NCERT Class 10 Maths Solutions: Triangles
Question:
All circles are ______.
A. congruent
B. similar
Concept in a Minute:
Similar figures are figures that have the same shape but may differ in size. Corresponding angles are equal and corresponding sides are in the same ratio. Congruent figures are figures that have the same shape and same size.
Explanation:
Two circles are always similar because they have the same shape. The ratio of their radii will be the constant of proportionality between them, even if the radii are different. For example, a circle with radius 2 cm and a circle with radius 4 cm are similar. Their corresponding angles (which are all 360 degrees in a circle) are equal, and the ratio of their circumferences (2 * pi * 2) / (2 * pi * 4) = 1/2, and the ratio of their radii is also 2/4 = 1/2. Congruent circles would have to have the same radius, which is not always the case for all circles. Therefore, all circles are similar.
The final answer is $\boxed{B}$.
Question:
All squares are ______.
A. similar
B. congruent
Concept in a Minute:
Understanding the definitions of similar and congruent geometric figures, specifically in relation to squares. Similar figures have the same shape but can differ in size. Congruent figures have the same shape and the same size. Squares are quadrilaterals with four equal sides and four right angles.
Explanation:
All squares have four equal sides and four right angles. This means that all squares have the same shape. However, squares can have different side lengths, meaning they can have different sizes. Therefore, all squares are similar because they share the same shape. They are not necessarily congruent because their sizes might differ.
The final answer is $\boxed{A}$.
Question:
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB.
Concept in a Minute:
Properties of Parallelograms: Opposite sides are parallel and equal.
Similar Triangles: Two triangles are similar if their corresponding angles are equal (AA similarity criterion).
Explanation:
We are given a parallelogram ABCD, and a point E on the side AD produced. The line segment BE intersects CD at F. We need to show that triangle ABE is similar to triangle CFB.
To prove that ΔABE ∼ ΔCFB, we need to establish that their corresponding angles are equal. We can use the properties of a parallelogram to identify equal angles.
Step 1: Identify parallel lines.
Since ABCD is a parallelogram, we know that AD || BC and AB || DC.
Step 2: Use the property of parallel lines intersected by a transversal.
Consider the transversal BE intersecting the parallel lines AB and DC. When a transversal intersects two parallel lines, alternate interior angles are equal. Therefore, ∠ABE = ∠CFB.
Step 3: Identify another pair of equal angles.
Consider the transversal AD (extended to E) intersecting the parallel lines AB and DC.
Since AD || BC, and BE is a transversal, we can look at other angle relationships.
Let’s re-examine the angles. We need two pairs of equal angles for AA similarity.
Consider the parallel lines AB and DC, and the transversal BE. We have already identified ∠ABE = ∠CFB (alternate interior angles).
Now consider the parallel lines AD and BC.
The line segment AE is an extension of AD. So, AE is parallel to BC.
The transversal BE intersects AE and BC.
Therefore, the alternate interior angles are ∠AEB and ∠CBF. So, ∠AEB = ∠CBF.
Step 4: Apply the AA similarity criterion.
In ΔABE and ΔCFB:
1. ∠ABE = ∠CFB (Alternate interior angles, since AB || DC and BE is a transversal)
2. ∠AEB = ∠CBF (Alternate interior angles, since AE || BC and BE is a transversal)
Since two pairs of corresponding angles are equal, by the AA similarity criterion, ΔABE ∼ ΔCFB.
Question:
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that $ABPQ=ADPM$.
Concept in a Minute:
Similar Triangles: If two triangles are similar, their corresponding angles are equal and the ratio of their corresponding sides is constant.
Medians: A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
Proportionality: The ratio of corresponding sides of similar triangles is equal to the ratio of their corresponding medians.
Explanation:
Given that triangle ABC is similar to triangle PQR (ΔABC ~ ΔPQR).
This means that the ratio of corresponding sides is equal:
AB/PQ = BC/QR = AC/PR
Also, the corresponding angles are equal:
∠A = ∠P
∠B = ∠Q
∠C = ∠R
AD is the median of triangle ABC, so D is the midpoint of BC.
Therefore, BD = DC = BC/2.
PM is the median of triangle PQR, so M is the midpoint of QR.
Therefore, QM = MR = QR/2.
Now consider triangles ABD and PQM.
We have AB/PQ = BD/QM (since AB/PQ = BC/QR and BC/2 = BD, QR/2 = QM).
We also have ∠B = ∠Q (since ΔABC ~ ΔPQR).
