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Detailed Chapter 05 Geometry TN Board Solutions for Class 8 Maths
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Class 8 Maths Chapter 05 Geometry TN Board Solutions PDF
Question 1. In the figure, given that \( \angle 1 = \angle 2 \) and \( \angle 3 = \angle 4 \). Prove that \( \Delta MUG \equiv \Delta TUB \).
Answer:
| S. No. | Statements | Reasons |
|---|---|---|
| 1. | In \( \Delta MUG \) and \( \Delta TUG \) | |
| \( \angle 3 = \angle 4 \) | These angles are given as equal, so the sides opposite to them must also be equal. | |
| MU = TU | Side opposite to equal angles | |
| \( \angle 1 = \angle 2 \) | Given as equal | |
| 2. | UG = UB | The sides opposite to equal angles are equal in length. |
| 3. | \( \angle GUM = \angle BUT \) | These are vertically opposite angles, which means they are always equal. |
| 4. | \( \Delta MUG \equiv \Delta TUB \) | This congruence is proven by the SAS (Side-Angle-Side) criteria, using the results from points 1, 2, and 3. |
🎯 Exam Tip: When proving triangle congruence, clearly state which congruence rule (SSS, SAS, ASA, AAS, RHS) you are using and list the corresponding parts that satisfy it.
Question 2. From the figure, prove that \( \Delta SUN \sim \Delta RAY \).
Answer: We need to show that the corresponding sides of \( \Delta SUN \) and \( \Delta RAY \) are proportional. From the given figure, we have:
SN = 14
RA = 5
AY = 6
RY = 7
SU = 10
UN = 12
Now let's find the ratios of the corresponding sides:
\( \frac { SU }{ RA } = \frac { 10 }{ 5 } = \frac { 2 }{ 1 } \) ....(1)
\( \frac { UN }{ AY } = \frac { 12 }{ 6 } = \frac { 2 }{ 1 } \) ....(2)
\( \frac { SN }{ RY } = \frac { 14 }{ 7 } = \frac { 2 }{ 1 } \) ....(3)
From equations (1), (2) and (3), we see that the ratios of the corresponding sides are all equal:
\( \frac { SU }{ RA } = \frac { UN }{ AY } = \frac { SN }{ RY } = \frac { 2 }{ 1 } \)
Since all three pairs of corresponding sides are proportional, the triangles are similar. This means they have the same shape, even if their sizes are different.
\( \implies \Delta SUN \sim \Delta RAY \)
In simple words: We checked the lengths of the sides of both triangles. When we divided the length of each side from the first triangle by the length of the matching side from the second triangle, we got the same answer every time. This shows that the two triangles are similar, meaning they look alike but might be different sizes.
🎯 Exam Tip: To prove triangle similarity using SSS (Side-Side-Side) criterion, always calculate the ratios of all three corresponding sides and ensure they are equal. Clearly match the corresponding vertices for correct ratio formation.
Question 3. The height of a tower is measured by a mirror on the ground at R by which the top of the tower's reflection is seen. Find the height of the tower. If \( \Delta PQR \sim \Delta STR \)
Answer: We are given that \( \Delta PQR \) is similar to \( \Delta STR \). This means their corresponding sides are proportional. The reflection principle shows that the angles of incidence and reflection are equal, which makes the triangles similar.
From the property of similar triangles, the ratios of corresponding sides are equal:
\( \frac { PQ }{ ST } = \frac { QR }{ TR } = \frac { PR }{ SR } \)
We can use the ratio for the height and the distances along the ground:
\( \frac { PQ }{ ST } = \frac { QR }{ TR } \)
Let \( h \) be the height of the tower (PQ). From the figure, ST = 8 ft, QR = 60 ft, and TR = 10 ft.
\( \frac { h }{ 8 } = \frac { 60 }{ 10 } \)
Now, we solve for \( h \):
\( h = \frac { 60 }{ 10 } \times 8 \)
\( h = 6 \times 8 \)
\( h = 48 \text{ feet} \)
Therefore, the height of the tower is 48 feet.
