The Triangle and its Properties - Class 7 - Mathematics
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Examples - The Triangle and its Properties | NCERT | Mathematics | Class 7
Find angle $x$ in Fig 6.11.
To find the angle $x$, you can use the property that the sum of the interior angles of a triangle equals $180^\circ$, or you can use the fact that the exterior angle of a triangle is equal to the sum of the two opposite interior angles.
In this case, since the exterior angle is given as $110^\circ$ and one of the interior opposite angles is $50^\circ$, you can find $x$ by the latter method:
$$ x + 50^\circ = 110^\circ $$
Solving for $x$, we get:
$$ x = 110^\circ - 50^\circ = 60^\circ $$
Therefore, the value of the angle $x$ is $60^\circ$.
In the given figure (Fig 6.18) find $\mathrm{m} \angle \mathrm{P}$.
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Sign up nowIs there a triangle whose sides have lengths $10.2 \mathrm{~cm}, 5.8 \mathrm{~cm}$ and $4.5 \mathrm{~cm}$ ?
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Sign up nowThe lengths of two sides of a triangle are $6 \mathrm{~cm}$ and $8 \mathrm{~cm}$. Between which two numbers can length of the third side fall?
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Sign up nowDetermine whether the triangle whose lengths of sides are $3 \mathrm{~cm}, 4 \mathrm{~cm}$, $5 \mathrm{~cm}$ is a right-angled triangle.
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Sign up now$\Delta \mathrm{ABC}$ is right-angled at $\mathrm{C}$. If $\mathrm{AC}=5 \mathrm{~cm}$ and $\mathrm{BC}=12 \mathrm{~cm}$ find the length of $\mathrm{AB}$.
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Sign up nowExercise 6.1 - The Triangle and its Properties | NCERT | Mathematics | Class 7
In $\Delta \mathrm{PQR}, \mathrm{D}$ is the mid-point of $\overline{\mathrm{QR}}$.
$\overline{\mathrm{PM}}$ is
$\mathrm{PD}$ is
Is $\mathrm{QM}=\mathrm{MR}$ ?
$\overline{\mathrm{PM}}$ is the altitude, because $\angle \mathrm{~PMR}$ is $90^\circ$
$\mathrm{PD}$ is the median, because $D$ is the mid-point of $\mathrm{QR}$
No, $\mathrm{QM}$ is not equal to $\mathrm{MR}$ because $D$ is the mid-point of $QR$, so $\mathrm{QD}$ is equal to $\mathrm{DR}$
Draw rough sketches for the following:
(a) In $\triangle \mathrm{ABC}, \mathrm{BE}$ is a median.
(b) In $\triangle \mathrm{PQR}, \mathrm{PQ}$ and $\mathrm{PR}$ are altitudes of the triangle.
(c) In $\triangle X Y Z, Y L$ is an altitude in the exterior of the triangle.
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Sign up nowVerify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
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Sign up nowExercise 6.2 - The Triangle and its Properties | NCERT | Mathematics | Class 7
Find the value of the unknown exterior angle $x$ in the following diagrams:
(i) One interior angle is 50 degrees, second is 70 degrees. find the value of x which is an exterior angle opposite these interior angles. explain with the rules applied.
(ii) One interior angle is 65 degrees, second is 45 degrees. find the value of x which is an exterior angle opposite these interior angles. explain with the rules applied.
(iii) One interior angle is 30 degrees, second is 40 degrees. find the value of x which is an exterior angle opposite these interior angles. explain with the rules applied.
(iv) One interior angle is 60 degrees, second is 60 degrees. find the value of x which is an exterior angle opposite these interior angles. explain with the rules applied.
(v) One interior angle is 50 degrees, second is 50 degrees. find the value of x which is an exterior angle opposite these interior angles. explain with the rules applied.
(vi) One interior angle is 30 degrees, second is 60 degrees. find the value of x which is an exterior angle opposite these interior angles. explain with the rules applied.
To find the value of the unknown exterior angle $x$ in the given diagrams with diverse pairs of interior angles, we will apply the fundamental rule that the sum of the measures of the interior angles and the exterior angle (opposite these interior angles) of a triangle is equal to $180^\circ$.
