| |
|
|
| |
|
Question (1):
ABC is a triangle in which AB=AC. D is a point on BC produced. Prove that AD>AB.
 |
Answer:
 AB=AC D is a point on BC produced.
AD>AB
AB=AC (given)



 |
|
Question (2):
In the adjoining diagram, name the shortest and the longest side. B is the greatest angle.
 |
Answer:
 Angle A is the least angle.
 |
|
Question (3):

 |
Answer:


 ( exterior angle = sum of int. opposite angle)




 |
|
Question (4):
 
 |
Answer:

 

 
 Combining relation (i) and (ii), we can write

 |
|
Question (5):


 |
Answer:
 

 

 






 |
|
Question (6):


 |
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)


 |
|
Question (7):

 |
Answer:



 (corollary on theorem on exterior angle of a triangle)
 
 |
|
Question (8):
ABCD is a quadrilateral in which AB=BC, AD=DC and AD>AB. 
 |
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

  |
|
Question (9):
In the adjoining figure, AGH is an isosceles triangle with base GH. Prove that KH>KG.
 |
Answer:
In D AGH, AG=AH
KH > KG
In D AGH, AG=AH ( given)







 |
|
Question (10):
In the adjoining figure, LM is the base of isosceles D KLM. Prove that KM>XL.
 |
Answer:
In D KLM , KL=KM
KM > XL
KL > XL (theorem on inequality KLM is an isosceles triangle) But KL = KM \ KM > XL |
|
Question (11):
In D ABC, AD is a median of BC. Prove that AB+AC>2AD.
 |
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)
  |
|
Question (12):
In the adjoining figure PQ=PS. Prove that PS
 |
Answer:
PQ=PS
PS


,





 |
|
Question (13):
In the adjoining figure AB>AC. Show that AB>AD.
 |
Answer:
AB>AC
AB>AD
AB>AC (given)
 (Converse of theorem on inequality) In D ABD,



 |
|
Question (14):
Show that in a right triangle the hypotenuse is the longest side.
 |
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. |
|
Question (15):
In the adjoining figure PQRS is a quadrilateral. PQ is the longest side and RS is its shortest side. Prove that .
 |
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

 |
|
Question (16):

 |
Answer:

AB>BD
 (Exterior angle is > interior opposite angle)
 
 
AB>BD (theorem on inequality) |
|
Question (17):
PQRS is a quadrilateral. Prove that (i) PQ+QR+RS+SP>PR+QS (ii) PQ+QR+RS+SP<2(PR+QS)
 |
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). |
|
Question (18):
In the adjoining figure AB=AC. Prove that BD>CD.
 |
Answer:
AB=AC
BD>CD
AB=AC ( given )
 (angles opposite to equal sides are equal)


 (side opposite to greater angle) |
|