In figure- AO - OC - BO - OD -1-2 and AB -5 cm- Find the value of DC -

In AOB and COD, we have ∠ AOB = ∠ COD [Vertically opposite angles] AO OC = OB ODex [Given] So, by SAS criterion of similarity, we have AOB ∼ COD ⇒AO OC = BO O ...

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Standard X

Mathematics

Relation between Areas and Sides of Similar Triangles

In figure, AO...

Question

In figure, $\frac{A O}{O C} = \frac{B O}{O D} = \frac{1}{2}$ and AB = 5cm. Find the value of DC.

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Solution

In $△ A O B a n d △ C O D$ , we have
$∠ A O B = ∠ C O D$ [Vertically opposite angles]
$\frac{A O}{O C} = \frac{O B}{O D} [G i v e n]$
So, by SAS-criterion of similarity, we have
$△ A O B \sim △ C O D$
$\Rightarrow \frac{A O}{O C} = \frac{B O}{O D} = \frac{A B}{D C}$
$\Rightarrow \frac{1}{2} = \frac{5}{D C} [∵ A B = 5 c m]$
$\Rightarrow$ DC = 10cm

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Q. In given figure,

\frac{A O}{O C} = \frac{B O}{O D} = \frac{1}{2}

and AB=5 cm. Find the value of DC.

In the figure, ABCD is a trapezium in which, AB || DC and $\frac{A O}{O C} = \frac{B O}{O D} = \frac{1}{2} a n d A B = 5 c m$ . Find the value of CD .

In the figure, $\frac{A O}{O C} = \frac{B O}{O D} = \frac{1}{2} a n d A B = 5 c m$ . Find the value of CD .

Figure 2.

In the given figure, ABCD is a trapezium in which
(i) AB || CD
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(iii) AB = 5 cm

Find the value of CD .

Q. Find

x

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and

O D = x c m .

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In figure, AOOC=BOOD=12 and AB = 5cm. Find the value of DC.

In △AOB and △COD, we have ∠AOB=∠COD [Vertically opposite angles] AOOC=OBOD[Given] So, by SAS-criterion of similarity, we have △AOB∼△COD ⇒AOOC=BOOD=ABDC ⇒12=5DC[∵AB=5cm] ⇒ DC = 10cm

In figure, $\frac{A O}{O C} = \frac{B O}{O D} = \frac{1}{2}$ and AB = 5cm. Find the value of DC.