Here's how we calculate $x$ in each case, applying the rule:
First Pair (50 degrees and 70 degrees):
Sum of interior angles: $50 + 70 = 120$ degrees
So, to find (x), we calculate:
$$180 - 120 = 60^\circ$$
Thus, $x = 60^\circ$.Second Pair (65 degrees and 45 degrees):
Sum of interior angles: $65 + 45 = 110$ degrees
So, to find (x), we calculate:
$$180 - 110 = 70^\circ$$
Thus, $x = 70^\circ$.Third Pair (30 degrees and 40 degrees):
Sum of interior angles: $30 + 40 = 70$ degrees
So, to find (x), we calculate:
$$180 - 70 = 110^\circ$$
Thus, $x = 110^\circ$.Fourth Pair (60 degrees and 60 degrees):
Sum of interior angles: $60 + 60 = 120$ degrees
So, to find (x), we calculate:
$$180 - 120 = 60^\circ$$
Thus, $x = 60^\circ$.Fifth Pair (50 degrees and 50 degrees):
Sum of interior angles: $50 + 50 = 100$ degrees
So, to find (x), we calculate:
$$180 - 100 = 80^\circ$$
Thus, $x = 80^\circ$.Sixth Pair (30 degrees and 60 degrees):
Sum of interior angles: $30 + 60 = 90$ degrees
So, to find (x), we calculate:
$$180 - 90 = 90^\circ$$
Thus, $x = 90^\circ$.
For every pair, we used the principle that the sum of the interior angles and the adjacent exterior angle in a triangle always sums up to $180^\circ$.
Find the value of the unknown interior angle $x$ in the following figures:
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Sign up nowExercise 6.3 - The Triangle and its Properties | NCERT | Mathematics | Class 7
Find the value of the unknown $x$ in the following diagrams:
To solve each of these questions, we'll use the property that the sum of the interior angles of a triangle is $180^\circ$
(i) $$x + 50^\circ + 60\circ = 180^\circ$$
$$ x = 70^\circ$$
(ii) $$x + 30^\circ + 90\circ = 180^\circ$$
$$ x = 60^\circ$$
(iii) $$x + 30^\circ + 110\circ = 180^\circ$$
$$ x = 40^\circ$$
(iv) $$x + x + 50\circ = 180^\circ$$
$$ x = \frac{130^\circ}{2} = 65^\circ$$
(v) $$x + x + x = 180^\circ$$
$$ x = \frac{180^\circ}{3} = 60^\circ$$
(vi) $$x + 2x + 90^\circ = 180^\circ$$
$$ x = \frac{90^\circ}{3} = 30^\circ$$
Find the values of the unknowns $x$ and $y$ in the following diagrams:
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Sign up nowExercise 6.4 - The Triangle and its Properties | NCERT | Mathematics | Class 7
Is it possible to have a triangle with the following sides?
(i) $2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$
(ii) $3 \mathrm{~cm}, 6 \mathrm{~cm}, 7 \mathrm{~cm}$
(iii) $6 \mathrm{~cm}, 3 \mathrm{~cm}, 2 \mathrm{~cm}$
To determine if a triangle with the given side lengths is possible, we use the Triangle Inequality Theorem. This theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This must be true for all three combinations of sides.
Mathematically, for sides $a$, $b$, and $c$, the following inequalities must hold true:
$a + b > c$
$a + c > b$
$b + c > a$
Let's apply these conditions to the given sets of side lengths:
For (i) $2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$:
$2 + 3 > 5$ (False)
$2 + 5 > 3$ (True)
$3 + 5 > 2$ (True)
For (ii) $3 \mathrm{~cm}, 6 \mathrm{~cm}, 7 \mathrm{~cm}$:
$3 + 6 > 7$ (True)
$3 + 7 > 6$ (True)
$6 + 7 > 3$ (True)
For (iii) $6 \mathrm{~cm}, 3 \mathrm{~cm}, 2 \mathrm{~cm}$:
$6 + 3 > 2$ (True)
$6 + 2 > 3$ (True)
$3 + 2 > 6$ (False)
Using the Triangle Inequality Theorem, we find that:
(i) The sides $2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$ cannot form a triangle, as one of the conditions fails.
