MATHS M.C.Q

Let:
$a = 20\ cm, b = 16\ cm$ and $c = 12\ cm$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{26+16+12}{2}=24\text{cm}$
By Heron's formula, we have:
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{24(24-20)(24-16)(24-12)}$
$=\sqrt{24\times4\times8\times12}$
$=\sqrt{6\times4\times4\times4\times4\times6}$
$=6\times4\times4$
$=96\text{cm} ^2$

The ratio of the sides is $12 : 17 : 25$
Perimeter $= 540\ cm$
Let the sides of the triangle be $12x, 17x$ and $25x.$
Hence,
$12x + 17x + 25x = 540cm$
$54x = 540cm$
$x = 10$
Therefore,
$a = 12x = 12 × 10 = 120$
$b = 17x = 17 × 10 = 170$
$c = 25x = 25 × 10 = 250$
$\text{Semi}-\text{perimeter, s}= \frac{540}{2}= 270\text{cm}$
Using Heron’s formula:
$\text{A}=\sqrt{8(8-\text{a})(8-\text{b})(8-\text{c})}$
$=\sqrt270{(270-120)(270-170)(270-250)}$
$=\sqrt(270\times150\times100\times20)$
$= 9000 \text{ sq.}\text{cm}$

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

If side of a square is $a\ cm$
Then, its diagonal $=\sqrt{2}\text{a}\text{cm}$
But diagonal $=12\sqrt{2}\text{cm}$
$\Rightarrow\sqrt{2}\text{a}=12\sqrt{2}$
$⇒ a = 12\ cm$
$⇒ $ Perimeter of a square $= 4a = 4 × 12 = 48\ cm$
Now, perimeter of an equilateral triangle with side $x = 3x\ cm$
But perimeter of equilateral triangle $=$ Perimeter of square
$⇒ 3x = 48$
$⇒ x = 16\ cm$
Now, Area of equilateral $\triangle=\frac{\sqrt{3}\text{x}^2}{4}=\frac{\sqrt{3}}{4}\times16\times16=64\sqrt{3}\text{cm}^2$
Hence, correct option is $(d).$

Let each of the equal sides be $x \ cm$. Then,
$\text{x}+\text{x}+5=11$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}=3\text{cm}$
$\text{s}=\frac{3+3+5}{2}=\frac{11}{2}\text{cm}$
$\text{Area}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{sc})}$
$=\sqrt{\frac{11}{2}\big(\frac{11}{2}-3\big)\big(\frac{11}{2}-3\big)\big(\frac{11}{2}-5\big)}$
$=\sqrt{\frac{11}{2}\times\frac{5}{2}\times\frac{5}{2}\times\frac{1}{2}}$
$=\frac{5}{4}\sqrt{11}\text{cm}^2$

$a = 325\ m, b = 300\ m, c = 125\ m$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{325+300+125}{4}=375\text{m}$
$s - a = 50\ m, s - b = 75\ m, s - c = 250\ m$
Area $=\sqrt{\text{s}}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})$
$=\sqrt{375\times50\times75\times250}$
$=\sqrt{15\times25\times25\times2\times3\times25\times25\times10}$
$=\sqrt{25\times25\times25\times25\times30\times30}$
$=25\times25\times30$
$=18750\text{m}^2$

Area of equilateral $\triangle\text{ i.e., } 9\sqrt{3}=\frac{\sqrt{3}}{4}(\text{side})^2$
$\Rightarrow(\text{side})^2=\frac{9\sqrt{3}\times4}{\sqrt{3}}=36$
$\therefore\ \text{side}=\pm\sqrt{36}=6\text{cm}$

$\text{s}=\frac{56+60+52}{2}=\frac{168}{2}=84\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{84(84-56)(84-60)(84-52)}$
$=\sqrt{84\times28\times24\times32}$
$=\sqrt{12\times7\times7\times4\times12\times2\times16\times2}$
$=12\times7\times2\times2\times4$
$=1344\text{cm}^2$

Area of $\triangle=\frac12\text{Base}\times\text{Height}$
The smallest side is $11\ m$
$\Rightarrow\text{Area}=\frac12\times11\times\text{Height}\dots(1)$
Area by Heron's Formula $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$\text{s}=\frac{11+60+61}{2}=66\text{m}$
$\Rightarrow\text{Area}=\sqrt{66\times55\times6\times5}=330\text{m}^2$
From eq. $(1)$
$330=\frac12\times11\times\text{Height}$
$\Rightarrow\text{Height}=\frac{2\times330}{11}=60\text{m}$
Hence, correct option is $(d).$

