Maths M.C.Q

4. 6cm

Solution:

Area of equilateral $\triangle\text{ i.e., } 9\sqrt{3}=\frac{\sqrt{3}}{4}(\text{side})^2$

$\Rightarrow(\text{side})^2=\frac{9\sqrt{3}\times4}{\sqrt{3}}=36$

$\therefore\ \text{side}=\pm\sqrt{36}=6\text{cm}$

1. 5.196cm²

Solution:

Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$

$=\frac{\sqrt{3}}{4}\big(2\sqrt{3}\big)^2=3\sqrt{3}=3\times1.732$

$=5.196\text{cm}^2$

2. 24cm

Solution:

Given, area of an equilateral triangle $=16\sqrt{3}\text{cm}^2$

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text{side})^2$

$\frac{\sqrt{3}}{4}(\text{side})^2={16}\sqrt{3}$

$\Rightarrow(\text{side})^2=64$

$\Rightarrow\text{sides}=8\text{cm}$

[taking positive square root because side is always positive]

Perimeter of an equilateral triangle = 3 × Side= 3 × 8 = 24cm

Hence, the perimeter of an equilateral triangle is 24cm.

2. Rs. 2.16

Solution:

Since, the edges of a triangular are a = 6cm, b = 8cm and c = 10cm

Now, semi-perimeter of a triangular board.

$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$

$=\frac{6+8+10}{2}=\frac{24}{2}=12\text{cm}$

Now, area of a triangular board $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$=\sqrt{12(12-6)(12-8)(12-10)}$

$=\sqrt{12\times6\times4\times2}$

$=\sqrt{(12)^2\times(2)^2}$

 $=12\times2=24\text{cm}^2$

Since, the cost of painting for area 1cm² = Rs. 0.09

$\therefore$ Cost of paint for area 24cm² = 0.09 × 24 = Rs. 2.16

Hence, the cost of a triangular board is Rs. 2.16

3. $24\sqrt{5}\text{cm}$

Solution:

Sides of the triangle are 35cm, 54cm and 61cm

$\text{s}=\frac{35+54+61}{2}=75\text{cm}$

Area of $\triangle=\sqrt{75(75-35)(75-54)(75-61)}$

$=\sqrt{75\times40\times21\times14}$

$=\sqrt{5\times5\times3\times2\times2\times2\times5\times3\times7\times7\times2}$

$=5\times3\times2\times2\times7\sqrt{5}$

$=420\sqrt{5}\text{cm}^2$

Now, longest altitude will be the perpendicular on the smallest side of the triangle from the opposite vertex.

$\therefore$ Length of longest altitude $=\frac{2(\text{Area of }\triangle)}{35}$

$=\frac{2\times420\sqrt{5}}{35}=24\sqrt{5}\text{cm}$

3. 1344cm²

Solution:

Since, the three sides of a triangle are a = 56cm, b = 60cm and c = 52cm

Then, semi-perimeter of a triangle,

$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{56+60+52}{2}=\frac{168}{2}=84\text{cm}$

Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]

$=\sqrt{84(84-56)(84-60)(84-52)}$

$=\sqrt{4\times7\times3\times4\times7\times4\times2\times3\times4\times4\times2}$

$=\sqrt{(4)^6\times(7)^2\times(3)^2}$

$=(4)^3\times7\times3=1344\text{cm}^2$

Hence, the area of triangle is 1344cm²

1. $\sqrt{32}\text{cm}$

Solution:

ABC is an isosceles right triangle. We have,

$\text{AB}=\text{BC}=\text{a cm}$

Area of $\triangle=\frac{1}{2}\text{base}\times\text{Height}$

$\Rightarrow\ 8=\frac{1}{2}\times\text{a}\times\text{a}$ $\big[\because\ \text{AB}=\text{BC}=\text{a cm}\big]$

$\Rightarrow\ \text{a}^2=16$ $\therefore\ \text{a}=+\sqrt{16}=4\text{cm}$

Using Pythagoras theorem,

Hypotenuse $\text{AC}=\sqrt{4^2+4^2}$

$=\sqrt{16+16}=\sqrt{32}\text{cm}$

1. $\sqrt{15}\text{cm}^2$

Solution:

Here, $\text{s}=\frac{4+4+2}{2}=5\text{cm}$

Area of $\triangle=\sqrt{5(5-2)(5-4)(5-4)}$

$=\sqrt{5\times3\times1\times1}=\sqrt{15}\text{cm}^2$

4. $100\sqrt{3}\text{m}^2$

Solution:

Perimeter of triangle = 3a

Now, $3\text{a}=60$

$\Rightarrow\text{ a}=60\div3=20\text{m}$

Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$

$=\frac{\sqrt{3}}{4}\times(20)^2=100\sqrt{3}\text{m}^2$