Practise set 4.2

 QUE 

Using the property ab = akbk$\frac{\mathrm{<br>}}{\mathrm{<br>}}$, fill in the blanks substituting proper numbers in the following.

(i) 57 = ....28 = 35.... = ....3.5            (ii) 914= 4.5.... = ....42 = ....3.5

ANSWER :

(i)
28 = 7 × 4, 35 = 5 × 4, 3.5 = 7 × 0.5

Now,

57 = 5×47×4 = 5×77×7 = 5×0.57×0.5⇒57 = 2028 = 3549 = 2.53.5$$\begin{gathered} \frac{5}{7} = \frac{5\times 4}{7\times 4} = \frac{5\times 7}{7\times 7} = \frac{5\times 0.5}{7\times 0.5} \\ \Rightarrow \frac{5}{7} = \frac{\underline{\mathbf{20}}}{28} = \frac{35}{\underline{\mathbf{49}}} = \frac{\underline{\mathbf{2}\mathbf{.}\mathbf{5}}}{3.5} \end{gathered}$$
(ii)
4.5 = 9 × 0.5, 42 = 14 × 3, 3.5 = 14 × 0.25

Now,

914=9×0.514×0.5=9×314×3=9×0.2514×0.25⇒914= 4.57 = 2742 = 2.253.5

QUE :(2) Find the following ratios. 

(i) The ratio of radius to circumference of the circle.

 (ii) The ratio of circumference of circle with radius r to its area. 

 (iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm. 

(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. 

Find the ratio of its perimeter to area. 

ANSWER :

(i) The ratio of radius to circumference of the circle:

The circumference of a circle is given by the formula C = 2πr, where "r" is the radius.

Ratio = Radius / Circumference = r / (2πr) = 1 / (2π)

(ii) The ratio of circumference of a circle with radius "r" to its area:

The area of a circle is given by the formula A = πr^2.

Ratio = Circumference / Area = 2πr / πr^2 = 2 / r

(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm:

In a square, the diagonal (d) and the side length (s) are related by the Pythagorean theorem: d^2 = s^2 + s^2 = 2s^2.

Ratio = Diagonal / Side = d / s = √(2s^2) / s = √2

(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area:

Perimeter of a rectangle = 2 * (length + width)

Area of a rectangle = length * width

Given: Length = 5 cm, Width = 3.5 cm.

Perimeter = 2 * (5 + 3.5) = 2 * 8.5 = 17 cm

Area = 5 * 3.5 = 17.5 cm^2

Ratio = Perimeter / Area = 17 / 17.5 = 0.9714

QUE (3) Compare the following pairs of ratios

(i) 5√3, 37√$\frac{\sqrt{5}}{3}, \frac{3}{\sqrt{7}}$      (ii)  35√57√,  63√125√$\frac{3\sqrt{5}}{5\sqrt{7}}, \frac{\sqrt{63}}{\sqrt{125}}$     (iii)   518 , 17121$\frac{5}{18} , \frac{17}{121}$    

(iv)  80√48√ , 45√27√$\frac{\sqrt{80}}{\sqrt{48}} , \frac{\sqrt{45}}{\sqrt{27}}$     (v) 9.25.1 ,  3.47.1

ANSWER :

(i)
5–√×7–√=5×7−−−−√=35−−√3×3=9=81−−√$$\begin{gathered} \sqrt{5}\times \sqrt{7}=\sqrt{5\times 7}=\sqrt{35} \\ 3\times 3=9=\sqrt{81} \end{gathered}$$
Now, 35−−√<81−−√$\sqrt{35}<\sqrt{81}$

∴5√3<37√$\therefore \frac{\sqrt{5}}{3}<\frac{3}{\sqrt{7}}$

(ii)

35–√×125−−−√=35–√×25×5−−−−−√=35–√×55–√=7557–√×63−−√=57–√×9×7−−−−√=57–√×37–√=105$3\sqrt{5}\times \sqrt{125}=3\sqrt{5}\times \sqrt{25\times 5}=3\sqrt{5}\times 5\sqrt{5}=75$

$5\sqrt{7}\times \sqrt{63}=5\sqrt{7}\times \sqrt{9\times 7}=5\sqrt{7}\times 3\sqrt{7}=105$
Now, 75 < 105

∴35√57√<63√125√$\therefore \frac{3\sqrt{5}}{5\sqrt{7}}<\frac{\sqrt{63}}{\sqrt{125}}$

(iii)
5×121=60518×17=306$$\begin{gathered} 5\times 121=605 \\ 18\times 17=306 \end{gathered}$$
Now, 605 > 306

∴518>17121$\therefore \frac{5}{18}>\frac{17}{121}$

(iv)
80−−√×27−−√=16×5−−−−−√×9×3−−−−√=45–√×33–√=1215−−√45−−√×48−−√=9×5−−−−√×16×3−−−−−√=35–√×43–√=1215−−√$\sqrt{80}\times \sqrt{27}=\sqrt{16\times 5}\times \sqrt{9\times 3}=4\sqrt{5}\times 3\sqrt{3}=12\sqrt{15}$

$\sqrt{45}\times \sqrt{48}=\sqrt{9\times 5}\times \sqrt{16\times 3}=3\sqrt{5}\times 4\sqrt{3}=12\sqrt{15}$
Now, 1215−−√=1215−−√$12\sqrt{15}=12\sqrt{15}$

∴80√48√=45√27√$\therefore \frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{45}}{\sqrt{27}}$

(v)
9.2×7.1=65.325.1×3.4=17.34$$\begin{gathered} 9.2\times 7.1=65.32 \\ 5.1\times 3.4=17.34 \end{gathered}$$
Now, 65.32 > 17.34

∴9.25.1>3.47.1

QUE (4) 

(i)  □ABCD is a parallelogram. The ratio of  ∠A and  ∠ B of this parallelogram is 5 : 4. Find the measure of  ∠B.

 (ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 

3 : 5. Find their present ages.

ANSWER :

(i) 
Quadrilateral ABCD is a parallelogram.

Let the measure of ∠$\angle$A and ∠$\angle$B be 5x and 4x, respectively.

Now,

∠$\angle$A + ∠$\angle$B = 180º          (Adjacent angles of a parallelogram are supplementary)

∴ 5x + 4x = 180º

⇒ 9x = 180º

⇒ x = 20º

∴ Measure of ∠$\angle$B = 4x = 4 × 20º = 80º

Thus, the measure of ∠$\angle$B is 80º.

(ii)
Let the present ages of Albert and Salim be 5x years and 9x years, respectively.

5 years hence,

Age of Albert = (5x + 5) years

Age of Salim = (9x + 5) years

It is given that five year hence, the ratio of their ages will be 3 : 5.

∴5x+59x+5=35⇒25x+25=27x+15⇒27x−25x=25−15⇒2x=10⇒x=5
∴ Present age of Albert = 5x = 5 × 5 = 25 years

Present age of Salim = 9x = 9 × 5 = 45 years

Thus, the present age of Albert is 25 years and the present age of Salim is 45 years.