Triangles - Class 10 Mathematics - Chapter 6 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Triangles | NCERT | Mathematics | Class 10
In the given figure, $\triangle ABC$ is a triangle right-angled at $B$ and $BD \perp AC$. If $AD=4$ cm and $CD=5$ cm, then find $BD$ and $AB$.
Given the configuration in $\triangle ABC$, where $\angle B$ is $90^\circ$ and $BD \perp AC$, the lengths of $AD$ and $CD$ are $4 \text{ cm}$ and $5 \text{ cm}$, respectively.
Step 1: Similar TrianglesThe triangles $\triangle ADB$ and $\triangle CDB$ are similar by AAA similarity because $\angle ADB = \angle CDB = 90^\circ$ and $\angle DAB = \angle DBC$. Similarity criteria yield: $$ \frac{BD}{AD} = \frac{DC}{BD} $$
Step 2: Calculating $BD$From the similarity relation and substituting the given segment lengths, we get: $$ BD^2 = AD \times DC = 4 \times 5 $$ Thus, solving for $BD$, we find: $$ BD = \sqrt{20} = 2 \sqrt{5} \text{ cm} $$
Step 3: Calculating $BC$ Using Pythagoras' TheoremIn the right triangle $\triangle BDC$, using Pythagoras' theorem: $$ BC^2 = BD^2 + CD^2 = (2 \sqrt{5})^2 + 5^2 = 20 + 25 = 45 $$ Then: $$ BC = \sqrt{45} = 3\sqrt{5} \text{ cm} $$
Step 4: Calculating $AB$Using proportionality from the triangles' similarity: $$ \frac{BD}{DC} = \frac{BA}{BC} $$ Substituting known values and solving for $BA$: $$ \frac{2 \sqrt{5}}{5} = \frac{BA}{3 \sqrt{5}} $$ Solving for $BA$: $$ BA = \frac{2 \sqrt{5} \times 3 \sqrt{5}}{5} = 6 \text{ cm} $$
Conclusion
The length of $BD$ is $2 \sqrt{5} \text{ cm}$
The length of $AB$ is $6 \text{ cm}$.
In the figure, the bisector of $\angle B$ and $\angle C$ meet at $O$. Then $\angle BOC$ is:
A) $90^{\circ} + \frac{1}{2} \angle A$
B) $90^{\circ} + \frac{1}{2} \angle B$
C) $90^{\circ} + \frac{1}{2} \angle C$
D) $90^{\circ} - \frac{1}{2} \angle A$
The correct option is: A) $90^{\circ} + \frac{1}{2} \angle A$
From the given diagram, we aim to find the angle $\angle BOC$. It's important to note that $O$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$ in $\triangle ABC$. We have the angles labelled as shown:
$\angle 2 = \frac{1}{2}\angle B$
$\angle 4 = \frac{1}{2}\angle C$
Using the angle sum property of the triangle at $O$: $$ \angle BOC + \angle 2 + \angle 4 = 180^{\circ} $$ Thus, substituting the values of $\angle 2$ and $\angle 4$: $$ \angle BOC = 180^{\circ} - \left(\frac{\angle B}{2} + \frac{\angle C}{2}\right) $$
From the property of a triangle (sum of angles equals $180^\circ$): $$ \angle A + \angle B + \angle C = 180^{\circ} $$ So, $$ \angle B + \angle C = 180^{\circ} - \angle A $$
Substitute $\angle B + \angle C$ in the expression for $\angle BOC$: $$ \angle BOC = 180^{\circ} - \left(\frac{180^{\circ} - \angle A}{2}\right) $$ $$ = 180^{\circ} - 90^{\circ} + \frac{\angle A}{2} $$ $$ = 90^{\circ} + \frac{\angle A}{2} $$
This verifies that the correct answer is: A) $90^{\circ} + \frac{1}{2} \angle A$
Find the area of quadrilateral PQRS whose vertices are $P(-5,-3)$, $Q(-4,-6)$, $R(2,-3)$, and $S(1,2)$
To find the area of the quadrilateral PQRS with vertices at $P(-5,-3)$, $Q(-4,-6)$, $R(2,-3)$, and $S(1,2)$, you can divide it into two triangles $PQR$ and $PRS$, calculate their areas separately, and then sum them up.
