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
$48\sqrt{3}\text{cm}^2$
- ✓
$64\sqrt{3}\text{cm}^2$
AnswerCorrect 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 $=12\sqrt{2}\text{cm}$
$\Rightarrow\sqrt{2}\text{a}=12\sqrt{2}$
$\Rightarrow a = 12cm$
$\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 $(d)$. View full question & answer→MCQ 21 Mark
The sides of a triangle are $11m, 60\ m$ and $61\ m$. The altitude to the smallest side is:
- A
$11\ cm$
- B
$66\ cm$
- C
$50\ cm$
- ✓
$60\ cm$
AnswerCorrect option: D. $60\ cm$
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 $(d)$.
View full question & answer→MCQ 31 Mark
The length of each side of an equilateral triangle of area $4\sqrt{3}\text{cm}^2,$ is:
AnswerCorrect option: A. $4\text{cm}$
If side of an equilateral triangle is $'a'$, then its
Area $=\frac{\sqrt{3}}{4}\text{a}^2$
Now, $\frac{\sqrt{3}}{4}\text{a}^2=4\sqrt{3}$
$\Rightarrow a^2=4^2$
$\Rightarrow a=4 \mathrm{~cm}$
Hence, correct option is $(a)$.
View full question & answer→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
$400\%$
AnswerCorrect 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, $\text{s}'=\frac{\text{a}'+\text{b}'+c'}{2}=\frac{2\text{a}+2\text{b}+2\text{c}}{2}=2\text{s}$
$\text{A}'=\sqrt{\text{s}'(\text{s},-\text{a}')(\text{s}'-\text{b}')(\text{s}'-\text{c}')}$
$=\sqrt{2\text{s}(2\text{s}-2\text{a})(2\text{s}-2\text{b})(2\text{s}-2\text{c})}$
$=4\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$\Rightarrow\text{A}' = 4\text{A}$
⇒ Increase in Area $=\frac{4\text{A}-\text{A}}{\text{A}}\times100\%=300\%$
Hence, correct optin is $(c)$.
View full question & answer→MCQ 51 Mark
The sides of a triangle are $7\ cm, 9\ cm$ and $14cm$. Its area is:
- ✓
$12\sqrt{5}\text{cm}^2$
- B
$12\sqrt{3}\text{cm}^2$
- C
$24\sqrt{5}\text{cm}^2$
- D
$63\text{cm}^2$
AnswerCorrect 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 $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{15\times8\times6\times1}$
$=\sqrt{5\times3\times4\times2\times3\times2}$
$=12\sqrt{5}\text{cm}^2$
Hence, correct option is $(a)$.
View full question & answer→MCQ 61 Mark
The sides of a triangle are $11\ cm, 15\ cm$ and $16\ cm$. The altitude to the largest side is:
AnswerCorrect option: D. $60\text{cm}$
$\text{s}=\frac{11+60+61}{2}=66\text{m}$
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 $(d)$.
View full question & answer→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
$450\text{cm}^2$
AnswerCorrect 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.
View full question & answer→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 $\triangle\text{ABC}?$
AnswerLet the sides of $\triangle\text{ABC}$ be $n, n + 1, n + 2$.
$\Rightarrow $ Perimeter $= n + n + 1 + n + 2$
$\Rightarrow (9 + 9 + 9) = 3n + 3$
$\Rightarrow 3n = 24$
$\Rightarrow n = 8cm$
Thus, the shortest side is $8\ cm$.
Hence, correct option is $(c)$.
View full question & answer→MCQ 91 Mark
If the length of a median of an equilateral triangle is $x\ cm$, then its area is:
AnswerCorrect 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 $(c)$. View full question & answer→MCQ 101 Mark
If the area of an isosceles right triangle is $8cm^2$, what is the perimeter of the triangle?
AnswerCorrect 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 $\triangle=\frac12\times\text{a}\times\text{a}$
$\Rightarrow8=\frac{\text{a}^2}{2}$
$\Rightarrow a^2 = 16$
$\Rightarrow a = 4cm$
$\Rightarrow$ Perimeter
$\Rightarrow\text{a}+\text{a}+\text{a}\sqrt{2}=4+4+4\sqrt{2}\text{cm}$
Hence, correct option is $(b)$.
