Maths M.C.Q

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

Solution:

$\text{s}=\frac{5+7+8}{2}=10\text{cm}$

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

$=\sqrt{10(10-5)(10-7)(10-8)}$

$=\sqrt{10\times5\times3\times2}$

$=10\sqrt{3}\text{ sq.cm}$

1. 48cm²

Solution:

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 = x² + 6²

⇒ 4 + 4x = 36

⇒ x = 8cm

Now, area of triangle $ = \frac{1}{2}\times \text{Base}\times\text{Height}$

$=\frac{1}{2}\times12\times8=48\text{cm}^2$

1. 5.196cm²

Solution:

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

Solution:

Given, Perimeter = 180cm

3a = 180 (Equilateral triangle)

a = 60cm

Semi-perimeter $\frac{180}{2}=90\text{cm}$

Now as per Heron’s formula,

$\text{A}=\sqrt{\text{a}(8-\text{a})(8-\text{b})(8-\text{c})}$

In the case of an equilateral triangle, a = b = c = 60cm

Substituting these values in the Heron’s formula, we get the area of the triangle as:

$\text{A}=\sqrt{90(90-60)(90-60)(90-60)} $

$ = \sqrt{(90\times{30}\times{30}\times{30})}$

$\text{A}=900\sqrt{3\text{cm}^2}$

4. 8

Solution:

Let the sides of $\triangle\text{ABC}$ be n, n + 1, n + 2

⇒ Perimeter = n + n + 1 + n + 2

⇒ (9 + 9 + 9) = 3n + 3

⇒ 27 = 3n + 3

⇒ 3n = 24

⇒ n = 8cm

Thus, the shortest side is 8cm.

2. Smallest

Solution:

Length of the perpendicular drawn on the longest side of a scale is $\triangle$ smallest.

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

4. 12x² sq.units.
   Solution:
   $\text{s}=\frac{5\text{x}+5\text{x}+8\text{x}}{2}=9\text{x}\text{ cm}$
   Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   $=\sqrt{9\text{x}(9\text{x}-5\text{x})(9\text{x}-5\text{x})(9\text{x}-8\text{x})}$
   $=\sqrt{9\text{x}\times4\text{x}\times4\text{x}\times\text{x}}$
   $=12\text{x}^2\text{ sq.cm}$

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.

Solution:

Height of isoscale 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)

$=\frac{1}{2}\times\sqrt{100-36}$

$=\frac{1}{2}\times\sqrt{64}$

$=\frac{1}{2}\times8$

$=4\text{cm}$

2. Rs. 2.16
   Solution:
   Since, the edges of a triangular are a = 6cm, b = 8cm and c = 10cm
   Now, semi-perimeter of a triangular board.
   $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
   $=\frac{6+8+10}{2}=\frac{24}{2}=12\text{cm}$
   Now, area of a triangular board $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
   $=\sqrt{12(12-6)(12-8)(12-10)}$
   $=\sqrt{12\times6\times4\times2}$
   $=\sqrt{(12)^2\times(2)^2}$
   $=12\times2=24\text{cm}^2$
   Since, the cost of painting for area 1cm² = Rs. 0.09
   $\therefore$ Cost of paint for area 24cm² = 0.09 × 24 = Rs. 2.16
   Hence, the cost of a triangular board is Rs. 2.16

3. 3k

Solution:

Semiperimeter of scalene triangle of side k, 2k and 3k $=\frac{\text{k}+2\text{k}+3\text{k}}{2}=3\text{k}$

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

Solution:

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

The smallest side is 11m

Area $=\frac{1}{2}\times11\times\text{Height}\ ....(\text{i})$

Area by Heron's Formula $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$

$\text{s}=\frac{11+60+61}{2}=66\text{m}$

Area $=\sqrt{66\times55\times6\times5}=330\text{m}^2$

From eq (i)

$330=\frac{1}{2}\times11\times\text{height}$

$\text{Height}=\frac{2\times330}{11}=60\text{m}$

3. 768m²

Solution:

According to the question,

Area of given quadrilateral $=\frac{1}{2}\times\text{Product of diagonals}$

$=\frac{1}{2}\times48\times32$

$=768\text{ sq.m}$

1. 0.003m²

Solution:

$\text{s}=\frac{5+12+13}{2}=15\text{cm}$

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

Solution:

Since diagonals of a rhombus bisect each other at right angle.

