$D A, C B, O M$ are each perpendicular to line segment $A B$ where $O$ is the point of intersection of $A C$ and $D B$ .If $O A = 2.4$ cm, $O C = 3.6$ cm then find the ratio of $A M : B M$

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Solution

Given: $A D ∥ O M ∥ B C$

By intercept theorem, $\frac{O A}{O C} = \frac{A M}{M B}$

$\Rightarrow \frac{2.4}{3.6} = \frac{A M}{M B}$

$\Rightarrow \frac{A M}{M B} = \frac{2}{3}$

$∴ A M : M B = 2 : 3$

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Similar questions

Q. In the given figure

D A ⊥ A B, C B ⊥ A B

and

O M ⊥ A B

. If

A O = 5.4

cm,

O C = 7.2

cm and

B O = 6

cm, then the length of

D O

is:

In $Δ A B C$ , AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:
(i) $Δ O M A \sim Δ O L C$
(ii) $\frac{O A}{O C} = \frac{O M}{O L}$

Q. In the given figure

D A ⊥ A B, C B ⊥ A B

and

O M ⊥ A B

. If AO = 5.4 cm, OC = 7.2 cm and BO = 6 cm, then the length of DO is:

Q. Fill in the blank :
A perpendicular bisector XY is constructed of line segment AB of length 2.4 cm. It meets line AB at O. Then OA = _____.

Q. In the trapezium ABCD, side AB

∥

side DC. Diagonals AC and DB intersect each other at O.

If AB = 15, DC = 10. Find

\frac{O C}{O A}

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DA,CB,OM are each perpendicular to line segment AB where O is the point of intersection of AC and DB.If OA=2.4cm,OC=3.6cm then find the ratio of AM:BM

Given:AD∥OM∥BCBy intercept theorem,OAOC=AMMB⇒2.43.6=AMMB⇒AMMB=23∴AM:MB=2:3

$D A, C B, O M$ are each perpendicular to line segment $A B$ where $O$ is the point of intersection of $A C$ and $D B$ .If $O A = 2.4$ cm, $O C = 3.6$ cm then find the ratio of $A M : B M$

Given: $A D ∥ O M ∥ B C$

By intercept theorem, $\frac{O A}{O C} = \frac{A M}{M B}$

$\Rightarrow \frac{2.4}{3.6} = \frac{A M}{M B}$

$\Rightarrow \frac{A M}{M B} = \frac{2}{3}$

$∴ A M : M B = 2 : 3$