View full question & answer→MCQ 111 Mark
The base and hypotenuse of a right triangle are respectively $5\ cm$ and $13\ cm$ long. Its area is:
- A
$25 \mathrm{~cm}^2$
- B
$28 \mathrm{~cm}^2$
- ✓
$30 \mathrm{~cm}^2$
- D
$40 \mathrm{~cm}^2$
AnswerCorrect option: C. $30 \mathrm{~cm}^2$
$\text{AB}=\sqrt{(13)^2-(5)^2}=12\text{cm}$
$\text{Area}=\frac{1}{2}\times\text{BC}\times\text{AB}=\frac12\times5\times12=30\text{cm}^2$
Hence, correct option is $(c)$. View full question & answer→MCQ 121 Mark
In the given figure, the ratio $AD$ to $DC$ is $3$ to $2$. If the area of $\triangle\text{ABC}$ is $40 \mathrm{~cm}^2$, what is the area of $\triangle\text{BDC}?$
- ✓
$16 \mathrm{~cm}^2$
- B
$24 \mathrm{~cm}^2$
- C
$30 \mathrm{~cm}^2$
- D
$36 \mathrm{~cm}^2$
AnswerCorrect option: A. $16 \mathrm{~cm}^2$
$\frac{\text{AD}}{\text{DC}}=\frac32$
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 \mathrm{~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).
View full question & answer→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$
AnswerCorrect 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=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD}=\frac{1}{2} \times 112 \times \mathrm{BD}=56 \times \mathrm{BD}$
$\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{50+78+115}{2}=120 \mathrm{~cm}$
$\mathrm{~s}-\mathrm{AB}=70 \mathrm{~cm}, \mathrm{~s}-\mathrm{BC}=42 \mathrm{~cm}, \mathrm{~s}-\mathrm{AC}=8 \mathrm{~cm}$
$\text { Area }=\sqrt{\mathrm{s}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$
$=\sqrt{120 \times 70 \times 42 \times 8}=1680 \mathrm{~cm}^2$
Now, $56 \times \mathrm{BD}=1680 \mathrm{~cm}^2$
$\Rightarrow \mathrm{BD}=\frac{1680}{56}=30 \mathrm{~cm}$
Hence, correct option is $(b)$. View full question & answer→MCQ 141 Mark
The sides of a triangular field are $325m, 300m$ and $125m$. Its area is:
- ✓
$18750 \mathrm{~m}^2$
- B
$37500 \mathrm{~m}^2$
- C
$97500 \mathrm{~m}^2$
- D
$48750 \mathrm{~m}^2$
AnswerCorrect option: A. $18750 \mathrm{~m}^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)$.
View full question & answer→MCQ 151 Mark
The area of an equilateral triangle with side $2 \sqrt{3} cm$ is
- A
$5.196 cm^2$
- B
$0.866 cm^2$
- C
$3.496 cm^2$
- D
$1.732 cm^2$
View full question & answer→MCQ 161 Mark
The edges of a triangular board are 6 cm, 8 cm and 10 cm long. The cost of painting it at the rate of 9 paise per $cm ^2$ is
View full question & answer→MCQ 171 Mark
The sides of a triangle are 56 cm, 60 cm and 52 cm . Area of the triangle is
- A
$1322 cm^2$
- B
$1311 cm^2$
- C
$1344 cm^2$
- D
$1392 cm^2$
View full question & answer→MCQ 181 Mark
The sides of a triangle are 35 cm, 54 cm and 61 cm respectivery. The length of its longest altitude is
- A
$16 \sqrt{5} cm$
- B
$10 \sqrt{5} cm$
- C
$24 \sqrt{5} cm$
- D
View full question & answer→MCQ 191 Mark
If the area of an equilateral triangle is $16 \sqrt{3} cm^2$, then its perimeter is
View full question & answer→MCQ 201 Mark
The length of each side of an equilateral triangle having an area of $9 \sqrt{3}cm^2$ is
View full question & answer→MCQ 211 Mark
The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm is
AnswerCorrect option: A. $\sqrt{15} cm^2$
View full question & answer→MCQ 221 Mark
The perimeter of an equilateral triangle is 60 m . The area is
