Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

## Practice Set 3.3 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles

Question 1.

Find the values of x and y using the information shown in the given figure. Find the measures of ∠ABD and ∠ACD.

Solution:

i. ∠ACB = 50° [Given]

In ∆ABC, seg AC ≅ seg AB [Given]

∴ ∠ABC ≅ ∠ACB [Isosceles triangle theorem]

∴ x = 50°

ii. ∠DBC = 60° [Given]

In ABDC, seg BD ≅ seg DC [Given]

∴ ∠DCB ≅ ∠DBC [Isosceles triangle theorem]

∴ y = 60°

iii. ∠ABD = ∠ABC + ∠DBC [Angle addition property]

= 50° + 60°

∴ ∠ABD = 110°

iv. ∠ACD = ∠ACB + ∠DCB [Angle addition property]

= 50° + 60°

∴ ∠ACD = 110°

∴ x = 50°, y = 60°,

∠ABD = 110°, ∠ACD = 110°

Question 2.

The length of hypotenuse of a right angled triangle is 15. Find the length of median on its hypotenuse.

Solution:

Length of hypotenuse = 15 [Given]

Length of median on the hypotenuse = \(\frac { 1 }{ 2 }\) x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]

= \(\frac { 1 }{ 2 }\) x 15 = 7.5

∴ The length of the median on the hypotenuse is 7.5 units.

Question 3.

In ∆PQR, ∠Q = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS).

Solution:

i. PQ = 12, QR = 5 [Given]

In APQR, ∠Q = 90° [Given]

∴ PR^{2} = QR^{2} + PQ^{2} [Pythagoras theorem]

= 25 + 144

∴ PR^{2} =169

∴ PR = 13 units [Taking square root of both sides]

ii. In right angled APQR, seg QS is the median on hypotenuse PR.

∴ QS = \(\frac { 1 }{ 2 }\)PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]

= \(\frac { 1 }{ 2 }\) x 13

∴ l(QS) = 6.5 units

Question 4.

In the given figure, point G is the point of concurrence of the medians of ∆PQR. If GT = 2.5, find the lengths of PG and PT.

Solution:

i. In ∆PQR, G is the point of concurrence of the medians. [Given]

The centroid divides each median in the ratio 2 : 1.

PG : GT = 2 : 1

∴ PG = 2 x 2.5

∴ PG = 5 units

ii. Now, PT = PG + GT [P – G – T]

= 5 + 2.5

∴ l(PG) = 5 units, l(PT) = 7.5 units

**Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.3 Intext Questions and Activities**

Question 1.

Can the theorem of isosceles triangle be proved by doing a different construction? (Textbook pg. no.34)

Solution:

Yes

Construction: Draw seg AD ⊥ seg BC.

Proof:

In ∆ABD and ∆ACD,

seg AB≅ seg AC [Given]

∠ADB ≅ ∠ADC [Each angle is of measure 90°]

seg AD ≅ seg AD [Common side]

∴ ∆ABD ≅ ∆ACD [Hypotenuse side test]

∴ ∠ABD ≅ ∠ACD [c.a.c.t.]

∴ ∠ABC ≅ ∠ACB [B-D-C]

Question 2.

Can the theorem of isosceles triangle be proved without doing any construction? (Textbook pg, no.34)

Solution:

Yes

Proof:

In ∆ABC and ∆ACB,

seg AB ≅ seg AC [Given]

∠BAC ≅ ∠CAB [Common angle]

seg AC ≅ seg AB [Given]

∴ ∆ABC ≅ ∆ACB [SAS test]

∴ ∠ABC ≅ ∠ACB [c. a. c. t.]

Question 3.

In the given figure, ∆ABC is a right angled triangle, seg BD is the median on hypotenuse. Measure the lengths of the following segments.

i. AD

ii. DC

iii. BD

From the measurements verify that BD = \(\frac { 1 }{ 2 }\)AC. (Textbook pg. no. 37)

Solution:

AD = DC = BD= 1.9 cm

AC = AD + DC [A – D – C]

= 1.9 + 1.9

= 2 x 1.9 cm

∴ AC = 2 x BD

∴ BD = \(\frac { 1 }{ 2 }\) AC