EDUCATION THROUGH SELFHELP IS OUR MOTTO RAYAT SHIKSHAN

“EDUCATION THROUGH SELFHELP IS OUR MOTTO” .

. . RAYAT SHIKSHAN SANSTHA’S

. VITA HIGHSCHOOL, VITA

TEACHER NAME SHRI. GHADAGE A. P. B. Sc B. Ed

SUBJECT GEOMETRY

PROPERTIES OF AREAS OF TWO TRIANGLE

. The ratio of areas of two triangles is equal to the ratio of products of their bases and the corresponding heights. P A B C Q D A(∆ABC) …………… A(∆PQR) = BC X AD …………… QS X PT S T

Having equal bases The ratio of the areas of two triangles having equal bases is equal to the ratio of their corresponding heights. A A(∆ABC) …………… = C D E B A(∆EBC) AB …… ED

Having equal height The ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding heights A A(∆ABC) …………… B E C D A(∆ACD) = BC ------- CD

Having equal height & base Areas of two triangles having equal bases and equal heights are EQUAL A(∆ABC) D A B C = A(∆DBC)

Basic praportionality theorem If a line is parallel to one side of a triangle which intersects the other two sides in two distinct points, Then the line divides those side in proportion. P M Q N R

Given - In ∆PQR line l ║ QR. Line l intersect the side PQ and PR in the points M and N respectively. TO P M N PROVE – PM ………… MQ Q = PN ……. . NR R Construction- Draw seg. QN and Seg. RM

P. roof - ∆PMN & ∆QMN have equal heights & with common vertex N A(∆PMN) A(∆QMN) ………. . . . PM MQ = ……………………. 1 ∆PMN & ∆RMN have equal heights with common vertex M A(∆PMN) PN ……………… A(∆RMN) = ………… NR A(∆QMN) = A(∆RMN) P I M Q ………………… 2 …… between lines MN & QR ………. 3 From 1 , 2 And 3 PM A(∆PMN) …………. . = A(∆QMN) N R A(∆PMN) ……………. A(∆RMN) PM PN ……… = ……. . MQ NR ………. From 1, 2, 4 4

Property of an angle bisector of a triangle In a triangle, the angle bisector divide the side opposite to the angle in the ratio of the remaining sides. C in ∆ABC ray CE is the bisect of angle C AE …………. EB = CA ………… CB A E B

. AREAS. OF SIMILAR TRIANGLES

. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides A P Given∆ABC ~ ∆PQR To prove – A(∆ABC) ……………. . = A(∆PQR) B 2 BC ……. . QR 2= Construction - Draw seg. AM and Seg. PN M C Q 2 2 AB ……. 2 = QR ……. . . side BC side QR AC PR 2 N R

. A Proof- ∆ABC~∆PQR AB …… BC AC ……. = ……. . = PQ QR PR ( c. s. s. t )…. 1 ~ PQR ( c. a. s. t ) ABC ----~ PQN ABM ---…………. . 2 --in ∆ABM and ∆PQN __ __ ABM ~ PQN , AMB ~ PNQ ………. . 3 ∆AMB ~ ∆PNQ ………( AA test for similarity) AM MB AB ( c. s. s. t ) ……. . =. . . . = ……. . PN NQ QR B ------ AM …. . AB BC …. . = = …. . PN PQ QR A(∆ABC) ……. . . . BC x AM ………… BC …. . AM …. . . = 2 2 A(∆ABC) AB AC …………… …. = = A(∆PQR) PQ 2 PR 2 N BCx …. . BC = …. . BC 2 …. . 2 QR QR QR x = = QR x PN QR PN 2 A(∆ABC) BC …………… = …. . 2 similarly we can prove that A(∆PQR) QR A(∆PQR) P Q (from 1 and 3 ) M C R

. THEOREM OF

. In a right angle triangle, square of hypotenuse is equal to the sum of the squares of the remaining two sides. Given- in ∆ABC angle B=90 To prove – A AC = AB + BC D Construction – Draw perpendicular BD from vertex B, to the side AC A-D-C B C

. Proof- in ∆ABC angle B=90 seg. BD ┴ hypotenuse AC by similarity in right angled triangles, ∆ABC~∆ADC~ ∆BDC now , ∆ABC ~ ∆ADC AB AC = ………. . AD AB 2 AB = AC x AD ………… 1 ∆ABC ~ ∆BDC BC = DC ………. . DC BC 2 BC = AC x DC …………. 2. . . A ………. . D from 1 and 2 2 AB + 2 BC = AC x AD + AC x DC = AC x ( AD + DC ) = AC x AC 2 2 2 AB + BC = AC B C

Theorem of 30 – 60 – 90 Triangle If the angle of a triangle are 30 -60 -90, then The side opposite to 30 is half of the hypotenuse ― √ 3 time hypotenuse And that opposite to 60 is. . 2 2

Theorem of 45 – 90 Triangle If the angle of a triangle are 45 -45 -90, then Each of the perpendicular sides is 1 . . ― √ 2 times the hypotenuse

APOLLONIUS’ THEOREM In triangle ABC, if seg. AD is a median, then 2 2 AB + AC = 2 AD + 2 BD B ……………. . A E D C