- A
$10 \sqrt{3} m^2$
- B
$15 \sqrt{3} m^2$
- C
$20 \sqrt{3} m^2$
- ✓
$100 \sqrt{3} m^2$
AnswerCorrect option: D. $100 \sqrt{3} m^2$
View full question & answer→MCQ 231 Mark
If an isosceles right triangle has area $8 cm^2$, then the length of its hypotenuse is
- ✓
$\sqrt{32} cm$
- B
$\sqrt{48} cm$
- C
$\sqrt{24} cm$
- D
AnswerCorrect option: A. $\sqrt{32} cm$
View full question & answer→MCQ 241 Mark
The sides of a triangle are 45 cm, 60 cm and 75 cm. The length of the altitude drawn to the longest side from its opposite vertex is
Answer(d) 36 cm
Let $s$ be the semi-perimeter of the triangle. Then,
$2 s=(45+60+75) cm \Rightarrow s=90$
Let $A$ be the area of the triangle. Then,
$\begin{array}{l}\quad A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{90(90-45)(90-60)(90-75)} \\
\Rightarrow \quad A=\sqrt{90 \times 45 \times 30 \times 15}=\sqrt{2 \times 3^2 \times 5 \times 3^2 \times 5 \times 3 \times 2 \times 5 \times 3 \times 5}=2 \times 3^3 \times 5^2 cm^2=1350 cm^2 \\
Also, \quad A=\frac{1}{2}(75 \times \text { Altitude drawn to the longest side }) \\
\Rightarrow \quad 1350=\frac{1}{2} \times 75 \times \text { Altitude drawn to the longest side } \\
\Rightarrow \quad \text { Altitude drawn to the longest side }=36 cm .\end{array}$
View full question & answer→MCQ 251 Mark
In a scalene triangle, one side exceeds the other two sides by 4 cm and 5 cm respectively and the perimeter of the triangle is 36 cm . The aren of triangle is
- A
$63 cm^2$
- B
$9 \sqrt{10} cm^2$
- C
$18 \sqrt{10} cm^2$
- ✓
$12 \sqrt{21} cm^2$
AnswerCorrect option: D. $12 \sqrt{21} cm^2$
(d) $12 \sqrt{21} cm^2$
Let the sides of the triangle be $a, a-4$ and $a-5$. It is given that perimeter of the triangle is 36 cm .
$\therefore \quad a+a-4+a-5=36 \Rightarrow 3 a=45 \Rightarrow a=15$
Thus, the lengths of the sides are $15 cm, 11 cm$ and 10 cm .
We have, $2 s=36 \Rightarrow s=18$
The area of the triangle is given by
$\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{18(18-15)(18-11)(18-10)}=12 \sqrt{21} cm^2$
View full question & answer→MCQ 261 Mark
Each side of a triangle is multiplied with the sum of the squares of the other two sides. If the sum of all such possible results is 6 times the product of the sides, then the triangle must be
Answer(a) equilateral
Let $a, b, c$ be the lengths of the sides of the triangle. According to the question
$\begin{array}{ll}& a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)=6 a b c \\
\Rightarrow & a(b-c)^2+b(c-a)^2+c(a-b)^2=0 \\
\Rightarrow & a(b-c)=0, b(c-a)=0, c(a-b)=0 \quad\left[\because a(b-c)^2 \geq 0, b(-a)^2 \geq 0 \text { and } c(a-b)^2 \geq 0\right] \\
\Rightarrow & b=c, c=a \text { and } a=b \Rightarrow a=b=c\end{array}$
So, triangle $A B C$ is equilateral.
View full question & answer→MCQ 271 Mark
The area of a right triangle is $28 cm^2$. If one of its perpendicular sides exceeds the other by 10 cm . then the length of the longest of the perpendicular is
Answer(b) 14 cm
Let the perpendicular sides be of length $x cm$ and $(x+10) cm$ respectively. Then,
$\text { Area }=28 cm^2 \Rightarrow \frac{1}{2} x(x+10)=28 \Rightarrow x^2+10 x-56=0 \Rightarrow(x+14)(x-4)=0\Rightarrow x=4$
Thus, the length of the longest perpendicular side is $(x+10) cm =14 cm$.
