Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2

Miscellaneous and Practice Problems

Question 1.
Identify the centroid of ΔPQR.

Solution:
In ΔPQR, PT = TR ⇒ QT is a median from vertex Q.
QS = SR ⇒ PS is a median from vertex P.
QT and PS meet at W and therefore W is the centroid of ΔPQR.

Question 2.
Name the orthocentre of ΔPQR.

Solution:
∠P = 90°
This is a right triangle
∴ orthocentre = P [∴ In right triangle orthocentre is the vertex containing 90°]

Question 3.
In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.

Solution:
Given A is the midpoint of YZ.
∴ ZA = AY
G is the centroid of ΔXYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2 : 1
\(\frac{XG}{GA}\) = \(\frac{2}{1}\)
\(\frac{XG}{3}\) = \(\frac{2}{1}\)
XG = 2 × 3
XG = 6 cm
XA = XG + GA
= 6 + 3 ⇒ XA = 9 cm

Challenging Problems

Question 4.
Find the length of an altitude on the hypotenuse of a right angled triangle of legs of length 15 feet and 20 feet.
Solution:

Since ∠B = 90°
Using pythagoras theorem
AC² = AB² + BC² = 20² + 15²
= 400 + 225
AC² = 62⁵
AC² = 25⁵
AC = 25
Area of a triangle ΔABC = \(\frac{1}{2}\) × 15 × 20 = 150 feet²
Again Area of ΔABC = \(\frac{1}{2}\) × AC × BD
150 = \(\frac{1}{2}\) × 25 × BD
BD = \(\frac{2 × 150}{25}\) = \(\frac{300}{25}\)
BD = 12 feet
∴ Length of the altitude on the hypotenuse of the right angled triangle is 12 feet.

Question 5.
If I is the incentre of ΔXYZ, ∠IYZ = 30° and ∠IZY = 40°, find ∠YXZ.

Solution:
Since I is the incentre of ΔXYZ
∠IYZ = 30° ⇒ ∠IYX = 30°
∠IZY = 40° ⇒ ∠IZX = 40°
∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°
∠XYZ = 60°
lll ly ∠XYZ = ∠XZI + ∠IZY = 40° + 40°
∠XYZ = 80°
By angle sum property of a triangle
∠XZY + ∠XYZ + ∠YXZ = 180°
80° + 60° + ∠YXZ = 180°
140° + ∠YXZ = 180°
∠YXZ = 180° – 140°
∠YXZ = 40°

Question 6.
In ΔDEF, 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 FD = ?
(iv) If OE = 36, then EP = ?

Solution:
Given DN, EO, FM are medians.
∴ FN = EN
DO = FO
EM = DM
(i) If DE = 44, then
DM = \(\frac{44}{2}\) = 22
DM = 22

(ii) If PD = 12, PN = ?
\(\frac{PD}{PN}\) = \(\frac{2}{1}\)
\(\frac{12 }{PN}\) = \(\frac{2}{1}\) ⇒ PN = \(\frac{12}{2}\) = 6
PN = 6

(iii) If DO = 8, then
FD = DO + OF = 8 + 8
FD = 16

(iv) If OE = 36,
then \(\frac{EP}{PO}\) = \(\frac{2}{1}\)
\(\frac{EP}{2}\) = PO
OE = OP + PE
36 = \(\frac{PE}{2}\) + PE
36 = \(\frac{PE}{2}\) + \(\frac{2PE}{2}\)
36 = \(\frac{3PE}{2}\)
PE = \(\frac{36 × 2}{3}\)
PE = 24