Chapter 3.4, Problem 26P

(a)

To determine

To Find: the lengths of side AB and CD in the given trapezium.

Answer to Problem 26P

  $\text{30}\text{cm}\text{and}\text{20}\text{cm}$

Explanation of Solution

Given:

A trapezium ABCD is given with side AB is $\frac{3}{2}$ times longer the CD.

AD is equal to the side BC and are $13\text{cm}$ long .

Perimeter of the trapezium is $\text{76}\text{cm}$ .

Calculation:

Figure is given below.

  

Since, AB is $\frac{3}{2}$ times longer the CD therefore.

  $AB=\frac{3}{2}CD$ …… (1)

Perimeter is.

  $\begin{array}{l}AB+BC+CD+AD=76\\ \frac{3}{2}CD+13+CD+13=76\\ \frac{3}{2}CD+CD=76-26\\ \frac{3}{2}CD+CD=50\\ \frac{5}{2}CD=50\\ CD=\frac{50\times 2}{5}\\ CD=20\end{array}$

From equation (1).

  $\begin{array}{l}AB=\frac{3}{2}CD\\ AB=\frac{3}{2}\times 20\\ AB=30\end{array}$

Therefore, the answer is $\text{30}\text{cm}\text{and}\text{20}\text{cm}$ .

(b)

To determine

To Find; the area of trapezium ABCD.

Answer to Problem 26P

  $300{\text{cm}}^2$

Explanation of Solution

Given;

  $AB=30\text{cm},BC=13\text{cm},CD=20\text{cm}\&AD=13\text{cm}$ .

Height of trapezium is $12\text{cm}\text{.}$

Calculation; Figure is given below.

  

Area of trapezium is.

  $\begin{array}{l}=\frac{1}{2}\times \text{Height}\times \left(\text{Sum of base lengths}\right)\\ =\frac{1}{2}\times h\times \left(AB+CD\right)\\ =\frac{1}{2}\times 12\times \left(30+20\right)\\ =\frac{1}{2}\times 12\times 50\\ =300{\text{cm}}^2\end{array}$

Therefore, the answer is $300{\text{cm}}^2$ .