View full question & answer→MCQ 281 Mark
If the sum of any two sides of a triangle exceeds the third by 6 cm, the area of the triangle is
- A
$12 \sqrt{3} cm^2$
- B
$18 \sqrt{3} cm^2$
- C
$15 \sqrt{3} cm^2$
- ✓
$9 \sqrt{3} cm^2$
AnswerCorrect option: D. $9 \sqrt{3} cm^2$
(d) $9 \sqrt{3} cm^2$
Let s be the semi-perimeter of a tringle of sides a cm, b cm and c cm. Then, 2s - a + b + c. It is given that
$a+b=c+6, b+c=a+6$ and $c+a=b+6$
$\Rightarrow a+b - c=6, b+c - a=0$ and $c+a-b=0$
$\Rightarrow (a+b+c)-2 c=6,(a+b+c) - 2 a=6$ and $(a+b+c)-2 b=6$
$\Rightarrow 2 s-2 c=6,2 s-2 a=6$ and $2 s - 2 b=6 \Rightarrow s-a=3, s-b=3$ and $s - c = 3$
Adding these three, we obtain
$3 s-(a+b+c)=9 \Rightarrow 3 s-2 s=9 \Rightarrow s=9$
So , area $A$ of the triangle is given by
$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{9 \times 3 \times 3 \times 3} cm^2=9 \sqrt{3} cm^2$
View full question & answer→MCQ 291 Mark
In a triangle, the average of any two sides is 6 cm more than half of the third side. The area of the triangle is
- A
$64 \sqrt{3} cm^2$
- B
$48 \sqrt{3} cm^2$
- C
$72 \sqrt{3} cm^2$
- ✓
$36 \sqrt{3} cm^2$
AnswerCorrect option: D. $36 \sqrt{3} cm^2$
(d) $36 \sqrt{3} cm^2$
Let the lengths of three sides of the triangle be $a, b, c$ (in cms). It is given that
$\quad$$\frac{a+b}{2}=\frac{c}{2}+6, \frac{b+c}{2}=\frac{a}{2}+6$ and $\frac{c+a}{2}=\frac{b}{2}+6$
$\begin{array}{ll}\Rightarrow & a+b-c=12, b+c-a=12 \text { and } c+a-b=12
\\ \Rightarrow & (a+b+c)-2 c=12,(b+c+a)-2 a=12 \text { and }(c+a+b)-2 b=12
\\ \Rightarrow & 2 s-2 c=12,2 s-2 a=12 \text { and } 2 s-2 b=12, \text { where } 2 s=a+b+c
\\ \Rightarrow & s-c=6, s-a=6 \text { and } s-b=6
\\ \Rightarrow & (s-a)+(s-b)+(s-c)=6+6+6 \Rightarrow 3 s-(a+b+c)=18 \Rightarrow 3 s-2 s=18 \Rightarrow s=18\end{array}$
So, area $A$ of the triangle is given by
$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{18 \times 6 \times 6 \times 6} cm^2=36 \sqrt{3} cm^2$
View full question & answer→MCQ 301 Mark
Area of a right-angled triangle is $6 cm^2$ and its perimeter is 12 cm. Then length of its hypotenuse, is
Answer(a) 5 cm
Let the lengths of the perpendicular sides of the right-angled triangle be a cm and b cm. Then, its hypotenuse is of length $\sqrt{a^2+b^2} cm$. It is given that the perimeter of the triangle is 12 cm and area is $6 cm^2$.
$\therefore \quad \frac{1}{2} a b=6 \Rightarrow a b=12$$\quad$$\ldots$ (i)
and, $a+b+\sqrt{a^2+b^2}=12$
$\Rightarrow a+b+\sqrt{(a+b)^2-2 a b}=12$
$\Rightarrow x+\sqrt{x^2-2 \times 12}=12$, where $x=a+b$
$\Rightarrow \sqrt{x^2-24}=(12-x)$
$\Rightarrow x^2-24=(12-x)^2 \Rightarrow x^2-24=144-24 x+x^2 \Rightarrow 24 x=168 \Rightarrow x=7 \Rightarrow a+b=7$$\quad$...(ii)
$\therefore \quad Hypotenuse =\sqrt{a^2+b^2}=\sqrt{(a+b)^2-2 a b}=\sqrt{7^2-2 \times 12}=\sqrt{25}=5 cm$.