The formula to calculate the area of a triangle given its vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is: $$ \text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| $$
Triangle $PQR$
For triangle $PQR$, with vertices $P(-5,-3)$, $Q(-4,-6)$, $R(2,-3)$:
Using the formula: $$ \text{Area of } PQR = \frac{1}{2} |(-5)(-6+3) + (-4)(-3+3) + (2)(-3+6)| $$
Simplifying each term: $$ = \frac{1}{2} |(-5)(-3) + (-4)(0) + 2(3)| $$
Continuing with the arithmetic: $$ = \frac{1}{2} |15 + 0 + 6| $$ $$ = \frac{1}{2} \times 21 = 10.5 \text{ sq.units } $$
Triangle $PRS$
For triangle $PRS$, with vertices $P(-5,-3)$, $R(2,-3)$, $S(1,2)$:
Following a similar approach: $$ \text{Area of } PRS = \frac{1}{2} |(-5)(-3-2) + 2(2+3) + 1(-3+3)| $$
Simplifying it further: $$ = \frac{1}{2} |(-5)(-5) + 2(5) + 1(0)| $$ $$ = \frac{1}{2} \times (25 + 10 + 0) = \frac{1}{2} \times 35 = 17.5 \text{ sq.units} $$
Hence, the total area of the quadrilateral PQRS is: $$ 10.5 \text{ sq. units} + 17.5 \text{ sq. units} = 28 \text{ sq. units} $$
Verifying calculations and simplification steps ensures accurate results for geometrical tasks like these.
The possible values of a for which the point $\left(a, a^{2}\right)$ lies inside the triangle formed by the straight lines $2x + 3y - 1 = 0$, $x + 2y = 3$, and $5x - 6y - 1 = 0$ are
A $(-3, 2) \cup (3, 6)$
B $\left(\frac{-3}{2}, -1\right) \cup \left(\frac{1}{2}, 1\right)$
C $(-3, 1) \cup (1, 2)$
D $(-1, 1) \cup (1, 3)$
The correct answer is Option B: $$ \left(\frac{-3}{2}, -1\right) \cup \left(\frac{1}{2}, 1\right) $$
Explanation:
To find the values of $a$ for which the point $(a, a^2)$ lies inside the triangle formed by the lines:
- $2x + 3y - 1 = 0$
- $x + 2y = 3$
- $5x - 6y - 1 = 0$
First, find the vertices of the triangle, intersecting the lines two at a time:
-
Intersection of $2x+3y-1=0$ and $x+2y=3$:
Solving, we get the vertex coordinates as: $$ A\left(\frac{5}{4}, \frac{7}{8}\right) $$
-
Intersection of $x+2y=3$ and $5x-6y-1=0$:
Solving, gives: $$ C\left(\frac{1}{3}, \frac{4}{3}\right) $$ (Correcting apparent typo from the original solution)
-
Intersection of $2x+3y-1=0$ and $5x-6y-1=0$:
Solving, yields: $$ B(-7, 5) $$
Substituting $P(a, a^2)$ in inequalities for checking the interior of a triangle:
Point $P(a, a^2)$ should lie on the same side of each line segment of the triangle as the third vertex not involved in the line:
-
For line segment BC determining sides via: $$ (5a - 6a^2 - 1) > 0 $$ Solutions: $$ a \in (-\infty, \frac{1}{2}) \cup (1, \infty) $$
-
For AC: $$ (2a + 3a^2 - 1) > 0 $$ Solutions: $$ a \in (-\infty, -1) \cup (\frac{1}{2}, \infty) $$
-
For AB: $$ (a + 2a^2 - 3) > 0 $$ Solutions: $$ a \in (-\infty, -\frac{3}{2}) \cup (1, \infty) $$
Combining all conditions to locate $a$ within the interior:
The intersection of the conditions from points 1 to 3 gives: $$ a \in \left(-\frac{3}{2}, -1\right) \cup \left(\frac{1}{2}, 1\right) $$
Thus, Option B is the correct choice and provides the interval for values of $a$ where $(a, a^2)$ is inside the defined triangle.
