A geometry problem by Ajay Sambhriya

Geometry Level 3

We are given that

  • ∠ A B D = ∠ C D B = ∠ P Q D = 9 0 ∘ \angle ABD = \angle CDB = \angle PQD = 90^\circ .
  • A B : C D = 3 : 1 AB: CD = 3:1 .

What is the ratio C D : P Q CD: PQ ?

5 : 4 5:4 5 : 3 5:3 3 : 2 3:2 4 : 3 4:3

Ajay Sambhriya by Ajay Sambhriya , 27, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Use the formula 1 x = 1 a + 1 b \dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b} where: x x is the center length, a a and b b are the other lengths. We have

1 P Q = 1 A B + 1 C D \dfrac{1}{PQ}=\dfrac{1}{AB}+\dfrac{1}{CD}

Substituting, we have

1 P Q = 1 3 + 1 1 \dfrac{1}{PQ}=\dfrac{1}{3}+\dfrac{1}{1}

P Q = 3 4 PQ=\dfrac{3}{4}

It follows that,

C D : P Q = 1 : 3 4 CD:PQ=1:\dfrac{3}{4}

C D : P Q = 4 : 1 \boxed{CD:PQ=4:1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...