Maharashtra Board Class 8 Maths Chapter 4 Altitudes and Medians of a Triangle Set 4.1 Solutions

Question 1. In ALMN, ___ is an altitude and ___ is a median, (write the names of appropriate segments.)
ℹ️ चित्र व्याख्या (Diagram Explanation): LMN त्रिभुज का एक चित्र है, जिसमें L से भुजा MN पर एक शीर्षलंब LX खींचा गया है, और L से भुजा MN पर एक माध्यिका LY खींची गई है।
Answer: In ALMN, seg LX is an altitude and seg LY is a median.
In simple words: An altitude is a perpendicular line segment from a vertex to the opposite side, while a median connects a vertex to the midpoint of the opposite side. In the given triangle, LX is perpendicular to MN, making it an altitude, and LY goes to the midpoint of MN, making it a median.

🎯 Exam Tip: Distinguish between altitudes and medians by remembering that altitudes involve perpendicularity and medians involve midpoints. Clearly labeling segments on diagrams helps in answering such questions.

Question 2. Draw an acute angled ΔPQR. Draw all of its altitudes. Name the point of concurrence as 'O'.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक न्यूनकोण त्रिभुज PQR का चित्र है, जिसके तीनों शीर्षलंब PA, QB और RC त्रिभुज के अंदर बिंदु O पर प्रतिच्छेद करते हुए दिखाए गए हैं। बिंदु O त्रिभुज का लंबकेंद्र है।
Answer: The diagram above shows an acute-angled triangle PQR with its altitudes drawn. The point of concurrence 'O' is the orthocentre, which lies inside the triangle.
In simple words: For an acute-angled triangle, all three altitudes meet at a single point called the orthocentre, which is always located inside the triangle.

🎯 Exam Tip: For acute-angled triangles, ensure all altitudes are drawn accurately from each vertex to the opposite side, forming a 90-degree angle. The orthocentre should always be within the triangle for this type.

Question 3. Draw an obtuse angled ΔSTV. Draw its medians and show the centroid.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक अधिककोण त्रिभुज STV का चित्र है, जिसमें शीर्षों से विपरीत भुजाओं के मध्यबिंदुओं तक खींची गई माध्यिकाएँ SU, TP, और VR त्रिभुज के अंदर बिंदु G पर प्रतिच्छेद कर रही हैं। बिंदु G त्रिभुज का केंद्रक है।
Answer: The diagram above illustrates an obtuse-angled triangle STV with its medians SU, TP, and VR drawn. The point of concurrence, G, is the centroid, located within the triangle.
In simple words: The medians of any triangle, including an obtuse-angled one, are line segments connecting each vertex to the midpoint of the opposite side. They always intersect at a point called the centroid, which is always inside the triangle.

🎯 Exam Tip: When drawing medians, accurately find the midpoint of each side before connecting it to the opposite vertex. The centroid (point of concurrence) will always be inside the triangle, regardless of its type.

Question 4. Draw an obtuse angled ΔLMN. Draw its altitudes and denote the ortho centre by 'O'.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक अधिककोण त्रिभुज LMN का चित्र है। इसमें शीर्षलंब LX और MY खींचे गए हैं, और तीसरे शीर्षलंब को प्राप्त करने के लिए भुजाओं को बढ़ाया गया है। सभी शीर्षलंब (या उनकी विस्तारित रेखाएँ) त्रिभुज के बाहर बिंदु O पर प्रतिच्छेद करते हैं, जो त्रिभुज का लंबकेंद्र है।
Answer: The provided diagram shows an obtuse-angled triangle LMN with its altitudes drawn. For an obtuse triangle, the altitudes might fall outside the triangle, requiring extensions of the sides to find their point of intersection. The orthocentre 'O' is located outside the triangle.
In simple words: For an obtuse-angled triangle, the orthocentre (where altitudes meet) lies outside the triangle, and you may need to extend the sides to draw all three altitudes and find their common intersection point.

