MATHS M.C.Q

$\text{s}=\frac{5+12+13}{2}=15\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{15(15-5)(15-12)(15-13)}$
$=\sqrt{15\times10\times3\times2}$
$=30\text{ sq.cm}$
$=0.003\text{ sq.m}$

$\text{s}=\frac{5+7+8}{2}=10\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{10(10-5)(10-7)(10-8)}$
$=\sqrt{10\times5\times3\times2}$
$=10\sqrt{3}\text{ sq.cm}$

Let the sides of $\triangle\text{ABC}$ be $n, n + 1, n + 2.$
$⇒$ Perimeter $= n + n + 1 + n + 2$
$⇒ (9 + 9 + 9) = 3n + 3$
$⇒ 3n = 24$
$⇒ n = 8\ cm$
Thus, the shortest side is $8\ cm.$
Hence, correct option is $(c).$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}\text{a}^2$
$=\frac{\sqrt{3}}{4}\times4\times4$
$=4\sqrt{3}\text{cm}^2.$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2=16\sqrt{3}$
$⇒ ($Side$)^2 = 64$
$⇒$ Side $= 8\ cm$
Perimeter of equilateral triangle $= 3\ ×$ side
$= 3 × 8 = 24\ cm.$

Area of rhombus $=\frac{1}{2}\times\text{Product of diagonal}$
$\Rightarrow200=\frac{1}{2}\times10\times\text{d}_2$
$\Rightarrow\text{d}_2=\frac{200\times2}{10}=40\text{cm}$

Given: Side $=2\sqrt3\text{cm}$
We know that, area of equilateral triangle $=\bigg(\frac{\sqrt3}{4}\bigg)\text{a}^2$ square units.
$=\bigg(\frac{\sqrt3}{4}\bigg)​​​​​​(2\sqrt3)^2=\bigg(\frac{\sqrt3}{4}\bigg)(12)$
$=3\sqrt3=3(1.732)=5.196\text{cm}^2$

Let the height of the isosceles triangle be $x\ cm$
Then length of equal side $= (x + 2)cm$
Since altitude of isosceles triangle bisects the base.
Then, in a right-angled triangle,
$(x + 2)^2 = x^2 + 6^2$
$⇒ 4 + 4x = 36$
$⇒ x = 8\ cm$
Now, area of triangle $ = \frac{1}{2}\times \text{Base}\times\text{Height}$
$=\frac{1}{2}\times12\times8=48\text{cm}^2$

Let $ABC$ be a triangle in which sides $AB = 35\ cm, BC = 54\ cm$ and $CA = 61\ cm$
Now semi-perimeter of a triangle,
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{35+54+61}{32}=\frac{150}{2}=75\text{cm}$
$\Big[\therefore$ semiperimeter, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}\Big]$
$\because$ Area of $\triangle\text{ABC}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]
$=\sqrt{75(75-35)(75-54)(75-61)}$
$=\sqrt{75\times40\times21\times14}$
$=\sqrt{25\times3\times4\times2\times5\times7\times3\times7\times2}$
$=5\times2\times2\times3\times7\sqrt{5}$
$=420\sqrt{5}\text{cm}^2$
Also, Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{AB}\times\text{Altitude}$
$\Rightarrow\frac{1}{2}\times35\times\text{CD}$
$\Rightarrow\text{CD}=\frac{420\times2\sqrt5}{35}$
$\therefore\text{CD}=24\sqrt5$
Hence, the length of altitude is $24\sqrt5\text{cm}.$

$\text{s}=\frac{6+8+10}{2}=12\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{12(12-6)(12-8)(12-10)}$
$=\sqrt{12\times6\times4\times2}$
$=24\text{ sq.cm}$
Therefore, the cost of painting it at the rate of $Rs. 0.09$ per sq.cm $= 24 × 0.09 = Rs 2.16$

Given: $a = 6\ cm, b = 8\ cm, c = 10\ cm.$
$\text{s}=(\frac{6+8+10}{2})=12\text{cm}$
Hence, by using Heron’s formula, we can write:
$\text{A}=\sqrt{12(12-6)(12-8)(12-10)}=\sqrt{(12)(6)(4)(2)}=\sqrt{576=24\text{cm}^2}$
$\text{Therefore, the cost of painting at a rate of 9 paise per cm}^2=24\times9\text{paise}=\text{Rs.2.16}$

