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
Answers
please show diagram then we solve this question
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