In simple words: Because the mirror creates similar triangles, we can use the given lengths to find the tower's height. We set up a simple division problem with the known heights and distances, then solved for the unknown height.
🎯 Exam Tip: For problems involving similar triangles, correctly identify the corresponding sides before setting up the ratios. A common error is mixing up the order of vertices.
Question 4. Find the length of the support cable required to support the tower with the floor.
Answer: In the given figure, we have a right-angled triangle formed by the tower, the floor, and the support cable. The length of the support cable is the hypotenuse of this triangle. We can use the Pythagoras theorem to find its length. The Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Let \( x \) be the length of the support cable.
From the figure, the height of the tower is 20 ft and the distance from the base of the tower to where the cable is anchored on the floor is 15 ft.
According to Pythagoras theorem:
\( x^2 = 20^2 + 15^2 \)
\( x^2 = 400 + 225 \)
\( x^2 = 625 \)
To find \( x \), we take the square root of 625:
\( x^2 = 25^2 \)
\( x = 25 \text{ ft} \)
Therefore, the length of the support cable required to support the tower with the floor is 25 ft.
In simple words: We used the Pythagoras theorem, which works for triangles with a right angle. We know the height of the tower and how far the cable reaches on the ground. By squaring these two numbers, adding them, and then finding the square root, we found the length of the cable.
🎯 Exam Tip: Always remember that the Pythagoras theorem \( a^2 + b^2 = c^2 \) only applies to right-angled triangles, where 'c' is always the hypotenuse (the longest side).
Question 5. Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Give reason.
Answer: A TV screen size (like 25 inches) refers to the length of its diagonal. We can think of the screen as a right-angled triangle, where the height and width are the two shorter sides, and the diagonal is the hypotenuse. We can use the Pythagoras theorem to find the width of the screen.
Let the sides of the right-angled triangle (the TV screen) be:
Height (a) = 7 inches
Diagonal (b) = 25 inches (This is the hypotenuse)
Width (c) = ?
According to the Pythagoras theorem: \( b^2 = a^2 + c^2 \)
Substitute the known values:
\( 25^2 = 7^2 + c^2 \)
\( 625 = 49 + c^2 \)
To find \( c^2 \), subtract 49 from 625:
\( c^2 = 625 - 49 \)
\( c^2 = 576 \)
To find \( c \), take the square root of 576:
\( c = \sqrt{576} \)
\( c = 24 \text{ inches} \)
So, the width of the TV screen is 24 inches. Now we compare this to the cabinet width.
The TV cabinet is 20 inches wide.
Since the TV screen is 24 inches wide, and the cabinet is only 20 inches wide, the TV will not fit into the cabinet.
In simple words: First, we used a math rule called Pythagoras theorem to find out how wide the TV screen is, using its diagonal and height. The TV's width turned out to be 24 inches. Then, we checked if 24 inches would fit into a 20-inch cabinet. It won't fit because the TV is wider than the cabinet.
🎯 Exam Tip: Always draw a simple diagram if one isn't provided for geometry problems; it helps visualize the right-angled triangle and correctly apply the Pythagoras theorem. Clearly label your knowns and unknowns.
Challenging Problems
Question 6. In the figure, \( \angle TMA = \angle IAM \) and \( \angle TAM = \angle IMA \). P is the midpoint of MI and N is the midpoint of AI. Prove that \( \Delta PIN \sim \Delta ATM \).
Answer: We need to prove that \( \Delta PIN \) is similar to \( \Delta ATM \). For triangles to be similar, their corresponding angles must be equal, or their corresponding sides must be proportional. We will use the Angle-Angle-Angle (AAA) similarity criterion.
