5. D, E and F are mid-points of side BC, CA and AB
of AARC Show that ratio of the areas of ADET
AABC is 1:4
6. Two triangles BAC and BDC right angled And
D are drawn on the same side of BC. AC and De
intersect at P. Prove that AP x PC - DP X PB
7. Two poles of height a metres and b metres are p mer
apart. Prove that the height of the point of intersection
of the lines joining the top of each pole to the foot of
the opposite pole is given by
ab
a+b
+ b
Answers
Step-by-step explanation:
5) GIVEN :- ∆ABC and D , E and F mid - points of AB , BC and CS respectively
TO PROVE:- Area of ∆ DEF and Area of ∆ABC is 1:4
PROOF:- We know that lines joining mid - points of two sides of Triangle is parallel to the 3rd side.
In ∆ABC ,
D , F are mid - points of AB and AC respectively
DF||BC
so , DF||BE also .......1
Similarly ,
E , F are mid - points of BC and AC respectively
EF ||AB
Hence , EF || DB .......2
From 1 and 2
DF ||BE and FE || DB
Therefore, opposite sides of quadrilateral is parallel DBEF is a parallelogram
Now we know that,
in parallelogram , opposite sides are equal
Hence < DFE =< ABC ........3
Similarly,
we can prove. DECF is a parallelogram
In a parallelogram, opposite angles are equal
Hence , <DFE =<ABC .......4
Now,in ∆EDF = ∆ ABC
< DFE = < ABC ( FROM 3 ))
<EDF = <ACB ( FROM 4 ))
By using AA similarity criterion
∆DEF ~∆ ABC
We know thatif two Triangles are similar, the ratio of their area is always equal to the squareof the ratio of their corresponding side
area of∆ DEF = DE2 / area of ∆ ABC = AC2
ar∆ DEF =FC2 / ar∆ ABC = AC 2
area of ∆DEF = (AC)2/2 / area of ∆ABC = (AC)2
area of ∆DEF = (AC)2/4 / area of∆ ABC = (AC)2
area of ∆DEF = 1/4 / area of∆ ABC = 1
Hence , Area of ∆DEF = 1
Area of ∆ABC = 4
Ratio of area of ∆DEF and ∆ABC = 1:4
6. In △APB and △DPC, we have
∠A=∠D=90
∘
and
∠APB=∠DPC [Vertically opposite angles]
Thus, by AA-criterion of similarity, we have
△APB∼△DPC
⇒
DP
AP
=
PC
PB
⇒ AP×PC=DP×PB [Hence proved]
7. Let AB and CD be two poles of heights a metres and b metres respectively such that the poles are p metres apart i.e.AC=p metres. Suppose the lines AD and BC meet at O such that OL=h metres.
Let CL=x and LA=y. Then, x+y=p.
In △ABC and △LOC, we have
∠CAB=∠CLO [Each equal to 90
∘
]
∠C=∠C [Common]
∴ △CAB∼△CLO [By AA-criterion of similarity]
⇒
CL
CA
=
LO
AB
⇒
x
p
=
h
a
⇒ x=
a
ph
...........(i)
In △ALO and △ACD, we have
∠ALO=∠ACD [Each equal to 90
∘
]
∠A=∠A [Common]
∴ △ALO∼△ACD [By AA-criterion of similarity]
⇒
AC
AL
=
DC
OL
⇒
p
y
=
b
h
⇒ y=
b
ph
[∵ AC=x+y=p]........(ii)
From (i) and (ii), we have
x+y=
a
ph
+
b
ph
⇒ p=ph(
a
1
+
b
1
) [∵ x+y=p]
⇒ 1=h(
ab
a+b
)
⇒ h=
a+b
ab
metres
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is
a+b
ab
metres.