{"id":581486,"date":"2022-02-23T10:50:42","date_gmt":"2022-02-23T10:50:42","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=581486"},"modified":"2022-02-25T09:25:56","modified_gmt":"2022-02-25T09:25:56","slug":"maharashtra-board-solutions-class-9-maths-part-2-chapter-3-3-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-3-3-triangles\/","title":{"rendered":"Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 3.3- Triangles"},"content":{"rendered":"\n

Class 9: Maths Chapter 3.3 solutions. Complete Class 9 Maths Chapter 3.3 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 3.3- Triangles<\/h2>\n\n\n\n

Maharashtra Board 9th Maths Chapter 3.3, Class 9 Maths Chapter 3.3 solutions<\/p>\n\n\n\n

Question 1.
Find the values of x and y using the information shown in the given figure. Find the measures of \u2220ABD and \u2220ACD.
\"Maharashtra
Solution:<\/strong>
i. \u2220ACB = 50\u00b0 [Given]
In \u2206ABC, seg AC \u2245 seg AB [Given]
\u2234 \u2220ABC \u2245 \u2220ACB [Isosceles triangle theorem]
\u2234 x = 50\u00b0<\/p>\n\n\n\n

ii. \u2220DBC = 60\u00b0 [Given]
In ABDC, seg BD \u2245 seg DC [Given]
\u2234 \u2220DCB \u2245 \u2220DBC [Isosceles triangle theorem]
\u2234 y = 60\u00b0<\/p>\n\n\n\n

iii. \u2220ABD = \u2220ABC + \u2220DBC [Angle addition property]
= 50\u00b0 + 60\u00b0
\u2234 \u2220ABD = 110\u00b0<\/p>\n\n\n\n

iv. \u2220ACD = \u2220ACB + \u2220DCB [Angle addition property]
= 50\u00b0 + 60\u00b0
\u2234 \u2220ACD = 110\u00b0
\u2234 x = 50\u00b0, y = 60\u00b0,
\u2220ABD = 110\u00b0, \u2220ACD = 110\u00b0<\/p>\n\n\n\n

Question 2.
The length of hypotenuse of a right angled triangle is 15. Find the length of median on its hypotenuse.
\"Maharashtra
Solution:<\/strong>
Length of hypotenuse = 15 [Given]
Length of median on the hypotenuse = 12 x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= 12 x 15 = 7.5
\u2234 The length of the median on the hypotenuse is 7.5 units.<\/p>\n\n\n\n

Question 3.
In \u2206PQR, \u2220Q = 90\u00b0, PQ = 12, QR = 5 and QS is a median. Find l(QS).
\"Maharashtra
Solution:<\/strong>
i. PQ = 12, QR = 5 [Given]
In APQR, \u2220Q = 90\u00b0 [Given]
\u2234 PR2<\/sup> = QR2<\/sup> + PQ2<\/sup> [Pythagoras theorem]
= 25 + 144
\u2234 PR2<\/sup> =169
\u2234 PR = 13 units [Taking square root of both sides]<\/p>\n\n\n\n

ii. In right angled APQR, seg QS is the median on hypotenuse PR.
\u2234 QS = 12PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= 12 x 13
\u2234 l(QS) = 6.5 units<\/p>\n\n\n\n

Question 4.
In the given figure, point G is the point of concurrence of the medians of \u2206PQR. If GT = 2.5, find the lengths of PG and PT.
\"Maharashtra
Solution:<\/strong>
i. In \u2206PQR, G is the point of concurrence of the medians. [Given]
The centroid divides each median in the ratio 2 : 1.
PG : GT = 2 : 1
\"Maharashtra
\u2234 PG = 2 x 2.5
\u2234 PG = 5 units<\/p>\n\n\n\n

ii. Now, PT = PG + GT [P \u2013 G \u2013 T]
= 5 + 2.5
\u2234 l(PG) = 5 units, l(PT) = 7.5 units<\/p>\n\n\n\n

Question 1.
Can the theorem of isosceles triangle be proved by doing a different construction? (Textbook pg. no.34)
\"Maharashtra
Solution:<\/strong>
Yes
Construction: Draw seg AD \u22a5 seg BC.
Proof:
In \u2206ABD and \u2206ACD,
seg AB\u2245 seg AC [Given]
\u2220ADB \u2245 \u2220ADC [Each angle is of measure 90\u00b0]
seg AD \u2245 seg AD [Common side]
\u2234 \u2206ABD \u2245 \u2206ACD [Hypotenuse side test]
\u2234 \u2220ABD \u2245 \u2220ACD [c.a.c.t.]
\u2234 \u2220ABC \u2245 \u2220ACB [B-D-C]<\/p>\n\n\n\n

Question 2.
Can the theorem of isosceles triangle be proved without doing any construction? (Textbook pg, no.34)
\"Maharashtra
Solution:<\/strong>
Yes
Proof:
In \u2206ABC and \u2206ACB,
seg AB \u2245 seg AC [Given]
\u2220BAC \u2245 \u2220CAB [Common angle]
seg AC \u2245 seg AB [Given]
\u2234 \u2206ABC \u2245 \u2206ACB [SAS test]
\u2234 \u2220ABC \u2245 \u2220ACB [c. a. c. t.]<\/p>\n\n\n\n

Question 3.
In the given figure, \u2206ABC is a right angled triangle, seg BD is the median on hypotenuse. Measure the lengths of the following segments.<\/strong>
i. AD
ii. DC
iii. BD
From the measurements verify that BD = 12AC. (Textbook pg. no. 37)
Solution:
<\/strong>AD = DC = BD= 1.9 cm
AC = AD + DC [A \u2013 D \u2013 C]
= 1.9 + 1.9
= 2 x 1.9 cm
\u2234 AC = 2 x BD
\u2234 BD = 12 AC<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 3.3- Triangles<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 3.3- Triangles PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 9 Maths :<\/strong><\/h2>\n\n\n\n

Part 2<\/strong><\/p>\n\n\n\n