In AB and BC legs of ABC triangle are points S and R, respectively, such that AS=3/7 of AB and BR=3/8 of BC. Find the area of SBR triangle if the area of ABC is 126 sq.cm

2. Okay, I understand what you mean. (English is not my native language also.)

You are saying,
In AB and BC (not CD) legs of ABC triangle are points S and R, respectively, such that AS=3/7 of AB and BR=3/8 of BC. Find the area of SBR triangle if the area of ABC is 126 sq.cm.

Since only sides are given with values, and the area of triangle ABC, only, and since triangles ABC and SBR have the same angle B, then we will use the formula for the area of any triangle,
A = (1/2)ab*sin(theta)
where
theta is the angle between sides a and b.

In triangle ABC,
Let x = AB
And y = BC
So,
Area of ABC = (1/2)(x)(y)sinB = 126
xy*sinB = 252 ------------------------(i)

----------------------------------
In triangle SBR,

SB = (4/7) of AB because AS is (3/7) of AB
So,
SB = (4/7)x

BR = (3/8) of BC
BR = (3/8)y

Area of SBR
= (1/2)[(4/7)x][(3/8)y]sinB
= [12/(2*56)](xy)sinB
= (3/28)(xy)sinB

Substitute into that the xy*sinB = 252,

= (3/28)(252)

3. Hello, blertta!

In AB and CD legs of ABC triangle are picked up two points S and R
where: .$\displaystyle AS\,=\,\frac{3}{7}(AB),\;BR\,=\,\frac{3}{8}(BC)$

Find the area of $\displaystyle \Delta SBR$ if the area of $\displaystyle \Delta ABC = 126$ cm².
Code:
                C
*
*  *
5/8 *     *
*        *
R o           *
* **           *
*   *   *         *
3/8 *     *      *       *
*       *         *     *
*         *            *   *
B * * * * * * o * * * * * * * * * A
4/7     S        3/7

$\displaystyle \Delta SBR$'s height is $\displaystyle \frac{3}{8}$ of $\displaystyle \Delta ABC$'s height.

$\displaystyle \Delta SBR$'s base is $\displaystyle \frac{4}{7}$ of $\displaystyle \Delta ABC$'s base.

Therefore: .$\displaystyle \Delta SBR \;=\;\left(\frac{3}{8}\right)\left(\frac{4}{7}\rig ht)(126) \;=\;27\text{ cm}^2$