🎯 Exam Tip: When drawing altitudes for an obtuse-angled triangle, remember that two altitudes may fall outside the triangle. Carefully extend the sides to locate their point of concurrence, the orthocentre, which will also be outside the triangle.

Question 5. Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज XYZ का चित्र है। इसकी माध्यिकाएँ XR, YP और ZS त्रिभुज के अंदर बिंदु G पर प्रतिच्छेद कर रही हैं। बिंदु G त्रिभुज का केंद्रक है।
Answer: The diagram illustrates a right-angled triangle XYZ with its medians XR, YP, and ZS. The point of concurrence, G, is the centroid, which is always found within the triangle, even for a right-angled one.
In simple words: In any right-angled triangle, the medians connect vertices to the midpoints of opposite sides and intersect at the centroid, which is always located inside the triangle.

🎯 Exam Tip: For any type of triangle, including right-angled ones, always locate the midpoints of each side precisely to draw the medians correctly. The centroid will consistently be an interior point of the triangle.

Question 6. Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समद्विबाहु त्रिभुज PQR का चित्र है। इसमें माध्यिकाएँ PA, QB, RC बिंदु G पर प्रतिच्छेद कर रही हैं और शीर्षलंब PS, QT, RU बिंदु O पर प्रतिच्छेद कर रहे हैं। बिंदु G (केंद्रक) और O (लंबकेंद्र) दोनों एक ही रेखा PS पर स्थित हैं, जो भुजा QR की लंब समद्विभाजक भी है।
Answer: The diagram depicts an isosceles triangle PQR with its medians and altitudes drawn. The point of concurrence of medians (centroid, G) and the point of concurrence of altitudes (orthocentre, O) both lie on the same line segment PS. This line PS is also the perpendicular bisector of seg QR.
In simple words: In an isosceles triangle, the centroid and orthocentre both lie on the median/altitude drawn from the vertex angle to the base. This line also acts as the perpendicular bisector of the base.

🎯 Exam Tip: For isosceles triangles, the median and altitude from the vertex angle to the base coincide. This alignment means the centroid and orthocentre will always lie on this specific line, a key property to remember for constructions and observations.

Question 7. Fill in the blanks. Point G is the centroid of ΔABC.
(i) If I(RG) = 2.5, then I(GC) = ___
(ii) If I(BG) = 6, then I(BQ) = ___
(iii) If I(AP) = 6, then I(AG) = ___ and I(GP) = ___
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC का चित्र है, जिसमें माध्यिकाएँ AR, BQ और CP बिंदु G पर प्रतिच्छेद कर रही हैं। बिंदु G त्रिभुज का केंद्रक है।
Answer: The centroid of a triangle divides each median in the ratio 2:1.
(i) Point G is the centroid and seg CR is the median.
\( \frac{I(GC)}{I(RG)} = \frac{2}{1} \)
\( \implies \frac{I(GC)}{2.5} = \frac{2}{1} \) ...[ \(I(RG) = 2.5\)]
\( \implies I(GC) \times 1 = 2 \times 2.5 \)
\( \implies I(GC) = 5 \)
(ii) Point G is the centroid and seg BQ is the median.
\( \frac{I(BG)}{I(GQ)} = \frac{2}{1} \)
\( \implies \frac{6}{I(GQ)} = \frac{2}{1} \) ...[ \(I(BG) = 6\)]
\( \implies 6 \times 1 = 2 \times I(GQ) \)
\( \implies \frac{6}{2} = I(GQ) \)
\( \implies 3 = I(GQ) \) i.e. \(I(GQ) = 3 \)
Now, \(I(BQ) = I(BG) + I(GQ) \)
\( \implies I(BQ) = 6 + 3 \)
\( \implies I(BQ) = 9 \)
(iii) Point G is the centroid and seg AP is the median.
\( \frac{I(AG)}{I(GP)} = \frac{2}{1} \)
\( \implies I(AG) = 2 I(GP) \) .....(i)
Now, \(I(AP) = I(AG) + I(GP) \) ... (ii)
\( \implies I(AP) = 2I(GP) + I(GP) \) ... [From (i)]
\( \implies I(AP) = 3I(GP) \)
\( \implies 6 = 3I(GP) \) ...[ \(I(AP) = 6\)]
\( \implies \frac{6}{3} = I(GP) \)
\( \implies 2 = I(GP) \) i.e. \(I(GP) = 2 \)
\(I(AP) = I(AG) + I(GP) \) ...[from (ii)]
\( \implies 6 = I(AG) + 2 \)
\( \implies I(AG) = 6 - 2 \)
\( \implies I(AG) = 4 \)
In simple words: The centroid of a triangle divides each median into two parts, with the segment from the vertex being twice as long as the segment to the midpoint of the opposite side (2:1 ratio). This property allows us to calculate unknown segment lengths if one part or the total length of the median is given.