By the Side-Angle-Side (SAS) similarity criterion, if two sides in one triangle are proportional to two sides in another triangle and the included angles are equal, then the triangles are similar.
Therefore, ΔABD ~ ΔPQM.
Since ΔABD ~ ΔPQM, the ratio of their corresponding sides is equal:
AB/PQ = BD/QM = AD/PM
We already know that AB/PQ = BC/QR.
And BD = BC/2 and QM = QR/2.
So, BD/QM = (BC/2) / (QR/2) = BC/QR.
Therefore, we have:
AB/PQ = AD/PM
We are asked to prove ABPQ = ADPM.
This can be rewritten as AB/PQ = AD/PM.
Which is exactly what we have derived from the similarity of triangles ABC and PQR and their respective medians AD and PM.
Thus, ABPQ = ADPM is proven.
Question:
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $AOBO=CODO$
Concept in a Minute:
Similar Triangles: Triangles are similar if their corresponding angles are equal. If two triangles are similar, the ratio of their corresponding sides is equal.
Properties of Parallel Lines: When a transversal intersects two parallel lines, alternate interior angles are equal and vertically opposite angles are equal.
Explanation:
To show that $AO \cdot BO = CO \cdot DO$, we will utilize the properties of similar triangles formed within the trapezium.
1. Identify Similar Triangles: Since AB || DC, consider the diagonals AC and BD as transversals.
* $\angle OAB = \angle OCD$ (alternate interior angles)
* $\angle OBA = \angle ODC$ (alternate interior angles)
* $\angle AOB = \angle COD$ (vertically opposite angles)
Therefore, triangle AOB is similar to triangle COD (by AAA similarity criterion).
2. Use Ratios of Corresponding Sides: Because triangle AOB ~ triangle COD, the ratio of their corresponding sides is equal:
$\frac{AO}{CO} = \frac{BO}{DO} = \frac{AB}{CD}$
3. Rearrange the Equation: From the equality of the first two ratios, we have:
$\frac{AO}{CO} = \frac{BO}{DO}$
Cross-multiplying this equation gives:
$AO \cdot DO = CO \cdot BO$
Rearranging this slightly to match the required form:
$AO \cdot BO = CO \cdot DO$
Hence, $AO \cdot BO = CO \cdot DO$ is shown.
Question:
E and F are points on the sides PQ and PR, respectively, of a ΔPQR. For the following case, state whether EF || QR:
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Concept in a Minute:
The question is about testing if a line segment EF is parallel to the base QR of a triangle PQR, where E and F are points on sides PQ and PR respectively. This can be solved using the converse of the Basic Proportionality Theorem (also known as the Thales’s Theorem). The theorem states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Explanation:
We are given a triangle PQR, with points E on PQ and F on PR.
We are provided with the lengths of the segments:
PE = 4 cm
QE = 4.5 cm
PF = 8 cm
RF = 9 cm
According to the converse of the Basic Proportionality Theorem, EF || QR if and only if PE/QE = PF/RF.
Let’s check if this condition holds true for the given values.
Calculate the ratio PE/QE:
PE/QE = 4 cm / 4.5 cm
To simplify this ratio, we can multiply the numerator and denominator by 10 to remove the decimal:
PE/QE = 40/45
Now, we can divide both the numerator and the denominator by their greatest common divisor, which is 5:
PE/QE = 8/9
Calculate the ratio PF/RF:
PF/RF = 8 cm / 9 cm
PF/RF = 8/9
Now, compare the two ratios:
PE/QE = 8/9
PF/RF = 8/9
Since PE/QE = PF/RF, the condition for the converse of the Basic Proportionality Theorem is satisfied.
Therefore, EF is parallel to QR.
Answer: Yes, EF || QR
Question:
E and F are points on the sides PQ and PR, respectively, of a ΔPQR. For the following case, state whether EF || QR.
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Concept in a Minute:
The question is about testing the converse of the Basic Proportionality Theorem (also known as Thales’s Theorem). The theorem states that if a line divides two sides of a triangle proportionally, then the line is parallel to the third side. To solve this, we need to check if the ratio of the segments on one side is equal to the ratio of the segments on the other side.
Explanation:
We are given a triangle PQR and points E on side PQ and F on side PR. We need to determine if the line segment EF is parallel to the side QR.
According to the converse of the Basic Proportionality Theorem, if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
In ΔPQR, if PE/EQ = PF/FR, then EF || QR.