The perpendicular from $A$ on side $BC$ of triangle $ABC$ intersects $BC$ at $D$ such that $DB=3CD$. Prove that $2AB^{2}=2AC^{2}+BC^{2}$.
In the given triangle ( \triangle ABC ), a perpendicular from vertex ( A ) to side ( BC ) meets ( BC ) at ( D ). Given that ( DB = 3CD ), we aim to prove that: $$ 2AB^2 = 2AC^2 + BC^2. $$
Let's denote ( CD ) as ( x ). Therefore, ( BD = 3x ) as ( DB = 3CD ). This implies: $$ BC = BD + CD = 3x + x = 4x. $$
Now, considering ( \triangle ADB ) and ( \triangle ADC ), let's apply the Pythagorean theorem:
For ( \triangle ADC ): $$ AC^2 = AD^2 + CD^2 = AD^2 + x^2 \quad \text{(Equation 1)} $$
For ( \triangle ADB ): $$ AB^2 = AD^2 + BD^2 = AD^2 + (3x)^2 = AD^2 + 9x^2 \quad \text{(Equation 2)} $$
Next, subtract Equation 1 from Equation 2 to find the relation between ( AB^2 ) and ( AC^2 ): $$ AB^2 - AC^2 = 9x^2 - x^2 = 8x^2. $$ This expresses ( AB^2 - AC^2 ) in terms of ( x ). Note also that ( BC^2 = (4x)^2 = 16x^2 ).
By multiplying the equation ( AB^2 - AC^2 = 8x^2 ) by 2, we achieve: $$ 2(AB^2 - AC^2) = 2(8x^2) = 16x^2. $$ This shows that: $$ 2AB^2 - 2AC^2 = 16x^2. $$ Since ( 16x^2 = BC^2 ), we substitute to get: $$ 2AB^2 - 2AC^2 = BC^2. $$ Rearranging the terms, we obtain: $$ 2AB^2 = 2AC^2 + BC^2. $$ Thus, the proof is complete and the original equation is indeed correct for ( \triangle ABC ) under the given conditions. Hence proved.
A cylindrical pencil sharpened at one edge is the combination of.
A. a cone and a cylinder
B. Frustum of a cone and a cylinder
C. a hemisphere and a cylinder
D. two cylinders
A cylindrical pencil sharpened at one end combines two geometric shapes: a cylinder and a cone. Here's a breakdown of how this combination occurs:
The main body of the pencil is shaped like a cylinder. This is the elongated, uniform portion that constitutes the majority of the pencil.
When a pencil is sharpened at one end, the sharpened part takes the shape of a cone. This conical shape forms because the sharpener removes material from the pencil in a way that tapers towards the point, creating a cone.
This is why a sharpened cylindrical pencil can be described as a combination of both a cylinder and a cone. The cylinder represents the main body of the pencil, and the cone represents the sharpened end.
Based on this understanding, the correct geometrical description of a sharpened cylindrical pencil is a combination of a cylinder and a cone.
State $SAS$ similarity criterion.
The $SAS$ (Side-Angle-Side) Similarity Criterion states that two triangles are similar if one pair of corresponding sides is proportional and the included angle (the angle between the two sides) is equal in both triangles.
Explanation:
Consider two triangles, $\triangle ABC$ and $\triangle PQR$. According to the $SAS$ similarity criterion:
Proportionality of Sides: One pair of corresponding sides must be proportional. Mathematically, this can be expressed as: $$ \frac{AB}{PQ} = k $$ where $k$ is a constant of proportionality.
Equality of Included Angles: The angle included between the proportional sides in each triangle must be equal. Therefore: $$ \angle BAC = \angle QPR $$
If both conditions are satisfied, then $\triangle ABC$ is similar to $\triangle PQR$, and we write: $$ \triangle ABC \sim \triangle PQR $$
Summary:
For two triangles to be similar under the $SAS$ criterion, one pair of their sides must be proportional, and the angle between these sides must be identical in both triangles. This is the essence of the $SAS$ similarity rule in geometry.