View full question & answer→MCQ 311 Mark
The perimeter of a right angled triangle is 72 cm and its area is $216 cm^2$. The sum of the lengths of its perpendicular sides is
Answer(c) 42 cm
Let the lengths of base and perpendicular be $a cm$ and $b cm$ respectively. Then, its hypotenuse is of length $\sqrt{a^2+b^2}$. It is given that the perimeter is 72 cm and area is $216 cm^2$.
$\therefore \quad a+b+\sqrt{a^2+b^2}=72$$\quad$$\ldots$ (i)
and, $\quad \frac{1}{2} a b=216 \Rightarrow a b=432$$\quad$$\ldots$ (ii)
Now, $\quad a+b+\sqrt{a^2+b^2}=72$
$\Rightarrow(a+b)+\sqrt{(a+b)^2-2 a b}=72$
$\Rightarrow(a+b)+\sqrt{(a+b)^2-2 \times 432}=72$
$\Rightarrow x+\sqrt{x^2-864}=72$$\Rightarrow \sqrt{x^2-864}=(72-x)$
$\Rightarrow x^2-864=(72-x)^2$
$\Rightarrow x^2-864=72^2-144 x+x^2 \Rightarrow 144 x=5184+864 \Rightarrow 144 x=6048 \Rightarrow x=42$$\Rightarrow a+b=42 cm$.
View full question & answer→MCQ 321 Mark
If a kite in the shape of an isosceles triangle of base 8 cm and each equal side 6 cm is to be made, the area of paper required to make the kite, is
- A
$10 \sqrt{2} cm^2$
- ✓
$8 \sqrt{5} cm^2$
- C
$10 \sqrt{2} cm^2$
- D
$8 cm^2$
AnswerCorrect option: B. $8 \sqrt{5} cm^2$
(b) $8 \sqrt{2} cm^2$
Here, $a=8 cm$ and $b=6 cm$.
$\therefore \quad \text { Area of kite }=\frac{a}{4} \sqrt{4 b^2-a^2}=\frac{8}{4} \sqrt{4 \times 36-64}=2 \sqrt{80} cm^2=8\sqrt{5} cm^2$
View full question & answer→MCQ 331 Mark
$A D$ is a median of triangle $A B C$ and area of $\triangle A D C=15 cm^2$, then ar $(\triangle A B C)$ is
- A
$15 cm^2$
- B
$22.5 cm^2$
- ✓
$30 cm^2$
- D
$37.5 cm^2$
AnswerCorrect option: C. $30 cm^2$
(c) $30 cm^2$
A median of a triangle divides it into two triangles of equal area. Therefore,
$\operatorname{ar}(\triangle A B C)=2 \operatorname{ar}(\triangle A D C)=2 \times 15 cm^2=30 cm^2$
View full question & answer→MCQ 341 Mark
The area of a right angled triangle is $240 cm^2$ and side other than hypotenuse is 30 cm , the perimeter of the triangle, is
Answer(b) 80 cm
Let the lengths of two sides, other than hypotenuse, of right triangle be $a cm$ and $b cm$. It is given that $a=30 cm$. Then,
$\text { Area }=240 cm^2 \Rightarrow \frac{1}{2} a b=240 \Rightarrow 30 b=480 \Rightarrow b=16 cm$
Applying Pythagoras theorem, we obtain
$\begin{aligned}& \text { Hypotenuse }=\sqrt{a^2+b^2}=\sqrt{30^2+16^2}=\sqrt{900+256}=\sqrt{1156} cm=34 cm \\
\therefore \quad & \text { Perimeter }=(30+16+34) cm=80 cm .\end{aligned}$
View full question & answer→MCQ 351 Mark
The perimeter of an isosceles right triangle having area $100 cm^2$ is
- A
$20 \sqrt{2} cm$
- ✓
$20(\sqrt{2}+1) cm$
- C
- D
$(10+\sqrt{2}) cm$
AnswerCorrect option: B. $20(\sqrt{2}+1) cm$
(b) $20(\sqrt{2}+1) cm$
Let $a$ be the hypotenuse and $b$ be the length of each remaining side of the triangle. Then,
$a^2=b^2+b^2 \Rightarrow a^2=2 b^2$
It is given that the area of the triangle is $100 cm^2$.
