MCQ(1M)

MCQ 11 Mark

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $12\sqrt{2}\text{ cm},$ then area of the triangle is:

A
$24\sqrt{2}\text{ cm}^2$
B
$24\sqrt{3}\text{ cm}^2$
C
$48\sqrt{3}\text{ cm}^2$
✓
$64\sqrt{3}\text{ cm}^2$

Answer

Correct option: D.

$64\sqrt{3}\text{ cm}^2$

If side of a square is a $\ cm$
Then, its diagonal $=\sqrt{2}\text{a}\text{ cm}$
But diagonal $=12\sqrt{2}\text{ cm}$
$\Rightarrow\sqrt{2}\text{a}=12\sqrt{2}$
$\Rightarrow a = 12\ cm$
$\Rightarrow$ Perimeter of a square $= 4a = 4 \times 12 = 48\ cm$
Now, perimeter of an equilateral triangle with side $x = 3x \ cm$
But perimeter of equilateral triangle $ =$ Perimeter of square
$\Rightarrow 3x = 48$
$\Rightarrow x = 16\ cm$
Now, Area of equilateral $\triangle$
$=\frac{\sqrt{3}\text{x}^2}{4}=\frac{\sqrt{3}}{4}\times16\times16=64\sqrt{3}\text{ cm}^2$
Hence, correct option is $(d)$.

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MCQ 21 Mark

The sides of a triangle are $11\ m, 60\ m$ and $61\ m$. The altitude to the smallest side is:

A
$11\ cm$
B
$66\ cm$
C
$50\ cm$
✓
$60\ cm$

Answer

Correct option: D.

$60\ cm$

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

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MCQ 31 Mark

The length of each side of an equilateral triangle of area $4\sqrt{3}\text{cm}^2,$ is:

✓
$4\ \text{cm}$
B
$\frac{4}{\sqrt{3}}\ \text{cm}$
C
$\frac{\sqrt{3}}{4}\ \text{cm}$
D
$3\ \text{cm}$

Answer

Correct option: A.

$4\ \text{cm}$

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

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MCQ 41 Mark

If every side of a triangle is doubled, then increase in the area of the triangle is:

A
$100\sqrt{2}\%$
B
$200\%$
✓
$300\%$
D
$400\%$

Answer

Correct option: C.

$300\%$

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

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MCQ 51 Mark

The sides of a triangle are $7\ cm, 9\ cm$ and $14\ cm$. Its area is:

✓
$12\sqrt{5}\text{ cm}^2$
B
$12\sqrt{3}\text{ cm}^2$
C
$24\sqrt{5}\text{ cm}^2$
D
$63\text{ cm}^2$

Answer

Correct option: A.

$12\sqrt{5}\text{ cm}^2$

Let $a = 7\ cm, b = 9\ cm, c = 14\ cm$
Semi $-$ perimeter $= \text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{7+9+14}{2}=15\text{ cm}$
$s - a = 15 -7 = 8\ cm, s - b = 15 - 9 = 6\ cm$ and $s - c = 15 - 14 = 1\ cm$
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 $(a).$

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MCQ 61 Mark

The sides of a triangle are $11\ cm, 15\ cm$ and $16\ cm$. The altitude to the largest side is:

A
$30\sqrt{7}\text{ cm}$
B
$\frac{15\sqrt{7}}{2}\text{ cm}$
✓
$\frac{15\sqrt{7}}{2}\text{ cm}$
D
$30\text{ cm}$

Answer

Correct option: C.

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

$\text{s}=\frac{11+60+61}{2}=66\text{ m}$
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 $(d)$.

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MCQ 71 Mark

The sides of a triangle are $16\ cm, 30\ cm, 34\ cm$. Its area is:

A
$225\text{ cm}^2$
✓
$225\sqrt{3}\text{ cm}^2$
C
$225\sqrt{2}\text{ cm}^2$
D
$450\text{ cm}^2$

Answer

Correct option: B.

$225\sqrt{3}\text{ cm}^2$

Let $a = 16\ cm, b = 30\ cm, c = 34\ cm$
Semi $-$ perimeter of a triangle $=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{16+30+34}{2}=40$
Now $, s - a = 24\ cm, s - b = 10\ cm$ and $s - c = 6\ cm$
By Heron's formula.
Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{40\times24\times10\times6}$
$=\sqrt{4\times10\times4\times6\times10\times6}$
$=\sqrt{4^2\times10^2\times6^2}$
$=4\times10\times6$
$= 240\text{ cm}^2$
Note: Correct option not given.

