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71. The perimeter of an isosceles triangle is equal to 14 cm and the lateral side is to the base in the ratio 5 : 4. The area of the triangle is :

A. 21 cm²

B. $0.5 \sqrt{21}$ cm²

C. $1.5 \sqrt{21}$ cm²

D. $2 \sqrt{21}$ cm²

Discuss

Solution:

Let the sides of the triangle be 5x, 5x and 4x cm respectively
Then,
5x + 5x = 4x + 14
⇒ 14 x = 14
⇒ x = 1
So, a = 5 cm, b = 5 cm, c = 4 cm

\begin{aligned} s = \frac{a  +  b  +  c}{2} \\ = (\frac{14}{2}) c m \\ = 7 c m \end{aligned}

(s - a) = 2 cm, (s - b) = 2 cm, (s - c) = 3 cm
∴ Area of the triangle :

\begin{aligned} = \sqrt{7 \times 2 \times 2 \times 3} \\ = 2 \sqrt{21} c m^{2} \end{aligned}

72. The areas of two equilateral triangles are in the ratio 25 : 36. Their altitudes will be in the ratio :

A. 25 : 36

B. 36 : 25

C. 5 : 6

D. $\sqrt{5}$ : $\sqrt{6}$

Discuss

Solution:

Let the length of sides of the two triangles be a₁ and a₂ respectively and their altitudes be h₁ and h₂ respectively.
Then,

\begin{aligned} \Leftrightarrow \frac{\frac{\sqrt{3}}{4} a_{1}^{2}}{\frac{\sqrt{3}}{4} a_{2}^{2}} = \frac{25}{36} \\ \Rightarrow {(\frac{a_{1}}{a_{2}})}^{2} = {(\frac{5}{6})}^{2} \\ \Rightarrow \frac{a_{1}}{a_{2}} = \frac{5}{6} \end{aligned}

And

\begin{aligned} \Leftrightarrow \frac{\frac{1}{2} \times a_{1} \times h_{1}}{\frac{1}{2} \times a_{2} \times h_{2}} = \frac{25}{36} \\ \Rightarrow \frac{5}{6} \times \frac{h_{1}}{h_{2}} = \frac{25}{36} \\ \Rightarrow \frac{h_{1}}{h_{2}} = \frac{25}{36} \times \frac{6}{5} \\ \Rightarrow \frac{h_{1}}{h_{2}} = \frac{5}{6} \end{aligned}

73. A diagonal of a rhombus is 6 cm. If its area is 24 cm² then the length of each side of the rhombus is :

A. 5 cm

B. 6 cm

C. 7 cm

D. 8 cm

Discuss

Solution:

\begin{aligned} \frac{1}{2} \times 6 \times d = 24 \\ \Rightarrow d = \frac{24}{3} \\ \Rightarrow d = 8 c m \\ O A = 4 c m and O B = 3 c m \\ ∴ A B = \sqrt{{(O A)}^{2} + {(O B)}^{2}} \\ = \sqrt{4^{2} + 3^{2}} \\ = 5 c m \end{aligned}

74. The magnitude of the area of a circle is seven times that of its circumference. What is the circumference (in units) of the circle ?

A. 88 units

B. 132 units

C. 616 units

D. Cannot be determined

Discuss

Solution:

\begin{aligned} π R^{2} = 7 \times (2 π R) \\ \Rightarrow R = 14 \\ ∴ Circumference : \\ = (2 \times \frac{22}{7} \times 14) units \\ = 88 units \end{aligned}

75. A small disc of radius r is cut out from a disc of radius R. The weight of the disc which now has a hole in it, is reduced to $\frac{24}{25}$ of the original weight. If R = xr, what is the value of x ?

A. 4

B. 4.5

C. 24

D. 25

Discuss

Solution:

Since weight of the disc is proportional to its area, we have :

\begin{aligned} π (R^{2} - r^{2}) = \frac{24}{25} π R^{2} \\ \Rightarrow R^{2} - r^{2} = \frac{24}{25} R^{2} \\ \Rightarrow r^{2} = \frac{1}{25} R^{2} \\ \Rightarrow R^{2} = 25 r^{2} \\ \Rightarrow R = 5 r \end{aligned}

76. The area of the largest circle, that can be drawn inside a rectangle with side 18 cm by 14 cm, is :

A. 49 cm²

B. 154 cm²

C. 378 cm²

D. 1078 cm²

Discuss

Solution:

Radius of the required circle :

\begin{aligned} = (\frac{1}{2} \times 14) c m \\ = 7 c m \end{aligned}

Area of the circle :

\begin{aligned} = (\frac{22}{7} \times 7 \times 7) c m^{2} \\ = 154 c m^{2} \end{aligned}

77. If the radius of a circle is increased by 75%, then its circumference will increase by :

A. 25%

B. 50%

C. 75%

D. 100%

Discuss

Solution:

Let original radius be R cm
Then, original circumference =

2 π r

cm
New radius :

\begin{aligned} = (175 % of R) c m \\ = (\frac{175}{100} \times R) c m \\ = \frac{7 R}{4} c m \end{aligned}

New circumference :

\begin{aligned} = (2 π \times \frac{7 R}{4}) c m \\ = \frac{7 π R}{2} c m \end{aligned}

Increase in circumference :

\begin{aligned} = (\frac{7 π R}{2} - 2 π R) c m \\ = \frac{3 π R}{2} c m \end{aligned}

Increase % :

\begin{aligned} = (\frac{3 π R}{2} \times \frac{1}{2 π R} \times 100) % \\ = 75 % \end{aligned}

78. A plate on square base made of brass is of length x cm and width 1 mm. The plate weights 4725 gm. If 1 cubic cm cm of brass weight 8.4 grams, then the value of x is :

A. 75

B. 76

C. 72

D. 74

Discuss

Solution:

Given length and width of a square base plate of brass is x cm and 1 mm
Volume of the plate of square base = Area of base × height

\begin{aligned} = x^{2} \times \frac{1}{10} \\ = \frac{x^{2}}{10} c u . c m . \end{aligned}

According to the question,

\begin{aligned} \Rightarrow \frac{x^{2}}{10} \times 8.4 = 4725 \\ \Rightarrow x^{2} = \frac{4725 \times 10}{8.4} \\ \Rightarrow x^{2} = 5625 \\ \Rightarrow x = \sqrt{5625} = 75 c m \end{aligned}

79. What would be the area of a rectangle whose area is equal to the area of a circle of radius 7 cm ?

A. 77 cm²

B. 154 cm²

C. 184 cm²

D. 180 cm²

Discuss

Solution:

Radius of circle = 7cm
Given area of rectangle :
= Area of circle
=

\frac{22}{7} \times 7 \times 7

= 154 cm²

80. The perimeter of a rectangle is 60 metres. If its length is twice its breadth, then its area is :

A. 160 m²

B. 180 m²

C. 200 m²

D. 220 m²

Discuss

Solution:

Let the breadth of the rectangle be x metres
Then, length of the rectangle = 2x metres
⇒ 2(2x + x) = 60
⇒ 6x = 60
⇒ x = 10
So, length = 20 m, breadth = 10 m
∴ Area = (20 × 10) m² = 200 m²

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