We start by calculating the area of the trapezoid.
This can be calculated as the sum of the area of the triangle and the rectangle.
The triangle has a base of 10-7=3 and a height of 4, and the rectangle has sides 4 and 7, so we can calculate the trapezoid area as:
![A_{}=A_t+A_r=(3\cdot4)/(2)+4\cdot7=6+28=34\operatorname{cm}^2]()
The area of the trapezoid is 34 cm^2, as the area of the triangle is 6 cm^2 and the area of the rectangle is 28 cm^2.
Now, if the bigger triangle has also an area of 34 cm^2, we can write the area as:
![\begin{gathered} A=(b\cdot h)/(2)=(10\cdot h)/(2)=5h=34 \\ h=(34)/(5)=6.8\operatorname{cm} \end{gathered}]()
To have the same area, the height of the triangle has to be h=6.8 cm.
Answer:
The area of the trapezoid is the sum of the area of one rectangle with area 28 cm^2 and one triangle with area 6 cm^2. The sum of this areas is 34 cm^2. For the triangle, area = 34 cm^2, so h=6.8 cm.