Triangles

Hi Nancy 

Please see if the question is complete. 

I thing there is a typographical error , the length of RT ( 6cm ) is already mentioned in the  question . 

Two medians PS and RT of ∆PQR intersect at G at right angles. If PS =9CM and  RT =6 cm,then length of RS is

a) 10

B) 6

C)5

D)3

Centroid divides the medians in 2 : 1 .

So PG : GS = 2 : 1

GS = 3 cm , PG = 6 cm

Again RG : GT = 2 : 1

GT = 2cm , RG = 4 cm

∆GSR is a right angled ∆ .

So , RS² = RG² + GS²

RS ² = 3² + 4²

RS = √25 = 5 .

Since centroid divides the median in 2 : 1 , 

BO : OY = 2 : 1 

OY = 6/2 = 3. 

OY = OX = 3.

Option (B)  

Hello Richa , 

Here , OC = Altitude on AB = Sum of diameters of the circles + radius of the largest circle 

OC = 9 + 6 + 2 + 2/3 +..... = 9 + ( 6)/ ( 1-1/3) = 18 

Area of the Triangle ABC = 1/2 x OC x AB 

1/2 x 18 x 18 = 162 sq units . 

Hello Samyak, 

PFA the solution . 

samyak

Hello Richa , 

In triangles ADC and BDE 

Angle A = Angle B 

Angle D = Angle D  (Each 90)

Hence , triangle ADC similar triangle BDE 

AD/CD = BD/ED 

8/4 = 6/ ED 

ED = 3 units . 

Hello Aniket , 

PFA the solution 

x = 22.5 

Alternate Approach : 

Sides: a, a  and 1 

Two cases:
(i) a is an even integer  , sum of two even integer and 1 is always odd 

(ii) a is an odd integer , 

odd integer  + odd integer  + 1 = an odd integer 

so Option C is false . 

Using Pythagoras Theorem : Altitude of the triangle = sqrt {a^2 - (1/2)^2} = 

Area = 1/2 x Base x Altitude 

=> sqrt{4a^2 -1}/4

4a^2 is a perfect square, so sqrt{4a^2 -1} is irrational 

So area is always an irrational Number . 

Triangle CAD ~ Triangle ABD 

CD/AD = AD/BD 

AD² = CD × BD 

16  = CD × 6 

CD = 16/6 

BC = 8/3 + 6 = 26/3 

Option (A) 

We know , 

Area of a triangle = Semiperimeter × inradius of the triangle 

So , T = ( a + b + c)/2 × r

According to the question 

(a + b +  c) = ( a + b + c)/2 × r

r = 2 units.