$\therefore \quad \frac{1}{2} b \times b=100 \Rightarrow b^2=200 \Rightarrow b=10 \sqrt{2} cm$
Putting $b^2=200$ in $a^2=2 b^2$, we obtain
$\begin{array}{ll}& a^2=400 \Rightarrow a=20 cm \\
\therefore \quad & \text { Perimeter }=a+2 b=(20+20 \sqrt{2}) cm=20(\sqrt{2}+1) cm\end{array}$
View full question & answer→MCQ 361 Mark
If the perimeter of an isosceles triangle is 32 cm and the ratio of the equal side to its base is $3: 2$, then area of the triangle is
- A
$16 \sqrt{2} cm^2$
- B
$20 \sqrt{2} cm^2$
- C
$30 \sqrt{2} cm^2$
- ✓
$32 \sqrt{2} cm^2$
AnswerCorrect option: D. $32 \sqrt{2} cm^2$
(d) $32 \sqrt{2} cm^2$
Let $a cm$ be the base and each equal side be $b cm$. Then,
$a+2 b=32$$\quad$[Given] ...(i)
It is given that $\frac{b}{a}=\frac{3}{2} \Rightarrow 2 b=3 a \Rightarrow b=\frac{3 a}{2}$.
Putting $b=\frac{3 a}{2}$ in (i), we obtain
$\begin{array}{ll}& a+3 a=32 \Rightarrow a=8 cm \\
\therefore \quad & b=12 cm\end{array}$
Let $A$ be the area of the triangle. Then,
$A=\frac{a}{4} \sqrt{4 b^2-a^2}=\frac{8}{4} \sqrt{4 \times 12^2-8^2} cm^2=2 \sqrt{512} cm^2=32 \sqrt{2} cm^2$
View full question & answer→MCQ 371 Mark
The area of an isosceles triangle with base 8 cm and each equal side is of length 6 cm , is
- A
$4 \sqrt{5} cm^2$
- B
$6 \sqrt{5} cm^2$
- ✓
$8 \sqrt{5} cm^2$
- D
$16 \sqrt{5} cm^2$
AnswerCorrect option: C. $8 \sqrt{5} cm^2$
(c) $8 \sqrt{5} cm^2$
The area $A$ of an isosceles triangle with base $a$ and each equal side $b$ is given by
$A=\frac{a}{4} \sqrt{4 b^2-a^2}$
Here, $a=8 cm$ and $b=6 cm$.
$\therefore \quad A=\frac{8}{4} \sqrt{4 \times 6^2-8^2}=2 \sqrt{144-64}=8 \sqrt{5} cm^2$
View full question & answer→MCQ 381 Mark
If each side of an equilateral triangle of area A is doubled, then the area of new triangle is
Answer(b) $4 A$
Let $a$ be the length of each side of the triangle. Then, $\Lambda=\frac{\sqrt{3}}{4} a^2$.