Given: $(s - a) (s - b) (s - c) = 13200\ m$ and $s = 132\ m$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{13200\times132}$
$=1320\text{ sq.m}$

Area of right angle triangle $= 20\ sq. m$
$\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height} = 20 $
$\Rightarrow\frac{1}{2}\times\text{Base}\times4=20$
$\Rightarrow\text{Base}=10\text{cm}$
Then, Hypotenuse $=\sqrt{10^2+4^2}=2\sqrt{29}\text{m}$
If the altitude drawn to the hypotenuse of a right-angle triangle, then the length of required altitude $=\frac{10\times4}{2\sqrt{29}}=\frac{20}{\sqrt{29}}\text{cm}$

Area of triangle with sides $a, b$ and $c.$
$(\text{A})=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
New sides are $2a, 2b$ and $2c$
$\text{s}'=\frac{2\text{a}+2\text{b}+2\text{c}}{2}=\text{a}+\text{b}+\text{c}=2\text{s}\ ....(\text{i})$
New area $=\sqrt{\text{s}'(\text{s}'-2\text{a})(\text{s}'-2\text{b})(\text{s}'-2\text{c})}$
$=\sqrt{2\text{s}(2\text{s}-2\text{a})(2\text{s}-2\text{b})(2\text{s}-2\text{c})} [$From eq.$(i)]$
$=4\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=4\text{A}$
Therefore, the new area will be four times the old area

Let the sides be $3x, 4x$ and $5x.$
Then according to quesiton, $3x + 4x + 5x = 84$
$⇒ 12x = 84$
$⇒ x = 7$
Therefore, the sides are $3 × 7 = 21\ cm, 4 × 7 = 28\ cm$ and $5 × 7 = 35\ cm$
$\text{s}=\frac{21+28+35}{2}=42\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{42(42-21)(42-28)(42-35)}$
$=\sqrt{42\times21\times14\times7}$
$21\times7\times2=294\text{ sq.cm}$

Arrea of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
$=\frac{\sqrt{3}}{4}(25)^2$
$=\frac{625\sqrt{3}}{4}\text{ sq.cm}$

Since diagonals of a rhombus divide it into $4$ triangles of equal area. Therefore,
Area of rhombus $= 4\ × $ Area of triangle
$= 4 × 125 = 500\  sq.cm$

Given: $s - a = 8\ cm, s - b = 7\ cm$ and $s - c = 5\ cm$
Adding all equations,
$s - a + s - b + s - c = 8 + 7 + 5$
$\Rightarrow3\text{s}-(\text{a+b+c})=20\Big[\text{s}=\frac{\text{a+b+c}}{2}\Big]$
$\Rightarrow3\text{s}-2\text{s}=20$
$\Rightarrow\text{s}=20\text{cm}$

Since in the quadrilateral $ABCD,$ the diagonals are perpendicular.
Area of quadrilateral $=\frac{1}{2}\times$ Product of diagonals
$=\frac{1}{2}\times12\times8=48\text{sq}.\text{cm}.$

If side of an equilateral triangle is $'a',$ then its
Area $=\frac{\sqrt{3}}{4}\text{a}^2$
Now, $\frac{\sqrt{3}}{4}\text{a}^2=4\sqrt{3}$
$⇒ a^2 = 4^2$
$⇒ a = 4\ cm$
Hence, correct option is $(a).$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
$=\frac{\sqrt{3}}{4}\times2\sqrt{3}\times2\sqrt{3}$
$=3\sqrt{3}$
$=3\times1.732=5.196\text{sq.cm}$

Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
Area of $\triangle\text{ABC}=\frac{1}{2}\times12\times5=30\text{cm}^2$

$\text{d}_2=\frac{\text{Area}\times2}{\text{d}_1}$
$=\frac{96\times2}{12}$
$=16\text{cm}$
Length of side of rhombus $=\sqrt{6^2+8^2}=10\text{cm}$
perimeter of rhombus $= 4\ × $ side
$= 4 × 10 = 40\ cm$

According to question, in given right-angled triangle $AB$ and $AC$ are Base and Perpendicular respectively.
$\text{Area}=\frac{1}{2}\times4\times4$
$= 8\ cm^2$.

