Question

ABC and DBC are two Triangles on same base and parallel lines L and M. if ab = 3 cm BC = 5 cm angle A = 90 degree. find the area of triangle DBC? ​

Answer 1

Area of $\Delta DBC = 6 \ cm^2$

Step-by-step explanation:

• Suppose, both triangles are as shown in the figure (attached).

• $\Delta ABC$ will be right angle triangle. Because, $\angle A = 90^o$

• So, we can apply Pythagoras theorem in $\Delta ABC.$

Therefore, we wrtite as,

• $AC^2 + AB^2 = BC^2$

        $AC^2 + 3^2 = 5^2$

        $AC^2 = 25 - 9$

        $AC^2 = 16$

        $AC = \pm4$

     Since it is a length. so,

       AC = 4 cm

Now, we can understand that;

• Area of $\Delta ABC = \dfrac{1}{2} \times base \times height$

• Area of $\Delta ABC = \dfrac{1}{2} \times 3 \times 4$

• Area of $\Delta ABC$ = 6 sq cm.

• Also, we know that the two triangles on same base and parallel lines have equal area.

• Area of $\Delta ABC$ = Area of $\Delta DBC$

• Thus, Area of $\Delta DBC = 6 \ cm^2$

Answer 2

Answer:

6cm²1

Step-by-step explanation:

Area of

Step-by-step explanation:

Suppose, both triangles are as shown in the figure (attached).

will be right angle triangle. Because,

So, we can apply Pythagoras theorem in

Therefore, we wrtite as,

       

       

       

       

    Since it is a length. so,

      AC = 4 cm

Now, we can understand that;

Area of

Area of

Area of  = 6 sq cm.

Also, we know that the two triangles on same base and parallel lines have equal area.

Area of  = Area of

Thus, Area of dbc= 6cm²