Question

PQRS is a square of sides 2 cm & ST = 2 cm. Also, PT=RT. What is the area of ∆PST?
2 cm2
√3 cm2
√2 cm2
1/√2 cm2

Answer 1

please show diagram then we solve this question

Answer 2

Answer:

the area of ∆PST = √2 cm²

Step-by-step explanation:

PQRS is a square of sides 2 cm & ST = 2 cm. Also, PT=RT. What is the area of ∆PST?

2 cm2

√3 cm2

√2 cm2

1/√2 cm2

As PT = RT => point T lies on Diagonal SQ

and O is the intersection point of diagonals

Area of ΔPST = Area of ΔPOS  + Area of ΔPOT

PR = QS = Digonal of square = sideof square√2 = 2√2 cm

SO = SQ/2 = 2√2/2 = √2 cm

in ΔPOS Sides are OS = OP = √2 cm  & PS = 2 cm

Diagonals of square are perpendicular

∠O = 90°

=> area of ΔPOS = (1/2)(OP)(OS) = (1/2)(√2)(√2) = 1 cm²

in ΔPOT Sides are  OP = √2 cm  & OT = ST - OS = 2 - √2 cm

=> area of ΔPOT = (1/2)(OP)(OT) = (1/2)(√2)(2-√2) = √2 - 1 cm²

Area of ΔPST = Area of ΔPOS  + Area of ΔPOT

=> Area of ΔPST = 1 + √2 - 1 = √2 cm²

the area of ∆PST = √2 cm²