Practice Set 4.1 8th Std Maths Answers Chapter 4 Altitudes and Medians of a Triangle
Question 1
Maharashtra Board Solution
In ∆LMN, ___ is an altitude and __ is a median, (write the names of appropriate segments.)
Solution & Step-by-Step Answer:
In ∆LMN, seg LX is an altitude and seg LY is a median.
Question 2
Maharashtra Board Solution
Draw an acute angled ∆PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
Draw an obtuse angled ∆STV. Draw its medians and show the centroid.
Solution & Step-by-Step Answer:
Question 4
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Draw an obtuse angled ∆LMN. Draw its altitudes and denote the ortho centre by ‘O’.
Solution & Step-by-Step Answer:
Question 5
Maharashtra Board Solution
Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
Solution & Step-by-Step Answer:
Question 6
Maharashtra Board Solution
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Solution & Step-by-Step Answer:
The point of concurrence of medians i.e. G and that of altitudes i.e. O lie on the same line PS which is the perpendicular bisector of seg QR.
Question 7
Maharashtra Board Solution
Fill in the blanks. Point G is the centroid of ∆ABC. i. If l(RG) = 2.5, then l(GC) = ___ ii. If l(BG) = 6, then l(BQ) = ____ iii. If l(AP) = 6, then l(AG) = ___ and l(GP) = ___.
Solution & Step-by-Step 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. ∴ ∴ ……[∵ l(RG) = 2.5] ∴ l(GC) × 1 = 2 × 2.5 ∴ l(GC) = 5
ii. Point G is the centroid and seg BQ is the median.
∴
∴ …..[∵ l(BG) = 6]
∴ 6 × 1 = 2 × l(GQ)
∴ = l(GQ)
∴ 3 = l(GQ)
i.e. l(GQ) = 3
Now, l (BQ) = l(BG) + l(GQ)
∴ l(BQ) = 6 + 3
∴ l(BQ) = 9
iii. Point G is the centroid and seg AP is the median.
∴
∴ l(AG) = 2 l(GP) …..(i)
Now, l(AP) = l(AG) + l(GP) … (ii)
∴ l(AP) = 2l(GP) + l(GP) … [From (i)]
∴ l(AP) = 3l(GP)
∴ 6 = 3l(GP)..[∵ l(AP) = 6]
∴ = l(GP)
∴ 2 = l(GP)
i.e. l(GP) = 2
l(AP) = l(AG) + l(GP) …[from (ii)]
∴ 6 = l(AG) + 2
∴ l(AG) = 6 – 2
∴ l(AG) = 4
Maharashtra Board Class 8 Maths Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 Intext Questions and Activities
Question 1
Maharashtra Board Solution
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)
Solution & Step-by-Step Answer:
Step 1: Draw a line l and a point P lying outside it. Step 2: By placing a set-square on line l, draw a perpendicular to the line from point P.
Question 2
Maharashtra Board Solution
Draw an acute angled ∆ABC and all its altitudes. Observe the location of the orthocentre. (Textbook pg. no. 20)
Solution & Step-by-Step Answer:
Point O is the orthocentre. Orthocentre lies in the interior of ∆ABC.
Question 3
Maharashtra Board Solution
Draw a right angled triangle and draw all its altitudes. Write the point of concurrence. (Textbook: pg, no. 20)
Solution & Step-by-Step Answer:
Point Q is the orthocentre. The point of concurrence of altitudes PQ, QR and QS is Q.
Question 4
Maharashtra Board Solution
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)
Solution & Step-by-Step Answer:
i. Point O is the orthocentre.
ii. Yes, all the altitudes intersect at point O in the exterior of ∆PQR.
Question 5
Maharashtra Board Solution
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)
Solution & Step-by-Step Answer:
i. Right angled triangle: ii. Obtuse angled triangle: iii. Acute angled triangle:
Question 6
Maharashtra Board Solution
Draw a sufficiently large ∆ABC. Draw medians; seg AR, seg BQ and seg CP of ∆ABC. Name the point of concurrence as G. Measure the lengths of segments from the figure and fill in the boxes in the following table.
l(AG) = | l(GR) = | l(AG): l(GR) = |
l(BG) = | l(GQ) = | l(BG): l(GQ) = |
l(CG) = | l(GP) = | l(CG): l(GP) = |
Observe that all of these ratios are nearly 2 : 1 (Textbook pg. no. 21)
Solution & Step-by-Step Answer:
l(AG) = 2.9 | l(GR) = 1.4 | l(AG): l(GR) = |
l(BG) = 2.4 | l(GQ) = 1.2 | l(BG): l(GQ) = |
l(CG) = 2.8 | l(GP) = 1.4 | l(CG): l(GP) =
Question 7
Maharashtra Board Solution
As shown in the given figure, a student drew ∆ABC 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)
Solution & Step-by-Step Answer:
Draw seg AP ⊥ seg PE and seg EQ ⊥ seg QC. Side AP || side EQ and AC is their transversal. ∴ ∠PAE ≅ ∠QEC …(i) [Corresponding angles] In ∆ APE and ∆ EQC, ∠PAE ≅ ∠QEC …[From (i)] ∠APE ≅ ∠EQC … [Each angle is of measure 90°] side PE ≅ side QC …. [Perpendicular distance between parallel lines] ∴ ∆ APE ≅ ∆ EQC … [By AAS test] ∴ AE = EC … [Corresponding sides of congruent triangles] ∴ E is the midpoint of AC. ∴ seg BE is the median. Similarly, seg CF is the median. Since, the medians of a triangle are concurrent. ∴ G is the centroid of ∆ABC.
Question 8
Maharashtra Board Solution
Draw an equilateral triangle. Find its circumcentre (C), incentre (I), centroid (G) and orthocentre (O). Write your observation. (Textbook pg. no. 22)
Solution & Step-by-Step Answer:
From the figure, circumcentre (C), incentre (I), centroid (G) and orthocentre (O) of an equilateral triangle are the same.
Question 9
Maharashtra Board Solution
Draw an isosceles triangle. Locate its centroid, orthocentre, circumcentre and incentre. Verify that they are collinear. (Textbook pg. no. 22)
Solution & Step-by-Step Answer:
From the figure, centroid (G), orthocentre (O), circumcentre (C) and incentre (I) of an isosceles triangle lie on the same line AD. ∴ they are collinear.
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