🎯 Exam Tip: Always remember the 2:1 ratio for the centroid dividing a median. This fundamental property is crucial for solving problems involving segment lengths related to the centroid. Clearly state the ratio and apply it correctly in calculations.

Maharashtra Board Class 8 Maths Chapter 4 Altitudes And Medians Of A Triangle Practice Set 4.1 Intext Questions And Activities

Question 1. Draw a line. Take a point outside the line. Draw a perpendicular from the point to the line with the help of a set-square (Textbook pg. no, 19)
Answer:
(i) Step 1: Draw a line I and a point P lying outside it.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक रेखा 'l' और उसके बाहर स्थित एक बिंदु 'P' का चित्रण है।
(ii) Step 2: By placing a set-square on line I, draw a perpendicular to the line from point P.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह रेखा 'l' पर एक सेट-स्क्वायर रखकर बिंदु 'P' से रेखा पर लंब खींचने की प्रक्रिया को दर्शाता है। बिंदु 'P' से रेखा 'l' पर एक लंब खींचा गया है।
In simple words: To draw a perpendicular from a point to a line using a set-square, align one edge of the set-square with the line, then slide it until the other perpendicular edge touches the point, and draw the line.

🎯 Exam Tip: Practice using a set-square for perpendicular constructions regularly. Accuracy in aligning the set-square is key to drawing a precise perpendicular line, a skill often tested in geometry.

Question 2. Draw an acute angled ΔABC and all its altitudes. Observe the location of the orthocentre. (Textbook pg. no. 20)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक न्यूनकोण त्रिभुज ABC का चित्र है, जिसके तीनों शीर्षलंब AD, BE और CF त्रिभुज के अंदर बिंदु O (लंबकेंद्र) पर प्रतिच्छेद कर रहे हैं।
Answer: Point O is the orthocentre. Orthocentre lies in the interior of ΔABC.
In simple words: For an acute-angled triangle, the point where all three altitudes intersect, known as the orthocentre, is always located inside the triangle.

🎯 Exam Tip: When identifying the orthocentre in acute-angled triangles, visually confirm that the point of concurrence of all altitudes lies strictly within the triangle's boundaries. This is a defining characteristic.

Question 3. Draw a right angled triangle and draw all its altitudes. Write the point of concurrence. (Textbook: pg, no. 20)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR का चित्र है, जिसमें कोण Q समकोण है। इस त्रिभुज के शीर्षलंब PQ (QR पर), QR (PQ पर) और QS (PR पर) दिखाए गए हैं। सभी शीर्षलंब बिंदु Q पर मिलते हैं, जो त्रिभुज का लंबकेंद्र है।
Answer: Point Q is the orthocentre. The point of concurrence of altitudes PQ, QR and QS is Q.
In simple words: In a right-angled triangle, the orthocentre is always the vertex where the right angle is formed, because the two legs of the right triangle themselves act as altitudes.

