Question

in square ABCD, two points p and q
are on side AD and AB respectively
such that Ap =25 cm, PD=15 cm
and AQ: QB = 3:2 find the area
of ∆ BCR ?​

Answer 1

Answer:

256 sq units

Step-by-step explanation:

Answer 2

GIVEN :

square ABCD, two points p and q

are on side AD and AB respectively

AP =25 cm, PD=15 cm

AQ: QB = 3:2

TO FIND :

the area of ∆ BCR

SOLUTION :

◆AD = 25 +15 =40cm

◆Side of square

AB =BC =CD = AD =40cm

◆AQ :QB is 3:2.

◆Consider A on the origin (0,0)

◆Then , solving with ratio,

3+2 =5

40/5 =8.

◆Q is at , 8×3 =24 ; (24,0)Q

◆R is the intersection of PB , QC .

◆Slope of CQ ;

CB /BQ = 40/16 =5/2.

◆Equation of line ,

y =mx + c

◆Equation of CQ = 2y = 5x - 120

◆Equation of PB = 8y = -5x +200

◆ Solving both equation,

We get (x,y) of R

That is ,( 27.2, 8)

◆RT = x2 - x1 = 40 -27.2 = 12.8

◆Area of ∆BCR = (BC × RT )/2

= 40×12.8 /2 = 256 sq units.

ANSWER :

Area of ∆BCR = = 256 sq units.