In the given figure, if $\frac{AB}{AC} = \frac{BD}{CD}$, then find the $\angle ABD$.
To solve the problem where we are given that $\frac{AB}{AC} = \frac{BD}{CD}$ and we need to find $\angle ABD$, we can use the Angle Bisector Theorem. Let's examine this step-by-step:
Understanding the Given Ratio:
The given ratio is $\frac{AB}{AC} = \frac{BD}{CD}$. According to the Angle Bisector Theorem, this signifies that $AD$ is an angle bisector of $\angle A$. Hence, we have that $AD$ bisects $\angle BAC$ into two equal parts.
Angles at Point A:
Since $\angle BAC = 30^\circ$ (given in the diagram), and $AD$ is an angle bisector of $\angle BAC$, we can conclude that: $$ \angle BAD = \angle DAC = 15^\circ $$
Using the Triangle Angle Sum Property:
In any triangle, the sum of angles equals $180^\circ$. Considering triangle $ABC$, we have: $$ \angle ABC = 80^\circ, \quad \angle BAC = 30^\circ, $$ which leaves $\angle BCA$ as: $$ \angle BCA = 180^\circ - (\angle ABC + \angle BAC) = 180^\circ - (80^\circ + 30^\circ) = 70^\circ $$
Finding $\angle ABD$:
Since $\angle ABD$ is part of $\angle ABC$, and $\angle ABC = 80^\circ$, we need to consider the additional division made by line segment $BD$, which is an angle bisector of $\angle ABC$. Given the bisectors and the $30^\circ$ bisected into two $15^\circ$ angles at $\angle BAC$, $\angle ABD$ at the vertex B of triangle $ABC$ is: $$ \angle ABD = \angle ABC - \angle BAD = 80^\circ - 15^\circ = 65^\circ $$
Thus, $\angle ABD$ is 65 degrees. This reflects the amount $\angle BAD$ subtracts from $\angle ABC$, focusing on how the angle bisector influences this particular triangle geometry.
Prove that the area of an equilateral triangle described on a side of a right-angled isosceles triangle is half the area of the equilateral triangle described on its hypotenuse.
To prove the statement, let's first understand the geometry of the triangles involved.
Triangle Properties:
We start with a right-angled isosceles triangle where each of the two equal sides (let’s denote each as $a$) is perpendicular to the other.
This indicates that the hypotenuse will be $\sqrt{2}a$ as per the Pythagorean theorem since the triangle is isosceles. Thus, Hypotenuse, $c = \sqrt{2}a$.
Areas of Equilateral Triangles:
For an equilateral triangle with side length $s$, the area $A$ can be given by the formula: $$ A = \frac{\sqrt{3}}{4} s^2 $$
Step-by-Step Proof:
Equilateral Triangle on a Side ($T_1$):
For an equilateral triangle described on one side ($a$) of the right-angled isosceles triangle: $$ \text{Area of } T_1 = \frac{\sqrt{3}}{4} a^2 $$Equilateral Triangle on the Hypotenuse ($T_2$):
For another equilateral triangle described on the hypotenuse ($\sqrt{2}a$): $$ \text{Area of } T_2 = \frac{\sqrt{3}}{4} (\sqrt{2}a)^2 = \frac{\sqrt{3}}{4} \cdot 2a^2 = \frac{\sqrt{3}}{2} a^2 $$
Comparison of the Areas:
If we compare the areas of $T_1$ and $T_2$: $$ \text{Area of } T_2 = \frac{\sqrt{3}}{2} a^2 = 2 \left(\frac{\sqrt{3}}{4} a^2\right) = 2 \times \text{Area of } T_1 $$
Therefore, the area of the equilateral triangle described on the hypotenuse ($T_2$) is twice the area of the equilateral triangle described on a side ($T_1$). This means the area of $T_1$ is half the area of $T_2$: $$ \text{Area of } T_1 = \frac{1}{2} \text{Area of } T_2 $$
This successfully proves the statement that the area of an equilateral triangle described on a side of a right-angled isosceles triangle is half of the area of the equilateral triangle described on its hypotenuse.
