Question

The ratio of areas of triangles lies above the diagonal AC of the quadrilateral on the figure .
= 42:14
= 3:1
If MC = 7cm, then
AM =-------------------cm
What is the area of triangle AMB?
If DM = 5cm , then
BM = --------------------cm

Answer 1

Question:

The ratio of areas of triangles lies above the diagonal AC of the quadrilateral on the figure.

= 42:14

= 3:1

If MC = 7cm, then

AM =-------------------cm

What is the area of triangle AMB?

If DM = 5cm , then

BM = --------------------cm

Answer:

The correct answer is AM= 21 cm and BM= 10 cm.

Step-by-step explanation:

Given:-

The ratio of the area triangle lies above the diagonal AC of the quadrilateral in the figure is $42: 14 = 3: 1.$

$MC = 7 cm .$

$DM = 5 cm .$

Area $CMB = 28 cm^{2} .$

To Find :-

$AM = ?$

Area of ∆$AMB = ?$

$BM = ?$

Step 1

The ratio of the area of two similar triangles is equivalent to the square of the ratio of any pair of the complementary sides of the identical triangles.

⇒ Area ∆$AMD$ : Area ∆$AMC$ $= 42 : 14 : 3 : 1$

⇒ $(1/2) * AM * MD * sin \theta$ $:(1/2)*MC * MD sin (180 - \theta ) = 3 : 1$

Simplifying, the above equation, we get

⇒ $AM * sin \theta :$ $MC*sin \theta = 3 : 1$

⇒ $AM : MC = 3 : 1$

⇒ $\frac{AM}{MC} = \frac{3}{1}$

⇒ $\frac{AM}{7} =\frac{3}{1}$

we get,

⇒ $AM = 21 cm$

Step 2

similarly,

⇒ $Area AMB : Area CMB = AM : CM$

⇒ $Area AMB : 28 = 21 : 7$

Simplifying, the above equation

⇒ $Area AMB / 28 = 3/1$

⇒ $Area AMB = 28 * 3$

⇒ $Area AMB = 84 cm^{2} .$

Step 3

similarly,

⇒ $Area DMC : Area BMC = DM : BM$

Simplifying, the above equation

⇒ $14 : 28 = 5 : BM$

⇒ $(1/2) = 5/BM$

we get,

⇒ $BM = 10 cm$

Therefore, the correct answer is AM= 21 cm and BM= 10 cm.

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