| S. No. | Statements | Reasons |
|---|---|---|
| 1. | \( \angle TMA = \angle IAM \) \( \angle TAM = \angle IMA \) | These angles are given to be equal in the problem statement. |
| 2. | \( \angle ATM = \angle TIM \) | This angle is the remaining angle in the triangle. It is found using the angle sum property of triangles. |
| 3. | P is the midpoint of MI and N is the midpoint of IA. | This information is given in the problem, stating that P and N divide the sides into two equal parts. |
| \( \frac { IP }{ PM } = 1 \) \( \frac { IN }{ NA } = 1 \) | Since P and N are midpoints, they divide the segments into equal halves, making the ratio 1. | |
| 4. | \( \frac { IP }{ PM } = \frac { IN }{ NA } \) | From step 3, both ratios are equal to 1, so they are equal to each other. |
| 5. | PN || MA | Since P and N are midpoints and the ratio of segments is equal, PN is parallel to MA by the converse of the Midpoint Theorem. |
| 6. | \( \angle IPN = \angle IMN \) \( \angle INP = \angle IAM \) | Because PN is parallel to MA, these are corresponding angles and therefore equal. |
| 7. | In \( \Delta PIN \) and \( \Delta ATM \) | |
| (i) \( \angle IPN = \angle TAM \) (ii) \( \angle INP = \angle TMA \) (iii) \( \angle ATM = \angle PIN \) | We have established these angle equalities from steps 1, 2, and 6. For example, \( \angle TAM \) from step 1 is equal to \( \angle IPN \) from step 6. | |
| 8. | \( \Delta PIN \sim \Delta ATM \) | Since all three corresponding angles are equal, the triangles are similar by the AAA (Angle-Angle-Angle) similarity criterion. |
🎯 Exam Tip: When proving similarity, always aim to use the simplest criteria (AAA or SAS similarity). Clearly list corresponding angles or proportional sides to support your chosen criterion.
Question 7. In the figure, if \( \angle FEG = \angle 1 \) then, prove that \( DG^2 = DE \cdot DF \).
Answer: We need to prove that \( DG^2 = DE \cdot DF \). This type of relationship usually comes from similar triangles. We will show that \( \Delta DGF \) is similar to \( \Delta DEG \). If we can prove this, then the ratio of their corresponding sides will be equal, which will lead to the desired equation. The crucial first step is to establish angle equalities based on the given information and geometric properties.
| S. No. | Statements | Reasons |
|---|---|---|
| 1. | \( \angle FEG = \angle 1 \) \( \implies \angle DEG = 180^\circ - \angle FEG \) | The angles \( \angle FEG \) and \( \angle DEG \) form a linear pair, meaning they lie on a straight line and add up to 180 degrees. |
| 2. | In \( \Delta DFG \): \( \angle FDG + \angle DFG = \angle 1 \) | The exterior angle \( \angle 1 \) of \( \Delta DFG \) is equal to the sum of its two interior opposite angles. |
| \( \implies \angle FDG + \angle DFG = \angle FEG \) | Since \( \angle 1 = \angle FEG \) (given). | |
| 3. | In \( \Delta DFG \), \( \angle DGF = 180^\circ - (\angle FDG + \angle DFG) \) | The sum of angles in any triangle is 180 degrees. |
| \( \implies \angle DGF = 180^\circ - \angle FEG \) | Substitute \( (\angle FDG + \angle DFG) \) with \( \angle FEG \) from step 2. | |
| 4. | \( \angle DGF = \angle DEG \) | Comparing with step 1, both \( \angle DGF \) and \( \angle DEG \) are equal to \( 180^\circ - \angle FEG \). This is a crucial step for similarity. |
| 5. | \( \angle EDG = \angle GDF \) | This angle is common to both \( \Delta DEG \) and \( \Delta DGF \). |
| 6. | In \( \Delta DGF \) and \( \Delta DEG \) | |
| \( \angle DGF = \angle DEG \) (from step 4) \( \angle FDG = \angle EDG \) (common angle from step 5) | We have found two pairs of equal corresponding angles. | |
| Remaining angle, \( \angle DFG = \angle DGE \) | If two angles of two triangles are equal, the third angle must also be equal. | |
| 7. | \( \Delta DGF \sim \Delta DEG \) | The triangles are similar by the AAA (Angle-Angle-Angle) similarity criterion, using the equalities established in step 6. |
| 8. | \( \frac { DG }{ DE } = \frac { GF }{ EG } = \frac { DF }{ DG } \) | When triangles are similar, the ratios of their corresponding sides are proportional. |
| 9. | \( \frac { DG }{ DE } = \frac { DF }{ DG } \) | We take the first and third parts of the proportion from step 8. |
| 10. | \( DG \cdot DG = DE \cdot DF \) \( DG^2 = DE \cdot DF \) | Cross-multiply the terms from step 9 to get the final relationship. |
🎯 Exam Tip: When proving a geometric relationship like \( x^2 = yz \), it often points to using similar triangles where \( x \) is a common side in the ratio. Carefully establish angle equalities first.
