Answer:
i)Let the girl of height be at point D on the ground from the lamp post after 4 seconds. Therefore,
AD=1.2 m/sec ×4 sec =4.8 m =480 cm
Suppose the length of the shadow of the girl be x cm when she at position D. Therefore,
BD=x cm
In ∆BDE and ∆BAC,
∠BDE=∠BAC (90°)
∠DBE=∠ABC (Common)
Thus, ∆BDE∼∆BAC (AA similarity)
BC
BE
=
AC
DE
=
AB
BD
( Corresponding sides are proportional )
360
90
=
480+x
x
⇒
4
1
=
480+x
x
⇒4x=480+x
⇒4x−x=480
⇒3x=480
⇒x=
3
480
=160
Hence length of her shadow after 4 seconds is 160 cm.
ii)(i) triangle AEP sim triangle CDF
Given: triangle ABC
and, altitude AD and CE of triangle intersects each other at the point P.
To Prove: triangle AEP sim triangle CDP
Proof:
In triangle AEP and triangle CDP
angle AEP = angle CDP
angle APE = angle CPD
Hence, triangle AEP sim triangle CDP
Hence proved
D
B
E
(Since AD & CE are altitudes)
(Vertically opposite angles)
(AA Similarity)
iii) Let x be the digit at unit’s place and y be the digit at ten’s place.
Since y is at ten’s place, then the number formed is 10y+x.
By reversing the digits, it becomes 10x+y.
As the difference of the numbers is 18, so,
(10y+x)−(10x+y)=18
9(y−x)=18
y−x=2 .... (1)
As the sum of digits is 8, so,
x+y=8 .... (2)
On adding equations (1) and (2), we get
2y=10⇒y=5
Putting this in (2), we get x=8−5=3
x=3,y=5
Hence, number =10y+x=10×5+3=53.
Step-by-step explanation:
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