Triangles - Class 9 Mathematics - Chapter 7 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Triangles | NCERT | Mathematics | Class 9
How many triangles can be drawn having its angles as $53^{\circ}, 64^{\circ}$, and $63^{\circ}$?
A) A Unique triangle
B) Two triangles
C) Infinitely many triangles
D) None of these
The correct answer is C. Infinitely many triangles
We start by calculating the sum of the given angles: $$ 53^{\circ} + 64^{\circ} + 63^{\circ} = 180^{\circ} $$ Since the sum of the angles is $180^{\circ}$, which is the required sum for the angles of any triangle, we conclude that a triangle with these angles can indeed exist.
Furthermore, as these angle measures do not depend on the side lengths, and since triangle sides can vary while maintaining the same angle measures (by the property of similar triangles), infinitely many triangles can be constructed. These triangles will be similar in shape but can differ in size, leading to the conclusion that option C is correct.
The midpoints of an equilateral triangle of side $10 \mathrm{~cm}$ are joined to form a triangle. What type of triangle is obtained by joining the midpoints?
A. Equilateral triangle of side $5 \sqrt{3} \mathrm{~cm}$
B. Isosceles triangle, in which the equal sides measure $5 \sqrt{2} \mathrm{~cm}$
C. Isosceles triangle, in which the unequal side measures $5 \sqrt{2} \mathrm{~cm}$
D. Equilateral triangle of side $5 \mathrm{~cm}$
The correct answer is Option D:
Equilateral triangle of side $5 , \text{cm}$
Consider an equilateral triangle $ABC$ with each side measuring $10 , \text{cm}$. Let $D$, $E$, and $F$ be the midpoints of sides $AB$, $AC$, and $BC$ respectively.
By the midpoint theorem:
$DE \parallel BC$ and $DE = \frac{1}{2} BC$ -------------------- (I)
$DF \parallel AC$ and $DF = \frac{1}{2} AC$ -------------------- (II)
$EF \parallel AB$ and $EF = \frac{1}{2} AB$ -------------------- (III)
Since triangle $ABC$ is equilateral, $AB = BC = AC$, we have: $$ DE = DF = EF = \frac{1}{2} \times 10 , \text{cm} = 5 , \text{cm} $$
Thus, triangle $DEF$ is equilateral, with each side measuring $5 , \text{cm}$.
A triangle is formed by joining the coordinates $B(0,0)$, $C(8,0)$, and $A(4,8)$. $DE$ is formed where $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. $DC$ and $BE$ are joined to intersect at $F$. Then, the area of $\triangle DCB = \frac{n}{2} \times$ the area of $\triangle EBC$, where $n$ is an integer. What is the value of $n$?
Let's analyze the triangle $ABC$ where:
$A(4,8)$
$B(0,0)$
$C(8,0)$
Finding $D$ and $E$:
$D$ is the midpoint of $AB$. Using the midpoint formula $D\left(\frac{x_A + x_B}{2},\frac{y_A + y_B}{2}\right)$, $D$ is: $$ D\left(\frac{4+0}{2},\frac{8+0}{2}\right) = (2,4) $$
$E$ is the midpoint of $AC$. Again applying the midpoint formula, $E$ is: $$ E\left(\frac{4+8}{2},\frac{8+0}{2}\right) = (6,4) $$
Key Properties:
According to the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle (here, $DE$) is parallel to the third side ($BC$) and half its length. That gives us $DE \parallel BC$.
Implication on Area:
Because $DE$ is parallel to $BC$ and both $\triangle DCB$ and $\triangle EBC$ share the same height from $A$ perpendicular to $BC$, the base line of both triangles is effectively split in half by $DE$.
This implies that the areas of $\triangle DCB$ and $\triangle EBC$ are equal because both triangles have the same height and half the length of $BC$ as base.
Using the Given Ratio:
Given that the area of $\triangle DCB$ is $\frac{n}{2} \times$ the area of $\triangle EBC$, since the areas are equal, we can deduce: $$ \text{Area of }\triangle DCB = \text{Area of }\triangle EBC \Rightarrow \frac{n}{2} = 1 $$
Solving for $n$, we see $n = 2$.
Hence, the value of $n$ is $2$.
The point where the perpendicular bisectors of a triangle intersect is called as:
A) Circumcentre
B) Bisector
C) Perpendicular Bisector
D) Co-interior points
The point where the perpendicular bisectors of a triangle intersect is known as the Circumcentre.
If $\triangle ABC \sim \triangle DEF$ such that $AB=12$ cm and $DE=14$ cm, find the ratio of areas of $\triangle ABC$ and $\triangle DEF$.