In a $\triangle ABC$, $BD \perp CA$ and $CE \perp AB$. Prove that $\triangle ABD \sim \triangle ACE$.
To prove that triangles $\triangle ABD$ and $\triangle ACE$ are similar, we will use the Angle-Angle (AA) similarity criterion which states that two triangles are similar if they have two pairs of corresponding angles congruent.
Here are the steps to establish the similarity:
Identify Right Angles:
Since $BD \perp CA$, $\angle ADB$ is a right angle, thus $\angle ADB = 90^\circ$.
Since $CE \perp AB$, $\angle AEC$ is also a right angle, thus $\angle AEC = 90^\circ$.
Common Angle:
Both triangles $\triangle ABD$ and $\triangle ACE$ share the angle at vertex $A$, denoted as $\angle BAC$.
Use of AA Criterion:
We have identified that $\angle ADB = \angle AEC = 90^\circ$.
We also know that $\angle BAC$ is common to both triangles.
With these two pairs of congruent angles, by the AA criterion, triangles $\triangle ABD$ and $\triangle ACE$ are similar.
In conclusion, using the Angle-Angle similarity criterion, we have successfully shown that $\triangle ABD \sim \triangle ACE$.
\begin{tabular}{|l|l|l|l|l|l|l|} \hline Number of sides ($n$) & 3 & 4 & 5 & 6 & 7 & 8 \ \hline Number of diagonals ($d$) & 0 & 2 & 5 & 9 & $p$ & $q$ \ \hline \end{tabular}
For a polygon, $d$ and $n$ are related as $d = An^{2} + Bn$, the values of $A$ and $B$ are: A) $A = \frac{1}{2}, B = \frac{3}{2}$ B) $A = \frac{1}{2}, B = -\frac{3}{2}$ C) $A = \frac{1}{2}, B = \frac{3}{2}$ D) $A = -\frac{1}{2}, B = -\frac{3}{2}$
The problem involves determining the coefficients $A$ and $B$ in the formula relating the number of sides $n$ of a polygon to the number of its diagonals $d$: $$ d = An^2 + Bn $$
We start by using given data points for polygons with specific values for $n$ and $d$:
For $n = 3$ (a triangle), there are no diagonals, so $d = 0$.
For $n = 4$ (a square), there are 2 diagonals, so $d = 2$.
By substituting these points into the formula, we get two equations:
For $n = 3$, $d = 0$: $$ 0 = 9A + 3B $$
For $n = 4$, $d = 2$: $$ 2 = 16A + 4B $$
To simplify, let's divide the second equation by 2: $$ 1 = 8A + 2B $$
Now, we can express equation 1 and the simplified version of equation 2 as:
$9A + 3B = 0$
$8A + 2B = 1$
To eliminate $B$, we can multiply the second equation by 3 and subtract the first equation: $$ (24A + 6B) - (9A + 3B) = 3 - 0 $$ $$ 15A + 3B = 3 $$ $$ 15A = 3 $$ $$ A = \frac{3}{15} = \frac{1}{5} \quad (\text{Incorrect algebraic manipulation}) $$
Correction: To solve correctly, solving the simplified versions straightforwardly, equate and solve:
$9A + 3B = 0$
$8A + 2B = 1$
Solving the equations for A and B:
Multiply the second equation by 1.5 to align the coefficients with those of the first equation: $$ 12A + 3B = 1.5 $$
Subtract the first modified equation from this: $$ 12A + 3B - (9A + 3B) = 1.5 - 0 $$ $$ 3A = 1.5 $$ $$ A = 0.5 $$
Substitute $A = 0.5$ back into the first equation: $$ 9(0.5) + 3B = 0 $$ $$ 4.5 + 3B = 0 $$ $$ 3B = -4.5 $$ $$ B = -1.5 $$
Thus, the correct values are: $$ A = \frac{1}{2}, B = -\frac{3}{2} $$
The correct option matching these values of $A$ and $B$ is B) $A = \frac{1}{2}, B = -\frac{3}{2}$.
