(8)
since ΔPRS is right isosceles then PR = RS
let PR = RS = x, then using Pythagoras' identity on the triangle
x² + x² = (3√2)²
2x² = 18 ( divide both sides by 2 )
x² = 9 ( take the square root of both sides )
x = 3
that is RS = PR = 3 cm
MNPQ is a square hence PQ = [tex]\sqrt{25}[/tex] = 5
and QR = PQ - 3 = 5 - 3 = 2 cm
area of rectangle RSTQ = 3 × 2 = 6 cm²
(9)
Since one of the diagonals is an altitude , then right triangle is formed
let one side be x then the other side is x - 2
perimeter = (2 × length ) + (2 × width ) = 2x + 2(x - 2 ) = 4x - 4
now given perimeter = 40, then
4x - 4 = 40 ( add 4 to both sides )
4x = 44 ( divide both sides by 4 )
x = 11
hence sides ( legs of right triangle ) are 11 and 9
Using Pythagoras' identity on the right triangle with hypotenuse (x) being the altitude
x = √(11² + 9²) = √(121 + 81) = √202 ≈ 14.21 in ( to 2 dec. places )
