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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 = 62^{5}

AC² = 25^{5
}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