Question

HELP I AM BEGING YOU

Answer 1

(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 = $\sqrt{25}$ = 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 )