In right-angled triangle $\triangle APQ$, the measure of $\angle APC$ is:
A. $15^{\circ}$ B. $30^{\circ}$ C. $45^{\circ}$ D. $60^{\circ}$
To solve for the measure of $\angle APC$ in the right-angled triangle $\triangle APQ$, we can start by analyzing the given angles in the diagram.
In a triangle, the sum of all internal angles is always $180^{\circ}$. Therefore, in $\triangle APQ$: $$ \angle PAQ + \angle AQP + \angle APQ = 180^{\circ} $$
From the diagram:
$\angle PAQ = 45^{\circ} + 15^{\circ} = 60^{\circ}$
$\angle AQP = 90^{\circ}$ (since $\triangle APQ$ is a right-angled triangle at Q)
Now, plugging in these values: $$ 60^{\circ} + 90^{\circ} + \angle APQ = 180^{\circ} $$ $$ 150^{\circ} + \angle APQ = 180^{\circ} $$ $$ \angle APQ = 180^{\circ} - 150^{\circ} = 30^{\circ} $$
Therefore, the measure of $\angle APC$ is indeed $\boldsymbol{30^\circ}$, which aligns with option B.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle $ABC$, whose vertices are $A(0, -1)$, $B(2, 1)$, and $C(0, 3)$.
To solve this problem, we begin by finding the coordinates of the midpoints of the sides of triangle $ABC$. The vertices of $ABC$ are given as $A(0, -1)$, $B(2, 1)$, and $C(0, 3)$.
Step 1: Calculate the Midpoints
The midpoints of the sides of the triangle are calculated using the midpoint formula: $$ \text{Midpoint}((x_1, y_1), (x_2, y_2)) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) $$
Midpoint $P$ of $AB$: $$ P\left(\frac{0 + 2}{2}, \frac{-1 + 1}{2}\right) = (1, 0) $$
Midpoint $Q$ of $BC$: $$ Q\left(\frac{2 + 0}{2}, \frac{1 + 3}{2}\right) = (1, 2) $$
Midpoint $R$ of $CA$: $$ R\left(\frac{0 + 0}{2}, \frac{3 + (-1)}{2}\right) = (0, 1) $$
Step 2: Use the Area Formula
With the coordinates of the vertices $P(1, 0)$, $Q(1, 2)$, and $R(0, 1)$, we apply the formula for the area of a triangle given its vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$: $$ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| $$
Substituting the values, we have: $$ \text{Area} = \frac{1}{2} \left| 1(2-1) + 1(1-0) + 0(0-2) \right| = \frac{1}{2} \left| 1 + 1 + 0 \right| = \frac{1}{2} \times 2 = 1 $$
Therefore, the area of the triangle $PQR$ formed by joining the midpoints of the sides of triangle $ABC$ is $1$ square unit.
If $\mathrm{AD}$ and $\mathrm{PM}$ are medians of triangles $\mathrm{ABC}$ and $\mathrm{PQR}$, respectively, where $\triangle ABC \sim \triangle PQR$, prove that:
$$ \frac{AB}{PQ} = \frac{AD}{PM} $$
To prove that $$\frac{AB}{PQ} = \frac{AD}{PM}$$ where $\triangle ABC \sim \triangle PQR$, and $\mathrm{AD}$ and $\mathrm{PM}$ are medians of their respective triangles, consider the properties of similar triangles and properties of medians.
Step-by-Step Proof:
Using Similarity of Triangles:Triangle $ABC$ is similar to triangle $PQR$. By the property of similar triangles, corresponding sides are proportional. Thus: $$ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} $$
Property of Medians:Medians divide the triangles into two smaller triangles of equal area, hence:
Median $AD$ divides $BC$ into two equal segments, $BD$ and $DC$.
Median $PM$ divides $QR$ into two equal segments, $PQ$ and $QM$.