Given: $(s - a)(s - b)(s - c) = 13200m$ and $s = 132m$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{13200\times132}$
$1320\text{m}^2$

The ratio of the sides is $3: 5: 7$
Perimeter $= 300\ cm$
Let the sides of the triangle be $3x, 5x$ and $7x.$
Hence,
$3x + 5x + 7x = 300\ cm$
$15x = 300\ cm$
$x = 20$
Therefore,
$a = 3x = 3 × 20 = 60$
$b = 5x = 5 × 20 = 100$
$c = 7x = 7 × 20 = 140$
$\text{semiperimeter,s}=\frac{300}{2}=150\text{cm}$
Using Heron’s formula:
$\text{A}=\sqrt{\text{s}(\text{s}-{a})(\text{s}-{b})(\text{s}-{c})}$
$=\sqrt{150(150-60)(150-100)(150-40)}$
$=\sqrt{(150\times90\times50\times10)}$
$=1500\sqrt{3\text{sq.cm}}$

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

Since triangle $ABC$ is an isosceles triangle.
Hence, height $= 30\ cm$
$⇒$ Area of triangle $ABC =\frac{1}{2}\times\text{AB}\times\text{BC}=\frac{1}{2}\times30\times30=450\text{cm}^2$

Adding given three equaitons,
$2 x+2 y+2 z=24 \Rightarrow x+y+z=12$
Therefore, $\text{s}=\frac{12}{2}=6\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{6(6-\text{x})(6-\text{y})(6-\text{z})}$
$=\sqrt{6(12-6-\text{x})(12-6-\text{y})(12-6-\text{z})}$
$=\sqrt{6(\text{y}+\text{z}-6)(\text{x}+\text{z}-6)(\text{x}+\text{y}-6)}$
$=\sqrt{6(9-6)(8-6)(7-6)}$
$=\sqrt{6\times3\times2\times1}$
$=6\text{ sq.m}$

Let the two adjacent sides of the parallelogram be $a = 74\ cm, b = 40\ cm$
Let the length of diagonal be $c = 102\ cm$
These two sides and the diagonal forms a triangle
semi perimeter, $\text{s}= \frac{(\text{a} + \text{b} + \text{c})}{2}$
$\text{s}=\frac{(74+40+102)}{2}$
$=\frac{216}{2}$
$=108\text{cm}$
By Heron's formula, we have area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
Area of triangle $=\sqrt{108(108-74)(108-40)(108-102)}$
$=1224\text{cm}^2$
therefore, area of parallelogram $= 1224 × 2$
$= 2448\ sq. cm$

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 $=\frac{1}{2}\times\text{Base}\times\text{Height}$
$=\frac{1}{2}\times8=48\text{sq.cm}.$

$\text{s}=\frac{1+2+2}{2}=\frac{5}{2}\text{m}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{\frac{5}{2}(\frac{5}{2}-1)(\frac{5}{2}-2)(\frac{5}{2}-2)}$
$=\sqrt{\frac{5}{2}\times\frac{3}{2}\times\frac{1}{2}\times\frac{1}{2}}$
$=\frac{\sqrt{15}}{4}\text{ sq.m}$

Here, the base and height of the triangle are $10\ cm$ and $10\ cm,$ respectively.
Thus, we have
Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
$=\frac{1}{2}\times10\times10$
$=50\text{cm}^2$

$\text{Area}=\frac{1}{2}\times\text{Base}\times\text{Height}$
$=\frac{1}{2}\times8\times10$
$=40\text{cm}^2$

Let $a = 7\ cm, b = 9\ cm, c = 14\ cm$
Semi-perimeter $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{7+9+14}{2}=15\text{cm}$
$s - a = 15 - 7 = 8\ cm, s - b = 15 - 9 = 6\ cm$ and $s - c = 15 - 14 = 1\ cm$
Are of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{15\times8\times6\times1}$
$=\sqrt{5\times3\times4\times2\times3\times2}$
$=12\sqrt5\text{cm}^2$

Let $a = 7\ cm, b = 9\ cm, c = 14\ cm$
Semi-perimeter = $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{7+9+14}{2}=15\text{cm}$
$s - a = 15 -7 = 8\ cm, s - b = 15 - 9 = 6\ cm$ and $s - c = 15 - 14 = 1\ cm$
Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{15\times8\times6\times1}$
$=\sqrt{5\times3\times4\times2\times3\times2}$
$=2\sqrt{5}\text{cm}^2$
Hence, correct option is $(a).$