Find the area of the given quadrilateral by using area of triangles

$\frac{1}{2} A r [A B D] + \frac{1}{2} A r [B C D] = 12 c m^{2}$

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$A r [A B D] = 6 c m^{2}$

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$A r [A B D] = 12 c m^{2}$

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$N o n e o f t h e a b o v e$

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Solution

The correct option is A
$\frac{1}{2} A r [A B D] + \frac{1}{2} A r [B C D] = 12 c m^{2}$

$A r [A B C D] = \frac{1}{2} A r [A B D] + \frac{1}{2} A r [B C D]$
$= \frac{1}{/ 2} \times [/ 4^{2} \times 3] + \frac{1}{/ 2} \times [/ 4^{2} \times 3]$
$= 6 + 6$
$= 12 c m^{2}$

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Similar questions

Area of a quadrilateral can be found by splitting it into triangles.

Q. What is the minimum area of quadrilateral ABCD, if the area of a pair of diagonally opposite triangles (Formed by joining the diagonals of the quadrilateral) is 12 and 27 respectively?

In the figure shown, area of quadrilateral ABCD is equal to the area of _______.(Given AC parallel to BE)

Q. In the given figure

P Q R S

is a cyclic quadrilateral

P Q

and

S R

produced meet at

T

. Find area of quadrilateral

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if area of

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(given Triangle TPS is similar to TRQ)

If a quadrilateral is split into two triangles by drawing a diagonal then the area of the quadrilateral is equal to the sum of the areas of the triangles.

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Find the area of the given quadrilateral by using area of triangles

The correct option is A 12Ar[ABD]+12Ar[BCD]=12 cm2 Ar[ABCD]=12Ar[ABD]+12Ar[BCD] =1/2×[/42×3]+1/2×[/42×3] =6+6 =12 cm2

The correct option is A
$\frac{1}{2} A r [A B D] + \frac{1}{2} A r [B C D] = 12 c m^{2}$

$A r [A B C D] = \frac{1}{2} A r [A B D] + \frac{1}{2} A r [B C D]$
$= \frac{1}{/ 2} \times [/ 4^{2} \times 3] + \frac{1}{/ 2} \times [/ 4^{2} \times 3]$
$= 6 + 6$
$= 12 c m^{2}$