Consequently: $$ BD = DC \quad \text{and} \quad PQ = QM $$
Corresponding Medians in Similar Triangles:Since triangles $ABC$ and $PQR$ are similar and medians to corresponding sides are taken:
Medians $AD$ of $\triangle ABC$ corresponds to median $PM$ of $\triangle PQR$.
Considering the proportional sides, the medians are also proportionally divided: $$ \frac{AD}{AB} = \frac{AM}{AN} \text{ and } \frac{PM}{PQ} = \frac{PN}{PM} $$
But notice, as earlier stated median divides base into two equal segments, so $AM = \frac{1}{2}BC$ and $PN = \frac{1}{2}QR$.
Putting It Together:Substitute the equal segments from step 2: $$ \frac{AM}{AN} = \frac{1/2 BC}{1/2 QR} = \frac{BC}{QR} $$ By triangle similarity (step 1): $$ \frac{BC}{QR} = \frac{AB}{PQ} $$ Thus, the ratio of the medians: $$ \frac{AD}{PM} = \frac{BC}{QR} \quad (\because \text{medians are halves of their proportionally corresponding sides}) $$
Conclusion:$$ \frac{AB}{PQ} = \frac{AD}{PM} $$
Final Conclusion:
Since $\triangle ABC \sim \triangle PQR$, the ratio of any corresponding parts, including medians, will hold the same proportional relationship as the sides themselves, as proved.
In a $\triangle ABC$, $\angle A=x^{\circ}$, $\angle B=3x^{\circ}$, and $\angle C=y^{\circ}$. If $3y-5x=30$, prove that the triangle is right-angled.
In $\triangle ABC$, we are given the angles as follows:
$\angle A = x^\circ$
$\angle B = 3x^\circ$
$\angle C = y^\circ$
Additionally, the equation relating $x$ and $y$ is provided: $$ 3y - 5x = 30 $$
Our goal is to prove that triangle $\triangle ABC$ is a right-angled triangle.
From the properties of triangles, we know the sum of the angles in any triangle equals $180^\circ$, which gives us our first equation: $$ x + 3x + y = 180^\circ $$ $$ 4x + y = 180^\circ \quad \text{(Equation 1)} $$
Now, rewrite the equation relating $x$ and $y$ for clarity: $$ 3y - 5x = 30 \quad \text{(Equation 2)} $$
To solve for $x$ and $y$, let's manipulate these equations. First, multiply Equation 1 by 3: $$ 3(4x + y) = 3 \times 180 $$ $$ 12x + 3y = 540 \quad \text{(Modified Equation 1)} $$
Now, by subtracting Equation 2 from this modified Equation 1, we simplify to find $x$: $$ (12x + 3y) - (3y - 5x) = 540 - 30 $$ $$ 12x + 3y - 3y + 5x = 510 $$ $$ 17x = 510 $$ $$ x = \frac{510}{17} = 30^\circ $$
Substitute $x = 30^\circ$ back into Equation 2: $$ 3y - 5(30) = 30 $$ $$ 3y - 150 = 30 $$ $$ 3y = 180 $$ $$ y = \frac{180}{3} = 60^\circ $$
Now, substitute the values of $x$ and $y$ back into the expressions for $\angle A$, $\angle B$, and $\angle C$:
$\angle A = x = 30^\circ$
$\angle B = 3x = 3 \times 30^\circ = 90^\circ$
$\angle C = y = 60^\circ$
$\angle B$ is exactly $90^\circ$, making $\triangle ABC$ a right-angled triangle at $\angle B$. This completes our proof that $\triangle ABC$ is indeed a right-angled triangle.
$D, E, F$ are the mid-points of the sides $BC, CA$, and $AB$ respectively of a triangle. Then the ratio of the areas of $\triangle DEF$ and $\triangle ABC$ is
(A) $1: 4$
B $4: 1$
C $1: 2$
(D) $2: 1$
The correct option is A: $1: 4$.
Given: $D, E$, and $F$ are the mid-points of the sides $BC, CA$, and $AB$ respectively of $\triangle ABC$.
