Question

plz solve 5,6,7 question of this image.... i am lreparing for exams plz answer fast guys

Answer 1

5) you see in the fig,
let AB = CD =3x
AD = BC =2x
a/c to question,
area of ∆APD + area of ∆BPC =48 cm^2

if we sketch altitude from P to AD and P to BC then, we see sum of altitude =AB =CD

so, 1/2 altitude from P to AD × AD +1/2 altitude from P to BC × BC = 48 cm^2

1/2 (altitude from P to BC + altitude from P to AD ) x BC = 48 cm^2

1/2(length of CD) x BC = 48 cm^2

1/2 × 3x × 2x = 48 cm^2
x =4
so,
CD =AB = 3 × 4 = 12 cm
BC =AD = 2 × 4 = 8 cm
so ,
perimeter of ABCD =2(12+8) =40 cm

6) given,
PC : PA = 1:3
area of ∆BPC= 16 cm^2

ABCD is a parellogram,
now,
we see A join with C by straight line ,
if we draw altitude from D to AC and B to AC , both are equal because ABCD is Parellogram .
now,
ar ∆ADP/ ar∆BPC = 1/2 altitude of ADP x PA/1/2 altitude of BPC x PC

ar∆ADP/ar∆BPC =PA/PC
ar∆ADC/16 =3/1
ar∆ADC =48 cm ^2

7) let a , b , c are three sides of triangle .
given,
a + b + c = 36
altitude of these in ratio = 1 : 2 : 3
now,
let altitude from A to BC = x
altitude from B to AC = 2x
altitude from C to AB = 3x
now,
area of ∆ = 1/2 ( x ) a
area of ∆ = 1/2 ( 2x ) b
area of ∆ = 1/2 (3x ) c

hence,
a = 2b = 3c

let a =6k
b =3k
c=2k
so, 6k +3k +2k =36
11k =36
k=36/11

a =216/11
b=108/11
c=72/11
but this type of triangle doesn't possible
because b+c < a