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MATHS 1 Marks Question

The Triangles and Its Properties · Gujarat Board (GSEB) STD 7 (GSEB - English Medium). 12 questions.

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Question 11 Mark
Are these be the sides of a right triangle? In the case of right-angled triangles, identify the right angle.
$1.5 \ cm, 2 \ cm, 2.5 \ cm.$
Answer
$1.5 \ cm, 2 \ cm, 2.5 \ cm.$
We find that
$1.5^2+2^2=2.25+4=6.25=2.5^2$
Therefore, the given lengths can be sides of a right-angled triangle. Also, right angle will be in front of the side of $2.5\ cm$ measure.
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Question 21 Mark
Which of the following can be the sides of a right-angled triangle? $2 \ cm, 2\ cm, 5 \ cm$
Answer
$2 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$
$2^2=4$
$2^2=4$
$5^2=25$
$2^2+2^2=5^2 \ldots$ [The sum of the length of one side is not equal to the sum of the squares of the lengths; according to Pythagoras theorem]
$\therefore$ The given lengths cannot be the sides of a right-angled triangle.
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Question 31 Mark
Are these be the sides of a right triangle? In the case of right-angled triangles, identify the right angle.
$2.5 \ cm, 6.5 \ cm, 6 \ cm$
Answer
$2.5 \mathrm{~cm}, 6.5 \mathrm{~cm}, 6 \mathrm{~cm}$
We see that
$(2.5)^2+6^2=6.25+36=42.25=(6.5)^2$
The square of the length of one side is the sum of the squares of the lengths of the remaining two sides.
Hence, these are the sides of a right angled triangle.
Right angle will be in front of the side of $6.5 \ cm$ measure.
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Question 41 Mark
In $\triangle$ PQR the following figure, $D$ is the mid-point of $QR$, is $QM = MR?$
Answer
$QM \ne MR.$
If the Median and Altitude are at the same point, only then $QM = MR$, but in $\triangle PQR$, this is not the case.
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Question 71 Mark
$\triangle ABC$ is right-angled at $C$. If $AC = 5 \ cm$ and $BC = 12 \ cm$ find the length of $AB.$
Answer
Given that $\triangle A B C$ is right-angled at $C$.
By Pythagoras property,
$A B^2=A C^2+B C^2$
$=5^2+12^2$
$=25+144$
$=169=13^2$
$\text { or } A B^2=13^2$
So, $A B=13$ or the length of $A B$ is $13 \ cm .$
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Question 81 Mark
Determine whether the triangle whose lengths of sides are $3 \ cm, 4 \ cm, 5 \ cm$ is a right-angled triangle.
Answer
Here,
$3^2=3 \times 3=9$
$4^2=4 \times 4=16$
$5^2=5 \times 5=25$
We find $3^2+4^2=5^2$
Therefore, the triangle is right-angled.
As, we know that in any right-angled triangle, the hypotenuse happens to be the longest side. In this example, the side with length $5 \ cm$ is the hypotenuse.
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Question 91 Mark
The lengths of two sides of a triangle are $6 \ cm$ and $8 \ cm$. Between which two numbers can length of the third side fall?
Answer
We know that the sum of two sides of a triangle is always greater than the third.
Therefore, third side has to be less than the sum of the two sides.
The third side is thus, less than $8 + 6 = 14 \ cm.$
Also, the side cannot be less than the difference of the two sides.
Thus, the third side has to be more than $8 – 6 = 2 \ cm.$
Therefore, the length of the third side lies between $2 \ cm$ and $14 \ cm.$
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Question 101 Mark
Is there a triangle whose sides have lengths $10.2 \ cm, 5.8 \ cm$ and $4.5 \ cm?$
Answer
Suppose such a triangle is possible. Then the sum of the lengths of any two sides would be greater than the length of the third side. Let us check this.
Is $4.5 + 5.8 > 10.2?$ Yes
Is $5.8 + 10.2 > 4.5?$ Yes
Is $10.2 + 4.5 > 5.8?$ Yes
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Question 111 Mark
In the given figure, Find m$\angle P $
Answer
$m \angle P+47^{\circ}+52^{\circ}=180^{\circ}, \text { (By Angle Sum Property of a triangle ) }$
$\mathrm{m} \angle P=180^{\circ}-47^{\circ}-52^{\circ}$
$\mathrm{m} \angle P=180^{\circ}-99^{\circ}$
$\mathrm{m} \angle P=81^{\circ}$
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Question 121 Mark
Find angle $x$ in Fig.
Answer
Here,
Sum of interior opposite angles = Exterior angle
$\Rightarrow 50^\circ + x = 110^\circ $
$\Rightarrow x = 60^\circ $
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