🎯 Exam Tip: For right-angled triangles, quickly identify the vertex with the 90-degree angle; this vertex is the orthocentre. This shortcut saves time and ensures accuracy in such problems.

Question 4. (i) Draw an obtuse angled triangle and all its altitudes.
(ii) Do they intersect each other? Draw the lines containing the altitudes. Observe that these lines are concurrent. (Textbook pg. no. 20)
Answer:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक अधिककोण त्रिभुज PQR का चित्र है। इसके शीर्षलंब (या उनकी विस्तारित रेखाएँ) त्रिभुज के बाहर बिंदु O पर प्रतिच्छेद करती हैं।
(ii) Yes, all the altitudes intersect at point O in the exterior of ΔPQR.
In simple words: For an obtuse-angled triangle, the altitudes, when extended, intersect at a point called the orthocentre, which is always located outside the triangle.

🎯 Exam Tip: When constructing altitudes for an obtuse triangle, remember to extend the sides to find the point of concurrence. Correctly identifying the orthocentre outside the triangle demonstrates a clear understanding of its properties.

Question 5. Draw three different triangles; a right angled triangle, an obtuse angled triangle and an acute angled triangle. Draw the medians of the triangles. Note that the centroid of each of them is in the interior of the triangle. (Textbook pg. no. 21)
Answer:
(i) Right angled triangle:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQS का चित्र है, जिसकी माध्यिकाएँ PV, QU और SR त्रिभुज के अंदर बिंदु G पर प्रतिच्छेद कर रही हैं। बिंदु G त्रिभुज का केंद्रक है।
(ii) Obtuse angled triangle:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक अधिककोण त्रिभुज PQR का चित्र है, जिसकी माध्यिकाएँ PT, QU और RS त्रिभुज के अंदर बिंदु G पर प्रतिच्छेद कर रही हैं। बिंदु G त्रिभुज का केंद्रक है।
(iii) Acute angled triangle:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक न्यूनकोण त्रिभुज PQR का चित्र है, जिसकी माध्यिकाएँ PV, QU और SR त्रिभुज के अंदर बिंदु G पर प्रतिच्छेद कर रही हैं। बिंदु G त्रिभुज का केंद्रक है।
In simple words: The centroid of any triangle, regardless of whether it's acute, obtuse, or right-angled, is always located inside the triangle. It's the point where all three medians intersect.

🎯 Exam Tip: A key property of the centroid is its consistent location: it is always an interior point of any triangle. This observation can be used to quickly verify the correctness of median constructions.

Question 6. Draw a sufficiently large ΔABC. Draw medians; seg AR, seg BQ and seg CP of ΔABC. Name the point of concurrence as G. Measure the lengths of segments from the figure and fill in the boxes in the following table. Observe that all of these ratios are nearly 2 : 1 (Textbook pg. no. 21)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC का चित्र है, जिसकी माध्यिकाएँ AR, BQ और CP बिंदु G पर प्रतिच्छेद कर रही हैं। यह केंद्रक G को दर्शाता है।

In simple words: The centroid G divides each median in a 2:1 ratio. By measuring the segments created by the centroid on each median, we observe that the segment from the vertex to the centroid is approximately twice the length of the segment from the centroid to the midpoint of the opposite side.

🎯 Exam Tip: This hands-on activity reinforces the 2:1 centroid property. In exams, if asked to prove or verify this, accurate measurement and calculation are vital. Remember to round measurements appropriately if necessary.

