True False[1 Marks ]

Question 11 Mark

The area of $\triangle\text{ABC}$ is $8\ cm^2$ in which $AB = AC = 4\ cm$ and $\angle\text{A}=90^\circ$

Answer

$\text{Area of }\triangle=\frac{1}{2}\times\text{base}\times\text{height}$ $=\frac{1}{2}\times4\times4=8\text{cm}^2$

Hence, the given statement is true.

Question 21 Mark

The cost of levelling the ground in the form of a triangle having the sides $51 m, 37 m$ and $20 m$ at the rate of $Rs. 3$ per $m ^2$ is $Rs. 918.$

Answer

Let sides of a triangle be $a =51 m, b =37 m$ and $c =20 m$ Now, semi-perimeter of triangle,
$s =\frac{ a + b + c }{2}=\frac{51+37+20}{2}$ $=\frac{108}{2}=54 m$
$\therefore$ Area of a triangle $=\sqrt{ s ( s - a )( s - b )( s - c )}$ [by Heron's formula]
$=\sqrt{54(54-51)(54-37)(54-20)}=\sqrt{54 \times 3 \times 17 \times 34}=\sqrt{9 \times 3 \times 2 \times 3 \times 17 \times 17 \times 2}$
$=3 \times 3 \times 2 \times 17=306 m^2$
$\because$ Cost of levelling per $m ^2=Rs.3$
$\therefore$ Cost of leavelling per $306 m^2=3 \times 306= Rs. 918$

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Question 31 Mark

If the side of a rhombus is $10\ cm$ and one diagonal is $16\ cm$, the area of the rhombus is $96\ cm^2.$

Answer

Given, side of a rhombus PQRS is $10\ cm$ and one of the diagonal is $16cm.$ i.e., $PQ = QR = RS = SP = 10cm$ and $PR = 16cm$

In $\triangle\text{POQ},\text{ PO}^2=\text{OP}^2+\text{OQ}^2$
$[$by Pythagoras theorem$] [$since, the diagonal of rhombus bisects each other at $90^\circ ]$
$\Rightarrow OQ^2 = PQ^2 - OP^2$
$\Rightarrow OQ^2 = (10)^2 - (8)^2$
$\Rightarrow OQ^2 = 100 - 64 = 36$
$\Rightarrow OQ = 6cm [$taking positive square root because length is always positive$]$
$SQ =2 \times OP = 2 \times 6= 12cm$
$\text{Area of the rhombus}=\frac{1}{2}(\text{Product of diagonals})$
$=\frac{1}{2}(\text{OS}\times\text{PR})$
$=\frac{1}{2}\times12\times16$
$=96\text{cm}^2$

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Question 41 Mark

In a triangle, the sides are given as $11\ cm, 12\ cm$ and $13\ cm$. The length of the altitude is $10.25\ cm$ corresponding to the side having length $12\ cm.$

Answer

We have the length of the altitude corresponsing to the side having length $12cm$ $2\text{s}=11\text{cm}+12\text{cm}+13\text{cm}=36\text{cm}$
$\Rightarrow\text{s}=36\div2=18\text{cm}$
$\text{Area of }\triangle=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{18(18-11)(18-12)(18-13)}$
$=\sqrt{18\times7\times6\times5}$
$=\sqrt{2\times3\times3\times7\times2\times3\times5}$
$=2\times3\sqrt{105}$
$=6\sqrt{105}\text{cm}^2$
$\text{Length of altitude}=\frac{2\text{Area of }\triangle}{\text{Base}}$
$=\frac{2\times6\sqrt{105}}{12}=\sqrt{105}=10.25\text{cm}$
Hence, the given statement is true.

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Question 51 Mark

The area of the equilateral triangle is $20\sqrt{3}\text{cm}^2$ whose each side is 8cm.

Answer

False.

Solution:

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

$=\frac{\sqrt{3}}{4}(8)^2=\frac{\sqrt{3}}{4}\times64=16\sqrt{3}\text{cm}^2$

Hence, the given statement is false.

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Question 61 Mark

The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.

Answer

False.Solution:
We see a regular hexagon is divided into six equilateral triangles. So, the area of the regular hexagon is divided side ‘a’ is the sum of area of the six equilateral triangles with side a.

Hence, the given statement that the area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a, is false.

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Question 71 Mark

The area of a triangle with base $4\ cm$ and height $6\ cm$ is $24\ cm^2$

Answer

We know that, area of a triangle $=\frac{1}{2}(\text{Base}\times\text{Height})$
Here Base $= 4\ cm$ and Height $= 6\ cm$
$\therefore$ Area of a triangle $=\frac{1}{2}\times4\times6=12\text{ cm}^2$

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Question 81 Mark

The area of the isosceles triangle is $\frac{5}{4}\sqrt{11}\text{cm}^2,$ if the perimeter is $11\ cm$ and the base is $5\ cm$

Answer

Let equal of an isosceles triangle be $b.$
$\therefore$ Perimeter of a triangle, $2\text{s}=\text{b}+\text{b}+5$
$\big[\because\ 2\text{s}=\text{a}+\text{b}+\text{c}\big]$
$\therefore\ 11=2\text{b}+5$
$\Rightarrow2\text{b}=11-5$
$\Rightarrow2\text{b}=6$
$\Rightarrow\text{b}=\frac{6}{2}=3\text{cm}$
We know that, area of an isosceles triangle, $=\frac{\text{a}}{4}\sqrt{4\text{b}^2-\text{a}^2}$
Here, sides of triangle are $a = 5\ cm$ and $b = 3\ cm$
$\therefore$ Area of an isosceles triangle $=\frac{5\sqrt{4(3)^2-(5)^2}}{4}$
$=\frac{5\sqrt{4\times9-25}}{4}$
$=5\frac{\sqrt{36-25}}{4}$
$=\frac{5\sqrt{11}}{4}\text{cm}^2$

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Question 91 Mark

The base and the corresponding altitude of a parallelogram are $10\ cm$ and $3.5\ cm,$ respectively. The area of the parallelogram is $30 \mathrm{~cm}^2$.

Answer

The base of the parallelogram is $10\ cm$ and the corresponding altitude is $3.5\ cm$
Area of $\| \mathrm{gm}=$ base $\times$ corresponding altitude
$=10 \mathrm{~cm} \times 3.5 \mathrm{~cm}=35 \mathrm{~cm}^2$
Hence, the given statement is false.

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MATHS True False[1 Marks ]

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