$\text{OB}=\frac{24}{2}=12\text{cm}$ and $\text{OC}=\frac{10}{2}=5\text{cm}$

In triangle OBC,

$\text{BC}=\sqrt{12^2+5^2}=\sqrt{144+25}=13\text{cm}$

Perimeter of rhombus $=4\times\text{side}=4\times13=52\text{cm}$

4. 30cm²

Solution:

30cm²

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

Are of $\triangle\text{ABC}=\frac{1}{2}\times12\times5$

= 30cm²

3. 336cm²

Solution:

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

1. 18750m²

Solution:

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

$=\sqrt{375\times50\times75\times250}$

$=\sqrt{15\times25\times25\times2\times3\times25\times25\times10}$

$=\sqrt{\underline{25\times25}\times\underline{25\times25}\times\underline{30\times30}}$

$=25\times25\times30$

$= 18750\text{m}^2$

Hence, correct option is (a).

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

Solution:

Given: Base = 8cm and Hypotenuse = 10cm

$\text{Hence,hight}=\sqrt{(10)^2-(8)^2}=\sqrt{36=6\text{cm}}$

$\text{Therefore,area}=(\frac {1}{2})\times\text{b}\times\text{h} =(\frac{1}{2})\times8\times6=24\text{cm}^2$

1. 12cm, 6cm²

Solution:

Perimeter of triangle = 3 + 4 + 5 = 12cm

Now, $\text{s}=\frac{3+4+5}{2}=\text{6cm}$

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

Solution:

Area of equilateral triangle $=\frac{\sqrt3}{4}\times(\text{Side})^2$

$=\frac{\sqrt3}{4}\times(8)^2$

$=\frac{\sqrt3}{4}\times64$

$=16\sqrt3\text{cm}^2$

2. 36cm

Solution:

Let $\triangle\text{PQR}$ be an isoscale triangle and $\text{PX}\bot\text{QR}.$

Now,

Area of triangle = 48cm²

$\Rightarrow\frac{1}{2}\times\text{QR}\times\text{PX}=48$

$\Rightarrow\text{h}=\frac{96}{16}=6\text{cm}$

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 = 36cm

4. 43.3cm²

Solution:

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

$=\frac{1.732}{4}\times10\times10$

$=43.3\text{cm}^2$

1. 16cm²

Solution:

$\frac{\text{AD}}{\text{DC}}=\frac{3}{2}$

Let AD = 3x and DC = 2x

Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{AC}\times\text{BE}$ (BE = h)

$\Rightarrow40=\frac{1}{2}\times5\text{x}\times\text{h}$

$\Rightarrow80=5\times\text{h}$

$\Rightarrow\text{xh}=16\text{cm}^2$

Area of $\triangle\text{ABD}=\frac{1}{2}\times3\text{x}\times\text{h}$

$=3\times\frac{\text{h}}{2}=\frac{3}{2}\times16=24\text{cm}^2$

$\text{Area of }\triangle\text{BDC}=\text{Area of }\triangle\text{ABC}-\text{Area of }\triangle\text{ABD}$

$=40-24=16\text{cm}^2$

2. 48cm²

Solution:

Area of triangle $=\frac{1}{2}\times\text{base}\times\text{height}$

$=\frac{1}{2}\times12\times8=48\text{cm}$

1. 24cm

Solution:

Let:

a = 30cm, b = 24cm and c = 18cm

$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{30+24+18}{2}=36\text{cm}$

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 18cm.

