Question

In the figure , BC=7,BD=3,Write the ratio .A(∆ABD)/A(∆ABC) .​

Answer 1

Given:

In the figure, BC = 7, BD = 3

To find:

The ratio A(∆ABD)/A(∆ABC)

Solution:

Construction:

Draw AP ⊥ BD

i.e., AP is the height of Δ ABD as well as Δ ABC

We know,

$\boxed{\bold{Area\:of\:a\:triangle = \frac{1}{2}\times base\times height }}$

Therefore,

Area of Δ ABD = $\frac{1}{2} \times AP \times BD = \frac{1}{2} \times AP \times 3$

and

Area of Δ ABC = $\frac{1}{2} \times AP \times BC = \frac{1}{2} \times AP \times 7$

Now,

The ratio of A(∆ABD) and A(∆ABC) is,

= $\frac{\frac{1}{2} \times AP \times 3}{ \frac{1}{2} \times AP \times 7}$

= $\bold{\frac{3}{ 7} }$

Thus, the ratio of the areas of the triangles are:

$\boxed{\bold{\frac{Area(\triangle \:ABD)}{Area (\triangle ABC)} = \frac{3}{7} }}$.

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Also View:

In ΔABC, B-D-C and BD=7, BC=20 then find following ratios.

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In the figure, BC = 7 and BD = 3 then write the ratio of A(triangleADC):A(triangle ABC)​

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Answer 2

Given:

In the figure, BC = 7, BD = 3

To find:

The ratio A(∆ABD)/A(∆ABC)

Solution:

Construction:

Draw AP ⊥ BD

i.e., AP is the height of Δ ABD as well as Δ ABC

We know,

\boxed{\bold{Area\:of\:a\:triangle = \frac{1}{2}\times base\times height }}

Areaofatriangle=

2

1

×base×height

Therefore,

Area of Δ ABD = \frac{1}{2} \times AP \times BD = \frac{1}{2} \times AP \times 3

2

1

×AP×BD=

2

1

×AP×3

and

Area of Δ ABC = \frac{1}{2} \times AP \times BC = \frac{1}{2} \times AP \times 7

2

1

×AP×BC=

2

1

×AP×7

Now,

The ratio of A(∆ABD) and A(∆ABC) is,

= \frac{\frac{1}{2} \times AP \times 3}{ \frac{1}{2} \times AP \times 7}

2

1

×AP×7

2

1

×AP×3

= \bold{\frac{3}{ 7} }

7

3

Thus, the ratio of the areas of the triangles are:

\boxed{\bold{\frac{Area(\triangle \:ABD)}{Area (\triangle ABC)} = \frac{3}{7} }}

Area(△ABC)

Area(△ABD)

=

7

3

.