A) $\frac{25}{49}$
B) $\frac{49}{16}$
C) $\frac{36}{49}$
D) $\frac{49}{9}$
Given $\triangle ABC \sim \triangle DEF$, the correspondence of the sides implies that the ratio of any two corresponding sides equals the ratio of any other two corresponding sides. Therefore, using $AB$ and $DE$:
$$ \frac{AB}{DE} = \frac{12 \text{ cm}}{14 \text{ cm}} = \frac{12}{14} = \frac{6}{7} $$
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, we find the ratio of the areas as:
$$ \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle DEF} = \left(\frac{6}{7}\right)^2 = \frac{36}{49} $$
Thus, the correct option is C) $\frac{36}{49}$, which represents the ratio of the areas of $\triangle ABC$ to $\triangle DEF$.
In the following figure, the sides $AB$ and $BC$ and the median $AD$ of triangle $ABC$ are respectively equal to the sides $PQ$ and $QR$ and median $PS$ of the triangle $PQR$. Prove that $\triangle PQR$ is congruent.
To prove that $\triangle ABC$ is congruent to $\triangle PQR$, we use the given information and apply the congruence rules of triangles (SSS and SAS).
Given:
$AB = PQ$,
$BC = QR$,
$AD$ and $PS$ are medians of the triangles $\triangle ABC$ and $\triangle PQR$ respectively.
To Prove:
$\triangle ABC \cong \triangle PQR$
Proof:
As $AD$ and $PS$ are medians, they bisect $BC$ and $QR$ respectively. Therefore: $$ BD = DC \quad \text{and} \quad QS = SR $$
Since $BC$ is equal to $QR$, and both are bisected by their respective medians: $$ BD = \frac{1}{2}BC = \frac{1}{2}QR = QS $$
In $\triangle ABD$ and $\triangle PQS$: $$ \begin{aligned} AB &= PQ \quad \text{(Given)} \ BD &= QS \quad \text{(Proved above)} \ AD &= PS \quad \text{(Given medians)} \end{aligned} $$ Here, all three corresponding sides are equal.
By the SSS Congruence Rule:$$ \triangle ABD \cong \triangle PQS $$
Once two triangles are proven congruent by SSS, all their corresponding parts must be equal by CPCT (Corresponding Parts of Congruent Triangles). This includes $\angle ADB = \angle PSQ$.
Therefore, we can now look at the whole triangles $\triangle ABC$ and $\triangle PQR$ and state: $$ \begin{aligned} &AB = PQ \quad \text{(Given)} \ &\angle ADB = \angle PSQ \quad \text{(By CPCT from $\triangle ABD \cong \triangle PQS$)} \ &BC = QR \quad \text{(Given)} \end{aligned} $$
By the SAS Congruence Rule:$$ \triangle ABC \cong \triangle PQR $$
This completion of the proof shows that the triangles are congruent utilizing properties of medians and the congruence rules (SSS followed by SAS for the entire triangles).
A triangle can have at the most $\qquad$ right angle(s).
A) 1
B) 2
C) 3
D) 4
Solution
The correct answer is A) 1
In a triangle, a right angle measures exactly $90^\circ$.
However, the sum of the interior angles of any triangle always equals $180^\circ$. If a triangle had two right angles, their sum alone would be $90^\circ + 90^\circ = 180^\circ$, leaving no degrees for the third angle, which is impossible since each angle in a triangle must have a positive measure.
Therefore, a triangle cannot have more than one right angle.
Given $ABC$ is a right-angled triangle, find the length of $BD$ assuming $\triangle ABC \sim \triangle BDC$.
A. 8.6 units
B. 9.6 units
C. 6 units
D. 10 units
The correct answer is Option B: 9.6 units.
Given that $\triangle ABC \sim \triangle BDC$, the sides corresponding to these similar triangles are proportional. Thus, we have:
$$ \frac{AB}{BD} = \frac{AC}{BC} $$
Setting up the equation, we see:
$$ \frac{12}{BD} = \frac{AC}{16} $$
Now, using the Pythagorean theorem to find $AC$, we calculate:
$$ AC = \sqrt{BC^2 + AB^2} = \sqrt{16^2 + 12^2} = \sqrt{400} = 20 \text{ units} $$
Substitute $AC = 20$ into the ratio:
$$ \frac{12}{BD} = \frac{20}{16} $$
Solving for $BD$, we rearrange and simplify: $$ BD = \frac{12 \times 16}{20} = 9.6 \text{ units} $$
Thus, the length of $BD$ is 9.6 units.
