Maths M.C.Q

1. 12cm
   Solution:
   Area of rhombus $=\frac{1}{2}\text{d}_1\text{d}_2$
   $96=\frac{1}{2}\times16\times\text{d}_2$
   $\text{d}_2=\frac{96\times2}{16}$
   $\text{d}_2=6\times2=12\text{cm}$

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

Solution:

Here, $\text{s}=\frac{4+4+2}{2}=5\text{cm}$

Area of $\triangle=\sqrt{5(5-2)(5-4)(5-4)}$

$=\sqrt{5\times3\times1\times1}=\sqrt{15}\text{cm}^2$

3. $\frac{1}{2}\text{(Base}\times \text{height})$

3. 48cm²
   Solution:
   Since in the quadrilateral ABCD, the diagonals are perpendicular.
   Area of quadrilateral $=\frac{1}{2}\times$ Product of diagonals
   $=\frac{1}{2}\times12\times8=48\text{sq}.\text{cm}.$

2. 24cm
   Solution:
   Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
   $\Rightarrow\frac{\sqrt{3}}{4}(\text{Side})^2=16\sqrt{3}$
   ⇒ (Slide)² = 64
   ⇒ Side = 8cm
   Perimeter of equilateral triangle = 3 × side = 3 × 8 = 24cm

2. Rs 2.16
   Solution:
   Given: a = 6cm, b = 8cm, c = 10cm.
   $\text{s}=(\frac{6+8+10}{2})=12\text{cm}$
   Hence, by using Heron’s formula, we can write:
   $\text{A}=\sqrt{12(12-6)(12-8)(12-10)}=\sqrt{(12)(6)(4)(2)}=\sqrt{576=24\text{cm}^2}$
   $\text{Therefore, the cost of painting at a rate of 9 paise per cm}^2=24\times9\text{paise}=\text{Rs.2.16}$

3. 450cm²

Solution:

Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$

$=\frac{1}{2}\times30\times30=450\text{ sq.cm}$

3. $32\sqrt{2}\text{cm}^2$

Solution:

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 = 12cm

Therefore, the sides are 12cm, 12cm and 8cm

Area $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$=\sqrt{16(16-12)(16-12)(16-3)}$

$=\sqrt{16\times4\times4\times8}$

$=32\sqrt{2}\text{sq}.\text{cm.}$

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

Solution:

Area of regular hexagon $=\frac{3\sqrt{3}}{2}(\text{Side})^2$

$=\frac{3\sqrt{3}}{2}\times4\times4$

$=24\sqrt{3}\text{cm}^2$

4. 192cm²

Solution:

$\text{S}=\frac{(24+20+20)}{2}=32\text{cm}$

$\text{Area}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$=\sqrt{32(32-24)(32-20)(32-20)}$

$=192\text{sq}.\text{cm}.$

3. 24cm²

Solution:

Perpendicular $=\sqrt{10^2-8^2}=\sqrt{100-64}=6\text{cm}$

Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$

$=\frac{1}{2}\times8\times6=24\text{ sq.cm}$

1. 2560m²

Solution:

According to the question,

Area of the field $=\frac{1}{2}\times128\times17.3+\frac{1}{2}\times128\times22.7$

$=\frac{1}{2}\times128(17.3+22.7)$

$=2560\text{ sq.m}$

2. 30cm²
   Solution:
   Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
   Area of $\triangle\text{ABC}=\frac{1}{2}\times12\times5=30\text{cm}^2$

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
   $\frac{8}{4}\times\sqrt{4(6)^2-8^2}$
   $=\frac{8}{4}\times\sqrt{144-64}$
   $=\frac{8}{4}\times\sqrt{80}$
   $=\frac{8}{4}\times4\sqrt{5}$
   $=8\sqrt{5}\text{cm}^2$

3. 384m²

Solution:

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 $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$=\sqrt{48(48-40)(48-24)(48-32)}$

$=\sqrt{48\times8\times24\times16}$

$=\sqrt{24\times2\times8\times24\times8\times2}$

$=24\times8\times2$

$=384\text{m}^2$ 

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

3. $\frac{20}{\sqrt{29}}\text{cm}$
   Solution:
   Area of right angle triangle = 20 sq. m
   $\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height} = 20 $
   $\Rightarrow\frac{1}{2}\times\text{Base}\times4=20$
   $\Rightarrow\text{Base}=10\text{cm}$
   Then, Hypotenuse $=\sqrt{10^2+4^2}=2\sqrt{29}\text{m}$
   If the altitude drawn to the hypotenuse of a right-angle triangle, then the length of required altitude $=\frac{10\times4}{2\sqrt{29}}=\frac{20}{\sqrt{29}}\text{cm}$

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

1. $\frac{4}{\sqrt{3}}\text{cm}$
   Solution:
   If side of an equilateral triangle is 'a', then its
   Area $=\frac{\sqrt{3}}{4}\text{a}^2$
   Now, $\frac{\sqrt{3}}{4}\text{a}^2=4\sqrt{3}$
   ⇒ a² = 4²
   ⇒ a = 4cm
   Hence, correct option is (a).