Let the sides of $\triangle\text{ABC}$ be $n, n + 1, n + 2$
$⇒$ Perimeter $= n + n + 1 + n + 2$
$⇒ (9 + 9 + 9) = 3n + 3$
$⇒ 27 = 3n + 3$
$⇒ 3n = 24$
$⇒ n = 8\ cm$
Thus, the shortest side is $8\ cm.$

Perpendicular $=\sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\text{cm}$
Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Perpendicular}$
$=\frac{1}{2}\times8\times6$
$=24\text{cm}^2$

Perimeter of triangle with sides $a, b$ and $c$ is $P = a + b + c ....(i)$
New sides are $\frac{\text{a}}{2},\frac{\text{b}}{2},\frac{\text{c}}{2}$
New perimeter $=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{\text{P}}{2}. ($From eq.$(i))$
Decreased perimeter $=\text{P}-\frac{\text{P}}{2}=\frac{\text{P}}{2}$
$\%$ of decreased perimeter $=\frac{\frac{\text{P}}{2}}{\text{P}}\times100=50\%$

Since diagonals of a rhombus bisect each other at right angle.

$\text{OB}=\frac{24}{2}=12\text{cm}$ and $\text{OC}=\frac{10}{2}=5\text{cm}$
In triangle $OBC,$
$\text{BC}=\sqrt{12^2+5^2}=\sqrt{144+25}=13\text{cm}$
Perimeter of rhombus $= 4\ ×$ side $= 4 × 13 = 52\ cm$

Given: Area of equilateral triangle $= 9\sqrt3\text{cm}^2$
Hence, $\bigg({\frac{\sqrt3}{4}}\bigg)\text{a}^2=9\sqrt{3}$
$\text{a}^2=\frac{[(9\sqrt3)(4)]}{\sqrt3}$
$\text{a}^2=36$
$\text{a}=6\text{cm}$

Perimeter of equilateral triangle $= 60m$
$⇒ 3\ ×$ side $= 60m$
$⇒$ side $= 20m$
Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
$=\frac{\sqrt{3}}{4}20\times20$
$=100\sqrt{3}\text{sq}.\text{m}.$

Let the sides of the triangle be $5x \ cm, 12x \ cm$ and $13x\ cm.$
Perimeter $=$ Sum of all sides
Or, $150 = 5x + 12x + 13x$
Or, $30x = 150x$
Or, $x = 5$
Thus, the sides of the triangle are $5 × 5\ cm, 12 × 5\ cm$ and $13 × 5\ cm,$ i.e., $25\ cm, 60\ cm$ and $65\ cm$.
Now,
Let:
$a = 25\ cm, b = 60\ cm$ and $c = 65\ cm$
$\text{s}=\frac{150}{2}=75\text{cm}$
By using Heron's formula, we have:
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{75(75-25)(75-60)(75-65)}$
$=\sqrt{75\times50\times15\times10}$
$=\sqrt{15\times5\times5\times10\times15\times10}$
$=15\times5\times10$
$=750\text{cm}^2$

Let:
$a = 30\ cm, b = 24\ cm$ and $c = 18\ cm$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{30+24+18}{2}=36\text{cm}$
By applying Heron's formula, we get:
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{36(36-30)(36-24)(36-18)}$
$=\sqrt{36\times6\times12\times18}$
$=\sqrt{12\times3\times12\times6\times3}$
$=12\times3\times6$
$=216\text{cm}^2$
The smallest side is $18\ cm.$
Hence, the altitude of the triangle corresponding to $18\ cm$ is given by:
Area of triangle $ =216\ cm^2$
$\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height}=216$
$\Rightarrow\text{Height}=\frac{216\times2}{18}=24\text{cm}$