Let’s check the given values:
PE = 3.9 cm
EQ = 3 cm
PF = 3.6 cm
FR = 2.4 cm
Now, we calculate the ratio PE/EQ:
PE/EQ = 3.9 cm / 3 cm = 39/30 = 13/10 = 1.3
Next, we calculate the ratio PF/FR:
PF/FR = 3.6 cm / 2.4 cm = 36/24 = 6/4 = 3/2 = 1.5
Comparing the two ratios:
PE/EQ = 1.3
PF/FR = 1.5
Since PE/EQ ≠ PF/FR (1.3 ≠ 1.5), the line segment EF does not divide the sides PQ and PR in the same ratio.
Therefore, by the converse of the Basic Proportionality Theorem, EF is not parallel to QR.
Final Answer: No, EF is not parallel to QR.
Question:
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Concept in a Minute:
Similar Triangles. When two triangles have corresponding angles equal, they are similar. The ratio of corresponding sides in similar triangles is equal. In this problem, the sun’s rays are parallel, creating similar right-angled triangles formed by the vertical pole and its shadow, and the tower and its shadow.
Explanation:
Let the height of the vertical pole be $h_p$ and the length of its shadow be $s_p$.
Given: $h_p = 6$ m, $s_p = 4$ m.
Let the height of the tower be $h_t$ and the length of its shadow be $s_t$.
Given: $s_t = 28$ m.
We need to find $h_t$.
Since the sun’s rays are parallel at the same time, the angle of elevation of the sun is the same for both the pole and the tower. This means that the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow.
In similar triangles, the ratio of corresponding sides is equal.
Therefore, we can write the proportion:
$\frac{h_p}{s_p} = \frac{h_t}{s_t}$
Substitute the given values:
$\frac{6 \text{ m}}{4 \text{ m}} = \frac{h_t}{28 \text{ m}}$
Now, we solve for $h_t$:
$h_t = \frac{6}{4} \times 28$
$h_t = \frac{3}{2} \times 28$
$h_t = 3 \times 14$
$h_t = 42$ m.
Therefore, the height of the tower is 42 meters.
Question:
E and F are points on the sides PQ and PR, respectively, of a ΔPQR. For the following case, state whether EF || QR.
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Concept in a Minute:
This question involves the converse of the Basic Proportionality Theorem (BPT). The BPT states that if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally. The converse states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Explanation:
To determine if EF is parallel to QR, we need to check if the line segment EF divides the sides PQ and PR in the same ratio. According to the converse of the Basic Proportionality Theorem, if PE/EQ = PF/FR, then EF || QR.
We are given:
PQ = 1.28 cm
PR = 2.56 cm
PE = 0.18 cm
PF = 0.36 cm
First, calculate the lengths of EQ and FR:
EQ = PQ – PE = 1.28 cm – 0.18 cm = 1.10 cm
FR = PR – PF = 2.56 cm – 0.36 cm = 2.20 cm
Now, calculate the ratios PE/EQ and PF/FR:
PE/EQ = 0.18 cm / 1.10 cm
PF/FR = 0.36 cm / 2.20 cm
Let’s simplify these ratios to compare them:
PE/EQ = 18/110 = 9/55
PF/FR = 36/220 = 18/110 = 9/55
Since PE/EQ = PF/FR (both are equal to 9/55), according to the converse of the Basic Proportionality Theorem, EF is parallel to QR.
Therefore, EF || QR.
Question:
Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Concept in a Minute:
Basic Proportionality Theorem (BPT) and its Converse. BPT states that if a line parallel to one side of a triangle intersects the other two sides at distinct points, then it divides the two sides in the same ratio. The Converse of BPT states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. The midpoint of a line segment divides it in a 1:1 ratio.
Explanation:
Let ABC be a triangle. Let D be the midpoint of side AB and E be the midpoint of side AC. We need to prove that DE is parallel to BC.
Since D is the midpoint of AB, AD = DB. Therefore, AD/DB = 1.
Since E is the midpoint of AC, AE = EC. Therefore, AE/EC = 1.
From these two equations, we have AD/DB = AE/EC.
According to the Converse of the Basic Proportionality Theorem, if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
In triangle ABC, the line segment DE divides sides AB and AC in the same ratio (AD/DB = AE/EC).
Therefore, by the Converse of the Basic Proportionality Theorem, DE is parallel to BC.
Question:
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that $OAOC=OBOD$.
Concept in a Minute:
The key concept here is the similarity of triangles. Specifically, we will use the AA similarity criterion. This criterion states that if two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. Once triangles are proven similar, the ratio of their corresponding sides is equal.
