MATHS M.C.Q

3. $60\text{cm}^2$
   Solution:
   $\text{s}=\frac{13+13+24}{2}=25\text{cm}$
   Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   $\sqrt{25(25-13)(25-13)(25-24)}$
   $=\sqrt{25\times12\times12\times1}$
   $60\text{cm}^2$

2. $2\sqrt{3}\text{cm}$
   Solution:
   Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
   $\Rightarrow\frac{\sqrt{3}}{4}(\text{Side})^2=4\sqrt{3}$
   $\Rightarrow(\text{Side})^2=4^2$
   $\Rightarrow\text{Side}=4\text{cm}$
   Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
   $\Rightarrow4\sqrt{3}=\frac{1}{2}\times4\times\text{Height}$
   $\Rightarrow\text{Height}=2\sqrt{3}\text{cm}$

1. Becomes four times.
   Solution:
   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

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

2. 12 sq.cm
   Solution:
   Area of rhombus $=\frac{1}{2}\times\text{Product of diagonals}$
   $=\frac{1}{2}\times4\times6$
   $=12\text{ sq.cm}$

4. 20cm
   Solution:
   Given: s - a = 8cm, s - b = 7cm and s - c = 5cm
   Adding all equations,
   s - a + s - b + s - c = 8 + 7 + 5
   $\Rightarrow3\text{s}-(\text{a+b+c})=20\Big[\text{s}=\frac{\text{a+b+c}}{2}\Big]$
   $\Rightarrow3\text{s}-2\text{s}=20$
   $\Rightarrow\text{s}=20\text{cm}$

2. $\frac{3}{4}\sqrt{15}\text{cm}$
   Solution:
   $\text{s}=\frac{4+8+6}{2}=9\text{cm}$
   Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   $=\sqrt{9(9-4)(9-8)(9-6)}$
   $=\sqrt{9\times5\times1\times3}$
   $=3\sqrt{15}\text{ sq.cm}$
   Area of triangle taking base as longest side $=\frac{1}{2}\times8\times\text{h}$
   $\Rightarrow \frac{1}{2}\times8\times\text{h}=3\sqrt{15}$
   $\Rightarrow\text{h}=\frac{3}{4}\sqrt{15}\text{cm}$

2. $16\sqrt{3}\text{cm}^2$
   Solution:
   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. $1500\sqrt{3}\text{sq}.\text{cm}$
   Solution:
   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 = 300cm
   15x = 300cm
   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:
   $\text{A}=\sqrt{\text{s}(\text{s}-{a})(\text{s}-{b})(\text{s}-{c})}$
   $=\sqrt{150(150-60)(150-100)(150-40)}$
   $=\sqrt{(150\times90\times50\times10)}$
   $=1500\sqrt{3\text{sq.cm}}$

4. $24\sqrt{3}\text{cm}^2$

4. $8\sqrt5\text{cm}^2$
   Solution:
   Area of isosceles triangle $=\frac{\text{b}}{4}\sqrt{4\text{a}^2-\text{b}^2}$
   Here,
   a = 6cm and b = 8cm
   Thus, we have
   $\frac{8}{4}\times\sqrt{4(6)^2-8^2}$
   $=\frac{8}{4}\times\sqrt{144-64}$
   $=\frac{8}{4}\times\sqrt{80}$
   $=\frac{8}{4}\times4\sqrt{5}$
   $=8\sqrt{5}\text{cm}^2$

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 $\triangle\text{PQR}$ 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 $= (10 + 10 + 16)cm = 36\ cm$

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, $56 \times BD = 1680\ cm^2$
$\Rightarrow\text{BD}=\frac{1680}{56}=30 \ \text{cm}$

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

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

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

3. $40\text{cm}$
   Solution:
   $\text{d}_2=\frac{\text{Area}\times2}{\text{d}_1}$
   $=\frac{96\times2}{12}$
   $=16\text{cm}$
   Length of side of rhombus $=\sqrt{6^2+8^2}=10\text{cm}$
   perimeter of rhombus = 4 × side
   = 4 × 10 = 40cm

1. $\frac{\sqrt{15}}{4}\text{m}^2$

2. 2448 sq. cm

2. $16\sqrt{3}\text{cm}^2$

1.  $12\sqrt{5}\text{cm}^2$

3. $12\sqrt5\text{cm}^2$

4. $\frac{15\sqrt7}{4}\text{cm}$

2. $5\sqrt{3}\text{cm}$

2. $4\text{cm}$
   Solution:
   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 = 5cm and b = 6cm)
   $=\frac{1}{2}\times\sqrt{100-36}$
   $=\frac{1}{2}\times\sqrt{64}$
   $=\frac{1}{2}\times8$
   $=4\text{cm}$

Perimeter of Rhombus $= 20\ cm$

$\Rightarrow 4 \times side = 20\ cm \Rightarrow Side = 5\ \ cm$
Since diagonals of rhombus bisect each other at perpendicular, therefore
$\text{OC}=\frac{8}{2}=4 \ \text{cm}$
In triangle $OBC,$
$\text{OB}=\sqrt{5^2-4^2}=\sqrt{25-16}=3\text{cm}$
$\text{AC}=2\times3$
$​​​​​​​=6 \ \text{cm}$
Area of rhombus $=\frac{1}{2}\times\text{Product of diagonal}$
$=\frac{1}{2}\times8\times6=24\text{ sq.cm}$

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 $=\sqrt{8^2+6^2}=10 \ \text{cm}.$