Question

The triangle and the trapezoid have the same area. Base b2 is twice the length of base b1. What are the lenghts of the bases of the trapezoid. The triangle is 2lcm base and 6cm height. decompose the triangle

Answer 1

Answer:

The lengths of the bases of the trapezoid:

42/h cm and 84/h cm.

Step-by-step explanation:

The formula of an area of a triangle:

$A=\dfrac{bh}{2}$

b - base

h - height

We have b = 21cm, h = 6cm.

Substitute:

$A=\dfrac{(21)(6)}{2}=63\ cm^2$

The formula of an area of a trapezoid:

$A=\dfrac{b_1+b_2}{2}\cdot h$

b₁, b₂ - bases

h - height

We have b₁ = 2b₂, therefore b₁ + b₂ = 2b₂ + b₂ = 3b₂.

The area of a triangle and the area of a trapezoid are the same.

Therefore

$\dfrac{3b_2}{2}\cdot h=63$         multiply both sides by 2

$3b_2h=126$          divide both sides by 3

$b_2h=42$         divide both sides by h

$b_2=\dfrac{42}{h}$

$b_1=2b_2\to b_1=2\cdot\dfrac{42}{h}=\dfrac{84}{h}$

Answer 2

Answer:

The length of the bases of the trapezoid is $b_1=42H$ and $b_2=84H$      

Step-by-step explanation:

Given : The triangle and the trapezoid have the same area. Base $b_2$ is twice the length of base $b_1$.

To find : What are the lengths of the bases of the trapezoid. The triangle is 21 cm base and 6 cm height decompose the triangle?

Solution :

The area of the triangle is $A_t=\frac{1}{2}bh$

The triangle is 21 cm base and 6 cm height,

$A_t=\frac{1}{2}\times 21\times 6$

$A_t=63\ cm^2$

The area of the trapezoid is $A_T=\frac{b_1+b_2}{2}H$

The triangle and the trapezoid have the same area.

i.e. $63=\frac{b_1+b_2}{2}H$

Base $b_2$ is twice the length of base $b_1$

i.e. $b_2=2b_1$

Substitute,

$63=\frac{b_1+2b_1}{2}H$

$63=\frac{3b_1}{2}H$

$126=(3b_1)H$

$\frac{126}{3H}=b_1$

$b_1=42H$

Put in $b_2=2b_1$,

$b_2=2(42H)$

$b_2=84H$

The length of the bases of the trapezoid is $b_1=42H$ and $b_2=84H$