Hence, the altitude of the triangle corresponding to 18cm is given by:

Area of triangle = 216cm²

$\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height}=216$

$\Rightarrow\text{Height}=\frac{216\times2}{18}=24\text{cm}$

2. 40cm

Solution:

Area of rhombus $=\frac{1}{2}\times\text{Product of diagonal}$

$\Rightarrow200=\frac{1}{2}\times10\times\text{d}_2$

$\Rightarrow\text{d}_2=\frac{200\times2}{10}=40\text{cm}$

2. 8cm²
   Solution:
   According to question, in given right-angled triangle AB and AC are Base and Perpendicular respectively.
   $\text{Area}=\frac{1}{2}\times4\times4$
   = 8cm².

3. 1344cm²
   Solution:
   Since, the three sides of a triangle are a = 56cm, b = 60cm and c = 52cm
   Then, semi-perimeter of a triangle,
   $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{56+60+52}{2}=\frac{168}{2}=84\text{cm}$
   Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]
   $=\sqrt{84(84-56)(84-60)(84-52)}$
   $=\sqrt{4\times7\times3\times4\times7\times4\times2\times3\times4\times4\times2}$
   $=\sqrt{(4)^6\times(7)^2\times(3)^2}$
   $=(4)^3\times7\times3=1344\text{cm}^2$
   Hence, the area of triangle is 1344cm²

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

2. 24cm²

Solution:

$\text{Side}=\frac{20}{4}=5\text{cm}$

$\text{half diagnoal} =\sqrt{5^2 - 3^2} = 4\text{cm }$

$\text{diagnoal} = 4 \times 2 = 8 \text{cm} $

$\text{Area}=\frac{1}{2}\times6\times8=24\text{cm}^2$

1. 36cm

Solution:

Let $\triangle\text{PQR}$ be an isosceles triangle and PX $\perp$ QR

Now,

Area of triangle = 48cm²

$\Rightarrow\frac{1}{2}\times\text{QR}\times\text{PX}=48$

$\Rightarrow\text{h}=\frac{96}{16}=6\text{cm}$

Also,

$\text{QX}=\frac{1}{2}\times24=12\text{cm}$ and PX =12cm

$\text{PQ}=\sqrt{\text{QX}^2+\text{PX}^2}$

$\text{a}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{cm}$

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

2. 30cm

Solution:

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 = 70cm, s - BC = 42cm, s - AC = 8cm

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 × BD = 1680cm²

$\Rightarrow\text{BD}=\frac{1680}{56}=30\text{cm}$

Hence, correct option is (b).

1. 16cm²

Solution:

$\frac{\text{AD}}{\text{DC}}=\frac32$

Let AD = 3x and DC = 2x

Area of $\triangle\text{ABC}=\frac12\times\text{AC}\times\text{BC}$ (BE = h)

$\Rightarrow40=\frac12\times5\text{x}\times\text{h}$

⇒ 80 = 5xh

⇒ xh = 16cm² ....(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 $\triangle\text{ABD}=40-24=16\text{cm}^2$

Hence, correct option is (a).

4. 6cm²

Solution:

Since in a right-angled triangle, the circumcentre is the mid-point of the hypotenuse, then

Hypotenuse = 2 × 3 = 6cm

Now, Area of right-angled triangle $=\frac{1}{2}\times\text{Base}\times\text{Altitude}$

$=\frac{1}{2}\times6\times2=6\text{sq}.\text{cm}.$

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

Solution:

$\text{s}=\frac{11+60+61}{2}=66\text{m}$

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

$=\sqrt{21\times10\times6\times5}=30\sqrt{7}\text{cm}^2$

Also if we choose largest side and its Altitude, the area would be

$\text{A}=\frac12\times\text{largest side}\times\text{h}$

$330=\frac12\times11\times\text{Height}$

$\Rightarrow\text{Height}=\frac{2\times330}{11}=60\text{m}$

Hence, correct option is (d).

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

Solution:

Area of quadrilateral triangle $=\frac{\text{b}}{4}\sqrt{4\text{a}^2-\text{b}^2}$

Here,

a = 6cm and b = 8cm

Thus, we have:

$=\frac{8}{4}\times\sqrt{4(6)^2-8^2}$

$=\frac{8}{4}\times\sqrt{144-64}$

$=\frac{8}{4}\times\sqrt{80}$

$=\frac{8}{4}\times4\sqrt{5}$

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