Question

- In trapezium ABCD, AB and CD are the parallel sides.
The diagonals AC and BD intersect at X. The ratio of
the area of A AXB to the area of A CXD is
25 : 49. Find the ratio of the area of A AXD and the
area of ABCD.
​

Answer 1

Answer:

AXB:CXD

25:49

SIDE---5:7

25

35. 35

49

TOTAL AREA OF ABCD--144

AXD---35

RATIO - - 35/144

Answer 2

Answer:

The ratio of the area of ΔAXD and the area of trapezium ABCD

= $\frac{35x}{144x}$ =  $\frac{35}{144}$

Step-by-step explanation:

• In trapezium ABCD, AB is parallel to CD  ---------- (1)

• Diagonals AC and BD intersect at ‘X’

• In ΔAXB and ΔCXD,

∠AXB = ∠CXD (vertically opposite angles)

∠ABX = ∠CDX (alternate interior angles)

⇒ ΔAXB ~ ΔCXD (AA criterion for Similar triangles)

⇒ Area of ΔAXB / Area of ΔCXD = $AB^{2}$ / $CD^{2}$ = $\frac{25}{49}$

• Let the common factor of the above ration be x.

So, Area of ΔAXB is 25x.

Area of ΔCXD is 49x.

• Area of Similar Triangles Theorem

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Now $\frac{AB}{CD}$ = $\frac{AX}{CX}$ =  $\frac{BX}{DX}$ = $\sqrt{\frac{25}{49} }$ = $\frac{5}{7}$  

Now We have,       Area of ΔABX / Area of ΔADX =  $\frac{5}{7}$ [Appling Area of Similar Triangles Theorem]

• So, Area of ΔADX  = $\frac{7}{5}$ ×  Area of ΔABX =  $\frac{7}{5}$ × 25x = 35x .

and,  Area of ΔABX / Area of ΔBCX =  $\frac{5}{7}$

• So, Area of ΔBCX  = $\frac{7}{5}$ ×  Area of ΔABX =  $\frac{7}{5}$ × 25x = 35x .

• Area of trapezium ABCD =  Area of ΔABX + Area of ΔADX + Area of ΔCDX + Area of ΔBCX = 25x + 35x + 49x + 35x = 144x

• So, the ratio of the area of ΔAXD and the area of trapezium ABCD

       = $\frac{35x}{144x}$ =  $\frac{35}{144}$

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