# IGCSE Mathematics Paper-1: Specimen Questions with Answers 159 - 161 of 175

## Question number: 159

### Describe in Detail

Solve:

Solve:

(a) The perimeter,

(b) The area.

### Explanation

Here,

(a) A perimeter can find by adding the length of all shapes or boundaries.

So, here, perimeter =

Where, AB = 18 cm, BC = 14 cm, CD = 12 cm, DA = 14 cm.

So, Perimeter =

(b) Area of Trapezoid ABCD = , where, a and b = two parallel sides, h = height.

So, here, a = 12 cm, b = 18 cm and h = 12 cm.

So, Area A =

## Question number: 160

### Describe in Detail

Solve

Find out:

(a) The perimeter,

(b) The area.

### Explanation

Here,

(a) A perimeter can find by adding the length of all shapes or boundaries.

So, here, perimeter =

Where, AB = 10 cm, BC = 22 cm, CD = 10 – 6 = 4 cm, DE = 22 – 14 = 8 cm, EF = 6 cm, FA = 14 cm.

So, Perimeter =

(b) Area of ABCDEF = Area of ABGF (rectangle) + Area of EGCD (rectangle)

So, area of ABGF =

Now, area of EGCD =

So, .

## Question number: 161

### Describe in Detail

There is a sale of 60 % discount on a shop of clothes. Vicky bought shirts of in this sale.

(a) How much did Vicky paid?

(b) The next day, all items were sold at 70 % of the original price. How much more would Vicky have saved if he had waited until the next day to buy the shirts?

### Explanation

Here,

(a) Price of shirts is and sale is of 60 %.

So, he saved.

So, he paid **.**

(b) Next day, all items were sold at 70 % of the original price.

So, , so he will saved more than saved in 60 % discount.