Maths M.C.Q

$\text{Area}=25\sqrt{3}\text{sq}.\text{cm}$
$25\sqrt{3}=\frac{\sqrt{3}}{4}\text{a}^2$
$\text{a}^2=\frac{25\times4\times\sqrt{3}}{\sqrt{3}}=100$
$\text{Side}=\text{10cm}.$

Let each of the two equal sides of an isosceles right triangle be $a\ cm.$
Then, third side $=\text{a}\sqrt{2}\text{cm}$
Area of $\triangle=\frac12\times\text{a}\times\text{a}$
$\Rightarrow8=\frac{\text{a}^2}{2}$
$⇒ a^2 = 16$
$⇒ a = 4\ cm$
⇒ Perimeter $\Rightarrow\text{a}+\text{a}+\text{a}\sqrt{2}=4+4+4\sqrt{2}\text{cm}$
Hence, correct option is $(b).$

$\text{s}=\frac{5+8+11}{2}=12\text{m}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{12(12-5)(12-8)(12-11)}$
$=\sqrt{12\times7\times4\times1}$
$=4\sqrt{21}\text{m}^2$

$ABC$ 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 $\text{AC}=\sqrt{4^2+4^2}$
$=\sqrt{16+16}=\sqrt{32}\text{cm}$

$\text{S}=\frac{12+16+20}{2}=\frac{48}{2}=24\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{24(24-16)(24-12)(24-20)}$
$=\sqrt{24\times8\times12\times4}$
$=12\times8=96\text{ sq.cm}$

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

$\frac{\text{AD}}{\text{DC}}=\frac{3}{2}$
Let $AD = 3x$ and $DC = 2x$
Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{AC}\times\text{BE} (BE = h)$
$\Rightarrow40=\frac{1}{2}\times5\text{x}\times\text{h}$
$\Rightarrow80=5\times\text{h}$
$\Rightarrow\text{xh}=16\text{cm}^2$
Area of $\triangle\text{ABD}=\frac{1}{2}\times3\text{x}\times\text{h}$
$=3\times\frac{\text{h}}{2}=\frac{3}{2}\times16=24\text{cm}^2$
$\text{Area of }\triangle\text{BDC}=\text{Area of }\triangle\text{ABC}-\text{Area of }\triangle\text{ABD}$
$=40-24=16\text{cm}^2$

Perimeter of triangle $= 3 + 4 + 5 = 12\ cm$
Now, $\text{s}=\frac{3+4+5}{2}=6\text{cm}$
Area $=\sqrt{\text{s}(\text{s-a})(\text{s-b})(\text{s-c})}$
$=\sqrt{6(6-3)(6-4)(6-5)}$
$=\sqrt{6\times3\times2\times1}$
$=6\text{sq}.\text{cm}.$

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

Since in a right-angled triangle, the circumcentre is the mid-point of the hypotenuse, then
Hypotenuse $= 2 × 3 = 6\ cm$
Now, Area of right-angled triangle $=\frac{1}{2}\times\text{Base}\times\text{Altitude}$
$=\frac{1}{2}\times6\times2=6\text{sq}.\text{cm}.$

$\text{s}=\frac{5+8+11}{2}=12\text{m}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{12(12-5)(12-8)(12-11)}$
$=\sqrt{12\times7\times4\times1}$
$=4\sqrt{21}\text{ sq.m}$

Area of parallelogram $=$ Base $×$ Corresponding altitude
$= 32 × 6 = 192\ sq.m$

Let the side of equilateral $\triangle\text{ABC}$ be $a\ cm$
The median of equilateral triangle is its altitude drawn from $A$ to $BC. ($i.e. the height of $\triangle$ over Base $BC)$
$\Rightarrow\text{AD}=\text{a}\sin60^\circ$
$\Rightarrow\text{x}=\frac{\text{a}\sqrt{3}}{2} [AD = x ($given$)]$
$\Rightarrow\text{a}=\frac{2\text{x}}{\sqrt{3}}$
Area of equilateral $\triangle$ of side $\text{a}=\frac{\sqrt{3}\text{a}^2}{4}$
$=\frac{\sqrt3}{4}\Big(\frac{2\text{x}}{\sqrt3}\Big)^2$
$=\frac{\text{x}^2}{\sqrt3}$

