True False[1 Marks ]

Question 11 Mark

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 21 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 31 Mark

The base and the corresponding altitude of a parallelogram are 10cm and 3.5cm, respectively. The area of the parallelogram is 30cm².

Answer

False.

Solution:

The base of the parallelogram is 10cm and the corresponding altitude is 3.5cm.

Area of || gm = base × corresponding altitude

= 10cm × 3.5cm = 35cm²

Hence, the given statement is false.

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Question 41 Mark

The area of $\triangle\text{ABC}$ is 8cm² in which AB = AC = 4cm and $\angle\text{A}=90^\circ$

Answer

True.

Solution:

$\text{Area of }\triangle=\frac{1}{2}\times\text{base}\times\text{height}$

$=\frac{1}{2}\times4\times4=8\text{cm}^2$

Hence, the given statement is true.

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Question 51 Mark

The cost of levelling the ground in the form of a triangle having the sides 51m, 37m and 20m at the rate of Rs. 3per m² is Rs. 918.

Answer

True.

Solution:

Let sides of a triangle be a = 51m, b = 37m and c = 20m

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 m² = Rs. 3

$\therefore$ Cost of leavelling per 306 m² = 3 × 306 = Rs. 918

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Question 61 Mark

If the side of a rhombus is 10cm and one diagonal is 16cm, the area of the rhombus is 96cm².

Answer

True.
Solution:
Given, side of a rhombus PQRS is 10cm 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°]
⇒ OQ² = PQ² - OP²
⇒ OQ² = (10)² - (8)²
⇒ OQ² = 100 - 64 = 36
⇒ OQ = 6cm
[taking positive square root because length is always positive]
SQ =2 × OP = 2 × 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 71 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 81 Mark

The area of a triangle with base 4cm and height 6cm is 24cm²

Answer

False.

Solution:

We know that, area of a triangle $=\frac{1}{2}(\text{Base}\times\text{Height})$

Here Base = 4cm and Height = 6cm

$\therefore$ Area of a triangle $=\frac{1}{2}\times4\times6=12\text{cm}^2$

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Question 91 Mark

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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