Question

25. If the mean of five observations x, x+2, X+4. x+6. x+3 is 11 then find the value ofx.
26. The diagonals of a rhombus are 24 cm and 10 cm. Find its area and perimeter.​

Answer 1

Answer:

25.   x = 8

26.   Area = 120$cm^{2}$ , Perimeter = 52cm

Step-by-step explanation:

25.

mean = 11 -------( given )

⇒No. of observations = 5

⇒Mean = $\frac{ Sum of Observations }{ No. of Observations }$                              

⇒11 = $\frac{x + 2 + x + 4 + x + 6 + x + 3}{5}$

⇒11 = $\frac{ x + x + x + x + x + 2 + 4 + 6 + 3}{5}$

⇒11 = $\frac{5x + 15}{5}$

⇒11 = $\frac{5 ( x + 3 ) }{5}$

⇒11 = x + 3

⇒ x + 3 = 11

⇒ x = 11 - 3 = 8

∴ x = 8

∴ The value of x is 8

26.

Diagonals of Rhombus ⊥ bisect each-other ----( rhombus property) ---- (1)

⇒Taking rhombus ABCD

⇒AC = 24cm

⇒BD = 10cm

From (1),

⇒AO = CO, BO = DO

⇒AO + CO = AC

⇒AO + AO = 24cm

⇒2AO = 24

⇒AO = 24/2 = 12cm

∴ AO = 12cm = CO

⇒BO + DO = BD

⇒BO + BO = 10cm

⇒2BO = 10cm

⇒BO = 10/2 = 5cm

∴ BO = 5cm = DO

ΔAOB, ∠AOB = 90°

⇒$AB^{2} = AO^{2} +BO^{2}$

⇒$AB^{2} = 12^{2} + 5^{2} \\\\AB^{2} = 144 + 25\\\\AB^{2} = 169\\\\AB = \sqrt{169} \\\\AB = 13cm$

AB = BC = CD = AD = 13cm ---------( All sides of Rhombus are equal )

⇒Area Of Rhombus = $\frac{1}{2}$ × $d_{1}$ × $d_{2}$

⇒Ar (ABCD) = $\frac{1}{2}$ × 24 × 10

⇒Ar (ABCD) = 12 × 10

∴ Ar (ABCD) = 120$cm^{2}$

⇒Perimeter of Rhombus = 4 × s

⇒Perimeter (ABCD) = 4 × AB

⇒Perimeter (ABCD) = 4 × 13 = 52cm

∴ Perimeter (ABCD) = 52$cm$