In the figure- ABCD is a trapezium in which- AB DC and AO - OC - BO - OD -1-2 and AB -5 cm- Find the value of CD-A- 10 cmB- 15 cmC- 18 cmD- 20 cm

The correct option is A 10 cm In Δ AOB and Δ COD, ∠ AOB = ∠ COD nbsp; nbsp; nbsp; (vertically opposite angles) AO/OC=BO/OD nbsp; (given) ∴ AOB∼ CO ...

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Byju's Answer

Standard X

Mathematics

SAS Similarity

In the figure...

Question

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 .

10 cm

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15 cm

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18 cm

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20 cm

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Solution

The correct option is A
10 cm

In $Δ A O B$ and $Δ C O D$ ,

$∠ A O B$ = $∠ C O D$
(vertically opposite angles)

$\frac{A O}{O C} = \frac{B O}{O D}$ (given)

$∴ △ A O B \sim △ C O D$
(by SAS similarity criterion)

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

$\Rightarrow C D = 2 \times A B = 2 \times 5 = 10 c m$

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Similar questions

In the given figure, ABCD is a trapezium in which
(i) AB || CD
(ii) $\frac{A O}{O C} = \frac{B O}{O D} = \frac{1}{2}$
(iii) AB = 5 cm

Find the value of CD .

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

Q. In the adjoining figure,

A B C D

is a trapezium in which

A B ∥ D C

. The diagonals

A C

and

B D

intersect at

O

. Prove that

A O / O C = B O / O D

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

Figure 2.

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In the figure, ABCD is a trapezium in which, AB || DC and AOOC=BOOD=12 and AB=5 cm. Find the value of CD .

The correct option is A 10 cm In ΔAOB and ΔCOD, ∠AOB = ∠COD (vertically opposite angles) AOOC=BOOD (given) ∴△AOB∼△COD (by SAS similarity criterion) ⇒AOOC=BOOD=ABCD=12 ⇒CD=2×AB=2×5=10 cm

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 .