Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.8k views
in Triangles by (24.0k points)
closed by

A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the altitude x of the rectangle is half the base of the rectangle, then

(a) x = \(\frac{1}{2}h\)

(b) x = \(\frac{bh}{h+b}\)

(c) x = \(\frac{bh}{2h+b}\)

(d) x = \(\sqrt{\frac{hb}{2}}\)

1 Answer

+1 vote
by (23.5k points)
selected by
 
Best answer

(c)  x = \(\frac{bh}{2h+b}\)

Let AD be the height of the given triangle ABC, where AD = h and BC = b Also let height HG of rectangle EFGH equal \(x\) so that HE = GF = 2x 

Now ΔBGH ~ ΔBDA

⇒ \(\frac{BG}{BD} =\frac{HG}{AD}\)

⇒ \(\frac{BG}{BD} =\frac{x}{h}\) ⇒ BG = \(\frac{x}{h}\)BD            ....(i)

Also, ΔCFE ~ ΔCDA

⇒ \(\frac{Cf}{CD} =\frac{EF}{AD}\) = \(\frac{x}{h}\)

⇒ CF = \(\frac{x}{h}\)CD                         .....(ii)

BG + CF = BC – GF = b – 2x         …(iii) 

∴ From (i), (ii) and (iii)

b - 2x = \(\frac{x}{h}\) (BD + CD) ⇒ b - 2x = \(\frac{x}{h}\)BC = \(\frac{x}{h}\)b

⇒ bh – 2xh = xb ⇒ bh = xb + 2xh = x(b + 2h)

⇒  x = \(\frac{bh}{2h+b}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...