Question

A triangle ABC stands on a rectangle BCDE and AF is a perpendicular from A to DE. If $AB=6.5m,CD=5m,AF=7.5m$, then BC can't be less than

• A. 6 m
• B. 5.1 m
• C. 7.04 m
• D. 6.5 m

Answer 1

The correct option is C 7.04 m

Given Ab=6.5m, CD=5m and AF=7.5m

So Distance of line from A to BC i.e. AM=Af-CD=7.5-5=2.5m

So $B{M}^2=A{B}^2-A{M}^2$

$\to B{M}^2=(6.5{)}^2-(25{)}^2\to BM=\sqrt{42.25-6.25}=\sqrt{36}=6m$

As we know that

$A{B}^2=BM.BC\to {6.5}^2=6.BC\to 42.25=6.BC\to BC=\frac{42.25}{6}=7.04$