A-triangle-QRS-is-to-be-constructed-from-a-line-segment-of-lenght-15cm-Construct-the-triangle-using-the-division-of-the-line-segment-into-the-ratio-5-4-3-such-that-QS-and-RS-are-laegest-and-smallest

Question Number 171165 by MathsFan last updated on 09/Jun/22

A t r i a n g l e Q R S i s t o b e c o n s t r u c t e d f r o m

a l i n e s e g m e n t o f l e n g h t 15 c m .

C o n s t r u c t t h e t r i a n g l e u s i n g t h e

d i v i s i o n o f t h e l i n e s e g m e n t i n t o t h e

r a t i o 5 : 4 : 3 s u c h t h a t Q S a n d R S a r e

l a e g e s t a n d s m a l l e s t r a t i o r e s p e c t i v e l y .

c i r c u m s c r i b e t h e t r i a n g l e b y l o c a t i n g

t h e c i r c u m c e n t e r .

Answered by aleks041103 last updated on 09/Jun/22

F r o m t h e p r o b l e m s t a t e m e n t

w e c a n s a y t h a t :

Q S = 5 x, Q R = 4 x, R S = 3 x

B u t :

{Q S}^{2} = 25 x^{2} = 16 x^{2} + 9 x^{2} = {(4 x)}^{2} + {(3 x)}^{2}

\Rightarrow {Q S}^{2} = {Q R}^{2} + {R S}^{2}

\Rightarrow t h e △ Q R S i s a r i g h t t r i a n g l e w i t h

h y p o t e n u s e Q S .

I t i s k n o w n, t h a t t h e c e n t e r o f t h e

c i r c u m s c r i b e d c i r c l e i s i n t h e m i d d l e

o f t h e r i g h t {t r i a n g l e}^{'} s h y p o t e n u s e .

\Rightarrow T h e a n s w e r i s t h e m i d d l e o f Q S .

A s f o r t h e r a d i u s : R = \frac{Q S}{2} = \frac{5}{2} x

Q S + R S + Q R = 5 x + 3 x + 4 x = 12 x = 15

\Rightarrow x = \frac{15}{12} = \frac{5}{4} c m

\Rightarrow R = \frac{25}{8} c m = 3, 125 c m

Commented by MathsFan last updated on 09/Jun/22

t h a n k s