Question 8. The diagonals of the rhombus is 12 cm and 16 cm. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).
Answer: A rhombus is a four-sided shape where all four sides are equal in length. Its diagonals cross each other exactly in the middle and form a right angle (90 degrees). This creates four smaller right-angled triangles inside the rhombus. We can use the Pythagoras theorem to find the length of one side of the rhombus.
Let the diagonals be \( d_1 = 12 \text{ cm} \) and \( d_2 = 16 \text{ cm} \).
Since the diagonals bisect each other, half of each diagonal will be the legs of a right-angled triangle:
Half of \( d_1 = \text{AO} = \text{CO} = \frac { 12 }{ 2 } = 6 \text{ cm} \)
Half of \( d_2 = \text{BO} = \text{DO} = \frac { 16 }{ 2 } = 8 \text{ cm} \)
Consider any one of the right-angled triangles, for example, \( \Delta AOB \). The sides AO and BO are the legs, and the side AB (which is a side of the rhombus) is the hypotenuse.
By Pythagoras theorem, \( AB^2 = AO^2 + BO^2 \)
\( AB^2 = 6^2 + 8^2 \)
\( AB^2 = 36 + 64 \)
\( AB^2 = 100 \)
To find AB, take the square root of 100:
\( AB = \sqrt{100} \)
\( AB = 10 \text{ cm} \)
Since all sides of a rhombus are equal, each side is 10 cm.
The perimeter of a rhombus is \( 4 \times \text{side} \).
Perimeter = \( 4 \times 10 \text{ cm} = 40 \text{ cm} \)
Therefore, the perimeter of the rhombus is 40 cm.
In simple words: We used the fact that a rhombus's diagonals cut each other in half at a right angle, which makes small right-angled triangles. We found the lengths of the two shorter sides of one of these small triangles (half of each diagonal). Then, using Pythagoras theorem, we found the longest side of that triangle, which is also one side of the rhombus. Since all sides of a rhombus are equal, we multiplied this side length by 4 to get the total perimeter.
🎯 Exam Tip: Remember the key properties of a rhombus: all sides are equal, and diagonals bisect each other at right angles. These properties are essential for solving problems involving its sides and diagonals.
Question 9. In the figure, find AR.
Answer: We need to find the total length of AR. From the figure, we can see two right-angled triangles, \( \Delta AFI \) and \( \Delta RFI \), where \( I \) is the point where the altitude from F meets AR. (The OCR text shows \( \Delta AFI \) and \( \Delta FRI \), which is consistent with the diagram if F is the vertex and I is on AR). We can use the Pythagoras theorem in both triangles to find the segments AF and FR, then add them to get AR. However, the problem statement asks to find AR, and the solution uses \( \Delta AFI \) and \( \Delta FRI \) where I is an intermediate point. Let's assume the vertical line from F hits AR at I. From the diagram, \( AF = 25 \text{ ft} \), \( FI = 15 \text{ ft} \), \( RF = 17 \text{ ft} \).