Let $A^{\prime}$ be the area of the triangle whose each side is $2 a$. Then,
$A^{\prime}=\frac{\sqrt{3}}{4}(2 a)^2=4\left(\frac{\sqrt{3}}{4} a^2\right)=4 A$
View full question & answer→MCQ 391 Mark
If the area of an equilateral triansle is $81 \sqrt{3} cm^2$, then its semi - perimeter is
Answer(d) 27 cm
let a be the length of each side. Then,
$\begin{array}{l}\text { Area }=81 \sqrt{3} cm^2 \Rightarrow \frac{\sqrt{3}}{4} \times a^2=81 \sqrt{3} \Rightarrow a^2=81 \times 4 \Rightarrow a=18 cm \\
\text { Semi-perimeter }=\frac{3}{2} a=27 cm\end{array}$
View full question & answer→MCQ 401 Mark
The area of an equilateral triangle with perimeter 12 cm is
- A
$16 \sqrt{3} cm^2$
- B
$8 \sqrt{3} cm^2$
- ✓
$4 \sqrt{3} cm^2$
- D
$6 \sqrt{3} cm^2$
AnswerCorrect option: C. $4 \sqrt{3} cm^2$
(c) $4 \sqrt{3} cm^2$
Let the length of each side be $a cm$. Then,
$\begin{aligned}& \text { Perimeter }=12 cm \Rightarrow 3 a=12 cm \Rightarrow a=4 cm \\
\therefore & \text { Area }=\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4} \times 4^2 cm^2=4 \sqrt{3} cm^2\end{aligned}$
View full question & answer→MCQ 411 Mark
The area of an equilateral triangle with altitude $2 \sqrt{3} cm$ is
AnswerCorrect option: B. $4 \sqrt{3} cm^2$
(b) $4 \sqrt{3} cm^2$
The area of an equilateral triangle with altitude $p$ is $\frac{p^2}{\sqrt{3}}$. Here, $p=2 \sqrt{3} cm$. Therefore, area $A$ is given by
$A=\frac{p^2}{\sqrt{3}}=\frac{(2 \sqrt{3})^2}{\sqrt{3}}=4 \sqrt{3} cm^2$
View full question & answer→MCQ 421 Mark
The area of an equilateral triangle with side $6 \sqrt{3} cm$ is
- A
$27 cm^2$
- ✓
$27 \sqrt{3} cm^2$
- C
$18 \sqrt{3} cm^2$
- D
$54 \sqrt{3} cm^2$
AnswerCorrect option: B. $27 \sqrt{3} cm^2$
(b) $27 \sqrt{3} cm^2$
The area $A$ of an equilateral triangle with side $a$ unit is given by $A=\frac{\sqrt{3}}{4} a^2$.
Here, $a=6 \sqrt{3} cm$
$\therefore \quad A=\frac{\sqrt{3}}{4} \times(6 \sqrt{3})^2 cm^2=27 \sqrt{3} cm^2$
View full question & answer→MCQ 431 Mark
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $12 \sqrt{2} cm$, then area of the triangle is
- A
$24 \sqrt{2} cm^2$
- B
$24 \sqrt{3} cm^2$
- C
$48 \sqrt{3} cm^2$
- ✓
$64 \sqrt{3} cm^2$
AnswerCorrect option: D. $64 \sqrt{3} cm^2$
View full question & answer→MCQ 441 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
$400 \%$
AnswerCorrect option: C. $300 \%$
View full question & answer→MCQ 451 Mark
If the length of a median of an equilateral triangle is x cm, then its area, is
- A
$x^2$
- B
$\frac{\sqrt{3}}{2} x^2$
- ✓
$\frac{x^2}{\sqrt{3}}$
- D
$\frac{x^2}{2}$
AnswerCorrect option: C. $\frac{x^2}{\sqrt{3}}$
View full question & answer→MCQ 461 Mark
In Figure, the ratio of $A D$ to $D C$ is 3 to 2. If the area of $\triangle A B C$ is $40 cm^2$, what is the area of $\triangle B D C$?
- ✓
$16 cm^2$
- B
$24 cm^2$
- C
$30 cm^2$
- D
$36 cm^2$
AnswerCorrect option: A. $16 cm^2$
View full question & answer→MCQ 471 Mark
The lengths of the sides of $\triangle A B C$ are consecutive integers. It $\triangle A B C$ has the same perimeter as an equilateral triangle with a side of length 9 cm , what is the length of the shortest side of $\triangle A B C ?$
View full question & answer→MCQ 481 Mark
If the area of an isosceles right triangle is $8 cm^2$, what is the perimeter of the triangle?
- A
$8+\sqrt{2} cm^2$
- ✓
$8+4 \sqrt{2} cm^2$
- C
$4+8 \sqrt{2} cm^2$
- D
$12 \sqrt{2} cm^2$
AnswerCorrect option: B. $8+4 \sqrt{2} cm^2$
View full question & answer→MCQ 491 Mark
If the length of each edge of a regular tetrahedron is ' $a$ ', then its surface area is
AnswerCorrect option: C. $2 \sqrt{3} a^2$ sq. units
View full question & answer→MCQ 501 Mark
If the area of a regular hexagon is $54 \sqrt{3} cm^2$, then the length of its each side is
- A
- B
$2 \sqrt{3} cm$
- ✓
- D
$6 \sqrt{3} cm$
View full question & answer→