Step-by-step Explanation:
Since $D$ and $E$ are the mid-points of sides $BC$ and $AB$ respectively of $\triangle ABC$:
By the Mid-point theorem, $$DE \parallel BA$$ and $$DE = \frac{1}{2} \times BA$$
From the above, one can infer that $$DE \parallel FA$$ (considering $DE = FA$).
Similarly, considering $D$ and $F$ as mid-points of $BC$ and $AB$ respectively:
By the Mid-point theorem, $$DF \parallel CA$$, so $$DF \parallel AE$$.
Combining equations, it is evident that quadrilateral $AFDE$ turns out to be a parallelogram (both pairs of opposite sides are parallel).
Analogously, $BDEF$ also forms a parallelogram.
Examining the triangle relations:
In $\triangle DEF$ and $\triangle ABC$, note:
$$\angle FDE = \angle A$$ (opposite angles of parallelogram $AFDE$)
$$\angle DEF = \angle B$$ (opposite angles of parallelogram $BDEF$)
By the AA-similarity criterion, $\triangle DEF \sim \triangle ABC$.
For similar triangles, the ratio of areas is the square of the ratio of corresponding sides:
$$\frac{\text{Area}(\triangle DEF)}{\text{Area}(\triangle ABC)} = \left(\frac{DE}{AB}\right)^2$$
Given:
$$DE = \frac{1}{2} AB$$
Plugging the values in:
$$\frac{\text{Area}(\triangle DEF)}{\text{Area}(\triangle ABC)} = \left(\frac{1}{2}AB\right)^2 / AB^2 = \frac{1}{4}$$
Thus, the ratio of the areas is:
$$\text{Area}(\triangle DEF) : \text{Area}(\triangle ABC) = 1 : 4$$
The area of the triangle formed by the coordinate axes and a line is 6 square units and the length of its hypotenuse is 5 units. Find the equation of the line.
Let the coordinate axes be denoted by $X'OX$ and $YOY'$, and let the line intersect these axes such that $OA = a$ and $OB = b$. The line forms a triangle $AOB$ with the coordinate axes.
We know two things about this triangle:
The area of triangle $AOB$ is 6 square units.
The length of the hypotenuse of triangle $AOB$ is 5 units.
Step-by-Step
Deriving the Area:The area of the triangle can be expressed as: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times b = 6 $$ Therefore, we have: $$ ab = 12 $$
Hypotenuse Condition:Given that the length of the hypotenuse (line segment $AB$) is 5 units, we can use the Pythagorean theorem: $$ a^2 + b^2 = 25 $$
Solve the System of Equations:Substitute $b = \frac{12}{a}$ into the Pythagorean theorem equation: $$ a^2 + \left(\frac{12}{a}\right)^2 = 25 $$ Simplify and solve the resulting quadratic equation: $$ a^2 + \frac{144}{a^2} = 25 \ a^4 - 25a^2 + 144 = 0 $$ Let $p = a^2$: $$ p^2 - 25p + 144 = 0 $$ Factorizing: $$ (p - 16)(p - 9) = 0 $$
Find Values of $a$ and $b$:Thus, $p = 16$ or $p = 9$: $$ a^2 = 16 \quad \Rightarrow \quad a = 4 \quad (\text{since length is positive})\ a^2 = 9 \quad \Rightarrow \quad a = 3 $$ If $a = 4$, $b = \frac{12}{4} = 3$. If $a = 3$, $b = \frac{12}{3} = 4$.
Form the Equation of the Line:The general form of the line equation is: $$ \frac{x}{a} + \frac{y}{b} = 1 $$ Thus, the equations of the line can be: $$ \frac{x}{4} + \frac{y}{3} = 1 \quad \text{or} \quad \frac{x}{3} + \frac{y}{4} = 1 $$
Convert these to standard form: $$ 3x + 4y - 12 = 0 \quad \text{or} \quad 4x + 3y - 12 = 0 $$
Hence, the required equation of the line is: $$ \boxed{3x + 4y - 12 = 0 \text{ or } 4x + 3y - 12 = 0} $$
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