(ii) The sides $3 \mathrm{~cm}, 6 \mathrm{~cm}, 7 \mathrm{~cm}$ can form a triangle, as all conditions pass.
(iii) The sides $6 \mathrm{~cm}, 3 \mathrm{~cm}, 2 \mathrm{~cm}$ cannot form a triangle, as one of the conditions fails.
Take any point $\mathrm{O}$ in the interior of a triangle $\mathrm{PQR}$. Is
(i) $\mathrm{OP}+\mathrm{OQ}>\mathrm{PQ}$ ?
(ii) $\mathrm{OQ}+\mathrm{OR}>\mathrm{QR}$ ?
(iii) $\mathrm{OR}+\mathrm{OP}>\mathrm{RP}$ ?
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Sign up now$A M$ is a median of a triangle $A B C$.
Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CA}>2 \mathrm{AM}$ ?
(Consider the sides of triangles$\triangle \mathrm{ABM}$ and $\triangle \mathrm{AMC}$.)
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Sign up now$\mathrm{ABCD}$ is a quadrilateral.
Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>\mathrm{AC}+\mathrm{BD}$ ?
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Sign up nowThe lengths of two sides of a triangle are $12 \mathrm{~cm}$ and $15 \mathrm{~cm}$. Between what two measures should the length of the third side fall?
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Sign up nowExercise 6.5 - The Triangle and its Properties | NCERT | Mathematics | Class 7
$P Q R$ is a triangle, right-angled at $P$. If $P Q=10 \mathrm{~cm}$ and $P R=24 \mathrm{~cm}$, find $Q R$.
We have a right-angled triangle $PQR$ with $PQ = 10,\text{cm}$ and $PR = 24,\text{cm}$. We want to find the length of the hypotenuse $QR$. To do this, we can use Pythagoras' theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
Steps To Solve:
Identify the sides: Here, $PQ$ and $PR$ are the two sides that form the right angle, so they will be our $a$ and $b$ when applying Pythagoras' theorem. $QR$ is the hypotenuse.
Apply Pythagoras' theorem: The theorem states $a^2 + b^2 = c^2$, where $c$ is the length of the hypotenuse.
Substitute the given values: We substitute $PQ = 10,\text{cm}$ and $PR = 24,\text{cm}$ into the equation. This gives us $10^2 + 24^2 = QR^2$.
Solve for $QR$: We need to calculate the square root of the sum of the squares of $PQ$ and $PR$ to find $QR$.
Let's calculate $QR$:
$$QR^2 = (10)^2 + (24)^2$$ $$QR^2 = 100 + 576 = 676$$ $$QR = \sqrt{676}$$ $$QR = 26,\text{cm}$$
Therefore, the length of $QR$ is $26,\text{cm}$.
$\mathrm{ABC}$ is a triangle, right-angled at $\mathrm{C}$. If $\mathrm{AB}=25$ $\mathrm{cm}$ and $\mathrm{AC}=7 \mathrm{~cm}$, find $\mathrm{BC}$.
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Sign up nowA $15 \mathrm{~m}$ long ladder reached a window $12 \mathrm{~m}$ high from the ground on placing it against a wall at a distance $a$. Find the distance of the foot of the ladder from the wall.
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Sign up nowWhich of the following can be the sides of a right triangle?
(i) $2.5 \mathrm{~cm}, 6.5 \mathrm{~cm}, 6 \mathrm{~cm}$.
(ii) $2 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$.
(iii) $1.5 \mathrm{~cm}, 2 \mathrm{~cm}, 2.5 \mathrm{~cm}$.
In the case of right-angled triangles, identify the right angles.
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Sign up nowA tree is broken at a height of $5 \mathrm{~m}$ from the ground and its top touches the ground at a distance of $12 \mathrm{~m}$ from the base of the tree. Find the original height of the tree.