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MCQ 81 Mark

The lengths of the sides of $\triangle\text{ABC}$ are consecutive integers. It $\triangle\text{ABC}$ has the same perimeter as an equilateral triangle with a side of length $9\ cm,$ what is the length of the shortest side of $\triangle\text{ABC}?$

A
$4$
B
$6$
✓
$8$
D
$10$

Answer

Correct option: C.

$8$

Let the sides of $\triangle\text{ABC}$ be $n, n + 1, n + 2.$
$\Rightarrow$ Perimeter $= n + n + 1 + n + 2$
$\Rightarrow (9 + 9 + 9) = 3n + 3$
$\Rightarrow 3n = 24$
$\Rightarrow n = 8\ cm$
Thus, the shortest side is $8\ cm.$
Hence, correct option is $(c)$.

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MCQ 91 Mark

If the length of a median of an equilateral triangle is $x \ cm,$ then its area is:

A
$\text{x}^2$
B
$\frac{\sqrt{3}}{2}\text{x}^2$
✓
$\frac{\text{x}^2}{\sqrt{3}}$
D
$\frac{\text{x}^2}{2}$

Answer

Correct option: C.

$\frac{\text{x}^2}{\sqrt{3}}$

Let the side of equilateral $\triangle\text{ABC}$ be $a \ cm$
The median of equilateral triangle is its altitude drawn from $A$ to $BC$.
$\big($i.e. the height of $\triangle$ over Base $BC\big)$
$\Rightarrow\text{x}=\frac{\text{a}\sqrt{3}}{2}\ [AD = x($given$)]$
$\Rightarrow\text{a}=\frac{2\text{x}}{\sqrt{3}}$
Area of equilateral $\triangle$ of side a
$=\frac{\sqrt{3}\text{a}^2}{4}$
$=\frac{\sqrt{3}}{4}\Big(\frac{2\text{x}}{\sqrt{3}}\Big)^3$
$=\frac{\text{x}^2}{\sqrt{3}}$
Hence, correct option is $(c).$

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MCQ 101 Mark

If the area of an isosceles right triangle is $8\ cm^2$, what is the perimeter of the triangle?

A
$8+\sqrt{2}\text{cm}^2$
✓
$8+4{\sqrt{2}}\text{cm}^2$
C
$4+8{\sqrt{2}}{}\text{cm}^2$
D
$12\sqrt{2}\text{cm}^2$

Answer

Correct option: B.

$8+4{\sqrt{2}}\text{cm}^2$

Let each of the two equal sides of an isosceles right triangle be a $cm.$
Then, third side $=\text{a}\sqrt{2}\ \text{cm}$
Area of $\triangle=\frac12\times\text{a}\times\text{a}$
$\Rightarrow8=\frac{\text{a}^2}{2}$
$\Rightarrow a^2 = 16$
$\Rightarrow a = 4\ cm$
$\Rightarrow$ Perimeter $\Rightarrow\text{a}+\text{a}+\text{a}\sqrt{2}=4+4+4\sqrt{2}\ \text{cm}$
Hence, correct option is $(b).$

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MCQ 111 Mark

The base and hypotenuse of a right triangle are respectively $5\ cm$ and $13\ cm$ long. Its area is:

A
$25\ cm^2$
B
$28\ cm^2$
✓
$30\ cm^2$
D
$40\ cm^2$

Answer

Correct option: C.

$30\ cm^2$

$\text{AB}=\sqrt{(13)^2-(5)^2}=12\ \text{cm}$
$\text{Area}=\frac{1}{2}\times\text{BC}\times\text{AB}=\frac12\times5\times12=30\ \text{cm}^2$
Hence, correct option is $(c).$

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MCQ 121 Mark

In the given figure, the ratio $AD$ to $DC$ is $3$ to $2$. If the area of $\triangle\text{ABC}$ is $40\ cm^2$, what is the area of $\triangle\text{BDC}?$

✓
$16\ cm^2$
B
$24\ cm^2$
C
$30\ cm^2$
D
$36\ cm^2$

Answer

Correct option: A.

$16\ cm^2$

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

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MCQ 131 Mark

The sides of a triangle are $50\ cm$, $78\ cm$ and $112\ cm.$ The smallest altitude is:

A
$20\ cm$
✓
$30\ cm$
C
$40\ cm$
D
$50\ cm$

Answer

Correct option: B.

$30\ cm$

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, $56 \times BD = 1680\ cm^2$
$\Rightarrow\text{BD}=\frac{1680}{56}=30\text{cm}$
Hence, correct option is $(b)$.

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MCQ 141 Mark

The sides of a triangular field are $325m$, $300m$ and $125m$. Its area is:

✓
$18750m^2$
B
$37500m^2$
C
$97500m^2$
D
$48750m^2$

Answer

Correct option: A.

$18750m^2$

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

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