"How many triangles are possible if the length of two sides and the non-included angle are given, assuming that the given length is the minimum length required to form a triangle?
A. 1
B. 2
C. 3
D. 4"
Correct Answer: D. 4
When constructing triangles given two sides and a non-included angle, and assuming the given side length constitutes the minimum length necessary to form a triangle, it is indeed feasible to construct four distinct triangles.
This is due to the flexibility in positioning the non-included angle relative to the given sides. You can place the non-included angle in such a way that either:
- The angle opens upwards or downwards.
- The angle is drawn from either end of the baseline side.
Such configurations lead to the formation of four distinct triangles, given the initial conditions and considerations regarding the placement and orientation of the angle and sides.
In a triangle, the centroid is the point of intersection of:
A) altitudes.
B) perpendicular bisectors.
C) medians.
D) angular bisectors.
The correct answer is C) medians.
In a triangle, the centroid is specifically the point where the three medians intersect. A median of a triangle is the line segment connecting a vertex to the midpoint of the opposite side. Thus, the centroid divides each of the medians in a 2:1 ratio, where the larger segment is always adjacent to the vertex.
How many non-overlapping triangles can we make in an $n$-gon (polygon having $n$ sides) by joining the vertices? a) $n-1$ b) $n-2$ c) $n-3$ d) $n-4$
The number of non-overlapping triangles that can be formed in an $n$-gon (a polygon with $n$ sides) is given by the formula $n-2$. This represents that each non-overlapping triangle can be formed by selecting any three consecutive vertices from the $n$-gon. This approach ensures that the triangles do not overlap, and that every three consecutive vertices form exactly one triangle.
For a better understanding, we can see how it applies to polygons with several numbers of sides:
[ \begin{array}{|c|c|} \hline \text{No. of Sides} & \text{No. of Non-Overlapping Triangles} \ \hline 3 & 1 \ \hline 4 & 2 \ \hline 5 & 3 \ \hline 6 & 4 \ \hline \vdots & \vdots \ \hline n & n-2 \ \hline \end{array} ]
In conclusion, the correct answer for the number of non-overlapping triangles that can be formed by joining the vertices of an $n$-gon is $n-2$. Therefore, the answer choice corresponding to this is:
- b) $n-2$
53
State whether the following statement is True or False:
Pyramids do not have a diagonal.
Solution
True
Pyramids are polyhedra with a polygonal base and triangular faces converging at a common vertex. A diagonal is defined as a line segment connecting two non-adjacent vertices. In pyramids, such a combination of opposite vertices does not exist due to its geometrical structure. Therefore, pyramids have no diagonals.
The shape of $\mathrm{ClF}_3$ according to VSEPR model is
A Planar triangle
B T-shape
C Tetrahedral
D Square planar.
The correct answer is B: T-shape.
Chlorine trifluoride ($\mathrm{ClF}_3$) has a total of 5 regions of electron density around the central chlorine atom according to the Valence Shell Electron Pair Repulsion (VSEPR) model. This includes 3 bond pairs and 2 lone pairs. The arrangement of these electron pairs in space gives $\mathrm{ClF}_3$ a T-shaped molecular geometry.
The area of a triangle whose vertices are $(2,3)$, $(3,2)$, and $(1,8)$ is:
A) 1 sq. unit
B) 2 sq. units
C) 3 sq. units
D) 4 sq. units
To find the area of a triangle with vertices $(2,3)$, $(3,2)$, and $(1,8)$, we use the formula for the area based on its vertices' coordinates:
The formula for the area of a triangle given the coordinates of its vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is: $$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$
For our specific vertices:
$(x_1, y_1) = (2, 3)$
$(x_2, y_2) = (3, 2)$
$(x_3, y_3) = (1, 8)$
Substituting these into the formula, we have: $$ \text{Area} = \frac{1}{2} \left| 2(2 - 8) + 3(8 - 3) + 1(3 - 2) \right| $$
Expanding the terms: $$ \text{Area} = \frac{1}{2} \left| 2(-6) + 3(5) + 1(1) \right| $$ $$ \text{Area} = \frac{1}{2} \left| -12 + 15 + 1 \right| $$ $$ \text{Area} = \frac{1}{2} \left| 4 \right| $$
So, the area of the triangle is: $$ \text{Area} = \frac{1}{2} \times 4 = 2 \text{ sq. units} $$
Therefore, the correct answer is Option B) 2 sq. units.
A triangle has altitudes:
1
2
3
infinite
The correct option is C: 3
The perpendicular line segment from a vertex of a triangle to its opposite side is referred to as the altitude of the triangle. Importantly, a triangle has 3 altitudes.
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