Question 7. As shown in the given figure, a student drew ΔABC using five parallel lines of a notebook. Then he found the centroid G of the triangle. How will you decide whether the location of G he found, is correct. (Textbook pg. no. 21)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC का चित्र है, जिसे नोटबुक की पांच समांतर रेखाओं का उपयोग करके बनाया गया है। इसमें बिंदु G त्रिभुज के केंद्रक के रूप में दिखाया गया है।
Answer: Draw seg AP perpendicular seg PE and seg EQ perpendicular seg QC.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC और इसके केंद्रक G का चित्र है, जिसमें AP, EQ और QC जैसे सहायक निर्माण दर्शाए गए हैं, जो केंद्रक की स्थिति की पुष्टि करने के लिए उपयोग किए जाते हैं।
Side AP || side EQ and AC is their transversal.
\( \therefore \angle PAE \cong \angle QEC \) ...(i) [Corresponding angles]
In \( \triangle APE \) and \( \triangle EQC \),
\( \angle PAE \cong \angle QEC \) ...[From (i)]
\( \angle APE = \angle EQC \) ...[Each angle is of measure 90°]
side PE \( \cong \) side QC ...[Perpendicular distance between parallel lines]
\( \therefore \triangle APE \cong \triangle EQC \) ... [By AAS test]
\( \therefore AE = EC \) ...[Corresponding sides of congruent triangles]
\( \therefore E \) is the midpoint of AC.
\( \therefore \) seg BE is the median.
Similarly, seg CF is the median.
Since, the medians of a triangle are concurrent.
\( \therefore G \) is the centroid of \( \triangle ABC \).
In simple words: To verify if G is the correct centroid, we can use the property that the centroid divides each median in a 2:1 ratio. By drawing a median and checking this ratio for G, or by showing G lies at the intersection of all medians, its correctness can be confirmed. Using parallel lines helps demonstrate that a line segment like BE is indeed a median by showing E is a midpoint.

🎯 Exam Tip: When verifying geometric properties, always refer back to fundamental definitions and theorems. For centroids, demonstrating that the point lies on all three medians and divides them in a 2:1 ratio is the most robust method for verification.

Question 8. Draw an equilateral triangle. Find its circumcentre (C), incentre (I), centroid (G) and orthocentre (O). Write your observation. (Textbook pg. no. 22)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समबाहु त्रिभुज XYZ का चित्र है। इस त्रिभुज के परिकेंद्र (C), अंतःकेंद्र (I), केंद्रक (G) और लंबकेंद्र (O) सभी एक ही बिंदु पर स्थित हैं, जो समबाहु त्रिभुज की एक विशिष्ट विशेषता है।
From the figure, circumcentre (C), incentre (I), centroid (G) and orthocentre (O) of an equilateral triangle are the same.
In simple words: In an equilateral triangle, all four major centers-circumcentre, incentre, centroid, and orthocentre-coincide at a single point, reflecting the triangle's perfect symmetry.

🎯 Exam Tip: Remember this unique property for equilateral triangles: all four centers are the same point. This is a common theoretical question and can simplify constructions when dealing with equilateral triangles.

Question 9. Draw an isosceles triangle. Locate its centroid, orthocentre, circumcentre and incentre. Verify that they are collinear. (Textbook pg. no. 22)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समद्विबाहु त्रिभुज ABC का चित्र है, जिसमें माध्यिका AD है। इस त्रिभुज के केंद्रक (G), लंबकेंद्र (O), परिकेंद्र (C) और अंतःकेंद्र (I) सभी एक ही रेखा AD पर स्थित हैं, जो यह सत्यापित करता है कि वे संरेखीय हैं।
From the figure, centroid (G), orthocentre (O), circumcentre (C) and incentre (I) of an isosceles triangle lie on the same line AD.
\( \therefore \) they are collinear.
In simple words: In an isosceles triangle, the centroid, orthocentre, circumcentre, and incentre all lie on the same line, which is the altitude and median from the vertex angle to the base, demonstrating their collinearity.

🎯 Exam Tip: For isosceles triangles, the line connecting the vertex angle to the midpoint of the base (also an altitude and angle bisector) is crucial. All four major centers will lie on this line, showcasing a significant property of isosceles triangles.

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