Explanation:
We are given a trapezium ABCD where AB is parallel to DC. The diagonals AC and BD intersect at point O. We need to show that OA/OC = OB/OD.
Consider triangles ΔOAB and ΔOCD.
1. Angle OAB = Angle OCD (Alternate interior angles because AB || DC and AC is a transversal).
2. Angle OBA = Angle ODC (Alternate interior angles because AB || DC and BD is a transversal).
3. Angle AOB = Angle COD (Vertically opposite angles).
Since two angles of ΔOAB are equal to two angles of ΔOCD (e.g., Angle OAB = Angle OCD and Angle OBA = Angle ODC), we can conclude that ΔOAB is similar to ΔOCD by the AA similarity criterion.
Because the triangles are similar, the ratio of their corresponding sides are equal:
OA/OC = OB/OD = AB/CD
Therefore, OA/OC = OB/OD.
Question:
Give two different example of a pair of similar figures.
Concept in a Minute:
Similar figures are figures that have the same shape but can have different sizes. This means that their corresponding angles are equal and the ratio of their corresponding sides is constant.
Explanation:
To give two different examples of a pair of similar figures, we need to identify pairs of objects that satisfy the definition of similarity.
Example 1:
Consider two squares.
Let Square A have sides of length 2 units.
Let Square B have sides of length 4 units.
All angles in a square are 90 degrees, so all corresponding angles are equal.
The ratio of corresponding sides is 4/2 = 2. Since the ratio is constant for all sides, Square A and Square B are similar.
Example 2:
Consider two equilateral triangles.
Let Equilateral Triangle C have sides of length 3 units.
Let Equilateral Triangle D have sides of length 6 units.
All angles in an equilateral triangle are 60 degrees, so all corresponding angles are equal.
The ratio of corresponding sides is 6/3 = 2. Since the ratio is constant for all sides, Equilateral Triangle C and Equilateral Triangle D are similar.
Question:
Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Concept in a Minute:
Basic Proportionality Theorem (BPT) states that if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally. The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side. This question requires using BPT to prove a consequence related to the Midpoint Theorem.
Explanation:
Let the triangle be ABC. Let D be the midpoint of side AB and E be a point on side AC such that DE is parallel to BC. We need to prove that E is the midpoint of AC.
Since DE is parallel to BC, by the Basic Proportionality Theorem, we have:
AD/DB = AE/EC
Given that D is the midpoint of AB, we know that AD = DB.
Therefore, AD/DB = 1.
Substituting this into the BPT equation:
1 = AE/EC
This implies that AE = EC.
Thus, E is the midpoint of AC.
This proves that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Question:
Two polygons of the same number of sides are similar, if (a) their corresponding angles are ______ and (b) their corresponding sides are ______. (equal, proportional)
Concept in a Minute:
Similar polygons definition: Two polygons are similar if their corresponding angles are equal and the ratio of their corresponding sides is constant.
Explanation:
The question asks to fill in the blanks to complete the definition of similar polygons with the same number of sides. The definition of similar polygons requires two conditions to be met:
1. Their corresponding angles must be equal. This means that each angle in one polygon must have the same measure as the corresponding angle in the other polygon.
2. Their corresponding sides must be proportional. This means that the ratio of the lengths of any pair of corresponding sides must be the same for all pairs of corresponding sides.
Therefore, the blanks should be filled with “equal” and “proportional”.
The completed sentence is: “Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.”
Question:
S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.
Concept in a Minute:
To show that two triangles are similar, we can use the Angle-Angle (AA) similarity criterion. This criterion states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
Explanation:
We are given a triangle PQR and points S on PR and T on QR such that ∠P = ∠RTS. We need to show that ΔRPQ ∼ ΔRTS.
Step 1: Identify common angles.
Observe that both triangles ΔRPQ and ΔRTS share the angle ∠R. This means ∠PRQ = ∠SRT.
Step 2: Use the given information.
We are given that ∠P = ∠RTS.
Step 3: Apply the AA similarity criterion.
In ΔRPQ and ΔRTS:
1. ∠P = ∠RTS (Given)
2. ∠PRQ = ∠SRT (Common angle)
Since two angles of ΔRPQ are equal to two angles of ΔRTS, by the Angle-Angle (AA) similarity criterion, we can conclude that ΔRPQ ∼ ΔRTS.
Therefore, ΔRPQ is similar to ΔRTS.
Next Chapter: Acids, Bases and Salts
Refer Triangles Notes
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