True False[1 Marks ]

Question 11 Mark

Write True or False and justify your answer: The area of $\triangle\text{ABC}$ is $8\ cm^2$ in which $AB = AC = 4\ cm$ and $\angle\text{A}=90^\circ$

Answer

$\text{Area of }\triangle=\frac{1}{2}\times$ base $\times$ height
$=\frac{1}{2}\times4\times4$
$=8\text{ cm}^2$
Hence, the given statement is true.

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Question 21 Mark

Write True or False and justify your answer: 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 $\text{PQRS}$ is $10\ cm$ and one of the diagonal is $16\ cm.$
$i.e., PQ = QR = RS = SP = 10\ cm$ and $PR = 16\ cm$

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 $90^\circ ]$
$\Rightarrow OQ^2 = PQ^2 - OP^2$
$\Rightarrow OQ^2 = (10)^2 - (8)^2$
$\Rightarrow OQ^2 = 100 - 64 = 36$
$\Rightarrow OQ = 6\ cm$
$[$taking positive square root because length is always positive$]$
$SQ =2 \times OP = 2 \times 6= 12\ cm$
Area of the rhombus $=\frac{1}{2}$ 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 31 Mark

Write True or False and justify your answer:
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 41 Mark

Write True or False and justify your answer: 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 $=\frac{1}{2}$ Base $\times$ 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 51 Mark

Write True or False and justify your answer:
The area of the isosceles triangle is $\frac{5}{4}\sqrt{11}\text{cm}^2,$ if the perimeter is 11cm and the base is 5cm.

Answer

True.Solution:
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 = 5cm and b = 3cm $\therefore$ 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 61 Mark

Write True or False and justify your answer: The base and the corresponding altitude of a parallelogram are $10\ cm$ and $3.5\ cm$, respectively. The area of the parallelogram is $30\ cm^2.$

Answer

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

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Question 71 Mark

Write True or False and justify your answer:In a triangle, the sides are given as 11cm, 12cm and 13cm. The length of the altitude is 10.25cm corresponding to the side having length 12cm.

Answer

True.
Solution:
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 81 Mark

Write True or False and justify your answer: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 91 Mark

Write True or False and justify your answer: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, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{51+37+20}{2}$
$=\frac{108}{2}$
$=54\text{ m}$
$\therefore$ Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}\ [$by Heron's formula$]$
$=\sqrt{54(54-51)(54-37)(54-20)}$
$=\sqrt{54\times3\times17\times34}$
$=\sqrt{9\times3\times2\times3\times17\times17\times2}$
$=3\times3\times2\times17$
$=306\text{ m}^2$
$\because$ 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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