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

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

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

$\frac{\text{AD}}{\text{DC}}=\frac32$
Let $AD = 3x$ and $DC = 2x$
Area of $\triangle\text{ABC}=\frac12\times\text{AC}\times\text{BC} (BE = h)$
$\Rightarrow40=\frac12\times5\text{x}\times\text{h}$
$⇒ 80 = 5xh$
$⇒ xh = 16\ cm^2 ....(1)$
Now, Area of $\triangle\text{ABD}=\frac12\times3\text{x}\times\text{h}=\frac{3\times\text{h}}{2}=\frac{3}{2}\times16=24\text{cm}^2$
Area of $\triangle\text{BDC}=$ Area of $\triangle\text{ABC}-$ Area of $\triangle\text{ABD}=40-24=16\text{cm}^2$
Hence, correct option is $(a).$

The smallest altitude is $\perp$ drawn to the largest side of a $\triangle$ From opposite point. i. e. $BD$
Area of $\triangle=\frac12\times\text{AC}\times\text{BD}=\frac12\times112\times\text{BD}=56\times\text{BD}$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{50+78+115}{2}=120\text{cm}$
$s - AB = 70\ cm, s - BC = 42\ cm, s - AC = 8\ cm$
Area $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{120\times70\times42\times8}=1680\text{cm}^2$
Now, $56 × BD = 1680\ cm^2$
$\Rightarrow\text{BD}=\frac{1680}{56}=30\text{cm}$
Hence, correct option is $(b).$

Here, $\text{s}=\frac{4+4+2}{2}=5\text{cm}$
Area of $\triangle=\sqrt{5(5-2)(5-4)(5-4)}$
$=\sqrt{5\times3\times1\times1}=\sqrt{15}\text{cm}^2$

Let $\triangle\text{PQR}$ be a right-angled triangle and $\text{PQ}\bot\text{QR}.$
Now,
$\text{PQ}=\sqrt{\text{PR}^2-\text{QR}^2}$
$=\sqrt{50^2-48^2}$
$=\sqrt{2500-2304}$
$=\sqrt{196}$
$=14\text{cm}$
$\therefore$ Area of triangle $=\frac{1}{2}\times\text{QR}\times\text{PQ}\\=\frac{1}{2}\times48\times14=336\text{cm}^2$

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

Area of isoscale triangle $=\frac{\text{b}}{4}\sqrt{4\text{a}^2-\text{b}^2}$
Here,
$a = 13\ cm$ and $b = 24\ cm$
Thus, we have:
$=\frac{24}{4}\times\sqrt{4(13)^2-24^2}$
$=6\times\sqrt{676-576}$
$=6\times\sqrt{100}$
$=6\times10$
$=60\text{cm}^2$

Area of parallelogram $=$ Base $×$ corresponding height
$= 60 × 24 = 1440\ sq.cm.$

Let $\triangle\text{PQR}$ be an isosceles triangle and $PX \perp QR$
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,
$\text{QX}=\frac{1}{2}\times24=12\text{cm}$ and $PX =12\ cm$
$\text{PQ}=\sqrt{\text{QX}^2+\text{PX}^2}$
$\text{a}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{cm}$
$\therefore$ Perimeter $= (10 + 10 + 16)\ cm = 36\ cm$

Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
$=\frac{1}{2}\times30\times30=450\text{ sq.cm}$

Let each of the equal sides of given triangle be $x\ cm.$ Then the third side is $(32 - 2x)cm,$
According to quesiton, $\frac{\text{x}}{32-2\text{x}}=\frac{3}{2}$
According to quesiton, $x 32 - 2x = 32$
$⇒ 2x = 96 - 6x$
$⇒ 8x = 96$
$⇒ x = 12\ cm$
Therefore, the sides are $12\ cm, 12\ cm$ and $8\ cm$
Area $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{16(16-12)(16-12)(16-3)}$
$=\sqrt{16\times4\times4\times8}$
$=32\sqrt{2}\text{sq}.\text{cm.}$

Here, $\text{s}=\frac{\text{a+aab}}{2}=\frac{2\text{a+b}}{2}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{\frac{2\text{a+b}}{2}\Big(\frac{2\text{a+b}}{2}-\text{a}\Big)\Big(\frac{2\text{a+b}}{2}-\text{a}\Big)\Big(\frac{2\text{a+b}}{2}-\text{b}\Big)}$
$\sqrt{\frac{2\text{a+b}}{2}\Big(\frac{\text{b}}{2}\Big)\Big(\frac{\text{b}}{2}\Big)\Big(\frac{\text{2a-b}}{2}\Big)}$
$=\frac{\text{b}}{4}\sqrt{4\text{a}^2-\text{b}^2}.$

Area of equilateral triangle $9\sqrt{3}\text{sq}.\text{cm}$
$\Rightarrow\frac{\sqrt{3}}{4}(\text{Side})^2=9\sqrt{3}$
$⇒ ($Side$)^2 = 36$
$⇒$ Side $= 6\ cm.$