABC (or B) > ADC (or D)
In D ADC,
AD>DC (given)
AB>BC ( given)
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Question (7):
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Answer:
(corollary on theorem on
exterior angle of a triangle)
 
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Question (8):
ABCD is a quadrilateral in which AB=BC, AD=DC and AD>AB. 
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Answer:
In the quadrilateral ABCD, AB=BC, AD=DC and AD>AB
Join BD.
In D ABD, AD>AB ( given)
CD > BC
S since CD=AD and BC=AB and AD>AB
By adding (i) and (ii), we get
 
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Question (9):
In the adjoining figure, AGH is an isosceles triangle with base GH. Prove that KH>KG.
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Answer:
In D AGH, AG=AH
KH > KG
In D AGH,
AG=AH ( given)
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Question (10):
In the adjoining figure, LM is the base of isosceles D KLM. Prove that KM>XL.
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Answer:
In D KLM ,
KL=KM
KM > XL
KL > XL
(theorem on inequality KLM is an isosceles triangle)
But KL = KM
\ KM > XL
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Question (11):
In D ABC, AD is a median of BC. Prove that AB+AC>2AD.
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Answer:
AD is a median of D ABC. D is the midpoint of BC
AB+AC>2AD
Produce AD to E such that AD=DE. Join CE.
In triangles ABD and CED,
AD = DE ( by construction)
BD=DC ( D is the midpoint of BC)
(given)
 ( vertically opposite angles)
In D ACE,
AC+CE>AE (sum of any two sides of a
triangle is greater than the
third side)
 AC+AB>AE (Substituting AB for CE proved above)
But AE=AD+DE ( from the figure)
=AD+AD
=2AD
DE=AD (by construction)
 
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Question (12):
In the adjoining figure PQ=PS. Prove that PS
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Answer:
PQ=PS
PS
 ,
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Question (13):
In the adjoining figure AB>AC. Show that AB>AD.
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Answer:
AB>AC
AB>AD
AB>AC (given)
(Converse of theorem on inequality)
In D ABD,
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Question (14):
Show that in a right triangle the hypotenuse is the longest side.
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Answer:
In right triangle AC is the hypotenuse.
AC is the longest side.
In a triangle there can be only one right angle.
 AC>BC ...(i)
 AC>AB ...(ii)
From (i) and (ii), AC is the longest side.
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Question (15):
In the adjoining figure PQRS is a quadrilateral. PQ is the longest side and RS is its shortest side. Prove that  .
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Answer:
PQRS is a quadrilateral.
PQ is the longest side.
RS is the shortest side.
In D RPQ,
PQ>QR
PQ is the greatest side. (given)
In D SRP,
PS>SR
RS is the shortest side.
(i) + (ii)
In D PQS,PQ > PS (given)
PQ is the longest side.
In D SRQ, SRSR is the shortest side.
By adding (iii) and (iv), we get
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Question (16):
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Answer:
AB>BD
(Exterior angle is > interior opposite angle)
 
 
 AB>BD (theorem on inequality)
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Question (17):
PQRS is a quadrilateral. Prove that
(i) PQ+QR+RS+SP>PR+QS
(ii) PQ+QR+RS+SP<2(PR+QS)
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Answer:
PQRS is a quadrilateral. Diagonals PR and QS intersect at O.
(i) PQ+QR+RS+SP>PR+QS
(ii) PQ+QR+RS+SP<2(PR+QS)
In D PSR,
PS+SR>PR ...(i)
(Sum of any two sides > third side)
In DPQS,
PS+PQ>QS ...(ii)
(Sum of any two sides > third side)
In DQRS,
QR+SR>QS ...(iii)
(Sum of any two sides > third side)
InDPQR,
PQ+QR>PR ...(iv)
Add (i), (ii), (iii) and (iv), we get
2(PQ+QR+SR+OS)>2(PR+QS) ...(v)
 PQ+QR+SR+PS>PR+QS (divide both sides by 2)
PO + OQ > PQ
OQ+OR >QR
OR + OS > SR
OS + OP > SP
By adding,
 PQ+QR+SR+PS<2(PR+QS).
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Question (18):
In the adjoining figure AB=AC. Prove that BD>CD.
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Answer:
AB=AC
BD>CD
AB=AC ( given )
(angles opposite to equal sides are equal)
(side opposite to greater angle)
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