Height of equilateral triangle $=\frac{\sqrt{3}}{2}\times\text{Side}$
$\Rightarrow6=\frac{\sqrt{3}}{2}\times\text{Side}$
$\Rightarrow\text{Side}=\frac{12}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{12}{3}\times\sqrt{3}=4\sqrt{3}\text{cm}$
Now,
Area of equilateral triangle $=\frac{\sqrt{3}}{2}\times(\text{Side})^2$
$=\frac{\sqrt{3}}{4}\times(4\sqrt{3})^2$
$=\frac{\sqrt{3}}{4}\times48$
$=12\sqrt{3}\text{cm}^2$

$\text{s}=\frac{4+4+2}{2}=5\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{5(5-4)(5-4)(5-2)}$
$=\sqrt{5\times1\times1\times3}$
$=\sqrt{15}\text{ sq.cm}$

Let:
$a = 40m, b = 24m$ and $c = 32m$
$\text{s}=\frac{40+24+32}{2}=48\text{cm}$
By Heron's formula, we have
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{48(48-40)(48-24)(48-32)}$
$=\sqrt{48\times8\times24\times16}$
$=\sqrt{24\times2\times8\times24\times8\times2}$
$=24\times8\times2$
$=384\text{m}^2$

Sides of the triangle are $35\ cm, 54\ cm$ and $61\ cm$
$\text{s}=\frac{35+54+61}{2}=75\text{cm}$
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 $=\frac{2(\text{Area of }\triangle)}{35}$
$=\frac{2\times420\sqrt{5}}{35}=24\sqrt{5}\text{cm}$

Since, the three sides of a triangle are $a = 56\ cm, b = 60\ cm$ and $c = 52\ cm$
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 $1344\ cm^2$

$\text{s}=\frac{5\text{x}+5\text{x}+8\text{x}}{2}=9\text{x}\text{ cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{9\text{x}(9\text{x}-5\text{x})(9\text{x}-5\text{x})(9\text{x}-8\text{x})}$
$=\sqrt{9\text{x}\times4\text{x}\times4\text{x}\times\text{x}}$
$=12\text{x}^2\text{ sq.cm}$

Area of triangle $=\frac{1}{2}\times\text{base}\times\text{height}$
$=\frac{1}{2}\times12\times8=48\text{cm}$

Let $a = 16\ cm, b = 30\ cm, c = 34\ cm$
semi perimeter of a triangle $=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{16+30+34}{2}=40$
Now, $s - a = 24\ cm, s - b = 10\ cm$ and $s - c = 6\ cm$
By Heron's formula, we have area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{40\times24\times10\times6}$
$=\sqrt{4\times10\times4\times6\times10\times6}$
$=\sqrt{4^2\times10^2\times6^2}$
$=4\times10\times6$
$=240\text{cm}^2$

Cost of turfing a triangular field at the rate of $Rs. 45$ per $100 = Rs. 900$
$\frac{\text{Area}\times45}{100}=900$
$⇒$ Area $= 2000\ sq.cm$
According to question,
$2\ ×$ Base $= 5\ ×$ Height
$\Rightarrow\text{Base}=\frac{\text{Height}\times5}{2}$
Areo of triangle $= 2000$ sq.cm
$\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height}=2000$
$\Rightarrow\frac{1}{2}\times\frac{\text{Height}\times5}{2}\times\text{Height}=2000$
$⇒ ($Height$)^2 = 1600$
$⇒$ Height $= 40cm$

Height of isoscale triangle $=\frac{1}{2}\sqrt{4\text{a}^2-\text{b}^2}$
$=\frac{1}{2}\sqrt{4(5)^2-6^2} (a = 5\ cm$ and $b = 6\ cm)$
$=\frac{1}{2}\times\sqrt{100-36}$
$=\frac{1}{2}\times\sqrt{64}$
$=\frac{1}{2}\times8$
$=4\text{cm}$

Here, $\text{x}\sqrt{3}=\sqrt{48}$
$\Rightarrow\text{x}=\sqrt{16}$
Side $= 2x$
Area of equilateral triangle $=\frac{\sqrt{3}}{4}\text{(Side)}^2$
$=\frac{\sqrt{3}}{4}(2\text{x})^2$
$=\sqrt{3}\text{x}^2\text{ sq. cm}$
$=\sqrt{3}(\sqrt{16})^2$
$=16\sqrt{3}\text{cm}^2$