# CIE Mathematics Paper-1: Specimen Questions 159 - 160 of 175

## Question number: 159

Essay Question▾

### 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

Essay Question▾

### 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, .

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