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Sign up nowAngles $\mathrm{Q}$ and $\mathrm{R}$ of a $\triangle \mathrm{PQR}$ are $25^{\circ}$ and $65^{\circ}$. Write which of the following is true:
(i) $\mathrm{PQ}^{2}+\mathrm{QR}^{2}=\mathrm{RP}^{2}$
(ii) $\mathrm{PQ}^{2}+\mathrm{RP}^{2}=\mathrm{QR}^{2}$
(iii) $\mathrm{RP}^{2}+\mathrm{QR}^{2}=\mathrm{PQ}^{2}$
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Sign up nowFind the perimeter of the rectangle whose length is $40 \mathrm{~cm}$ and a diagonal is $41 \mathrm{~cm}$.
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Sign up nowThe diagonals of a rhombus measure $16 \mathrm{~cm}$ and $30 \mathrm{~cm}$. Find its perimeter.
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Sign up nowExtra Questions - The Triangle and its Properties | NCERT | Mathematics | Class 7
Two sides of a triangle have lengths of 5 units and 6 units respectively. Which of the following is a possible length for the third side?
A 11
B $\mathbf{12}$
C 13
D 10
The correct answer is D) 10.
According to the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Given side lengths of $5$ units and $6$ units, the combination possibilities for the third side ($x$), satisfying the triangle inequality, are:
$5 + 6 > x$
$5 + x > 6$
$6 + x > 5$
From the first inequality: $$ 5 + 6 > x \ 11 > x $$ $x$ must be less than $11$ units.
From the second and third inequalities, since they will always be true for $x$ in these scenarios (since adding any positive $x$ to $5$ or $6$ will exceed $5$ or $6$), the critical determination is made by the first inequality.
Thus, the possible values for $x$ are less than $11$ but more than 0. The available choice under $11$ from the given options that also complies with triangle side length rules is:
D) 10
All other options are either equal to or more than $11$, which would violate the triangle inequality theorem.
The opposite faces of a die always have a total of $\qquad$ on them.
A) 6
B) 7
C) 8
D) 9
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Sign up nowIn a $\triangle ABC$, $\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$. Then the ordered pair $(\angle A, \angle B)$ is equal to:
A $\left(45^{\circ}, 75^{\circ}\right)$
B $\left(75^{\circ}, 45^{\circ}\right)$
C $\left(105^{\circ}, 15^{\circ}\right)$
D $\left(15^{\circ}, 105^{\circ}\right)$
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Sign up nowWhich of the following can be the length of $BC$ required to construct the triangle $ABC$ such that $AC=6.7 \mathrm{~cm}$ and $AB=5.3 \mathrm{~cm}$?
A) $8 \mathrm{~cm}$
B) $5 \mathrm{~cm}$
C) $6 \mathrm{~cm}$
D) $9 \mathrm{~cm}$
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Sign up nowIn fig. $BD | CA$, $E$ is midpoint of $CA$ and $BD=\frac{1}{2}AC$. Then
(A) $\operatorname{ar}(ABC)=\frac{1}{2}\operatorname{ar}(DBC)$
(B) $\operatorname{ar}(ABC)=\operatorname{ar}(DBC)$
(C) $\operatorname{ar}(ABC)=2\operatorname{ar}(DBC)$
(D) $\operatorname{ar}(ABC)=\frac{5}{2}\operatorname{ar}(DBC)$
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Sign up nowA letter is drawn from the word TRIANGLE. The probability that it is a vowel is:
(A) 1/4 (B) 3/8 (C) 5/7 (D) 2/3
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Sign up now$\triangle ABC$ is an isosceles triangle, in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$. $\angle BCD$ is equal to $\qquad$
(A) $90^{\circ}$ (B) $45^{\circ}$ (C) $37^{\circ}$
(D) Cannot be determined
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Sign up nowA triangle can have all three angles greater than $60^{\circ}$.
A. True
B. False
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Sign up nowTwo triangles have the same base and are between the same parallels. The area of one triangle is $36 \mathrm{~cm}^{2}$. The area of the 2nd triangle is $\qquad$
(A) $18 \mathrm{~cm}^{2}$
(B) $36 \mathrm{~cm}^{2}$
(C) $48 \mathrm{~cm}^{2}$
(D) $54 \mathrm{~cm}^{2}$
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