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 $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{24(24-20)(24-16)(24-12)}$
$=\sqrt{24\times4\times8\times12}$
$=\sqrt{6\times4\times4\times4\times4\times6}$
$=6\times4\times4$
$=96\text{cm} ^2$

Let $\triangle\text{PQR}$ be an isoscale triangle and $\text{PX}\bot\text{QR}.$
Now,
Area of triangle = $48cm^2$
$\Rightarrow\frac{1}{2}\times\text{QR}\times\text{PX}=48$
$\Rightarrow\text{h}=\frac{96}{16}=6\text{cm}$
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 $= (10 + 10 + 16)\ cm = 36\ cm$

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 $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{48(48-40)(48-24)(48-32)}$
$=\sqrt{48\times8\times24\times16}$
$=\sqrt{24\times2\times8\times24\times8\times2}$
$=24\times8\times2$
$=384\text{m}^2$

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 = $216\ cm^2$
$\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height}=216$
$\Rightarrow\text{Height}=\frac{216\times2}{18}=24\text{cm}$

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 $=\frac{1}{2}\times\text{QR}\times\text{PQ}\\=\frac{1}{2}\times48\times14=336\text{cm}^2$

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:
$=\frac{24}{4}\times\sqrt{4(13)^2-24^2}$
$=6\times\sqrt{676-576}$
$=6\times\sqrt{100}$
$=6\times10$
$=60\text{cm}^2$