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
   $\Rightarrow3\text{s}-(\text{a+b+c})=20\Big[\text{s}=\frac{\text{a+b+c}}{2}\Big]$
   $\Rightarrow3\text{s}-2\text{s}=20$
   $\Rightarrow\text{s}=20\text{cm}$

3. 5.196cm²
   Solution:
   Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
   $=\frac{\sqrt{3}}{4}\times2\sqrt{3}\times2\sqrt{3}$
   $=3\sqrt{3}$
   $=3\times1.732=5.196\text{sq.cm}$

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 $=\frac{1}{2}\times8\times\text{h}$
   $\Rightarrow \frac{1}{2}\times8\times\text{h}=3\sqrt{15}$
   $\Rightarrow\text{h}=\frac{3}{4}\sqrt{15}\text{cm}$

3. $16\sqrt{3}\text{cm}^2$
   Solution:
   Here, $\text{x}\sqrt{3}=\sqrt{48}$
   $\Rightarrow\text{x}=\sqrt{16}$
   Side = 2x
   Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text{Side})^2$
   $=\frac{\sqrt{3}}{4}(2\text{x})^2$
   $=\sqrt{3}\text{x}^2\text{ sq.cm}$
   $=\sqrt{3}(\sqrt{16})^2=16\sqrt{3}\text{cm}$

1. 1320m²
   Solution:
   Given: (s - a)(s - b)(s - c) = 13200m and s = 132m
   Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   $=\sqrt{13200\times132}$
   $1320\text{m}^2$

1. 5.196cm²
   Solution:
   Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$
   $=\frac{\sqrt{3}}{4}\big(2\sqrt{3}\big)^2=3\sqrt{3}=3\times1.732$
   $=5.196\text{cm}^2$

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

Solution:

$\text{s}=\frac{1+2+2}{2}=\frac{5}{2}\text{m}$

Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$=\sqrt{\frac{5}{2}(\frac{5}{2}-1)(\frac{5}{2}-2)(\frac{5}{2}-2)}$

$=\sqrt{\frac{5}{2}\times\frac{3}{2}\times\frac{1}{2}\times\frac{1}{2}}$

$=\frac{\sqrt{15}}{4}\text{ sq.m}$

2. 2448 sq. cm

Solution:

Let the two adjacent sides of the parallelogram be a = 74cm, b = 40cm

Let the length of diagonal be c = 102cm

These two sides and the diagonal forms a triangle

semi perimeter, $\text{s}= \frac{(\text{a} + \text{b} + \text{c})}{2}$​​​​​​​

$\text{s}=\frac{(74+40+102)}{2}$

$=\frac{216}{2}$

$=108\text{cm}$

By Heron's formula, we have area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

Area of triangle $=\sqrt{108(108-74)(108-40)(108-102)}$

$=1224\text{cm}^2$

therefore, area of parallelogram = 1224 × 2

= 2448 sq. cm

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

Solution:

Area of quadrilateral triangle $=\frac{\sqrt{3}}{4}\times(\text{side})^2$

$=\frac{\sqrt{3}}{4}\times(8)^2$

$=\frac{\sqrt{3}}{4}\times64$

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

1.  $12\sqrt{5}\text{cm}^2$

Solution:

Let a = 7cm, b = 9cm, c = 14cm

Semi-perimeter = $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{7+9+14}{2}=15\text{cm}$

s - a = 15 -7 = 8cm, s - b = 15 - 9 = 6cm and s - c = 15 - 14 = 1cm

Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$=\sqrt{15\times8\times6\times1}$

$=\sqrt{5\times3\times4\times2\times3\times2}$

$=2\sqrt{5}\text{cm}^2$

Hence, correct option is (a).

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

Solution:

Height of equilateral triangle $=\frac{\sqrt{3}}{2}\times\text{Side}$

$=\frac{\sqrt{3}}{2}\times10$

$=5\sqrt{3}\text{cm}$

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)
   $=\frac{1}{2}\times\sqrt{100-36}$
   $=\frac{1}{2}\times\sqrt{64}$
   $=\frac{1}{2}\times8$
   $=4\text{cm}$

1. 10cm

Solution:

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 $=\sqrt{8^2+6^2}=10\text{cm}.$

3. 30cm²

Solution:

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. $40\text{cm}$

Solution:

$\text{d}_2=\frac{\text{Area}\times2}{\text{d}_1}$

$=\frac{96\times2}{12}$

$=16\text{cm}$

length of side of rhombus $=\sqrt{6^2+8^2}=10\text{cm}$

peimeter of rhombus = 4 × side

=4 × 10 = 40cm

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

Solution:

Perimeter of triangle = 3a

Now, $3\text{a}=60$

$\Rightarrow\text{ a}=60\div3=20\text{m}$

Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$

$=\frac{\sqrt{3}}{4}\times(20)^2=100\sqrt{3}\text{m}^2$

1. 2448 sq.cm

Solution:

Let the two adjacent sides of the parallelogram be a = 74cm, b = 40cm

Let the length of diagonal be c = 102cm

These two sides and the diagonal forms a triangle

semi perimeter, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$

$\text{s}=\frac{74+40+102}{2}$

$=\frac{216}{2}$

= 108cm

By Heron's formula, we have area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

Area of triangle $=\sqrt{108(108-74)(108-40)(108-102)}$

= 1224cm²

therefore, area of parallelogram = 1224 × 2

= 2448 sq.cm

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

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

Solution:

$\text{Side}=\frac{24}{3}=8\text{m}$

$\text{Area}=\frac{\sqrt{3}}{4}\times8\times8$

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