M.C.Q

Question 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:

$24\sqrt{2}\text{cm}^2$
$24\sqrt{3}\text{cm}^2$
$48\sqrt{3}\text{cm}^2$
$64\sqrt{3}\text{cm}^2$

Answer

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

Solution:

If side of a square is a cm
Then, its diagonal $=\sqrt{2}\text{a}\text{cm}$
But diagonal $=12\sqrt{2}\text{cm}$
$\Rightarrow\sqrt{2}\text{a}=12\sqrt{2}$
⇒ a = 12cm
⇒ Perimeter of a square = 4a = 4 × 12 = 48cm
Now, perimeter of an equilateral triangle with side x = 3x cm
But perimeter of equilateral triangle = Perimeter of square
⇒ 3x = 48
⇒ x = 16cm
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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Question 21 Mark

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

11cm
66cm
50cm
60cm

Answer

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).

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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, $\frac{\sqrt{3}}{4}\text{a}^2=4\sqrt{3}$
$\Rightarrow a^2 = 4^2$
$\Rightarrow a = 4\ cm$

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Question 41 Mark

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

$100\sqrt{2}\%$
$200\%$
$300\%$
$400\%$

Answer

$300\%$

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).

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Question 51 Mark

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

$12\sqrt{5}\text{cm}^2$
$12\sqrt{3}\text{cm}^2$
$24\sqrt{5}\text{cm}^2$
$63\text{cm}^2$

Answer

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

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).

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Question 61 Mark

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

$30\sqrt{7}\text{cm}$
$\frac{15\sqrt{7}}{2}\text{cm}$
$\frac{15\sqrt{7}}{2}\text{cm}$
$30\text{cm}$

Answer

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

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).

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Question 71 Mark

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

$225\text{cm}^2$
$225\sqrt{3}\text{cm}^2$
$225\sqrt{2}\text{cm}^2$
$450\text{cm}^2$

Answer

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

Solution:
Let a = 16cm, b = 30cm, c = 34cm
Semi-perimeter of a triangle $=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{16+30+34}{2}=40$
Now, s - a = 24cm, s - b = 10cm and s - c = 6cm
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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Question 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 9cm, what is the length of the shortest side of $\triangle\text{ABC}?$

Answer

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
⇒ 3n = 24
⇒ n = 8cm
Thus, the shortest side is 8cm.
Hence, correct option is (c).

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Question 91 Mark

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

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

Answer

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

Solution:

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
$\Rightarrow\text{a}+\text{a}+\text{a}\sqrt{2}$
$=4+4+4\sqrt{2}\text{ cm}$

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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.

$30\ cm^2$

$\text{AB}=\sqrt{(13)^2-(5)^2}$
$=12\text{ cm}$
Area $=\frac{1}{2}\times\text{BC}\times\text{AB}$
$=\frac12\times5\times12$
$=30\text{ cm}^2$

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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 $\triangle\text{ABC}=\frac12\times\text{AC}\times\text{BC}$ (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 $\triangle\text{ABD}=40-24=16\text{ cm}^2$

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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 $\triangle=\frac12\times\text{AC}\times\text{BD}=\frac12\times112\times\text{BD}=56\times\text{BD}$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{50+78+115}{2}=120\ cm$
$s - AB = 70\ cm, s - BC = 42\ cm, s - AC = 8\ cm$
Area $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{120\times70\times42\times8}=1680 \ cm^2$
Now, $56 \times BD = 1680\ cm^2$
$\Rightarrow\text{BD}=\frac{1680}{56}=30\ cm$
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 $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{375\times50\times75\times250}$
$=\sqrt{15\times25\times25\times2\times3\times25\times25\times10}$
$=\sqrt{\underline{25\times25}\times\underline{25\times25}\times\underline{30\times30}}$
$=25\times25\times30$
$= 18750 m^2$
Hence, correct option is $(a).$

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