MATHS M.C.Q

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

$\frac{\text{AD}}{\text{DC}}=\frac32$
Let $AD = 3x$ and $DC = 2x$
Area of $\triangle\text{ABC}=\frac12\times\text{AC}\times\text{BC}$ $(BE = h)$
$\Rightarrow40=\frac12\times5\text{x}\times\text{h}$
$\Rightarrow 80 = 5xh$
$\Rightarrow xh = 16\ cm^2 ....(1)$
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 $\triangle\text{ABD}$
$=40-24$
$=16 \ \text{cm}^2$

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

3. $\frac{15\sqrt{7}}{2}\text{cm}$

2. $8\sqrt{5}\text{cm}^2$

3. 9000 sq.cm
   Solution:
   The ratio of the sides is 12 : 17 : 25
   Perimeter = 540cm
   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:
   $\text{A}=\sqrt{8(8-\text{a})(8-\text{b})(8-\text{c})}$
   $=\sqrt270{(270-120)(270-170)(270-250)}$
   $=\sqrt(270\times150\times100\times20)$
   $= 9000 \text{ sq.}\text{cm}$

1. 1320 sq.m
   Solution:
   Explanation: Given,
   a = 122m
   b = 22m
   c = 120m
   $\text{Semi-perimeter, s}=\frac{(122+22+120)}{2}=132\text{m}$
   Using heron’s formula:
   $\text{A}=\sqrt{8(8-a)(8-b)(8-c)}$
   $=\sqrt{132(132-122)(132-22)(132-120)}$
   $=\sqrt{(132\times10\times110\times12)}$
   = 1320 sq.m

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

2. $4\sqrt{21}\text{m}^2$

3. 2cm

2. $6\sqrt{91}\text{cm}^2$
   Solution:
   $\text{s}=\frac{15+8+19}{2}=21\text{cm}$
   Area of triangle $=\sqrt{\text{s}(\text{s-a})(\text{s-b})(\text{s-c})}$
   $=\sqrt{21(21-15)(21-8)(21-19)}$
   $=\sqrt{21\times6\times13\times2}$
   $=6\sqrt{91}\text{sq}.\text{cm}.$

4. 60cm
   Solution:
   Area of $\triangle=\frac12\text{Base}\times\text{Height}$
   The smallest side is 11m
   $\Rightarrow\text{Area}=\frac12\times11\times\text{Height}\dots(1)$
   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).

1. $50\text{cm}^2$

1. $4\sqrt{21}\text{m}^2$

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

4. 6cm
   Solution:
   Given: Area of equilateral triangle $= 9\sqrt3\text{cm}^2$
   Hence, $\bigg({\frac{\sqrt3}{4}}\bigg)\text{a}^2=9\sqrt{3}$
   $\text{a}^2=\frac{[(9\sqrt3)(4)]}{\sqrt3}$
   $\text{a}^2=36$
   $\text{a}=6\text{cm}$

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

2. 24cm
   Solution:
   Given, area of an equilateral triangle $=16\sqrt{3}\text{cm}^2$
   Area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text{side})^2$
   $\frac{\sqrt{3}}{4}(\text{side})^2={16}\sqrt{3}$
   $\Rightarrow(\text{side})^2=64$
   $\Rightarrow\text{sides}=8\text{cm}$
   [taking positive square root because side is always positive]
   Perimeter of an equilateral triangle = 3 × Side= 3 × 8 = 24cm
   Hence, the perimeter of an equilateral triangle is 24cm.

2. $\sqrt{15}\text{cm}^2$

1. 36cm

3. $\frac{1}{2}\text{(Base x Height)}$

3. $300\%$
   Solution:
   $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2},\text{A}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   Now, if a' = 2a, b' = 2b and c' = 2c
   Then, $\text{s}'=\frac{\text{a}'+\text{b}'+c'}{2}=\frac{2\text{a}+2\text{b}+2\text{c}}{2}=2\text{s}$
   $\text{A}'=\sqrt{\text{s}'(\text{s},-\text{a}')(\text{s}'-\text{b}')(\text{s}'-\text{c}')}$
   $=\sqrt{2\text{s}(2\text{s}-2\text{a})(2\text{s}-2\text{b})(2\text{s}-2\text{c})}$
   $=4\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   $\Rightarrow\text{A}' = 4\text{A}$
   ⇒ Increase in Area $=\frac{4\text{A}-\text{A}}{\text{A}}\times100\%=300\%$
   Hence, correct optin is (c).