MATHS M.C.Q

1. $10\sqrt3\text{cm}^2$

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

3. $40\text{cm}$

4. $100\sqrt{3}\text{m}^2$

Side $=\frac{20}{5}=5 \ \text{cm}$
Half diagonal $=\sqrt{5^2-3^2}=4 \ \text{cm}$
Diagonal $=4\times2=8 \ \text{cm}$
Area $=\frac{1}{2}\times6\times8=24 \ \text{cm}^2.$

1. $60\text{cm}^2$

2. $9\sqrt3^{\text{cm}^2}$

4. $16\sqrt{3}\text{m}^2$

3. $10\sqrt{3}\text{cm}^2$

Let the height of the isosceles triangle be $x \ cm$
Then length of equal side $= (x + 2)cm$
Since altitude of isosceles triangle bisects the base.
Then, in a right$-$angled triangle,
$(x + 2)^2 = x^2 + 6^2$
$\Rightarrow 4 + 4x = 36$
$\Rightarrow x = 8\ cm$
Now, area of triangle $ = \frac{1}{2}\times \text{Base}\times\text{Height}$
$=\frac{1}{2}\times12\times8=48 \ \text{cm}^2$

Given: Side $=2\sqrt3 \ \text{cm}$
We know that, area of equilateral triangle $=\bigg(\frac{\sqrt3}{4}\bigg)\text{a}^2$ square units.
$=\bigg(\frac{\sqrt3}{4}\bigg)​​​​​​(2\sqrt3)^2$
$=\bigg(\frac{\sqrt3}{4}\bigg)(12)$
$=3\sqrt3$
$=3(1.732)$
$=5.196 \ \text{cm}^2$

2. $900\sqrt3\text{cm}^2$

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

4. 8

2. Smallest

3. $\frac{3}{16}\text{sq.}\text{cm}$

3. $24\sqrt{5}\text{cm}$
   Solution:
   Sides of the triangle are 35cm, 54cm and 61cm
   $\text{s}=\frac{35+54+61}{2}=75\text{cm}$
   Area of $\triangle=\sqrt{75(75-35)(75-54)(75-61)}$
   $=\sqrt{75\times40\times21\times14}$
   $=\sqrt{5\times5\times3\times2\times2\times2\times5\times3\times7\times7\times2}$
   $=5\times3\times2\times2\times7\sqrt{5}$
   $=420\sqrt{5}\text{cm}^2$
   Now, longest altitude will be the perpendicular on the smallest side of the triangle from the opposite vertex.
   $\therefore$ Length of longest altitude $=\frac{2(\text{Area of }\triangle)}{35}$
   $=\frac{2\times420\sqrt{5}}{35}=24\sqrt{5}\text{cm}$

3. 4cm.

3. 3k

3. 24 sq.cm
   Solution:
   side of rhombus $=\frac{\text{Perimeter}}{4}$
   $=\frac{20}{4}=5\text{cm}$
   diagonal $=2\sqrt{52-44}=2\sqrt{25-16}=2\times3=6\text{cm}$
   Area of rhombus $=\frac{1}{2}\times\text{Product of diagonal}$
   $=\frac{1}{2}\times8\times6=24\text{ sq.cm}$

2. 60m

Area of equilateral triangle $=\frac{\sqrt{3}\text{a}^2}{4}$ where $a =$ side of the triangle
$\sqrt{3}=\frac{\sqrt{3}\text{a}^2}{4} $
Solving
$a^2 = 4$
$a = 2\ cm$

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

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

1. 52cm

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

$a = 325m, b = 30m, c = 125m$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{325+300+125}{2}=375\text{m}$
$s - a = 50m, s - b = 75m, s - c = 250m$
Area $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{375\times50\times75\times250}$
$=\sqrt{15\times25\times25\times2\times3\times25\times25\times10}$
$=\sqrt{\underline{25\times25}\times\underline{25\times25}\times\underline{30\times30}}$
$=25\times25\times30$
$= 18750\text{m}^2$

1. $64\sqrt3\text{cm}^2$

1. $24\text{cm}^2$

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

3. $16\sqrt3\text{cm}^2$

2. $\frac{5}{4}\sqrt{11}\text{cm}^2$
   Solution:
   Let each of the equal sides be x cm. Then,
   $\text{x}+\text{x}+5=11$
   $\Rightarrow2\text{x}=6$
   $\Rightarrow\text{x}=3\text{cm}$
   $\text{s}=\frac{3+3+5}{2}=\frac{11}{2}\text{cm}$
   $\text{Area}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{sc})}$
   $=\sqrt{\frac{11}{2}\big(\frac{11}{2}-3\big)\big(\frac{11}{2}-3\big)\big(\frac{11}{2}-5\big)}$
   $=\sqrt{\frac{11}{2}\times\frac{5}{2}\times\frac{5}{2}\times\frac{1}{2}}$
   $=\frac{5}{4}\sqrt{11}\text{cm}^2$

$\text{Side}=\frac{20}{4}=5 \ \text{cm}$
$\text{half diagnoal} =\sqrt{5^2 - 3^2} = 4 \ \text{cm }$
$\text{diagnoal} = 4 \times 2 = 8 \ \text{cm} $
$\text{Area}=\frac{1}{2}\times6\times8=24 \ \text{cm}^2$

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 $= (10 + 10 + 16)cm$
$= 36\ cm$