First, find AI using \( \Delta AFI \):
\( AF^2 = AI^2 + FI^2 \)
\( 25^2 = AI^2 + 15^2 \)
\( 625 = AI^2 + 225 \)
\( AI^2 = 625 - 225 \)
\( AI^2 = 400 \)
\( AI = \sqrt{400} \)
\( AI = 20 \text{ ft} \)
Next, find IR using \( \Delta RFI \):
\( RF^2 = IR^2 + FI^2 \)
\( 17^2 = IR^2 + 15^2 \)
\( 289 = IR^2 + 225 \)
\( IR^2 = 289 - 225 \)
\( IR^2 = 64 \)
\( IR = \sqrt{64} \)
\( IR = 8 \text{ ft} \)
Finally, find the total length AR:
\( AR = AI + IR \)
\( AR = 20 \text{ ft} + 8 \text{ ft} \)
\( AR = 28 \text{ ft} \)
In simple words: We broke the problem into two smaller right-angled triangles. We used the Pythagoras theorem for each triangle to find the lengths of their bases. Then, we added these two base lengths together to find the total length of AR.
🎯 Exam Tip: When a complex shape can be divided into right-angled triangles, break it down and apply the Pythagoras theorem step-by-step. Make sure to clearly identify the hypotenuse and legs in each triangle.
Question 10. In \( \Delta DEF \), DN, EO, FM are medians and point P is the centroid. Find the following.
(i) If DE = 44, then DM = ?
(ii) If PD = 12, then PN = ?
(iii) If DO = 8, then PD = ? (The solution calculates FD for DO = 8)
(iv) If OE = 36 then EP = ?
Answer: In a triangle, medians connect a vertex to the midpoint of the opposite side. The centroid (point P) is where all three medians meet. The centroid divides each median in a 2:1 ratio, with the longer segment being from the vertex to the centroid and the shorter segment from the centroid to the midpoint of the side.
(i) If DE = 44, then DM = ?
DM is half of DE because M is the midpoint of DE (FM is a median).
\( DM = \frac { DE }{ 2 } \)
\( DM = \frac { 44 }{ 2 } \)
\( DM = 22 \)
So, DM is 22 units.
(ii) If PD = 12, then PN = ?
P is the centroid, so it divides the median DN in a 2:1 ratio, with DP being 2 parts and PN being 1 part.
\( \frac { DP }{ PN } = \frac { 2 }{ 1 } \)
Given PD = 12:
\( \frac { 12 }{ PN } = \frac { 2 }{ 1 } \)
To find PN, cross-multiply:
\( 2 \times PN = 12 \times 1 \)
\( 2 \times PN = 12 \)
\( PN = \frac { 12 }{ 2 } \)
\( PN = 6 \)
So, PN is 6 units.
(iii) If DO = 8, then FD = ?
(Note: The question asks for PD, but the solution calculates FD based on DO=8. We will follow the solution's calculation. EO is a median, so O is the midpoint of FD. This means DO = OF).
Since O is the midpoint of FD, then FD = DO + OF.
Given DO = 8, and since O is the midpoint, OF must also be 8.
\( FD = 8 + 8 \)
\( FD = 16 \)
So, FD is 16 units.
(iv) If OE = 36 then EP = ?
P is the centroid, so it divides the median OE in a 2:1 ratio, with OP being 2 parts and PE being 1 part. The total length OE is 3 parts (OP + PE = 2 parts + 1 part = 3 parts).
So, PE is \( \frac { 1 }{ 3 } \) of OE.
\( EP = \frac { 1 }{ 3 } \times OE \)
\( EP = \frac { 1 }{ 3 } \times 36 \)
\( EP = 12 \)
Alternatively, using the ratio: \( \frac { OP }{ PE } = \frac { 2 }{ 1 } \). Also, \( OE = OP + PE \).
Let PE = \( x \). Then OP = \( 2x \).
\( OE = 2x + x = 3x \)
Given OE = 36, so \( 3x = 36 \)
\( x = \frac { 36 }{ 3 } \)
\( x = 12 \)
Therefore, EP = 12 units.
In simple words: We used the special properties of medians and the centroid in a triangle. Medians connect a corner to the middle of the opposite side. The centroid is the point where they all cross, and it divides each median into two parts, with one part being twice as long as the other. We used these rules to find the missing lengths.
🎯 Exam Tip: Always remember the 2:1 ratio for the centroid dividing a median. The segment from the vertex to the centroid is twice as long as the segment from the centroid to the midpoint of the opposite side.
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TN Board Solutions Class 8 Maths Chapter 05 Geometry
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