Maths M.C.Q

Area of quadrilateral triangle $=\frac{\text{b}}{4}\sqrt{4\text{a}^2-\text{b}^2}$
Here,
$a = 6\ cm$ and $b = 8\ cm$
Thus, we have:
$=\frac{8}{4}\times\sqrt{4(6)^2-8^2}$
$=\frac{8}{4}\times\sqrt{144-64}$
$=\frac{8}{4}\times\sqrt{80}$
$=\frac{8}{4}\times4\sqrt{5}$
$=8\sqrt{5}\text{cm}^2$

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

Height of equilateral triangle $=\frac{\sqrt{3}}{2}\times\text{Side}$
$=\frac{\sqrt{3}}{2}\times10$
$=5\sqrt{3}\text{cm}$

$\text{s}=\frac{11+60+61}{2}=66\text{m}$
Area of $\triangle=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{21\times10\times6\times5}=30\sqrt{7}\text{cm}^2$
Also if we choose largest side and its Altitude, the area would be
$\text{A}=\frac12\times\text{largest side}\times\text{h}$
$330=\frac12\times11\times\text{Height}$
$\Rightarrow\text{Height}=\frac{2\times330}{11}=60\text{m}$
Hence, correct option is $(d).$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
$=\frac{1.732}{4}\times10\times10$
$=43.3\text{cm}^2$

Height of isosceles 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 $b = 6\ cm)$
$=\frac{1}{2}\times\sqrt{100-36}$
$=\frac{1}{2}\times\sqrt{64}$
$=\frac{1}{2}\times8$
$=4\text{cm}$

The smallest altitude is $\perp$ drawn to the largest side of a $\triangle$ from opposite point.
i.e. $BD$ Area of $\triangle=\frac{1}{2}\times\text{AC}\times​​\text{BD}=\frac{1}{2}\times112\times\text{BD}$
$=56\times\text{BD}$
$\text{s}=\frac{50+78+112}{2}=120\text{cm}$
$s - AB = 70\ cm, s - BC = 42\ cm, s - AC = 8\ cm$
Area $=\sqrt{\text{s}(\text{s}-\text{AB})-(\text{s}-\text{BC})(\text{s}-\text{AC})}$
$=\sqrt{120\times70\times42\times8}$
$=1680\text{cm}^2$
Now, $56 × BD = 1680\ cm^2$
$\Rightarrow\text{BD}=\frac{1680}{56}=30\text{cm}$

Let $\triangle\text{PQR}$ be an isoscale triangle and $\text{PX}\bot\text{QR}.$
Now,
Area of triangle $= 48\ cm^2$
$\Rightarrow\frac{1}{2}\times\text{QR}\times\text{PX}=48$
$\Rightarrow\text{h}=\frac{96}{16}=6\text{cm}$
Also,
$\text{QX}=\frac{1}{2}\times24=12\text{cm}\ \text{and}\ \text{PX}=12\text{cm}$
$\text{PQ}=\sqrt{\text{QX}^2+\text{PX}^2}$
$\text{a}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{cm}$
$\therefore$ Perimeter $= (10 + 10 + 16)\ cm = 36\ cm$

Area of triangle with sides $a, b, c (\text{A})=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
New sides are $2a, 2b$ and $2c$
Then $\text{s}' =\frac{2\text{a}+2\text{b}+2\text{c}}{2}=\text{a}+\text{b}+\text{c}$
$\Rightarrow\text{s}'=2\text{s}\ ...(\text{i})$
New area $=\sqrt{\text{s}'(\text{s}'-2\text{a})(\text{s}'-\text{2b})(\text{s}'-\text{2c})}$
$=\sqrt{2\text{s}(2\text{s}-2\text{a})(2\text{s}-2\text{b})(2\text{s}-2\text{c})}$
$=4\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=4\text{A}$
Increased area $= 4A - A = 3A$
$\%$ of increased area $=\frac{3\text{A}}{\text{A}}\times100=300\%$

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.
Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{40\times24\times10\times6}$
$=\sqrt{4\times10\times4\times6\times10\times6}$
$=\sqrt{4^2\times10^2\times6^2}$
$=4\times10\times6$
$= 240\text{cm}^2$
Note: Correct option not given.

Area of rhombus $=\frac{1}{2}\times$ Product of diagonal
$\Rightarrow96=\frac{1}{2}(16\times\text{d}_2)$
$\Rightarrow\text{d}_2=\frac{96\times2}{16}=12\text{cm}$
Since diagonals of a rhombus bisect each other at a right angle.
Therefore, the side of the rhombus is the hypotenuse of a triangle.
Side $=\sqrt{8^2+6^2}=10\text{cm}.$

Here, $\text{s}=\frac{12+16+20}{2}=24\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{24(24-12)(24-16)(24-20)}$
$=96\text{sq.cm}$

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

In the given triangle,
$\text{Base}(\text{BC})=\sqrt{13^2-12^2}=\sqrt{169-144}=5\text{cm}$
Area of triangle ABC $=\frac{1}{2}\times\text{BC}\times\text{AB}$
$=\frac{1}{2}\times5\times12$
$=30\text{ sq.cm}$

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

$\text{OB}=\frac{24}{2}=12\text{m}$
In triangle $OBC, OC =\sqrt{20^2-12^2}=\sqrt{400-144}=16\text{m}$
$\text{AC}=2\times16=32\text{m}$
Area of rhombus $=\frac{1}{2}\times\text{Product of diagonals}$
$=\frac{1}{2}\times24\times32$
$=384\text{ sq.m}$

Perimeter of triangle $= 3a$
Now, $3\text{a}=60$
$\Rightarrow\text{ a}=60\div3=20\text{m}$
Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$
$=\frac{\sqrt{3}}{4}\times(20)^2=100\sqrt{3}\text{m}^2$

Let each of the two equal sides of an isosceles right triangle be $a\ cm.$
Then, third side $=\text{a}\sqrt{2}\text{m}$
Area of $\triangle=\frac{1}{2}\times2\times2$
$\Rightarrow8=\frac{\text{a}^2}{2}$
$⇒ a^2 = 4\ cm$
$⇒ a = 4\ cm$
$⇒$ Perimeter
$\Rightarrow\text{a}+\text{a}+\text{a}\sqrt{2}=4+4+4\sqrt{2}$
$=8+4\sqrt{2}\text{cm}^2.$

$\text{S}=\frac{(24+20+20)}{2}=32\text{cm}$
$\text{Area}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{32(32-24)(32-20)(32-20)}$
$=192\text{sq}.\text{cm}.$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
$=\frac{\sqrt{3}}{4}(10)^2$
$=25\sqrt{3}\text{sq.cm}$

Area of rhombus $=\frac{1}{2}\text{d}_1\text{d}_2$
$96=\frac{1}{2}\times16\times\text{d}_2$
$\text{d}_2=\frac{96\times2}{16}$
$\text{d}_2=6\times2=12\text{cm}$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times(\text{side})^2$
$\Rightarrow\frac{\sqrt{3}}{4}\times(\text{side})^2=36\sqrt{3}$
$\Rightarrow(\text{Side})^2=144$
$\Rightarrow\text{Side}=12\text{cm}$
Now,
Perimeter $= 3 × 12 = 36\ cm$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}$ (Side)$^2$
$=\frac{1.732}{4}\times10\times10$
$=43.3\text{ sq.cm}$

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